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Multilinear Algebra

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Werner Greub
Multilinear
Igenra
2nd Edition
a
Springer-Verlag
New York Heidelberg
Berlin
I
Un iversitext
Werner Greub
Multiinear Algebra
2nd Edition
New York
Springer-Verlag
Heidelberg Berlin
Werner Greub
Department of Mathematics
University of Toronto
Toronto M5S 1A1
Canada
AMS Subject Classifications: 15-01, 15A75, 15A72
Library of Congress Cataloging in Publication Data
Greub, Werner Hildbert, 1925Multilinear algebra,
(Universitext)
Includes index.
1. Algebras, Linear.
I. Title.
QA 184.G74 1978
512'.5
78-949
ISBN-13:978-0-387-90284-5
e-ISBN-13:978-1-4613-9425-9
DO!: 10.1007/ 978-1-4613-9425-9
All rights reserved.
No part of this book may be translated or reproduced in any
form without written permission from Springer-Verlag.
© 1967 by Springer-Verlag Berlin Heidelberg
© 1978 by Springer-Verlag New York Inc.
987654321
Preface
This book is a revised version of the first edition and is intended as a
sequel and companion volume to the fourth edition of Linear Algebra
(Graduate Texts in Mathematics 23).
As before, the terminology and basic results of Linear Algebra are
frequently used without reference. In particular, the reader should be
familiar with Chapters 1-5 and the first part of Chapter b of that book,
although other sections are occasionally used.
In this new version of Multilinear Algebra, Chapters 1-5 remain essentially unchanged from the previous edition. Chapter b has been completely
rewritten and split into three (Chapters b, 7, and 8). Some of the proofs
have been simplified and a substantial amount of new material has been
added. This applies particularly to the study of characteristic coefficients
and the Pf of f ian.
The old Chapter 7 remains as it stood, except that it is now Chapter 9.
The old Chapter 8 has been suppressed and the material which it contained (multilinear functions) has been relocated at the end of Chapters 3,
S, and 9.
The last two chapters on Clifford algebras and their representations are
completely new. In view of the growing importance of Clifford algebras
and the relatively few references available, it was felt that these chapters
would be useful to both mathematicians and physicists.
In Chapter 10 Clifford algebras are introduced via universal properties
and treated in a fashion analogous to exterior algebra. After the basic
isomorphism theorems for these algebras (over an arbitrary inner product
space) have been established the chapter proceeds to a discussion of
finite-dimensional Clifford algebras. The treatment culminates in the complete classification of Clifford algebras over finite-dimensional complex
and real inner product spaces.
v
Preface
vi
The book concludes with Chapter 11 on representations of Clifford
algebras. The twisted adjoint representation which leads to the definition
of the spin-groups is an important example. A version of Wedderburn's
theorem is the key to the classification of all representations of the Clifford
algebra over an 8-dimensional real vector space with a negative definite
inner product. The results are applied in the last section of this chapter to
study orthogonal multiplications between Euclidean spaces and the existence of orthonormal frames on the sphere. In particular, it is shown that
the (n -1)-sphere admits an orthonormal k-frame where k
is the
Radon-Hurwitz number corresponding to n. A deep theorem of F. Adams
states that this result can not be improved.
The problems at the end of Chapter 11 include a basis-free definition of
the Cayley algebra via the complex cross-product analogous to the definition of quaternions in Section 7.23 of the fourth edition of Linear Algebra.
Finally, the Cayley multiplication is used to obtain concrete forms of some
of the isomorphisms in the table at the end of Chapter 10.
I should like to express my deep thanks to Professor J. R. Vanstone who
worked closely with me through each stage of this revision and who made
numerous and valuable contributions to both content and presentation. I
should also like to thank Mr. M. S. Swanson who assisted Professor
Vanstone and myself with the proof reading.
Toronto, April 1978
W. H. Greub
Table of Contents
Chapter 1
Tensor Products
Chapter 2
Tensor Products of Vector Spaces with
Additional Structure
41
Chapter 3
Tensor Algebra
60
Chapter 4
Skew-Symmetry and Symmetry in the Tensor Algebra
84
Chapter 5
Exterior Algebra
96
Chapter 6
Mixed Exterior Algebra
148
Chapter 7
Applications to Linear Transformations
174
Chapter 8
Skew and Skew-Hermitian Transformations
193
Chapter 9
Symmetric Tensor Algebra
209
Chapter 10
Clifford Algebras
227
Chapter 11
Representations of Clifford Algebras
260
Index
1
291
vii
Tensor Products
Throughout this chapter except where noted otherwise all vector spaces will be defined over a
fixed, but arbitrarily chosen, field T.
Multilinear Mappings
1.1. Bilinear Mappings
Suppose E, F and G are any three vector spaces, and consider a mapping
p : E x F -+ G.
'p is called bilinear if it satisfies the conditions
p(t x 1 + µx 2 , y) =
x1, x2EE, yeF,A,µeF,
p(x 1, y) + µup(x 2 , y)
xeE,y1,y2eF.
'p(x, Ay 1 + µY2) _ A p(x, Y 1) + µup(x, Y2)
Recall that if G = F, then 'p is called a bilinear function.
The set S of all vectors in G of the form '(x, y), x e E, y e F is not in
general a vector subspace of G. As an example, let E = F and G be respectively 2- and 4-dimensional vector spaces. Select a basis a 1, a2 in E
and a basis c (v = 1, ... , 4) in G and define the bilinear mapping 'p by
co(x, Y) =
'i 'c1 + 1i2C2 + 2i 'c3 + 2ii2C4
where x = c 1a 1 + 2a2 and y = ri' a 1 +
r12a2.
Then it is easy to see that
a vector
z=
VC
of G is contained in S if and only if the components satisfy the relation
-
= o.
1
1 Tensor Products
2
Since the vectors z 1 = 2c 1 + 2c2 + c3 + c4 and z2 = c 1 + c3 satisfy this
condition, while the vector z = z 1 - z2 = c 1 + 2c2 + c4 does not, it follows
that S is not a subspace of G.
We shall denote by Im p the subspace of G generated by S.
Now consider the set B(E, F; G) of all bilinear mappings of E x F into
G. By defining the sum of two bilinear mappings p1 and P2 by
('p1 + P2)(x, y) = P1(x, y) + P2(x, y)
and the mapping (hp) by
(A p) (x, y) =
x e E, y e F,
p(x, y)
F,
we can introduce a vector space structure in the set B(E, F; G). The space
B(E, F; F) of all bilinear functions in E x F will be denoted simply by
B(E, F).
1.2. Bilinear Mappings of Subspaces and Factor Spaces
Given a bilinear mapping p : E x F -+ G and a pair of subspaces E 1 c E,
F 1 c F, a bilinear mapping p 1: E 1 x F 1 -+ G is induced by
x1 e E1, Y1 e F1.
p1(x1, Y1) = P(x1, Y1)
p 1 is called the restriction of 'p to E 1 x F 1.
F be two direct decompositions of E and F
Let E = a Ea and F =
respectively and assume that for every pair (a, /3) a bilinear mapping
pa : Ea x Ffl -+ G is given. Then there exists precisely one bilinear mapping
gyp: E x F -+ G whose restriction to Ea x Ffl is
In fact, if ?Ca: E -+ Ea and
pfl : F -+ Ffl are the canonical projections, define 'p by
'(x, y) =
x
p
y e F.
a,13
Then the restriction of 'p to Ea x F13 is p.
Now let '1 and '2 be two bilinear mappings of E x F into G whose
Then it follows that
restrictions to Ea x F13 are
('p1 - '2)(x, y) = 'p1(x, y) - P2(xry)
=
x, p13 Y) = 0
x, p13 Y) a, f3
a, f3
whence Pi = 'p2 .
If E 1 c E and F 1 c F are subspaces and 'pr: E 1 x F 1 -+ G is a bilinear
mapping then there exists a (not uniquely determined) bilinear mapping
cP : E x F -+ G whose restriction to E 1 x F 1 is 'pr. To prove this choose
two subspaces E2 c E and F2 c F such that
E=E1QE2,
F=F1G F2
Multilinear Mappings
3
and define the bilinear mappings cpi; : E. x F; -+ G (i, j = 1,2) by P 1 1 = Pi
and cpt; = 0, (i, j) (1, 1). In view of the above remark there exists a bilinear
mapping p: E x F -+ G whose restriction to E. x F; is cot;.
Now suppose that p: E x F -+ G is bilinear, and for some subspaces
E 1 c: E and G 1 c: G c(x 1, y) E G 1 for every x1 E E 1, y e F. Let p: E -+ E/E 1
and it : G -+ G/G 1 be the canonical projections, and define a bilinear mapping
x F -+ G/G1
by
gP(Px, y) = irpCx, y)
px e E/E 1, y e F.
It is clear that P is a well-defined bilinear mapping. We say that P is the
bilinear mapping induced by gyp.
If also for some subspace F 1 c: F, c(x, y1) E G 1 for each x e E, y1 E F 1,
then cp(px, y1) = 0 for each px e E/E 1, y1 E F 1. Denoting the canonical
projection of F onto F/F1 by a we see that p induces a bilinear mapping
P : E/E 1 x F/F 1 -+ G/G 1
defined by
MCP x
cry) _i Cx,Y)
P xeEE
/ 1cry
Ye
1
1.3. Multilinear Mappings
, p), G. A mapping
x Ep -+ G is called p-linear if for every i (1 <_ i <_ p)
Suppose we are given p + 1 vector spaces Et (i = 1.
gyp: E1 x
(p(x 1, ... , x_ 1, ),x1
+ µyi, xi + 1, ... , xp) = p(x 1, ... , xt , ... , xp)
+ µup(x1, ..., y1, ... , x,,)
x,, yt e E., A,, t e F.
If G = F, then p is called a p-linear function.
As in the case p = 2 the subspace of G generated by the vectors
cp(x 1,
... , xp), xt E Et will be denoted by Im gyp. Let L(E 1, ..., E,,; G) be the
set of all p-linear mappings p: E 1 x
x Ep -+ G. Defining the linear
operations by
(co + I) (x 1, ... , x,,) = (p(x 1, ... , x p) + lU(x 1, ... , x,,)
and
(A p)(x1, ... , xp) = A p(x1, ... , xp)
we obtain a vector space structure in L(E 1, ... , Es,, G). The space of all px Ep will be denoted by L(E 1, ..., E,,).
linear functions in E 1 x
4
1 Tensor Products
PROBLEMS
1. Establish natural isomorphisms
B(E, F; G)
L(E; L(F; G))
L(F; L(E; G)).
2. Given a bilinear mapping (p : E x F --> G, define a mapping li : E x F -- G by
fiz = cp(ir1 z, ire z)
zEExF
where ir1: E x F -- E and ire : E x F --> F are the canonical projections. Show that
& satisfies the relations
'Y(z1 + z2) + 'Y(z1 - z2) = 2'&(z1) + 2'&(z2)
and
() _
3. Let E and F be arbitrary. Show that the mapping f3: L(E; F) x E -- F defined by
(gyp, x)
px is bilinear. Prove that Im f3 = F.
4. Let E, F, G be finite-dimensional real vector spaces with the natural topology.
Show that every bilinear mapping p : E x F --> G is continuous. Conclude that the
mapping L(E; F) x E -- F defined by (gyp, x) --> px is continuous.
5. Given a bilinear mapping qP : E x F -- G define the null-spaces N 1(cp) c E and
N2(q,) = F as follows:
N 1(qp) : {x; q (x, y) = 0)
for every y e F
N2(gp) : {y; q (x, y) = 0)
for every x e E
and
(a) Consider the induced bilinear mapping
P : E/N 1(q,) x F/N2(q) -- G
(cf. Section 1.2). Show that N 1(rp) = 0 and N 2(rp) = 0.
Given a linear map f : G -- H consider the bilinear mapping ii : E x F -- H
defined by li(x, y) = f ip(x, y). Show that
and
N 1(q,) c N
N2(ip) c N2(/i)
(b) Conversely, let ci : E x F -- H be a bilinear mapping such that N 1(ip) c N 1(ci)
and N2(gp) c N2(ci). Prove that there exists a linear map f : G -- H such that
li(x, y) = fp(x, y)
Consider the space L of linear maps f : G -- H satisfying this condition. Establish a
linear isomorphism
1
L -* L(G/Im gyp; H).
Conclude that f is uniquely determined by ii if and only if Im p = G.
The Tensor Product
5
6. Let E be a vector space and F be the space of all functions h : E -- r. Define a bilinear
mapping (p : L(E) x L(E) --> F by
(p(f, 9)(x) = f(x)g(x)
x e E.
Show that N1(q) = 0 and N2((p) = 0.
7. Let E, E* be a pair of dual spaces and assume that 4: E* x E -- r is a bilinear function such that
D(T*- lx*, Tx) = D(x*, x)
for every pair of dual automorphisms. Prove that iD(x*, x) = A<x*, x> where A is a
scalar.
The Tensor Product
1.4. The Universal Property
Let E and F be vector spaces and let ® be a bilinear mapping from E x F
into a vector space T. We shall say that ® has the universal property, if it
satisfies the following conditions:
®l : The vectors x ® y (x e E, y e F) generate T, or equivalently,
Im ®= T.
®2. If p is a bilinear mapping from E x F into any vector space H,
then there exists a linear map f:T -+ H such that the diagram
T
commutes.
The two conditions above are equivalent to the following single condition
®: To every bilinear mapping p: E x F -+ H there exists a unique
linear map f:T -+ H such that Diagram (1.1) commutes.
In fact, assume that ®1 and ®2 hold and let
T -+ H and 12: T -+ H
be linear maps such that
p(x, y) = fl(x ®x y)
and p(x, y) = f2(x ®x y)
Then we have
fi(x8y)=f2(x®y)
xEE,yeF.
Now ® 1 implies that f, = f2 and so f is uniquely determined by cp.
1 Tensor Products
6
Conversely, assume that Q holds. Then QZ is obviously satisfied. To
prove ®, let Ti be the subspace of T generated by the vectors x Q y with
x e E and y e F. Then Q determines a bilinear mapping gyp: E x F -+ Ti
such that
icp(x, y) = x Q y
xeE, yeF,
where i : Ti -+ T denotes the inclusion map. By Q there is a linear map
f : T -- Ti such that
p(x, y) = f (x Q y)
x e E, y e F.
Applying i to this relation we obtain
x®y
(i°f)(x®y) _
On the other hand, clearly,
l(x® y) = xQ y
xeE,yeF
where us the identity map of T. Now the uniqueness part of Q implies that
i of = 1. Thus i is surjective and so Ti = T This proves Q l .
EXAMPLE. Consider the bilinear mapping F x F -+ F given by Q y = Ay.
Since 1 Q y = y this map satisfies condition ®. To verify OZ, let gyp: F x
F -+ H be any bilinear mapping and define a linear map f : F -+ H by setting
f(y) _ (p(l, y)
Then we have for E F and y e F
y) _ A p(1, y) _ f (y) = f (;y) = f ( O y)
and so OZ is proved.
1.5. Elementary Properties
Before proving existence and uniqueness of bilinear mappings with the universal property we shall derive a few properties which follow directly from
the definition.
Thus we assume that Q : E x F -+ T is a bilinear mapping with the universal property.
Lemma 1.5.1. Let at (i = 1, ..., r) be linearly independent vectors in E and
let b. (i = 1, ..., r) be arbitrary vectors in F. Then the relation
atQbt = 0
implies that bt = 0 (i = 1,
..., r).
7
The Tensor Product
PROOF. Since the at are linearly independent we can choose r linear functions
f t in E such that
f'(a) = b
(i, j = 1, ..., r).
Now consider the bilinear function
r
(x, y) _
x e E, y e F
i= 1
where the g` are arbitrary linear functions in F. In view of QZ , there exists
a linear function h in T such that
h(x O Y) _
f `(x)g`(Y)
t
Then
a; O b; _
h
g`(bt)
f `(a;)g`(b;) _
Since L a; Q b; = 0, we obtain
g`(bt) = 0.
But the g` are arbitrary and so it follows that
(i = 1, ... , r).
b= O 0
Corollary. If a
0, then a Q b
0 and b
D
0.
be a basis of E. Then every vector z e T can be
Lemma 1.5.2. Let
written in the form
z = > ea Q ba ,
ba E F,
a
where only finitely many ba are different from zero. Moreover, the ba are
uniquely determined by z.
PROOF. In view of ®, z is a finite sum
z=
xv e E, Yv e F.
x O Yv
v
Now write
JaEF.
ea
a
Then we have
z = >e®yv = >e®yv = eaOba
v, c
a
v, cc
where
a
ba =
v Yv
v
1 Tensor Products
8
To prove uniqueness assume that
ea ® ba =
b, ba e F.
ea ® ba
a
a
Then
ea ® (ba - ba) = 0
a
and so Lemma 1.5.1 implies that ba = b'.
Lemma 1.5.3. Every nonzero vector z e T can be written in the form
r
z = > xl ® yl
xleE, yIeF,
i=1
where the x, (1, ... , r) are linearly independent and the yl (i = 1.
linearly independent.
,
r) are
xl ® yl where r is minimized.
PROOF. Choose a representation z =
If r = 1, it follows from bilinearity that x 1
0.
0 and y1
Now consider the case r > 2. If the vectors x, are linearly dependent we
may assume that
r- 1
i
xr =
i= 1
Then we have
r-1
r-1
r-1
r-1
>x,
!!',z =i=1> x1®y+
j ® (y i + `yr) = i=1 xi®yi
i=1
i=1
L!x1®y,. =
which contradicts the minimality of r. Thus the vectors x, are linearly independent. In the same way it follows that the vectors yl are linearly independent as well.
1.6. Uniqueness
Suppose that ® : E x F -+ T and c : E x F -+ T are bilinear mappings
with the universal property. Then there exists a linear isomorphism f:T
such that
f(x®y)=xy
xeE,yeF.
In fact, in view of ®2, we have linear maps
f:T-+T
and g: T -+ T
such that
f(x ® y) = x
y
T
9
The Tensor Product
and
g(x ®y) = x Q y
x e E, y e F.
These relations imply that
gf (x O Y) = x ® y and fg(x O Y) = x O Y
Now ®, shows that
gof=i and fig=i.
Thus f and g are inverse linear isomorphisms.
1.7. Existence
To prove existence consider the free vector space C(E x F) generated by
the set E x F (see Section 1.7 of Linear Algebra). Let N(E, F) denote the
subspace of C(E x F) generated by the vectors
(x + µx2, y) - (x 1, y) - µ(x2 , y)
and
(x, Y + µ)72) - (x, Y 1) - µ(x, Y2)
Set
T = C(E x F)/N(E, F)
and let it : C(E x F) - T be the canonical projection.
Now define a set map ® : E x F - T by
x O Y = ic(x, y).
We shall show that Q is a bilinear mapping and has the universal property.
In fact, since
ic( tx 1 + µx2, y) = A r(x 1, y) + µir(x2 , y),
it follows that
(x1 + ,1x2) Q y = Tr(x 1 + µx2, y)
= Ait(x 1, y) + µir(x 2 , y) = x1 O y + µx 2 ® y.
In the same way it is shown that Q is linear in y.
To prove ®,, observe that every vector z e T is a finite sum
vµ(Yµ)
z=t
xeE,yµeF.
v, µ
It follows that
yµ) _ it
vµxv 0 yµ
v,µ
v,µ
vµ(xv, yµ) = z.
v,µ
10
1
Tensor Products
To verify 02, consider a bilinear mapping U of E x F into a third vector
space H. Since the pairs (x, y) x e E, y e F form a basis for C(E x F) there
is a uniquely determined linear map
g:C(ExF)--H
such that
g(x, y) = U(x, Y)
The bilinearity of i'/ implies that N(E, F) c ker g. In fact, if
z = (x 1 + µx 2 , y) - (x1, y) - µ(x 2 , y)
is a generator of N(E, F), then
g(z) = g(tx 1 + µx2, y) - 2g(x 1, y) - ug(x2, y)
= U(x 1 + µx2, y) - A,4i(x 1, y) - µ4(x2 , y)
=0.
In a similar way it is shown that
g[(x, Y1 + µY2) - Ax, Y1) - µ(x, Y2)] = 0
and it follows that N(E, F) c ker g. Hence g induces a linear map
f : C(E x F)/N(E, F) -+ H
such that
foj=g.
In particular, it follows that
(f ° ®)(x, y) = f ir(x, y) = g(x, y) _ U(x, Y)
This shows that the bilinear mapping Q has the universal property.
Definition. The tensor product of two vector spaces E and F is a pair (T, ®),
where Q : E x F -+ T is a bilinear mapping with the universal property. The
space T, which is uniquely determined by E and F up to an isomorphism, is
also called the tensor product of E and F and is denoted by E Q F.
Now we show that the tensor product is commutative in the sense that
EOF FOE.
In fact, consider the bilinear mappings
cp:E x F-+FQE and t/i:F x E-+EQF
given by
p(x, y) = Y O x
and
U(y, x) = x Q y.
The Tensor Product
11
In view of 0 2, they induce linear maps f: E Q F -+ F O E and g: F Q E -+
E Q F such that
y Q x= f (x Q y) and x Q y= g(y O x)
for all x e E and y e F. These relations imply, in view of Ol, that g o f = i
and f o g = t. Thus f and g are inverse isomorphisms.
1.8. Reduction of Bilinear Mappings to Linear Maps
Fix E and F and let G be a third vector space. Then a linear isomorphism
L(EQF;G)=-B(E,F;G)
is defined by
1(f) = f° O
f e L(E Q F; G).
In fact, Q2 implies that 1 is surjective, since it states that any bilinear mapping gyp: E x F -+ G may be factored over the tensor product. To show that
1 is injective assume that f o Q = 0 for a certain linear map f : E Q F -+ G.
In view of Q 1 the space E Q F is generated by the products x Q y and hence
it follows that f = 0.
The correspondence between the bilinear mappings p: E x F -+ G and the
linear maps f : E Q F -+ G which is obtained by the above result is expressed
by the following commutative diagram:
E x F -- G
Proposition 1.8.1. Let cp: E x F -+ G be a bilinear mapping and f: E Q F -+ G
be the induced linear map. Then f is surjective if and only if p satisfies OQ l .
Moreoverf is injective [and only if p satisfis ®2.
PROOF. The first part follows immediately from the relation
Imp=Imf
To prove the second part assume that f is injective. Then the pair (Im gyp, 'p)
is a tensor product for E and F. Hence every bilinear mapping i'/ : E x F -+ K
induces a linear map g :Im p -+ K such that
U(x, y) =
y)
1f f is an extension of g to a linear map f : G -+ K it follows that
/i(x, y) = f p(x, y)
and hence p has the property 02.
12
1 Tensor Products
Conversely, assume that p satisfies ®2. Then the bilinear mapping
E x F -+ E® F induces a linear map h: G -+ E Q F such that
x O y = hto(x, y).
On the other hand, we have p(x, y) = f(x ® y) and it follows that
x®y=hf(x®y).
Hence h o f = i and so f is injective.
PROBLEMS
1. Consider the bilinear mapping f3 : r" x rm -- M" x m defined by
(b 1, ... ,x ( 1, ... ,
xn 1
.. ,
bb
Prove that the pair
/
bb
(M" x m, f3) is the tensor product of r" and rm.
n
2. Show that the bilinear mapping r" x E --> QE defined by
(b 1 . ... ,") ® x = (
1 x,
... ,"x)
is the tensor product.
3. Let S and T be two arbitrary sets and consider the vector spaces C(S) and C(T)
generated respectively by S and T (cf. Section 1.7 of Linear Algebra). Show that
C(S x T) is isomorphic to C(S) ® C(T).
4. Assuming that a® b
0, a e E, be F prove that
a®b=a'®b'
if and only if
a' = A.a and
b' = A-1 b
A E r, A
0.
5. Let A be a subfield of r and consider a vector space Eo over A. Then r ® Eo is again
a A-vector space. Define a scalar multiplication r x (r ® Eo) -- r ® Eo by
A,aer,xeEo
(a) Prove that this multiplication makes r ® Eo into a r-vector space E.
(b) Show that the restriction of this multiplication to A x Eo coincides with the
scalar multiplication in E.
(c) If {e2} is a basis of Eo prove that { 1 ® e2} is a basis of E.
(d) Let r = C and A = Ft Prove that E is isomorphic to the vector space E x E
constructed in Problem 5, §1, Chapter 1 of Linear Algebra.
6. With the notation of Problem 5 let po be a linear transformation of E.
(a) Prove that pr = i
is a linear transformation of E.
(b) For any polynomial f e r[t] prove that
(.f (q,e))r = .f (qpr)
(c) Find the minimum polynomial of pr in terms of the minimum polynomial of p.
(d) Show that pr is semisimple (nilpotent) if po is semisimple (nilpotent) and hence
construct the decomposition of pr into semisimple and nilpotent parts.
13
Subspaces and Factor Spaces
Subspaces and Factor Spaces
1.9. Tensor Products of Subspaces
Suppose that the bilinear mapping Q : E x F -+ T has the universal property
F. Let Q' denote the restricand consider two subspaces E 1 E and F 1
tion of Q to E 1 x F 1 and set T1 = Im Q'. We shall show that (T1, ®') is
the tensor product of E 1 and F 1.
Property Q i is immediate from the definitions. To verify ®2' let P1 : E 1 x
F 1 -+ H be a bilinear mapping. Extend P1 to a bilinear mapping p: E x F -+ H.
Since O has the universal property, there is a linear map
f:T -+ H
such that
x e E, y e F.
f(x O Y) = p(x, y)
This relation implies that
f(x1 0 Y1) = P(x1, Y1) = p1(x1, Y1)
x1 E E1, Y1 E
1,
and so P1 factors over 0.
1.10. Tensor Product of Factor Spaces
Again let E 1
E and F 1 c F be subspaces and set
T(E1, F1) = E1 ®F + E O F1
Define a bilinear mapping /3: E x F -+ (E Q F)/T (E 1, F 1) by
/3(x, y) = ic(x O Y),
where it denotes the canonical projection.
Since /3(x1, y) = 0 if x 1 E E 1, y e F and /3(x, y 1) = 0, if x e E, y l E F 1, /3
induces a bilinear mapping
J3:E/E1 x F/F1 -+ (E Q F)/T(E1, F1)
such that
/3(x, y) = /3(x, y)
x E E/E 1, y e F/F 1.
To prove that fi satisfies Q first notice that
Im fi = Im /3 = Im m = (E Q F)/T (E 1, F 1)
and so property Q i follows. To check ®2, let
E/E 1 x F/F 1 -+ H
be any bilinear mapping. Define a bilinear mapping p: E x F -+ H by setting
p(x, y) = (x, Y)
1 Tensor Products
14
Then there is a linear map f : E Q F -+ H such that
p(x, y) = f (x Q y)
x e E, y e F.
Moreover,
f (x 1 0 Y) = p(x 1, y) _ /i(0, j) = 0
x1eE1,yeF
and similarly,
xeE,yeF1.
f(x®Y1) = 0
Hence T (E 1, F 1) c: ker f, and so f induces a linear map
f:(E O F)/T(E1, F1) -+ H
such that
fo=f
It follows that
/ (x, j) _ p(x, Y) = f (x O Y) = fir(x O Y)
= f13(x, Y) = ffl(, y)
x E E/E 1, y E F/F 1
whence ,/, = f o J3. Thus JJ satisfies condition Q 2 and so the proof is complete.
The result obtained above shows that there is a canonical isomorphism
E/E 1 0 F/F 1 =- (E O F)/(E 1 0 F+ E O F 1)
PROBLEM
Let E1 and E2 be subspaces of E such that E = E1 + E2 and set E1 n E2 = F. Establish
an isomorphism
E/E1 OO E/E2
(E1 Ox E2)/(F OO E2 + E1 OO F).
Direct Decompositions
1.11. Tensor Product of Direct Sums
Assume that two families of linear spaces E, a e 1 and Ffl ,13 E J are given and
that for every pair (a, 13), (Ea Q Ffl , Q) is the tensor product of Ea and Ffl .
Then a bilinear mapping p of E = ®Ea and F = O Ffl into the direct sum
(= EE a, fl(Ea Q Ffl) is defined by
i (ica x O p fl Y)
(p(x, Y) _
a, fJ
where
ica : E -+ Ea ,
pfl : F -+ Ffl
Direct Decompositions
15
are the canonical projections and
Ea O F -+ G
are the canonical injections. It will be shown that the pair (G, gyp) is the tensor
product of E and F.
Condition Q 1 is trivially fulfilled. To verify Q 2 let :E x F -+ H be an
: Ea x Ffl -+ H by
arbitrary bilinear mapping. Define
y) = iU(la x, j y),
where
is : Ea -+ E and j : Ffl -+ F
are the canonical injections. Then
such that
U
(x, y) =
induces a linear map fag : Ea Q Ffl -+ H
xeEa,yeFf.
O Y)
Define a linear map f : G -+ H by
f=
where
-' Ea Q Ffl
are the canonical projections. Then it follows that
(f ° 'p)(, j) = f
x O P Y)
a, fJ
_
x O P Y)
a, fJ
=
Pfl Y)
=
a
=
Y)
Hence f ° P = i/i and so 02 is satisfied.
1.12. Direct Decompositions
Assume that the pair (E Q F, Q) is the tensor product of the vector spaces E
Ffl are
and F and that two direct decompositions E = a Ea and F =
given. It will be shown that E Q F is the direct sum of the subspaces Ea Q Ffl,
EOF = >Ea®Ff.
a, fJ
(1.2)
1 Tensor Products
16
In view of ®l the space E Q F is generated by the products x Q y;
xeE, yeF.
Since x = axa, xa a Ea and y =
yp E F it follows that
x®y = xa®x ys.
a, Ii
This equation shows that the space E Q F is the sum of the subspaces
Ea Q F.
To prove that the decomposition (1.2) is direct consider the direct sums
E = Qa Ea, F = Q F and G = pa, Ea Q F and let the injections is, J1,
ia and the projections ?Ca, pp, ira be defined as in the previous section. Then
if gyp: E x P -+ G is the bilinear mapping given by
p(i, Y) =
iap(ia x O p Y)
a, Ii
we have shown (in the previous section) that the pair (G, gyp) is the tensor
product of E and F.
Now consider the linear isomorphisms
f:E-+E and g:F-+F
defined by
fx=
and g y = >jpYp'
la xa
a
Ii
where
x=
xa , xa E Ea
and y =
a
Yp, Yp E F.
p
Define a bilinear mapping i/i : E x F -+ G by
/i(x, y) = P(f x, g y)-
In view of the factorization property there exists a linear map h : E Q F -+ U
such that
h(x O y) = Ji(x, y)
and hence
h(x O y) = P(f x, g y).
If x e Et and y e Fa it follows from the definition off, g, and p that
h(x O y) = p(f x, g y) = p(it x, jay)
=
l43(ia it x O ppja y) = lta(x 0 y)
a, p
But this equation shows that h maps every subspace Ea Q F of E Q F into
the subspace
of G. Since the decomposition
Q
O
G=
a, p
Direct Decompositions
17
is direct, the decomposition
EQF = Ea®Ff
a, fi
must be direct. This completes our proof.
Conversely, suppose that direct decompositions
E = > Ea ,
F = > Ffl ,
a
G = > Gafl
a, f3
13
and bilinear mappings Q : Ea x F13 -+ Gay are given such that the pair
is the tensor product of Ea and F13. Define a bilinear mapping
p : E x F -+ G by
p(x, y) = , xa O Y'
where
x=
xa
and y =
a
y13.
13
Then the pair (G, gyp) is the tensor product of E and F.
The condition Q 1 is obviously satisfied. To prove 02 let U : E x F -+ H
of U to
be an arbitrary bilinear mapping and consider the restriction
Ea x F13 . Then there exists a linear map fag : Gay -+ H such that
O
y13). Define a linear map f : G -+ H by
f (z) = L, f(z)
a,f
where z =
Then
zap, zap e G.
(f o p)(x, y) = f o(x, y)
=
Ox y13)
a, fi
=
y13) = U(x, y)
a, fi
whence f o P = U. Thus 02 is satisfied and the proof is complete.
1.13. Tensor Product of Basis Vectors
Suppose that (aa)a El and (b13) 13 E J are, respectively, bases of vector spaces E and
F. Then the products (aa O b13)a El, 1 E J form a basis of E Q F. To prove this,
let Ea, F13 denote the one-dimensional subspaces of E and F generated by as
in view of the result of the
Ea, F =
and b13 respectively. Then E =
previous section it follows that
EQF = >Ea®F13.
a, 13
1 Tensor Products
18
Now it was shown in Section 1.5 that as 0, bf 0 implies as Q bf 0.
On the other hand, Q 1 applied to Ea, Ffl and Ea Q Ffl gives that Ea Q Ffl is
spanned by the single element as Q bfl . Thus E Q F is the direct sum of the
one-dimensional subspaces generated by the products as Q bfl, and hence
these products form a basis of E Q F (see Lemma 1.5.1).
In particular, it follows from these remarks that if E and F have finite
dimensions, then E Q F has finite dimension, and
dim(E Q F) = dim E dim F.
(1.3)
1.14. Application to Bilinear Mappings
Let E and F be vector spaces with bases (x 3, E I and (yfl) E J respectively. Then
into
since xa Q y is a basis of E Q F, it follows that every set map of (xa Q
a third vector space G can be extended in a unique way to a linear map
f:E ® F - G
and every linear map f : E Q F - G is obtained in this way. In view of the
isomorphism
L(E Q F ; G)
B(E, F ; G),
it follows that every set map
(x«, yfi) - G
can be extended in a unique way to a bilinear mapping p : E x F - G and
every bilinear mapping p is obtained in this way. In particular, the space
Im 'p is generated by the vectors p(xa, yfl). This result implies that if E and F
have finite dimension, then
dim Im '
dim E dim F.
It is now easy to construct a basis of the space B(E, F; G) provided that the
dimensions of E and F are finite. Let xt (i = 1, ..., n), y; (j = 1, ..., m) and
(zy)y E K be bases of E, F, and G respectively. Then the products xt Q y; form
a basis of E Q F and hence the linear maps f yl (k = 1, ..., n; l = 1, ... , m;
y e K) defined by
f'(xi Ox y;) = 5 5 zy
form a basis of L(E Q F; G). Consequently, the bilinear mappings o given
by
p '(xi , y,) = bkb; zy
form a basis of B(E, F; G).
If G has finite dimension as well, it follows that
dim B(E, F; G) = dim E dim F dim G
Direct Decompositions
19
and so in particular
dim B(E, F) = dim E dim F.
1.15. Intersection of Tensor Products
Let E 1 and E2 be two subspaces of a vector space E. Then if F is a second
vector space,
(E1 QF)n(E2QF) = (E1 nE2)QF.
Clearly,
(E1 nE2)OFc (E1 OF)n(E20F)
To prove the inclusion in the other direction, let z be an arbitrary vector of
(E1 Q F) n (E2 Q F). If (bfl)fE, is a basis of F we can write
z=
u Q bfl,
ufi E E1
and z =
13
u Q bfl,
u E E2.
13
This yields
(u - v13) x0 b = o
e
and since the b13 are linearly independent we obtain u13 = v. Hence u13 E
E1 n E2 and z e (E1 n E2) Q F. This completes the proof of (1.4).
Next consider two subspaces E 1 c E and F 1 c F. Then
(E1 O F) n (E O F1) = E1 O F1
To prove (1.5), choose a subspace F' c F such that
F=F10+F'.
Then we have in view of Section 1.12 that
E1QF=E1QF1E E1QF'.
Intersecting with E Q F 1 and observing that E 1 Q F 1 c E Q F 1 we obtain
(E1 O F) n (E O F1) = (E1 O F1) O+ [(E1 O F') n (E1 O F1)] (1.6)
Now Formula (1.4) yields
(E1 Q F') n (E1 O F1) = E1 Q (F' n F1) = E1 Q 0 = 0
and thus (1.5) follows from (1.6).
1 Tensor Products
20
Finally let E 1, E2 and F 1, F2 be subspaces respectively of E and F. Then
(E 1 O F 1) n (E2 O F2) _ (E 1 n E2) O (F 1 n F2)
Clearly,
(E1 n E2) OO (F1 n F2) C (E1 Q F1) n (E2 Q F2).
Moreover, we have in view of (1.4) that
(E1 ® F1) n (E2 O F2) c: (E1 O F) n (E2 O F) _ (E1 n E2) O F.
In the same way it follows that
(E1 ®F1) n (E2 O F2) c: (E O F1) n (E O F2) = E 0 (F1 n F2).
Now formula (1.5) yields
(E 1 O F 1) n (E2 O F2)
(E 1 n E2) O (F 1 n F2)
This completes the proof of (1.7).
PROBLEMS
1. Let z e E O F, z 0, be any vector. Show by an explicit example that in general there
exist several representations of the form
z = x Qx y
x E E, y E F
where the
and the
representations
are respectively linearly independent. Given two such
r
s
i=1
j=1
z= >x®y and z= xJOyJ
prove that r = s.
2. Let p : E x F --> G be a bilinear mapping. Show that the following property is equivalent to ®2 Whenever the vectors xa E E and yp e F are linearly independent then so
are the vectors cp(xa, yp).
3. Let A 0 be an algebra of finite dimension and assume that the pair (A, /3) is a tensor
product for A and A where /3 denotes the multiplication. Prove that dim A = 1.
4. Let E be an arbitrary vector space and F be a vector space of dimension m. Establish
a (noncanonical) isomorphism
EE ...Q+ E=+EQx F.
m
5. Let E, E* and F, F* be two pairs of dual spaces of finite dimension. Consider the
bilinear mapping
fl: E x F -- B(E*, F*)
Linear Maps
21
given by
IJ,y(x*, Y*) _ <x*, x><Y*, y>
Prove that the pair (B(E*, F*), /3) is the tensor product of E and F.
Linear Maps
1.16. Tensor Product of Linear Maps
Given four vector spaces E, E', F, F' consider two linear maps
and t/i : F -- F'.
cP : E -- E'
Then a bilinear mapping E x F -+ E' Q F' is defined by
(x, y) -' cpx Q U y.
In view of the factorization property there exists a linear map
x:EQF-+E'®F'
such that
x(x O y) _ cpx O ty
(1.8)
and this map is uniquely determined by (1.8). The correspondence ((p, U) -+ x
defines a bilinear mapping
f3: L(E; E') x L(F; F') -+ L(E Q F; E' Q F').
Proposition 1.16.1. Let L(E; E') Q L(F; F') be the tensor product of L(E; E')
and L(F; F'). Then the linear map
f:L(E;E')QL(F;F')-+L(EQF;E'QF')
induced by the bilinear mapping fi is injectiue.
PROOF. Let w be an element such that f (w) = 0. If w
0 we can write
r
w= >Jp®
Pi E L(E, E'), thii E L(F, F'),
i= 1
where the co and i/ii are linearly independent.
Then
f(w) =
f3((p, th'1)
and hence f (w) = 0 implies that
r
q (x) O i(y) = 0
i= 1
for every pair x e E, y e F. Now choose a vector a e E such that p 1(a)
0.
1 Tensor Products
22
Let p >_ 1 be the maximal number of linearly independent vectors in the
set p 1(a), ... , cpr(a). Rearranging the cot we can achieve that the vectors
P 1(a), ... , p ,,(a) are linearly independent. Then we have
P
=
a)
J = p + 1, ... , r
i= 1
and Relation (1.9) yields
p
i=1
r
(a) O i(Y) +
j=p+ 1 \i=1
ji Pi(a) O
(Y) = 0
yeF;
r
p
i=1
p
(a)®
j=p+ 1
(Y) + Y' i(Y) = 0
y E F.
Since the vectors pi(a) (i = 1, ... , p) are linearly independent it follows that
r
i(Y) +
j=p+ 1
Y) = 0
i = 1, ... , p
= 0. This is in contradiction to our
for every y e F, i.e. i/ i +
=p+1
hypothesis that the maps l//; are linearly independent and hence f is injective.
Corollary I. The pair (Im /3, l3) is the tensor product of L(E; E') and L(F; F').
Corollary II. The bilinear mapping /3: L(E) x L(F) - L(E Q F) given by
fl(f, g)(x O Y) = f(x)g(y)
is such that (Im /3, /3) is the tensor product of L(E) and L(F).
Corollary III. If E and F have finite dimension the elements f3(cp, U) generate the
space
L(EQF;E'QF)
as will be shown in Section 1.27 and so the pair (L(E Q F; E' Q F'), /3) is the
tensor product of L(E; E') and L(F; F').
In general, however, this is not the case as the example below will show.
Nevertheless, by an abuse of language, we call
U) the tensor product of
U) = p Q U. Then formula (1.8)
the linear maps p and l// and write
reads
('p O U)(x O Y) = 'x O UY
In particular, if E = F = E' = F' and 'p = ui = t, then t Q l = 1.
23
Linear Maps
1.17. Example
Let E = F be a vector space with a countable basis and put E' = F' = F.
Then we have
L(E Q F; E' Q F') = L(E Q E)
L(E; E') = L(F; F') = L(E),
and the bilinear mapping /3 is given by
f3: (f g) -'f . g,
where
(f . g) (x O y) = f (x)g(y)
It will be shown that /3 does not satisfy the Condition Q l .
We associate with every linear function h in E Q E a subspace Eh c: E
(called the nullspace of the corresponding function) in the following way:
A vector xo E E is contained in Eh if and only if h(xo Q y) = 0 for every y e E.
Now assume that h e Im /3. Then the factor space E/Eh has finite dimension.
In fact, h can be written as a finite sum of the form
r
h(x O y) _
f (x)gt(y)i= 1
ker f we have
It follows that for any x e n
h(x O y) = 0
y E F;
i.e.,
r
Eh D flker.
i= 1
Now a result of Section 2.4 of Linear Algebra implies that dim E/Eh < r.
On the other hand, consider the linear function w on E Q E given by
w(x
x0
y) _
cv11v
V
where c and iv (v = 1, 2, ... ) are the components of x and y with respect to a
basis of E. It is easy to see that the nullspace Ew = 0 and hence dim E/Ew =
dim E = oo. Consequently, w is not contained in Im /3.
1.18. Compositions
Consider four linear maps
p : E --> E'
gyp' : E' --> E"
F -+ F'
t/i' : F' -+ F".
Then it is clear from the definition that
((P' O /J') 0 ((P O IJ) _ ((P' 0 (p) O (c/i' 0 i/i)
(1.10)
1 Tensor Products
24
Now assume that 'p and U are injective. Then there exist linear maps 0: E' -p E
and ,J: F' -+ F such that
(po(p=i and iiol'/=i.
Hence, formula (1.10) yields
(®
° (' O i/i) = (o gyp) O (c/iO i/i) = p _
showing that (p Q U is injective.
1.19. Image Space and Kernel
It follows immediately from the definition of (p Q U that
ImcppImiji.
In particular, if 'p and U are surjective, then so is 'p Q U. Now the formula
ker((ppU)=kercp® F+EpkerU
(1.12)
will be established. Consider the induced injective linear maps
i:E/ker cp -> E'
and
F/ker t/i -> F'.
Then
rp®E/ker'QF/keri//- E'QF'
is injective as well (Section 1.18). Let
m, :E -+ E/ker gyp,
ir2 : F -+ F/ker U,
and
U)
T(ker gyp, ker U) = ker 'p Q F + E Q ker 1i
be the canonical projections. According to Section 1.10 there exists a linear
isomorphism
g: E/ker cp Q F/ker 1i
(1.13)
E p F/T(ker p, ker U)
such that
9(irx O it2 Y) = ir(x O Y)
Now define a linear map
x
= (i O ) ° 9 -'
25
Linear Maps
Clearly x is injective. Moreover, if x e E and y e F are arbitrary we obtain
(x ° ir)(x O y) = (iii O /)g -1 g(irx O ice y)
= (pic x O i2 y
= (pXQi/iy
whence
x is injective, it follows that
ker((p Q U) = ker Tr = T(ker (p, ker U) = ker p Q F + E Q ker U.
PROBLEMS
1. Consider two linear maps (p : E - E' and /i : F - F'.
(a) Prove that p Q i/i is injective if and only if both mappings (p and /.i are injective.
(b) Assume that E and F have finite dimension. Prove that
r(q Qx /1) = r(ip)r(i/i)
where r denotes rank (see Section 2.34 of Linear Algebra).
2. Let E, F be two vector spaces of dimension n and m respectively and let p : E -p E,
/i : F - F be two linear transformations. Prove that
tr(q Q /i) = tr p tr /1
and
det(gp Q Ii) = (det q)m(det /,)n.
3. Consider a vector space E of dimension n. Given two linear transformations a and /3
of E let
L(E ; E) - L(E ; E)
be the linear transformation defined by
Q = a o Q o/
Q E L(E; E).
Show that
tr D = tr a tr /3
and
det D = det(a o f3)".
4. Let E and F be two finite-dimensional vector spaces and E', F' be arbitrary vector
spaces. Prove that the bilinear mapping
j3: L(E; E') x L(F; F') -> L(E Q F; E' Q F')
defined by l(q, 1i) = P Qx i/i is a tensor product.
1 Tensor Products
26
5. Let E be a finite-dimensional vector space and consider two commuting linear transformations (p, /i of E. Prove that
((P OO ')s = GPs OO 1s
and
(APO I)N=BPSOx'N+cPNOx
1N
(cf. Section 13.24 of Linear Algebra). Conclude that the tensor product of commuting
transformations is semisimple if and only if both transformations are semisimple.
Tensor Product of Several Vector Spaces
1.20. The Universal Property
Let Et (i = 1, ... , p) be any p vector spaces and let
Q:E1 x...xEp-T
be a p-linear mapping. This mapping is said to have the universal property if
it satisfies the following conditions:
Q l : The vectors x 1 ® O xp, (x, e Et) generate T.
®2: Every p-linear mapping p: E 1 x
can be written in the form
x E p -p H (H any vector space)
p(x 1, ... , x,,) = f (x 1 0 ... ® x,,)
where f : T -p H is a linear map.
The existence and uniqueness theorems are proved in the same way as in
the case p = 2 and we shall not repeat them.
Definition. The tensor product of the spaces Et (i = 1, ... , p) is a pair (T, Q)
x Ep -p T is a p-linear mapping with the universal
where Q : E1 x
property. T is also called the tensor product of the spaces Et and is denoted
by E 1 O ... O Ep .
If H is any vector space, then the correspondence 'p -+f expressed by the
commutative diagram
E1 x xEp- H
E1O...OEp
determines a linear isomorphism
L((E Q ... O E r); H) 4 L(E 1, ..., E; H).
1
27
Tensor Product of Several Vector Spaces
Proposition 1.20.1. Given three arbitrary spaces E 1, E2, E3 there exists a linear
isomorphism
f:E1 OE2 OE34(E1 OE2)OE3
such that
f(x®y®z) = (x®x y)Qx z.
PROOF. Consider the trilinear mapping
E 1 x E2 x E3 - (E 1 0 E2) O E3
defined by
(x, y, z)-(x®y)Oz.
In view of the factorization property, there is induced a linear map
f:E1 OE20E3 -(E1 OE2)OE3
such that
f (x Q y Q z) = (x xQ y) xQ z.
(1.14)
On the other hand, to each fixed z e E3 there corresponds a bilinear mapping
I3Z : E 1 x E2 - E 1 Q E2 Q E3 defined by
IZ(x, y) = x Q y Q z.
The mapping f3Z induces a linear map
gZ:EIOE2-E1OE20E3
such that
gZ(x®y) = x®y®z.
(1.15)
Define a bilinear mapping
fr: (E 1 0 E2) x E3 - E 1 0 E2 O E3
by
,&(u, z) = gZ(u)
u e E 1 Q E2, z E E3 .
(1.16)
Then /i induces a linear map
g:(E1 OE2)OE3-E1 OE20E3
such that
U(u, z) = g(u Q z)
u e E 1 Q E2, z e E3.
(1.17)
Combining (1.17), (1.16), and (1.15) we find
g((xOy)Oz) = U(xOy,z) = gZ(xOy) = xQyQz.
(1.18)
Equations (1.14) and (1.18) yield g f (x O y 0 z) = x O y O z and
fg((x O y) O z) = (x O y) O z showing that f is a linear isomorphism of
E 1 Q E2 Q E3 onto (E 1 Q E2) Q E3 and g is the inverse isomorphism.
1 Tensor Products
28
In the same way a linear isomorphism h : E Q E2 Q E3 --> E Q (E2 Q E3)
1
1
can be constructed such that
h(x Q y Qx z) = x 0 (y O z).
Hence, h o f -' is an isomorphism of (E1 Q E2) Q E3 onto E1 Q (E2 Q E3)
such that (x®y)QzHx®(y®z).
More generally, if Et (i = 1. , p + q) are p + q vector spaces then there
exists precisely one vector space isomorphism
f:(E1
...
Ep)
E1 O ... O Ep+q
(Ep+1 O ... O Ep+q)
such that
f ((x 1
®... ® xp) ® (xp + 1
... O x,+)) = x 1
®... ® x p + q -
It follows that there is a uniquely determined bilinear mapping
f3:(E1 Q - - - Q Ep) x (Ep+ 1 O ... Q Ep+q) -+ E1 O ... O Ep+q
such that
fJ(xl ®" . ® xp, xp+ 1 ® .. ® xp+q)
x1 Qx
Qx xp+q
.
and that (E 1 O
O Ep + q , /3) is the tensor product of E 1 ®... O Ep and
Ep+ 1 ® ... ® Ep+qThe theory developed for the case p = 2 carries over to the general case
in an obvious way, and the reader will have no difficulty in making (and
proving) the generalization himself. In particular, he should verify that if
ais a basis for Et (i = 1, ... , p) then the products ail Q - Q a p form a
basis for E1 O - O Es,, and then generalize the results of Section 1.14,
obtaining in the finite-dimensional case that
-
dim L(E1, ... , Ep; G) = dim E1
dim L(E 1,
..., E,,) =
dim E 1
-
-
- dim G
- dim E.
The tensor product of several linear maps can be defined in the same way
as in the case p = 2. If Bpi : Ei -+ Fi (i = 1, ... , p) are linear maps then there
exists precisely one linear map
such that
x(xl ® ... ® xp) = (Plxl
®... O opxp
xt e E,.
As in the case p = 2 we shall write x = P 1 Q - .. Q 'pr. As in that case, the
Q cop induces an injection
mapping (p1, ..., cP p) H P 1 Q
L(E1; F1) O ... O L(Ep; Fp) -+ L(E1 O ... O Es,; F1 O - . O Fr).
If
Gi
(i = 1, ... , p)
Dual Spaces
29
is another system of linear maps it follows from the above definition that
('l1
®---®'l
p)°(®---®
i
) = ('l1°
1)®'--®(4ip°
An argument similar to the one given for p = 2 shows that
Q...QIm(pP
O---® pp)=
and
P
Ei O - - ker Pt Q
ker(rpi O - - O Pp) _
-
O Ep-
i= 1
PROBLEM
Let EL be p vector spaces.
(a) Prove that x 1 Q
Q xp = 0 if and only if at least one x = 0.
(b) Assuming that x 1 O . O xp 0 prove that
x1O...Oxp=y1O...Oyp
if and only if y _
(i = 1, ... , p) and Al
Ap = 1.
Dual Spaces
1.21. Bilinear Mappings
Let two triples of vector spaces E, E', E" and F, F', F" be given and consider
two bilinear mappings
p : E x E' -+ E" and tai : F x F' -+ F".
Then there exists precisely one bilinear mapping
x:(E®F) x
such that
x(xOy,x'Oy')_ co(x,x')O'l,(y,y')
xeE, x'EE', yeF, y'EF'. (1.19)
Since the spaces E Q F and E' Q F' are generated by the products x Q y
and x' Q y' respectively it is clear that if x exists it is uniquely determined by
p and ,,Ii. To prove the existence of x consider the linear maps
f:E®E'-E" and g:FQF'-F"
induced by p and /, respectively. Then f Q g is a linear map of (E Q E') Q
(F Q F') into E" Q F". Now let
S:(EQF)O(E'OF')4(EOE')O(FOF')
1 Tensor Products
30
be the linear isomorphism defined by
S:(x®Y)O(x'OY')-+(x®x')O(YOY')
and define a bilinear mapping x by
x(u, u) _ (f ® g)S(u ® u)
u e E ® F, u e E' ® F'.
Then it follows that
x(x®Y, x'®Y') _ (f®g)((x®x')®(Y®Y'))
=f(x®x')®g(Y®Y')
_ P(x, x') ® U(Y, Y')
We shall denote x by p ® U, the justification for this notation being given in
problem 1 at the end of Section 1.26.
1.22. Bilinear Functions
In particular, every pair of bilinear functions 1 and 'P in E x E' and F x F'
induces a bilinear function 1 ® 'P in (E ® F) x (E' ® F') such that
(I ® `I')(x ® Y, x' ® Y') _'(x, x')`I'(Y, y').
We shall show that 1 ®'P is nondegenerate if and only if and 'P are both
nondegenerate.
Consider the linear maps
q: E -+ L(E')
l//: F -+ L(F')
x : E ® F -+ L(E' ® F')
which are determined by
IbY' _ t'(b, y')
Pa x' _ I (a, x')
x z' _ (I ® `I') (c, z') (1.20)
(Here sp(a), fi(b), and x(c) are denoted by Pa' 'i6, and xe). Then we have
p ® U : E ® F -+ L(E') ® L(F'). On the other hand, L(E') ® L(F') may be
considered as a subspace of L(E' ® F') (Corollary III to Proposition 1.16.1).
It will be shown that
(1.21)
where i denotes the injection of L(E') ® L(F') into L(E' ® F'). By definition
we have
('P
='Pa ® I'6
aEE,bEF
and thus
1('P ® tIJ)a ®6(x' ® Y') _ 'Pa x' ' b Y' _ 't'(a, x')`I'(b, y').
On the other hand it follows from (1.20) that
xa ®6(x' ® Y') _ ('Ia ® `I') (a ® b, x' ® Y') _ 't'(a, x')`I'(b, y')
Dual Spaces
31
and so we obtain
l(p ® 41)a®b = xa®b
aeE,beF
whence (1.21).
Since i is an injection, it follows from (1.12) that
ker p ®
ker x =
ker SIi.
Hence, the nullspaces NE(I), NF(tP), and NE ®F(' ®'F) (see Section 2.21
of Linear Algebra) are connected by the formula
NE ®F('t' ® "F) = NE(B) ® F + E O NF('f).
(1.22)
In the same way it is shown that
NE, ®F,(t' ® "F) =
F' + E' ® NF-("F).
(1.23)
Formulas (1.22) and (1.23) imply that 1 ® "F is nondegenerate if and only
if 1 and "F are nondegenerate.
Suppose now that E*, E and F*, F are two dual pairs and let both scalar
products be denoted by <p>. Then the above results show that there exists
precisely one bilinear function <,> in E* ® F*, E ® F such that
<x* O Y*, x O Y> = <x*, x>
<Y*, y>
(1.24)
and this bilinear function is again nondegenerate. In other words, if E*, E
and F*, F are dual pairs, then a duality between E* ® F* and E ® F is
induced.
Next, consider the special case F = E* and F* = E. Then we obtain a
scalar product in the pair E* ® E, E ® E* defined by
<x* ® x, y ® y*> = <x*, Y> <Y*, x>.
Since the mapping x* ® x F-* x ® x* is an isomorphism of E* ® E onto
E ® E*, this scalar product determines a scalar product in the pair E ® E*,
E ® E* given by
<x O x*, Y ®Y* = <x*, ,> 3,*, x>.
.
(1.25)
Hence the space E ® E* can be considered as dual to itself. It is clear, moreover, that this scalar product is symmetric.
Now suppose that E*, Et (i = 1, ... , p) are pairs of dual vector spaces
and let all the scalar products be denoted by <,>. As in the case p = 2 a scalar
® Ep such that
® Ep and E1 ®
product is induced between E1 ®
<x* 1 ® ... ® x*p, x 1
®... ® X,> = <x* 1, X1> ... <x*p, xp>,
1 Tensor Products
32
1.23. Dual Mappings
Let Et, E* and Ft, F* (i = 1,2) be four pairs of dual vector spaces, and let
co:E1-+E2
q*:Ei
i/i : F 1 -+ F2
,* : F i -- F2
E2
and
be two pairs of dual mappings. Then the mappings
p®iji:E1 ®F1 -'E20F2
and
are dual with respect to the induced scalar products.
In fact, let x 1 E E 1, y 1 E F 1, 4 E E, and y2 E F2 be arbitrarily chosen
vectors. Then we have
<4 O y, (px 1 O 'Y 1> = <4 (px 1> <YZ , '/'Y 1
= <(p*x2 xi><I1f*YZ, Y1
= <cP*x2 O U*YZ , xi O Y1
whence
((p O /j)* _ (p* O /j*.
1.24. Example
Consider the dual spaces E* = L(E) and F* = L(F). Then the induced
scalar product in L(E) Q L(F) and E Q F is given by
<f 0 9, x 0 Y> = f(x)9(y).
On the other hand, the space L(E Q F) is dual to E Q F with respect to the
scalar product
< h, x Q y> = h(x Q y)
h e L(E Q F).
(1.26)
Now consider the injection
i : L(E) Q L(F) -+ L(E Q F)
defined by
i(f O 9) (x O Y) = f(x)9(y).
Then formulas (1.26), (1.27), and (1.24) yield the relation
< i(f O 9), (x O )> = f(x)g(y) = <f O 9, x O Y
showing that the injection i preserves the scalar products.
(1.27)
Dual Spaces
33
1.25. Inner Product Spaces
An inner product in a vector space E is a nondegenerate symmetric bilinear
function ( , ) in E. So in particular, every Euclidean space is an inner product
space.
Now let E and F be inner product spaces and denote both inner products
by (, ). In view of Section 1.22, there is precisely one bilinear function ( , ) in
E Q F satisfying
(x1 Ox Y1 x2 O Y2) = (x1, x2) . (Y1' Y2)
Clearly, this rbHinear function is again symmetric. Moreover, it is nondegenerate as follows from Section 1.22. The inner product space E Q F
so obtained is called the tensor product of the inner product spaces E and F.
Now assume that E and F are Euclidean spaces of dimensions n and m
respectively. Choose orthonormal bases a (v = 1, ..., n) and bµ (u = 1, ... , m)
in E and F. Then we have
(a Q bµ, a Q bK) = Sv
S/lK .
Thus the products a O bµ form an orthonormal basis of E Q F. In particular,
E Q F is again a Euclidean space.
1.26. The Composition Algebra
Let E*, E be a pair of dual vector spaces. Define a multiplication in the space
E* Q E by setting
(x* O x) ° (Y* O Y) = <x*, Y>(Y* O x)
It is easy to verify that this multiplication makes E* Q E into an associative
(noncommutative) algebra called the composition algebra.
Next, consider the linear map T : E* Q E - L(E; E) given by
T(a* Q b)x = <a*, x>b
x E E.
Since
T [(ai O b1) ° (a2 O b2)] = T(ai O b1) ° T(a2 O b2),
T is an algebra homomorphism. We show that T is injective. In fact, assume
that T(z) = 0, z e E* Q E. Choose a basis {ej of E. Then z is a finite sum
r
z=
a* Q e where a* E E*
v= 1
(see Lemma 1.5.2). Then, for every x e E,
r
<a*, x>e,, = 0
1 Tensor Products
34
whence
a *, x> = 0
x E E.
This implies that a* = 0 (v = 1, ..., r) and so z = 0. Hence T is injective.
For the further discussion we distinguish two cases:
Case 1. dim E < oo. Set dim E = n. Then dim(E* Q E) = n2 = dim L(E; E)
and so T is a linear isomorphism (since it is injective). Thus, = T -'(l) is
the unit element of the composition algebra. It is called the unit tensor of E.
To obtain a more explicit expression for the unit tensor let {e*v},
(v = 1, ..., n) be a pair of dual bases for E* and E and consider the element
i e* Q Then we have
n
T
e*v Q
n
e) (x) =
v= 1
x>e = x
x E E,
v= 1
whence
n
Oev=.
v=1
e* Q e is independent of the choice of the dual
In particular, the sum
bases {e*v},
Next observe that the spaces E* Q E and L(E; E) are self-dual with respect
to the scalar products given by
<x* Qx x, y* OO Y> = <x*, y> <Y*, x>
and
a, i> = tr(a o /3)
a, /3 E L(E ; E)
respectively. A simple calculation shows that
< T(x* Q x), T (Y* O y)> = x* ® x, y* ® Y>
and so T preserves the scalar products.
Case 2. dim E = oo. Then the composition algebra does not have a unit
element. To prove this, assume that a is a unit element of E* U E. Let {ej
be a basis of E. Then a is a finite sum
r
a* Q e with a* E E*.
e=
v= 1
For x* E E* and x e E,
r
(a* x0
e o (x* xQ x) =
o (x* O x)
v= 1
r
_
r
<a* x> (x* O ev) _ >2v(x* O ev)
v= 1
v= 1
Finite-Dimensional Vector Spaces
35
where V = <a*, x>. Since a is a unit element, it follows that
r
v(x* O ev) = x* O x,
v= 1
whence
r
x=
v= 1
Thus the vectors e1, ..., er generate E and so E has finite dimension.
It follows from the result above that, whenever dim E = oo, the linear
map T is not an isomorphism and hence not surjective.
PROBLEMS
1. Given six vector spaces E, E', E" and F, F', F" consider the bilinear mapping
y:B(E,E';E") x
defined by
/i) : (x O Y, x' O Y') H q (x, x') O /i(Y, y').
Show that the pair (Im y, y) is the tensor product of B(E, E'; E") and B(F, F'; F").
2. Let E, E* and F, F* be two pairs of dual vectors spaces and consider subspaces
E and F 1 c F.
E1
(a) Show that a nondegenerate bilinear function is induced in
(E* Qx F*)/T(Ei, F±)
and
E1 Ox F1,
where
T(Ei, Fi) = Ei Ox F* + E* Ox Fi
by the scalar product in E* Q F*, E Q F.
(b) Prove the relation
(E1 Ox F1)-L = Ei Ox F* + E* Ox Ff.
3. Given a pair of dual transformations q : E -- E, p* : E*
transformation (p Q (p* is self-dual.
E* prove that the linear
Finite-Dimensional Vector Spaces
For the remainder of this chapter all vector spaces will be assumed to have
finite dimension.
1.27.
Let E and F be vector spaces of dimension n and m respectively. Then E Q F
has dimension nm (see Section 1.13).
1 Tensor Products
36
Proposition 1.27.1. Let gyp: E x F -+ T be a bilinear mapping where dim T =
nm. Then conditions ®i and ®2 for p are equivalent.
PROOF. Consider the induced linear map
f : E ®F -+ T.
Then, if 'p satisfies ®l, f is surjective, (Proposition 1.8.1). Since dim T =
nm = dim(E ® F), it follows that f is an isomorphism and so 'p satisfies
®2.
On the other hand, if 'p satisfies ®2, then f is injective (Proposition 1.8.1),
and hence it must be a linear isomorphism. Thus 'p satisfies ®l.
Next, let p: E -+ E' and l//: F -+ F' be linear maps where dim E' = n'
and dim F' = m'. It has been shown in Section 1.16 that the bilinear mapping
f3:L(E;E') x L(F;F`)-+L(EQF;E'QF')
given by 'p x i/i -+ 'p ® i/i satisfies ®2. In the finite-dimensional case we
have
dim L(E Q F; E' ® F') = (nm) (n'm') = (nn') (mm')
= dim L(E ; E') dim(F ; F').
Hence, by the proposition above, /3 satisfies ®1 as well. Thus /3 has the universal property and we may write
L(E ®F; E' ®F') = L(E; E') ®L(F; F').
This yields for E' = F' = r
L(E ® F; r) = L(E; r) ® L(F; r),
that is,
(E ® F)* = E* ® F*.
Thus the tensor product of linear functions f and g in E and F is the linear
function in E ® F given by
(f O g)(x O y) = f (x) . g(y)
x e E, y e F.
1.28. The Isomorphism T
Let E* be a dual space of E and consider the linear map
T : E* ®F-+ L(E; F)
given by
T(a* ® b)x = <a*, x>b
x e E.
Finite-Dimensional Vector Spaces
37
We show that T has the universal property. In fact, we have the commutative
diagram
E* Q F
L(E ; F)
L(E) Q L(r; F)
where a: E* - L(E) is the canonical isomorphism, /3F -+ L(r; F) is the
isomorphism given by f3() = y, y e F, e r, and O is the map defined in
Section 1.16. Since the bilinear mapping Q has the universal property
(Proposition 1.27.1), the same is true for T. Thus we may identify E* Q F with
L(E; F) under T.
A straightforward computation shows that
/joT(a*®b)= T(a* Q t/ib)
i/i e L(F; E)
= T(t/i*a* Q b)
U e L(F; E).
and
T(a* Q b) o
In particular, if a e E, a* e E*, b e F, b* e F*, then
T(b* O a) o T(a* Q b) = <b*, b>T (a* Q a).
(1.28)
Finally, consider the trace form tr : L(E, F) x L(F, E) -+ r given by
(gyp x U) -+
o U). We show that the operator T satisfies
tr(T(b* O a) o T (a* O b)) = <a* Q b, b* Q a> = <a*, aX b*, b>.
(1.29)
In fact, Formula (1.28) yields
tr(T(b* Q a) o T(a* Q b)) = <b*, b>tr T(a* Q a).
But since the linear map T(a* O a) is given by T(a* Q a)x = <a*, x>a,
we have
tr T(a* O a) = <a*, a>
(1.30)
and so (1.29) follows.
Formula (1.29) implies in particular, that the trace form is nondegenerate
(cf. Section 1.22).
1.29 The Algebra of Linear Transformations
To simplify notation we use the isomorphism T of the last section to identify
a* Q a with the corresponding linear transformation.
Consider the associative algebra A = A(E; E) of the linear transformations p : E --> E. A linear map cZ : A Q A -+ L(A ; A) is defined by (a, /3) H
cZ(a Q /3) where cZ(a Q /3) is the transformation defined by
cZ(oc O /3)(P = a ° P ° /3
(1.31)
1 Tensor Products
38
Proposition 1.29.1. cZ is an isomorphism.
PROOF. We recall that the space A is dual to itself with respect to the trace
form <U, p> = tr(1i o gyp). Now 1'et F : A x A -+ L(A ; A) be the linear map
defined by
F(a O
<a, p>/3
Then the mappings F and cZ are connected by the relation
S = F o Q,
(1.32)
where Q is the linear automorphism of L(A Q A) defined by
Q((a * O a) O (b* O b)) = (a * O b) O (b* O a)
To prove (1.32) it is sufficient to show that
O a) O (b* O b)) = F((a* O b) O (b* O a)).
(1.33)
Let p : E -+ E be an arbitrary linear transformation. Then we have, in view
of the results of Section 1.28, that
O a) O (b* O
(a* O a) ° (P ° (b* O b) = (a* O a) ° (b* O cob)
= <a*, cpb>b* Q a
= <<p*a*, b>b* Q a
and
F((a* O b) O (b* O a))co = <a* O b, p>b* Q a
= <<p*a*, b>b* Q a
whence (1.33).
In view of Section 1.28, F is a linear isomorphism. Since Q is a linear
automorphism of A Q A, relation (1.32) implies that S is an isomorphism.
Corollary. Suppose c and f3. (i = 1, ..., r) are elements of A such that the c
are linearly independent. Then the relation
al P
f3i = 0 for every (p E A
implies that f3i = 0 (i = 1, ..., r).
1.30. The Endomorphisms of A
Every linear automorphism a of E induces an endomorphism ha
algebra A given by
0 of the
hp =ao(poa-'.
It will now be shown that conversely, every nonzero endomorphism of the
algebra A is obtained in this way. In other words, every endomorphism
Finite-Dimensional Vector Spaces
39
0 of the algebra A can be written in the form hcp = a ° P ° x' where a
is a regular linear transformation of E.
h
Since the pair (L(A; A), SZ) is the tensor product of A and A, we can write
r
hcp= at°cp°fJ1
ai,f1EA,
i=1
where the ai and the /3i are linearly independent.
Now the relation
h(cp ° U) = hcp ° hi/i
implies that
i
°p°(i/i°f3i - fi°j°i/j°fj)
= 0.
j
Since the ai are linearly independent and p is an arbitrary element of A
it follows that (see the Corollary to Proposition 1.29.1)
- f3i°a;°l//°i3. = 0
(i= 1, ... , r).
This can be written in the form
- fi°a.)°ii°fj = 0
(1= 1, ... , r).
Now the linear independence of the elements fl; implies that
(i,j = 1,...,r).
For j
(1.34)
i we obtain from (1.34)
and for j = i
fi ° ai = 1.
These relations are compatible only if r = 1. Denoting a1 by a we obtain
hcp = ao (p°a-'.
It is easy to show that the element a is uniquely determined by h up to
a constant factor.
Our result shows, in particular, that every endomorphism h
0 of the
algebra A preserves the scalar product <iU, 'p> = tr(i1i ° gyp). In fact, for every
two elements (p, l// E A we have
<h(, hi/i> = tr(h( ° h U)
= trh(cp°I,I/) = tr(a°(°I,Ii°a-')
= tr((° iU) _
whence
<'p,'/'>
(p, I/(E A.
40
1 Tensor Products
PROBLEMS
1. Let a* E E* and b e F be two fixed vectors and consider the linear maps a* Q b : E --> F
and b Q a * : E*
F*. Prove that
b Q a* _ (a* Q b)*.
2. Let E, F be Euclidean spaces and consider the induced inner product in E Q F. Given
two linear transformations p : E -+ E, : F -+ F; prove that
(a) p Q i/i is a rotation if and only if p = AtE and _ A-1iF where 'CE and cF are
rotations of E, F and A 0 is a real number.
(b) p Q i/i is selfadjoint if and only if both transformations p and l.i are selfadjoint or
skew.
(c) p Q i/i is skew if and only if precisely one of the transformations is selfadjoint and
the other one is skew.
(d) p Q /i is normal if and only if both transformations are normal.
3. Let E be a real n-dimensional vector space and consider two regular transformations
p and /i of E. Given an orientation in E Q E prove that
(a) if n is even then p Q Ii preserves the orientation.
(b) if n is odd, then p Q /i preserves the orientation if and only if both mappings p
and c1' are orientation preserving or orientation reversing.
Tensor Products of Vector Spaces
with Additional Structure
Tensor Products of Algebras
2.1. The Structure Map
If A is an algebra, then the multiplication A x A -> A determines a linear
map µA:A Qx A -> A such that
µA(x O Y) = xy.
(2.1)
µA is called the structure map of the algebra A. (In this chapter the symbol µA
will be reserved exclusively for structure maps. Since such a notation appears
in no other chapters, there is no possibility of confusion.) Conversely, if A is a
vector space and µA: A Q A -p A is a linear map, a multiplication is induced
in A by
xY = µA(x O Y)
(2.2)
and so A becomes an algebra. The above remark shows that there is a 1-1
correspondence between the multiplications in A and the linear maps
µA:AOA-A.
Now let B be a second algebra and µB : B Q B -+ B be the corresponding
structure map. If gyp: A -+ B is a homomorphism we have
(ThµA(x O Y) = I B((PX O SPY) = I B((P
O pXx O Y)
whence
'P ° µA = µB ° ('P O 'P)
(2.3)
Conversely, every linear map gyp: A -p B which satisfies this relation is a
homomorphism.
41
42
2 Tensor Products of Vector Spaces with Additional Structure
2.2. The Canonical Tensor Product of Algebras
Let A and B be two algebras with structure maps µA and µB respectively.
Consider the flip-operator
S:(AQB)Q(AQB)--(AQA)Q(BQB)
defined by
S(x1 O Y1 ® x2 ® Y2) = x1 ® x2 ® Y1 ® Y2
Then a linear map
µA®B.(AQB)Q(AQB)-+AQB
defined by
µA ®B - (µA ®µB) ° S
determines an algebra structure in A ® B. The algebra A ® B is called the
canonical tensor product of the algebras A and B. It is easily checked that the
multiplication in A ® B satisfies
(2.5)
(x1 ® Y1)(x2 ® Y2) = xlx2 ® Y1Y2
This formula shows that a canonical tensor product of two associative
(commutative) algebras is again associative (commutative). If A and B have
unit elements IA and IB respectively then IA ® IB is the unit element of A ® B.
If B has a unit element IB we can define an injective linear mapping (p : A -+
AQBby
(px = x Q I B
x e A.
It follows from (2.5) that
(p(xx') = xx' ® I B = (x ®I B)(x' ® I B) = (ox (px',
x, x' e A,
i.e., (p preserves products and so it is a monomorphism.
2.3. Tensor Product of Homomorphisms
Let A 1, B 1, A2, B2 be algebras and suppose that
(P1 Al -' B1
(p2:A2 -+ B2
are homomorphisms. Then (i 1 ®(p2 is a homomorphism of the algebra
A 1 ® A2 into the algebra B 1 ® B2. In fact, since
((p l ®1 ®(P2 ®(p 2) ° S = S ° ((p 1 ®(P2 ®(Pl ® 'P2)
43
Tensor Products of Algebras
it follows that
0 cP2) ° µA1 ®A2 = [(cP1 ° µA1) 0 (cP2 ° µA2)] ° S
= [(µB1 °
O P2)] ° S
O P1)) O µB2 °
= [(µB1 O µB2) ° S] ° ((P1 ®42 ® Pi ®(P2)
0 (P2)]
= µB1 ®B2 ° [('P1 0 (P2) 0
This equation shows that P1 0 P2 is a homomorphism.
Now consider two involutions WA and WB in A and B respectively. Then
the mapping
W A ®B = WA 0 WB
is an involution in the algebra A 0 B.
2.4. Antiderivations
Let WA be an involution in the algebra A and SZA be an antiderivation with
respect to WA; i.e.,
SZA(x . y) = AX y + WAX
AY
(2.6)
In terms of the structure map, (2.6) can be rewritten as
A ° µA = µA °
0 l + WA 0 SZA)
(2.7)
To simplify notation we write
A O l+ WA 0 A = A ® A
and then (2.7) reads
A°µA =
Now let B be a second algebra and SZB be an antiderivation of B with respect
to an involution WB,
B ° µB = µB ° B ® B
Consider the linear maps
A®B 0 lA®B + WA®B 0 A®B
and
(A ®A) ®(B ®B) = SZA ®B 0 1B ®B + WA ®A 0 SZB ®B .
(2.12)
44
2 Tensor Products of Vector Spaces with Additional Structure
On the assumption that
So (A®B)®(A®B) =
°S
(2.13)
(this always holds if WA = 1A, WB = 'B), Equations (2.10), (2.11), (2.12), and
(2.13) imply that
µA ® B °
(A ® B) ® (A ® B) = (h A ® µB) ° S ° SZ(A ® B) ® (A ® B)
C UA ® µB) ° (A ® A) ®(B ® B) ° S
CUA ® JIB) ° (AA®A ® lB®B + WA®A
S
CUA ° A®A ® µB + µA ° WA®A ® µB ° B®B) ° S
= (cZA ° µA ®µB + WA ° JA QX cZB ° µB) ° S
= (cZA Q lB + WA Q SZB) ° (hA ® JIB) ° S
A®B °µA®B
This relation shows, in view of (2.8), that SZA ® B is an antiderivation in the
algebra A Q B with respect to the involution WA ® B
Tensor Products of G-Graded Vector Spaces
2.5. Poincare Series
Let E _ > E, a e G and F =
F, fi e H be respectively G- and H-
graded vector spaces. Then a (G O H)-gradation is induced in the space
EQFby
EQF= Ea®Ff.
(2.14)
a, fJ
If H = G, then (2.14) is a G-bigradation. The corresponding (simple) Ggradation is given by
EQF = (EQF)y,
(EQF)y =
Ea®FJ.
(2.15)
y
The space E Q F, together with its G-gradation is called the tensor product of
the G-graded spaces E and F.
It follows from (2.15) that for every two homogeneous elements x e Ea,
y e Ffl the element x Q y is homogeneous and
deg(x Q y) = deg x + deg y.
In particular, the linear isomorphism f : E Q F -+ F Q E given by
f(x® y) = y ® x
is homogeneous of degree zero. Moreover, we have as well that each (E Q F)y
is linearly generated by homogeneous decomposable elements of the form
x®y,xeEa,yeFf,a+/3= y.
45
Tensor Products of G-Graded Vector Spaces
Let E, E', F, F' be G-graded vector spaces, and consider homogeneous
linear maps
and
cP : E -- E'
/i : F -- F'
of degrees k and l respectively. Then (p ® U : E ® F -+ E' ® F' is homogeneous of degree k + 1. In fact, if x and y are homogeneous elements of
degree a and /3 respectively it follows that
deg((p ® t/ixx ® y) = deg((px ® ,y)
= deg cpx + deg ty
=a+k+f3+1
=(a+/3)+(k+1)
and hence p ® U is homogeneous of degree k + 1.
Now assume that G = 7l and that the gradations of E and F are positive
and almost finite. Then the Poincare series of E ® F is given by
dim(E ® F)k tk.
PE ®F(t) =
k
Since
E. ® Fj =
dim(E ® F)k = dim
i+j=k
dim Et dim Fj,
i+j=k
the above formula reads
PE ® F(t) =
dim Ei t` dim Ej tj = PE(t) PF(t)
k i+ j=k
showing that the Poincare series of E ® F is the product of the Poincare
series of E and F.
2.6. Tensor Products of Several G-Graded Vector Spaces
Let E. = a E", a e G be G-graded vector spaces. Then a G p-gradation is
induced in the space E = E 1 ® ® Ep by assigning the degree (a 1, ... , a p)
to the elements of E;1 ® ® E 7p. The corresponding simple G-gradation is
given by E = a E G Ea where
the sum being extended over all p-tuples (a 1, ... , a p) such that cc 1 +
+ ap = a. The space E together with this gradation is called a tensor product
for the G-graded spaces E,. It follows from the definitions that
deg(x 1 O ... O xp) = deg x i + ... + deg xp
46
2 Tensor Products of Vector Spaces with Additional Structure
for every p-tuple of homogeneous elements x1. As another immediate consequence of the definition we note that the isomorphism f: (E1 Q E2) Q E3 -+
E1 Q (E2 Q E3) defined by
f:(x1 0x2)Ox3 -+x1 0(x2 Ox3)
is homogeneous of degree zero.
It is easy to verify that if El, F1(i = 1, ... , p) are graded spaces, and if
1: E1 -+ F1 are homogeneous of degree kl then the map p1 O
O co is
homogeneous of degree p=1 kl.
Suppose now that G = 7l and that all the gradations of the E1 are positive
and almost finite. Then clearly the induced gradation in E is again positive
and almost finite. Moreover, the Poincare series of E is given by
PE(t) =
PE1(t)... P(t).
The proof is similar to that given for p = 2.
2.7. Dual G-Graded Spaces
Let E = >JizeG Ea, E* = >jzeG Ea and F = >fleG Ff, F* = >fleG F be two
pairs of dual G-graded vector spaces and consider the spaces
a, fJ
Ea Q F
E* Q F* =
a, fJ
as G-bigraded vector spaces. Then the induced scalar product between E Q F
and E* Q F* respects the G-bigradations. In fact, for any vectors x e Eat,
x * E E, y e Ff 1, y* E Fwe have
<x* O y*, x O y> = <x*, x><y*, y> = 0
unless a 1 = a2 and f 1 = I2 . As an immediate consequence we have that the
G-graded spaces E Q F, E* Q F* are dual G-graded spaces.
2.8. Anticommutative Tensor Products of Graded Algebras
Let A = L Ap and B = >q Bq be two graded algebras. Consider the anticommutative flip operator
Q:(AOB)O(AOB)-'(AOA)O(BOB)
defined by
Q(x®y®x'Oy')=(-1)' x®x'Oy®y',
47
Tensor Products of G-Graded Vector Spaces
where deg x' = p' and deg y = q. Then the linear map
µA®B:(AQ B)Q(AQB)-AQB
defined by
µA ®B =(µA OO 1 B) ° Q
determines an algebra structure in the graded vector space A Q B. The
resulting algebra, A Q B, is called the anticommutative tensor product or the
skew tensor product of A and B. The multiplication in A Q B is given by
(x O Y)(x' O Y') = (-1)1 'qxx' Q yy'
p' = deg x', q = deg y. (2.16)
If A and B are algebras without gradation, then by the tensor product of
A and B we shall mean the canonical tensor product (Section 2.2). If, in this
chapter, A and B are graded algebras, then by the tensor product of A and B
we shall mean the anticommutative tensor product. Observe that the underlying vector spaces of the algebras A Q B and A Q B coincide.
Now it will be shown that A Q B is a graded algebra. In fact, if x 1 E AP 1,
x2 E AP2 , y 1 E Bq 1, Y2 E Bq2 are arbitrary we have
(x1 O y1)(x2 O Y2) = (- 1)P2glx1x2 O YlY2
Since A and B are graded algebras, it follows that
deg(xlx2) = pl + p2
deg(Y1Y2) = q1 + q2
In view of the definition of the gradation in A Q B (Section 2.5) we obtain
that (x 1 O y 1)(x2 O Y2) is homogeneous of degree Pi + P2 + q1 + q2 and
hence A Q B is a graded algebra.
It is easy to verify that if C is a third graded algebra then the linear map
f:(AOB)QC-AQ(BQC)given by
f:(x®y)Ox z - xOx (y Ox z)
preserves products and hence is an isomorphism.
The anticommutative tensor product of two anticommutative graded
algebras is again an anticommutative algebra. In fact, let x E AP, y E Bq,
x' E A,,, and y' E Bq- be homogeneous elements. Then we have
(x O Y)(x' O Y') = ((
1)1qxx' O yy'
1)P'q+P'P+q"x'x O y'y
= (- 1)P'q + PP' + qq' + Pq'(x' ® y')(x O y)
= (, 1)(P +
q')(x' O Y')(x O Y)
The reader should observe that the canonical tensor product of anticommutative graded algebras is not in general an anticommutative algebra.
48
2 Tensor Products of Vector Spaces with Additional Structure
2.9. Homomorphisms and Antiderivations
Let C = >r Cr and D = > DS be two more graded algebras and assume that
cP : A - C,
t/i : B - D
are homomorphisms homogeneous of even degree k and l respectively. Then
(p ® U is a linear map homogeneous of degree k + 1. Moreover, we have
as is easily checked. By the same argument as that used in Section 2.3 it
follows that (p ® U is a homomorphism of A ® B into C ® D.
In particular, if WA and WB are involutions in A and B, homogeneous of
degree zero, then WA ® B = WA ®wB is an involution of degree zero in A ® B.
As a special case, suppose that WA and WB are the canonical involutions in
A and B. Then WA ® B is the canonical involution in A ® B (see Section 6.6 of
Linear Algebra).
Proposition 2.9.1. Let SZA and SZB be homogeneous antiderivations of odd degree
k. Then the mapping
AB = A ® 1B + WA
(where WA is the canonical involution) is an antiderivation, homogeneous of
degree k, in the graded algebra A ® B. If eA and eB are derivations of even
degree k then
eA ®B = eA OO LB + A ®0B
is a homogeneous derivation of degree k in A ® B.
PROOF. Clearly, SZA ® B is a homogeneous linear map of degree k. To show that
SZA ® B is an antiderivation we verify that
Q ° (A®B)®(A®B) = (A®A)®(B®B) ° Q
(2.17)
(see Section 2.4 for the notation). Then the same argument as that used in
Section 2.4 proves that SZA ® B is an antiderivation in A ® B. Similarly, to
prove the second part of the proposition we need only establish the formula
Q ° e(A®B)®(A®B) = e(A®A)®(B®B) ° Q
Now we proceed to the verification of (2.17) and (2.18).
Let
xeAp,
yEBq,
x'EAp-,
y'EBq-
(2.18)
49
Tensor Products of G-Graded Vector Spaces
be homogeneous elements. Then we obtain
Q (A ® B) ®(A ® B)(x OO Y OO X' O Y')
= Q[QAXOYOX'OY' + (-1)Px®QBYOx'OY'
+ (_ 1)P+,x ® y ® QA x' ® y' + (-1)P+q+P'x O y O x` O QBY']
(- 1)'Q x O x' O y O y' +(-1)P+q+(k+P')q.aG Q Ax' Q y Q y'
+ (_ 1)P+P (q+k)x O X' O BY O Y'
1)P+q+P,+P'qx
+ (-
O x' O y O By'
= (-1)" {QA x O x' + (_ 1)P + (k + 1)qx O A x'} O Y O Y'
+ (- 1)'(x O x') O {( -1)P+kP QBy Q y' +
Since k is odd we have
x O x' + (_ 1)P +(k + 1)qx O A x')
(-1)P+q+P
y Q QBy'}.
x O x' + (- 1)"x O A x')
=
= QA®A(x O x')
and
(-
1)P+kP
BY O Y' + (- 1)""y ® BY
=
= (-1)P
(-1)P+P
[ BY O Y' + (-1)y O BY]
+ P QB ®B(Y O Y')
Hence it follows that
Q
A ®B) ®(A ®B)(x OO Y OO x' O Y')
= (- 1)qP' LEA ®A(x O x')® Y O Y' + (-1)P + q (x O x')® B ®B(Y O Y')]
= (-
®A) ®(B ®B)(x OO x' O Y O Y')
= SZ(A ® A) ®(B ® B) Q(x OO Y Ox X' O Y')
whence (2.17).
Now let eA and eB be homogeneous derivations of even degree k. Then
we have
Q (A ® B) ® (A ® B)(x Ox Y OO X' OO Y')
= Q(OAx®YOx'OY' +
+ XOYOOAX'OY' + XOYOX'OOBY')
= (-1)P'q[OAXQx'Q y®y' + x®x'®y®0BY']
XOOBYOX'®Y'
+ (-1)q(P' + k)x ® eA x' O y O Y' + (-1)P' (q + k)x O X' O 0B Y O Y'
= (-1)P'q[OAx®x'Oy®y' +XOOAX'®y®Y'
+ XOX'OOBYOY' + XOx'OYOOBY'1
= (-1)P qe(A ®A) ®(B ®B)(x Ox x' Ox Y Ox Y')
= e(A ®A)
whence (2.18).
(B
B) Q(x OO Y OO x' OO Y')
2 Tensor Products of Vector Spaces with Additional Structure
50
Tensor Products of Differential Spaces
In Sections 2.10-2.17 the notations BE, B(E); ZE, Z(E); and HE, H(E) for the
boundary, cycle, and homology spaces of a differential space will be used interchangeably.
2.10. Tensor Products of Differential Spaces
Suppose that (E, aE) and (F, aF) are differential spaces. We wish to make
E ® F into a differential space. In order to do so we shall need an involution
w of E such that
(2.19)
Suppose we are given such an involution. Define D = E ®F by
aE®F = aE ®l + (U®
aF.
Then we have
aE ®F(x ® Y) = aE x ® y + wx ® 'FY .
(2.20)
From (2.19) we obtain that
aE®F=aE®l +waE®aF+aEw®aF+ l®aF
= (waE + aE w) ®aF
=0
and so (E ® F,
E ®F)
is a differential space. Formula (2.20) implies that
ZE ® ZF C
(2.21)
ZE®F.
Moreover
BE ® ZF C BE ®F
and
ZE ® BF
BE ®F '
(2.22)
In fact, if aE x e BE and y e ZF are arbitrary elements, then
EX ® Y = aE ®F(x ® Y)
Similarly, if x e ZE and aF y e BF, then
aE ®F(wx ® Y) = w2X ® 'FY = x ®aFY
It follows from Relations (2.21) and (2.22) that the bilinear mapping (E x F)
- E ® F induces a bilinear mapping p: HE x HF -+ HE ® F such that
(p(lrE Z 1,
F z2) = E ®F(Z 1 ® Z2)
Z1 e ZE, Z2 e ZF,
(2.23)
where ?rE, ?CF, and irE ®F are the canonical projections of the cycle spaces onto
the homology spaces.
51
Tensor Products of Differential Spaces
It is the purpose of this section to show that the pair (HE ® F' 'p) is the
tensor product of HE and HF. We first establish the formulas
ZE ®F = ZE ® ZF + BE ®F
(2.24)
(ZE ® ZF) n BE ®F = BE Q ZF + ZE Q BF.
(2.25)
Consider the linear operators D1: E Q F -+ E Q F and D2 : E Q F -+ E Q F
given by
D1 = aE Q
and D2 = a Q aF.
(2.26)
Then
D1 =D2=0,
D1D2+D2D1 =0,
(2.27)
and
D1+D2=D.
It follows from (2.26) and (1.11) that
ImDI=BEQF,
ImD2=EQBF,
and
Im(D 1 D2) = BE ® BF
whence, in view of (1.7),
Im(D1D2) = Im D1 n Im D2.
The kernels of D 1 and D2 are given by
ker D l= ZE Q F
ker D 2= E Q ZF
(cf. (1.12)) and so we obtain
ker D1 n ker D2 = ZE Q ZF.
Suppose now that z e ZE ® F is arbitrary. Then D 1 z = - D2 z and so
DIzeImD1 nImD2 = ImD1D2.
Let x e E O F be a vector such that D 1 D2 x= D 1 z. Then setting
y = z - (D1x + D2x)
we obtain
D1y = D1z - D1D2x = 0
and
D2y = D2z - D2D1x = -D1z + D1D2x = 0.
(2.28)
52
2 Tensor Products of Vector Spaces with Additional Structure
Thus
y E ker D 1 n ker D2 = ZF. ®ZF .
It follows that
Z=Dx+
yEBE®F+ZE®ZF
whence
ZE®F C ZE ® ZF + BE®F
Inclusion in the other direction is a consequence of (2.21) and so (2.24) is
proved.
Next, note that every element of BE ® F n (ZE ® ZF) can be written in the
form
Dx=Dix+D2x.
Then we obtain from (2.26) and (2.27)
D2(D1x) = D2Dx = 0
and
D1(D2x) = D1Dx = 0.
Hence
Dix e ker D2 n Im D1 = BE ® ZF
and
D2xekerD1 nImD2 = ZE®BF.
It follows that Dx E BE ® ZF + ZE ® BF ; i.e.,
BE ®F n (ZE ® ZF) C BE ® ZF + ZE ® BF
The inclusion in the other direction follows from (2.22) and hence (2.25)
is proved as well.
Now we are ready to prove that the pair (HE ® F, rp) is the tensor product
of HE and HF.
Let Q be the restriction of the canonical projection 1rE ®F . ZE ®F -' HE ® F
to the subspace ZE ® ZF. Then we have, in view of (2.24) and (2.25),
Im Q = HE®F
(2.29)
ker Q = BE ® ZF + ZE ® BF.
(2.30)
and
Consequently Q induces a linear isomorphism
Q : (ZE ® ZF)/Te(E, F) 4 HE ®F,
Tensor Products of Differential Spaces
53
where Ta(E, F) = BE ® ZF + ZE ® BF. Hence,
Qop=Q
(2.31)
where p denotes the canonical projection
p . ZE ® ZF -' (ZE ® ZF)/Te(E, F)
Consider the bilinear mapping
fJ: HE x HF -' (ZE ® ZF)/T,(E, F)
defined by
I (irE Z 1,
F Z2) = p(Z 1 ® Z2)
Z 1 E ZE, z2 E ZF.
(2.32)
Then Formulas (2.32), (2.31), and (2.23) yield
h'/(?rEZI, FZ2) = ap(Z1 ® Z2)
= Q(Z1 ® Z2)
= ?LE ®F(Z 1 ® Z2)
(p(?rEZ 1 ,
F Z2)
and so it follows that Th/i = rp. Hence we have the commutative diagram
F), i'/) is the tensor product for HE and HF
Since the pair (ZE ® ZF
(see (1.13)) and a is a linear isomorphism, it follows that the pair (HE ® F, co)
is also the tensor product of HE and HF. The result is restated in the
Kunneth theorem: Let (E, aE) and (F, aF) be two differential spaces and
(E ® F, aE ® F) be their tensor product. Then the pair (HE ® F, cp) is the tensor
product of HE and HF, where p : HE x HF -+ HE ® F denotes the bilinear
mapping induced by the bilinear mapping E x F -+ E ® F.
In view of the above theorem we may denote the mapping p by ®.
Then we have the relations
E®F(Z1 ® Z2) = EZ1
and
HE ®F = HE ® HF.
2 Tensor Products of Vector Spaces with Additional Structure
54
2.11. Tensor Products of Dual Differential Spaces
Suppose that (E, SE), (E*, aE) and (F, SF), (F*, aF) are two pairs of differential spaces dual with respect to scalar products <, > and suppose further
that w, w* is a pair of dual involutions in E and E* respectively. Then the
induced differential operators D, D* in E O F and E* O F* are given by
aF
and
D* = a*
i + w* Qx aF .
It follows from Section 1.23 that D and D* are again dual, and hence
(E O F, D) and (E* O F*, D*) are dual differential spaces.
As an immediate consequence we have that there is induced a bilinear
form 1 in HE* ® F* x HE ® F such that
1(pz*, irz) = <z*, z>
(2.33)
z* E ZE* ®F*, z E ZE ®F,
where
P ZE* ®F* -+HE* ®F*
and
7r : ZE ®F -+HE ® F
are the canonical projections. On the other hand, consider the bilinear
functions I 1 and 2 in HE* x HE and HF* x HF defined by
1 1(Plu*, ?r1u) = <u*, u>
u * E ZE*, u e ZE
= <U*, v)
U* E ZF*, v E ZF.
and
s,2(P2 U*, ?r2 U)
For the tensor product of the bilinear functions I 1 and '12 (see Section 1.22)
we obtain
(I O s2)(P1u* ® P2 u*, ir1u ® ir2 U) _
11(P1u*,
i 1u)12(P2
U*,
m2 U)
_
= (u* O u*, u O u>.
(2.34)
On the other hand, Formula (2.33) shows that
CP1u* O P2 u*, lt1u O ir2 u) = (p(u* O U*), ir(u O u))
= <u*
(2.35)
OQ u*, u Ox v>.
Comparing (2.34) and (2.35) we find that
2.
In particular, since I 1 and 1 2 are nondegenerate (see Section 1.22) so is 1.
In view of Section 6.9 of Linear Algebra the nondegeneracy of 1 is also derivable from the duality of D and D*.
Tensor Products of Differential Spaces
55
2.12. Graded Differential Spaces
Consider two graded differential spaces E = i E. and F = L F. We
shall assume that the operators aE and aF are homogeneous of odd degree k.
Then the canonical involution, w, defined by
wx=(-1)"x
xeEp
satisfies Condition (2.19). In fact,
(SEw + w aE)x = (- 1)p
1)P+k 3X = 0.
The induced differential operator D in E Q F is given by
D(x Q y) = aE x Q y+(-1)"x Q aF y x e E, y e F.
(2.36)
Clearly D is again homogeneous of degree k. In general, the induced
differential operator in the tensor product of graded differential spaces will
mean the operator defined with respect to the canonical involution, w, in E.
The gradations of E and F induce gradations in H(E) and H(F) respectively defined by
H(E) =
H.(E)
H.(E) = ic1Z1(E)
(2.37)
H (F)
H (F) = ir2 Z j(F).
(2.38)
i
and
H(F) =
J
Similarly, we have (in the induced simple gradation)
Hk(E Q F) = irZk(E Q F).
Hk(E O F)
H(E O F) =
(2.39)
k
Formulas (2.39), (2.37), and (2.38) yield in view of the Kunneth theorem
Hk(E O F) =
=
H1(E) O
k i+j=k
Hj(F)
Hi(E) xQ H (F).
(2.40)
Since
H.(E) O H(F) c H+ j(E
(E O F)
we obtain from (2.40) the Kunneth formula for graded differential spaces
H,(E) O H (F).
Hk(E O F) =
(2.41)
i+j=k
In particular, the gradation in H(E Q F) determined by the gradation
in E Q F coincides with the gradation obtained by identifying H(E Q F)
with the tensor product of the graded spaces H(E) and H(F).
2 Tensor Products of Vector Spaces with Additional Structure
56
Now assume that E and F are almost finite positively graded spaces.
Then so are H(E), H(F) and H(E Q F). Moreover, if
H(E)'
H(F)
and
PH(E ®F) are the corresponding Poincare polynomials, then
PH(E ®F) - PH(E) PH(F)
2.13. Dual Graded Differential Spaces
Suppose now that E* = (>t Et, 3) and F* = (L F;, 3) are two graded
differential spaces which are dual to the graded differential spaces E and F
respectively. Then aF, aE will be of degree - k and the canonical involution
w* of E* is dual to the canonical involution of E, (k = deg aE = deg SF).
It follows that the differential operator
D*=& ®i+(0*®a*
coincides with the differential operator in the graded space E* Q F* defined
in Section 2.11. Hence, (E Q F, D) and (E* Q F*, D*) are dual differential
spaces and graded spaces as well. Moreover, we have from Section 2.7
that E Q F and E* Q F* are dual graded spaces (i.e., the scalar product
respects the gradation). Thus these spaces are dual graded differential spaces.
PROBLEM
Let i? be a differential operator in a finite-dimensional vector space E. Define differential operators i1 and 2 in the space L(E; E) by
O1(p= 9
q)
and
a2(P = qP°a
(pEL(E;E)
and let H 1, H2 be the corresponding homology spaces. Prove that
H1
E* Qx H(E) and
H2
H(E*) ® E
(H(E) and H(E*) are the homology spaces of E and E*).
Tensor Products of Differential Algebras
2.14. The Structure Map of the Homology Algebra
Suppose (A, &4) is a differential algebra with respect to an involution cvA
(see Section 6.12 of Linear Algebra). Then the differential operator aA is an
antiderivation with respect to WA; i.e.,
aA(XY) = aA x y + WAX aA y
x, y e A.
(2.42)
Tensor Products of Differential Algebras
57
Introducing the differential- space (A ® A, aA ® A) we can rewrite (2.42) in
the form
µA aA ®A = aA µA ,
where µA denotes the structure map of A. Now consider the homology
algebra H(A) (see Section 2.10). The equation
z i, z2 E Z(A)
AZ1 AZ2 = A(Z1Z2)
shows that the structure map of H(A) is given by
µH(A) ° (?rA
7tA ° µZ(A)
(2.43)
where irA : Z(A) -p H(A) denotes the canonical projection and µZ(A) is the
structure map of the algebra Z(A). Since Z(A) is a subalgebra of A it is
clear that µZ(A) is the restriction of the structure map µA to the subspace
Z(A) ® Z(A) of A® A.
2.15. Tensor Products of Differential Algebras
Suppose now that (A, '3A) and (B, aB) are two differential algebras with
respect to involutions WA and c0B . Then the induced differential operator in
the space (A ® B, aA ® B), is given by
aA®B-aA®l +WA®
3B.
Now consider the canonical tensor product A ® B. Recall that WA ® B is an
involution in A ® B and that aA®B is an antiderivation with respect to
WA ® B Hence (A ® B, aA ® B) is a differential algebra, and so an algebra
structure is induced in H(A ® B). The structure map, µH(A ® B)' of H(A ® B)
is given by
µH(A ® B)
(irA ® B ® irA ® B) = irA ® B ° µZ(A ® B)'
(2.44)
where irA ® B : Z(A ® B) -+ H(A ® B) denotes the canonical projection, and
µZ(A ® B) is the structure map for the algebra Z(A ® B) (cf. (2.43)).
2.16. The Algebra H(A) p H(B)
It follows from the Ki nneth formula that the vector space H(A ® B) may
be considered to be the tensor product of the spaces H(A) and H(B),
H(A ® B) = H(A) ® H(B).
In this section it will be shown that H(A ® B) as an algebra is the canonical
tensor product of the algebras H(A) and H(B).
2 Tensor Products of Vector Spaces with Additional Structure
58
The structure map of the algebra H(A) ® H(B) is given by
µH(A) ®H(B) = (µH(A) ® µH(B))SH,
(2.45)
where SH denotes the flip operator for the pair H(A), H(B) (see Section 2.2).
It has to be shown that
µH(A ®B) = µH(A) ®H(B)
(2.46)
To simplify notation we set
µH(A) = Q,
µH(B) = '
µH(A ® B) = P
Then we obtain from (2.43) that
(Q ® t) (mA ® 1A ® ThB ® ThB) = Q(mA ® 2A) ® t(ThB
ThA µZ(A) ® T B µZ(B)
(ThA ® ThB) (µZ(A) ® / Z(B))
Next we observe that
SH(mA ® 2 B ® ThA ® ThB) _ (?LA ® 2 A ® 2 B ® TB)SZ,
(2.47)
where SZ is the flip operator for Z(A) and Z(B). The preceding two equations
yield
(Q ® t)SH(mA ® 2B ® ThA ® ThB) = (ThA ® ThB) (µZ(A) ® µZ(B))SZ .
(2.48)
But
CUz(A) 0 µZ(B))'SZ = µZ(A) ®Z(B)
and so (2.48) implies that
(Q ® t)SH(mA ® 2B ® ThA ® ThB) = (ThA ® ThB)µZ(A) ® Z(B)
(2.49)
On the other hand it follows from (2.44) that
P(2tA ® B ® ThA ® B) = ThA ® B µZ(A ® B)
Restricting this relation to the subspace Z(A) ® Z(B) ® Z(A) ® Z(B), and
observing that the restriction of ThA ®B to Z(A) ® Z(B) is ThA
(see
(2.23)), we obtain
P(2tA ® ThB ® ThA ® ThB) = (ThA ® ThB)µZ(A) ®Z(B)
(2.50)
Combining (2.49) and (2.50), we find that
((Q ®r)SH
P)(ThA ® ThB ® ThA ® ThB) = 0.
But
ThA ® ThB ® ?LA ® ?LB : Z(A) ® Z(B) ® Z(A) ® Z(B)
- H(A) ® H(B) ® H(A) ® H(B)
(2.51)
Tensor Products of Differential Algebras
59
is a surjective mapping, and so it follows from (2.51) that
(cr ® i)SH = P'
i.e.,
C UH(A) ® UH(B))SH = 1 H(A ® B)
(2.52)
Relations (2.45) and (2.52) yield (2.46).
2.17. Graded Differential Algebras
A, aA) and B = (>q Bq,'3B) be two graded differential algeLet A =
bras and assume that aA and aB are both antiderivations of odd degree k.
Consider the anticommutative tensor product A Q B. Then the structure
map of the algebra A Q B is given by
µA ® B = C UA ® µB) ° Q,
where Q denotes the anticommutative flip operator for A and B. It follows
from Section 2.8 that A Q B is a graded anticommutative algebra and, since
B) is
k is odd, A © B is an antiderivation of degree k. Hence (A Q B,
a graded differential algebra. It will be shown that H(A Q B) is the anticommutative tensor product of the graded algebras H(A) and H(B).
Let QH and QZ be the anticommutative flip operators for the pairs
H(A), H(B) and Z(A), Z(B). Then we have that
QH ° (irA ® ?CB ® irA ® iB) = (iA ® irA ® irB ® iB) ° QZ
where irA : Z(A) -+ H(A) and 1rB. Z(B) -+ H(B) denote the canonical pro-
jections. With this formula the proof coincides with the proof for the
analogous result in Section 2.16.
Tensor Algebra
3
In this chapter except where noted otherwise all vector spaces will be defined over an arbitrary
field r.
Tensors
3.1.
Definition. Let E be a vector space and consider for each p >_ 2 the pair
(®P E, 0 p) where
OpE=EQ...QE.
n
We extend the definition of Op E to the case p = 1 and p = 0 by setting
Q 1 E = E and 0° E = F. The paif (0" E, 0") is called the pth tensorial
power of E. The space Qp E is also called the pth tensorial power of E and
its elements are called tensors of degree p.
A tensor of the form x1 ® O xp, p
1, and tensors of degree zero are
called decomposable.
For every pair (p, q) there is a unique bilinear mapping
/3:Qp E X ®"E-+®E
+q
f3(x1 0 ... OO xp, xp+ 1 OO ... OO xp+q) = x1 0 ... 0 xp+q
x, E E.
(3.1)
Moreover, the pair (Op+q E, /3) is the tensor product of Op E and Oq E
(see Section 1.20). Hence we may write u Q u instead of /3(u, u) for u e Op E.
Then (3.1) reads
(x1 Ox ... Ox xp) Ox (xp + 1 Ox ... Ox xp + q) = x 1 Ox ... Ox x p + q
60
.
(3.2)
Tensors
61
The tensor u ® v is called the product of the tensors u and v. The product
(3.2) is associative, as follows from the definitions.
However, it is not commutative except for the case dim E = 1. (In fact, if
x e E and y e E are linearly independent vectors, then the products x ® y and
y ® x are also linearly independent and hence x ® y y ® x). Finally,
we notice that the product ® z (e ®° E = r, z e Op E) is the vector
in Op E obtained by multiplying the vector z by the scalar i In particular,
the scalar 1 acts as identity.
e l)
Let {eV}VEI be a basis of E. Then the products
form a basis of ®p E. (see Section 1.20). In particular, if E has finite dimension
®
and eV (v = 1, ..., n) is a basis of E then the products eVl ®
(v, = 1, ..., n) form a basis of ®p E and dim Op E = np (n = dim E).
Every tensor z e Op E can be uniquely written as a sum
:V1 ...., Vpev 1 O ... O e.
z=
(v)
The coefficients
.
VP
are called components of z with respect to the basis eV.
3.2. Tensor Algebra
Suppose that (Q p E, Q p) is a pth tensorial power of E (p = 0, 1,...) and
consider the direct sum
OCR
®E= Q ®p E.
p=o
The elements of ® E are the sequences
(zo , z 1, ...)
(p = 0, 1, ...)
zp e ®p E
such that only finitely many zp are different from zero in each sequence.
If ip: Qp E -+ ® E denotes the canonical injection we can write
®E=
i,
E.
p=o
Since the pair (ip p p E, i, Q p) is again a pth tensorial power of E we denote
the p-linear mapping i,® p by ®" and the vector space i, O Y' E by ®" E.
Then the above equation reads
®E= ®"E.
p=0
By assigning the degree p to the elements of Op E we obtain a positive
gradation in the space ® E.
3 Tensor Algebra
62
We now define a bilinear mapping
u, u,uue®E
(u,u)-+uu
by
ull =
uP Qx Vq
U=
P9
UP, U =
Vq.
q
P
This multiplication makes ® E into an associative (but noncommutative,
if dim E > 2) algebra in which the sequence (1, 0, ...) acts as a unit element.
It is clear from the definition that ® E is a positively graded algebra. ®E
is called a tensor algebra over the vector space E. From now on we shall
identify ®° E with r and ®' E with E. Then r and E are subspaces of
® E, and the elements of E, together with the scalar 1 generate (in the algebraic sense) the algebra ® E.
If E has finite dimension the gradation of ® E is almost finite and the
Poincare series of the graded space ® E is given by
P(t) =
nptp =
P=°
1
n = dim E.
1 - n t'
Remark : The reader should observe that if E 0 the bilinear mapping
/3:(® E) x (® E) -+ ® E which is defined by the multiplication is not a
tensor product. In fact, if /3pq denotes the restriction of /3 to (®P E) x (®q E),
we have
Im f3Pq = Im f3gP = ® p + q E.
Set E 1 = ®p E and F 1 = ®q E, p
q. Then
E1 n F1 = 0,
while j3(E 1 x F 1) = j3(F 1 x E 1). Hence if 3 were a tensor product it would
follow that E 1 = 0 or F 1 = 0, whence E = 0.
3.3. The Universal Property of ®E
Consider an arbitrary associative algebra A, with unit element e, and a
linear map ri : E -+ A. Then there exists precisely one homomorphism
h : ® E -+ A such that h(1) = e and h o i = ri ; i.e., such that the diagram
n
A
it
is commutative, where i denotes the injection of E into ® E. To define h
consider the p-linear mapping
ExxE-+A
Tensors
63
given by
(x 1, ..., x,,) - rix 1 ... ,ixP .
In view of 02 there exists a linear map h: OP E - A such that
hP(x 1 O ... O x,,) = yix 1 ...
P
We extend the definition of hP to the cases p = 1 and p = 0 by setting
ho(t) = to for all e r.
h1 = ri and
Then a homomorphism h : Q E - A is given by
hu =
u, e 0,, E, u =
h p uP
P
uP.
P
In fact, if u, v e Q E are decomposable, then it is clear that
h(uv) = hu hv.
Since every element in Q E is the sum of decomposable tensors, and h
is linear, it follows that h preserves products.
To show that h is uniquely determined by ri, we notice that the conditions
h o i=, and h(1) = e determine h in Q 1 E= E and in Q° E= r. But Q E
is generated by 1 and the vectors of E ; consequently, h is uniquely determined in Q E.
3.4. Universal Pairs
Now let U be an associative algebra with unit element 1 and E : E - U be a
linear map. We shall say that the pair (E, U) has the universal tensor algebra
property with respect to E if the following conditions are satisfied:
T1: The space Im E together with the unit element 1 generates U (in the
algebraic sense),
T2: If ri is a linear map of E into an associative algebra A with unit
element a then there is a homomorphism h : U - A such that h(1) = e
and the diagram
E
E
W
is commutative.
The properties T1 and T2 are equivalent to the property:
T: If ri is a linear mapping of E into an associative algebra A with unit
element, then there exists a unique homomorphism h : U - A such that
h(1) = e and Diagram (3.3) is commutative.
3 Tensor Algebra
64
It is clear that Tl and T2 imply T. Conversely, assume that the pair (E, U)
satisfies the condition T. Then T2 follows immediately. To prove Tl consider
the subalgebra V of U generated by Im E and the unit element. Then E can be
considered as a linear map of E into V (which, to avoid confusion, we denote
V such that h(1) = 1 and
by EV) and hence there is a homomorphism h : U
ho E = Ey . If j : V -+ U denotes the inclusion map we have E = J o Ey and
hence it follows that E = (j o h) o E. In diagram form we have
and hence the diagram
U
is commutative. On the other hand, we have the commutative diagram
E
U
where us the identity map of U. Since j o h is an endomorphism of the algebra
U, the uniqueness part of T implies that
joh = z.
Consequently, j is an onto map, and hence U = V. This proves T1.
We shall now prove the following
Uniqueness theorem. Let (E, U) and (E; U') be two universal pairs for E. Then
there exists precisely one isomorphism f : U -+ U' such that
fo =E'.
PROOF. In view of T there exist unique homomorphisms f : U -+ U' and
g : U' -+ U such that
fo
=
'
and goE'=E.
Hence, g of is an endomorphism of U which reduces to the identity in Im E.
Tensors
65
Since the space Im E generates U, it follows that g of = i. In the same way it
is shown that f o g = i', the identity map of U'. Hence, f is an isomorphism of
U onto U' and g = f -' . The uniqueness theorem is thereby proved.
Since (i, ®E) is a universal pair for E it follows from the above uniqueness
theorem that for every pair(, U) there is precisely one isomorphism f : ®E -+ U
such that f o i = E.
Since ®E is a graded algebra, a gradation is induced in U by the isomorphism f. The algebra U furnished with this gradation is a graded algebra
and f : ®E -+ U is a homogeneous isomorphism of degree zero.
In view of the uniqueness theorem, the universal algebra U is usually
called the tensor algebra over E and is denoted by ®E.
3.5. The Induced Homomorphism
Let p: E -+ F be a linear map from the vector space E to a second vector
space F. Then p extends in a unique way to a homomorphism
-+ ®F
such that 'p®(1) = 1. In fact, consider the linear map ri : E -+ ®F given by
ri = j o p (where j denotes the inclusion map) and apply the result of Section
3.3.
Clearly, the homomorphism'p® is homogeneous of degree zero. It follows
from the definition of 'o that
'pO (x 1 ® ... ® x,,) = (px 1 O ... O cpx p
xt e E.
Now let G be a third vector space, ® G be a tensor algebra over G and
U: F -+ G be a linear map. Then it is clear from the definitions that
° p)® =
o°
(3.4)
If F = E and us the identity map then ro is the identity map of ®E,
ro=l.
(3.5)
It follows from (3.4) and (3.5) that 'o is injective (surjective) whenever 'p is
injective (surjective). In fact, if 'p is injective there exists a linear map i/i : E - F
such that l// o 'p = u. Formulas (3.4) and (3.5) now imply that
7"0 ° 'p® = l® = l
and hence 'o is injective.
It is easy to see that
Im 'o = ®(Im 'p).
Hence 'o is surjective whenever 'p is.
3 Tensor Algebra
66
3.6. The Derivation Induced by a Linear Transformation
Let p be a linear transformation of E. Then p can be extended in a unique
we notice
way to a derivation e®(cp) in the algebra ®E. To construct
that for each p >_ 2 a p-linear mapping
x E- ®"E
Ex
p
is defined by
P
i= 1
This p-linear mapping induces a linear map
QPE - QPE such that
P
eo(p)(x 1 O ... O xp) = >x1
®... ®(pxt
®... ® x.
i= 1
We extend this definition to the case p = 1 and p = 0 by setting
Op) = p and
eo('p) = 0
to be the linear transformation of ®E into itself which
and define
extends the Op).
It remains to be shown that
v) =
is a derivation, i.e., that
v + u e®(p)v.
(3.6)
For the proof, we may assume (as before) that u and u are decomposable so
that
xieE.
and
Let us first assume that p >_ 1 and q >_ 1. We then obtain
e ®(p)(u u) = e ®(p)(x 1 ®. ® x p + q)
p+q
xl ®...®(pxi®x ...Ox xP+q
i= 1
P
x1 O ... O ® O xp+q
i= 1
p+q
x 1 0 ... O xp O ... O cpxi O ... O xp + q
+
i=p+ 1
= e®(p)u U + u e®(p)u.
Formula (3.6) remains true if p = 0 or q = 0. In the case p = 0, for instance,
we have
W®(p)U = e®(p)L v+
e®(p)u.
67
Tensors
Since the algebra ®E is generated by the elements of E and the unit
element, it follows that the extension of p into a derivation in the algebra ®E
is unique (see Section 5.6 of Linear Algebra).
Let U be a second linear transformation of E. Then
o®(2p + µt/i) _ Ao®(p) + uO®(i,U)
A, µ E r
(3.7)
and
o®( °
- t,U° q') = o®C(p) ° o®(I,U) - o®(U) ° o®(p)
(3.8)
Formula (3.7) follows immediately from the definition of o®. To prove
(3.8) we remark first that the operator on each side is a derivation (see Section
5.6 of Linear Algebra). Hence it is sufficient to show that
p) = eo('P) ° eo(n) - eo(U) °
But this is immediately clear since
e®('p) _ p and O®(U) _ fi.
3.7. Tensor Algebra Over a G-Graded Vector Space
Suppose E _ a Ea is a G-graded vector space and let ®E _ L ®"E be a
tensor algebra over E. Then a G-gradation is induced in each ®"E (p > 1) by
(®"E), where (®"E)a =
pE _
Ea 1
... O E.
al
a
It will be convenient to extend this gradation to ®°E = r by assigning the
degree 0 to the elements off. The induced G-gradation in the direct sum ®E
(®E)a where (®E)a = L (®"E). Then ®E becomes
is given by ®E _
a G-graded algebra. In fact, let
u = xa 1 0 ... O xap E (®E)a and v = yfl 1 0 ... O Yflq e (O E)
be two decomposable elements. Then
uv = xa O ... Q xap O Y 1 0 ... O Yflq E (O E)a +
1
and thus we have
deg(uv) = deg u + deg v.
The algebra ®E together with this gradation is called the G-graded tensor
algebra over E. If G = 7l and all the vectors of E are of degree one then the
induced gradation of ®E coincides with the gradation defined in Section 3.2.
3 Tensor Algebra
6g
PROBLEMS
1. Let u 1 = a 1 Q b 1 and u2 = a2 Q b2 be two decomposable tensors and assume that
u1
0. Prove that u1 + u2 is decomposable if and only if a2 = a1 or b2 = µb 1
(A, µ e r).
2. Assume that a 1 O
0. Prove that
O ap
a1®...®ap=b1Q...Qbp
if and only if
i = 1, ... , .. /., ... , 4,, =
b1 _
1 (A E r).
3. Use Formula (3.8) to prove that
tr(q Q /i) = tr (p tr /i.
Tensors Over a Pair of Dual Spaces
3.8.
Definition. Suppose that E and E* ire vector spaces, dual with respect to a
scalar product <, >, and let QE _ ®"E and ®E* _ ®PE* be tensor
algebras over E and E*, respectively. According to Section 1.22 there is
induced between x®PE and x®PE* for each p >_ 1 a unique scalar product
such that
<x* 1 O ... Q X*P, X1-® ... QX XP> _ <X* 1, X1>" . <X*P, XP>.
(3.9)
We extend the definition of <, > to the case p = 0 by setting
u E Q °E = r.
<A,, µ> _ Aµ
The scalar products between ®E and ®PE* can be extended in a unique
way to a gradation-respecting scalar product <, > between the spaces QE and
®E* (see Section 6.5 of Linear Algebra) and this scalar product is given by
<u* U\ _
<u*P UP>,
u*P, V = VP
u* =
P
P
Now assume that E has finite dimension and let
P
a*v be a pair of dual
bases for E and E*. Then the scalar product between the induced basis
vectors
O eVp and a*µ1 O
O e*µp is given by
O
<e*I O . . O e*µp' ev1 O
O ev,> _
(3.10)
.
This formula shows that the bases
Q
Q
and a*µ1 O
O e*µp are
again dual. It follows from (3.10) that the scalar product of two tensors
cv1.....
u=
(v)
VpeV1 O
O
and
u* _
µp a*µ1
(µ)
®... O e*µ"
69
Tensors Over a Pair of Dual Spaces
is given by
<u* u> _
V1, ..., Vp1
(V)
3.9. The Induced Homomorphism
Suppose that F, F* is a second pair of spaces dual with respect to a scalar
product (again denoted by <, >). Let p : E -+ F, p* : E* - F* be a dual pair of
linear maps. The homomorphism (gyp*)®: ®E* - ®F* induced by p* will
be denoted by p ®. Then we have
p®(y* 1 ®... O y*p) = (p* y* 1 ®... ® (p* y*P
y*i e F*. (3.11)
Now it will be shown that the homomorphisms
cP®: OE -+ OF and p®: ®E* - QF*
form a dual pair,
(3.12)
Let u e ®E and u* e ®F* be arbitrary elements. It has to be shown that
<U*, (p®u> = <(p®V*, u>
Since 'p® and 'p ® are both homogeneous of degree zero we may assume that
u and u* are both homogeneous of the same degree p. Moreover, we may
assume that u and v* are decomposable, u = x 1 ® ® xp and u* =
y* 1 ®
® y*p. Then Formulas (3.11), (3.8) and (3.9) yield
<'P®v*, u> = <(P®(y*1 x0 ... x0 y*p), x1
®... x0 xp>
= <(P* y* 1 ®... ® 'p* y*p, x 1 0 ... o xp>
= <<p*y* 1, x1> ...
_ < y*
1
<(p*y*P, xp> = <y* 1, 'x 1 > ... < y*p, cpxp>
®... ® y*p, 'x 1®... ® oxp> = <v*, o®u>
whence
<v*, P® u> = <(p®V*, u>
If G, G* is a third pair of dual spaces and
iU: F -+ G,
t/,* : F* - G*
is a second pair of dual mappings, we have in view of (3.12) and (3.4) that
( ° )® = [(/j° )*]® _ ((p* ° /J*)®_ (p*)®°(/,*)® ='p® ° 1//®
whence
3 Tensor Algebra
70
3.10. The Induced Derivation
Consider again a pair of dual mappings, p : E - E, p* : E* E- E*. The derivation in the algebra ®E* which is induced by p* will be denoted by 8®(p),
Then
O®(p) =
P
e®(p)(x* 1 0 ... O x*p) => x* 1 p ...
P*x*i
... O x*P.
(3.13)
i= 1
The derivations
and
again form a dual pair,
(3.14)
As in Section 3.9 we have to prove that
v* E QE*, u e QE
<e®(cp)v*, u> = <v*, 8®(cp)u>
and we may assume that v* and u are of the form
v* = x* 1 QX ... QX x*P
u = x 1 QX ... QX xp .
Then we obtain from (3.13) that
P
u> =
x*1 Q ...*x* ... ® x*P, x1 ® ... ® xp
i= 1
P
<x*1 xl> ...
xi> ... <x*p, xp>
i= 1
P
<x* 1, x1 > ... <x*i, Pxi> ... <x*p, xp>
i= 1
(x*i QX ... QX x*P,
P
x1 O ... (pxi ... O xp
i= 1
=
<v*, O®(p)u>
whence
<e®(p)v*, u> = <v*, 8®(p)u>
If i'/ : E - E, /i* : E* E- E* is a second dual pair of linear maps, then it
follows from (3.14) and (3.8) that
/,'p) = e®(/,*p* - (p*/,*) = e®(/,*)e®(p*) e®(i/i)e®(p)
- e®(p)e®(i/i)
71
Mixed Tensors
whence
PROBLEM
Let E, E* be a pair of dual n-dimensional vector spaces, and consider a tensor
c E* as follows: a vector x* e E* is to be contained in
Eif and only if
u e ®"E. Define a subspace
<u, x* Q v*> = 0 for every v* e ®"- lE*.
Show that u is decomposable if and only if dim E= p - 1.
Mixed Tensors
3.11.
Definition. Let E*, E be a pair of dual vector spaces and consider, for every pair
(p, q), p > 1, q >_ 1, the space
®(E*, E) _ (®PE*) O
(®"E).
Extend the definition of ®(E*, E) to the cases q = 0 and p = 0 by setting
®(E*, E) = ®PE* and Qq(E*, E) = QqE.
The elements of ®(E*, E) are called mixed tensors over the pair (E*, E) and
are said to be homogeneous of bidegree (p, q). The number p + q is called the
total degree.
A tensor of the form
w = xi Qx ...
xp Qx xl Qx ".
xq
x*EE*, x;EE,
is called decomposable.
The scalar product between E* and E induces a scalar product between
®(E*, E) and ®(E*, E) determined by
<u* p v, v* p u> _ <u*, u><v*, v>
(3.15)
(see Section 1.22). Thus any two spaces ®(E*, E) and ®(E*, E) are dual.
In particular, every space ®(E*, E) is self-dual. Finally note that
<zl, z2> = <z2, z1>
as follows from the definition.
z1e®(E*, E), z2 e ®(E*, E)
3 Tensor Algebra
72
3.12. The Mixed Tensor Algebra
The mixed tensor algebra over the pair E*, E is defined to be the canonical
tensor product of the algebras ®E* and QE (see Section 2.2). It will be
denoted by ®(E*, E),
®(E*, E) = (OE*) O (OE).
Thus ®(E*, E) is an associative (noncommutative) algebra with 1 Q 1 as
unit element. It is (algebraically) generated by the elements 1 p 1, x* Q 1,
and 1 p x with x* E E* and x e E.
Now let
iq : ®E - Q E
iP. ®PE * - ®E*,
and
iq : (p PE *) p (p qE) - ®(E*, E)
be the inclusion maps and identify the spaces ®)E*, ®"E, and ®(E*, E)
with their images under these maps. Then we have the direct decomposition
®(E*, E) =
(®PE*) O (®"E).
P, q
3.13. Contraction
Assume that p >_ 1 and q >_ 1. Fix a pair (i, j) with 1 <_ i <_ p and 1 < j < q
and consider the (p + q)-linear mapping
x E- ® : (E*,E)
x E* x E x
x
q
given by
..,xp,x1 ...xq)
= <xr, x;>x l 0 ... 0 x O ... 0 xP* 0 x 1 0 ... 0 x1
OX ... OX x.
q
In view of the universal property, 'I determines a linear map
C: ®(E*, E)
-
OP- i (E*, E).
C; is called the contraction operator with respect to the pair (i, j) and the
tensor FXw) is called the contraction of w with respect to (i, j). In particular,
Ci(x* Q x) = <x*, x>
x* E E*, x e E.
Now assume that E has finite dimension and let {e*v}, {ev} be a pair of
dual bases of E* and E. Then the products
evls ..., Vp
µ1s ..., µ9
= e*vl
O... O e*vp O eµl
O.. O eµ9
Mixed Tensors
73
form a basis of ®(E*, E). Thus every tensor w e ®(E*, E) can be written
in the form
w=
µ l s . µy eV l s : s Vp
Vls ' VP µ1s s µq
µ l s : µy E IT.
V1s
Vp
(v)(µ)
The scalars
,
; vp are called the components of w with respect to the basis
{eV}. Since
Ci.eV1s ...s Vp _ Vi eV1s ...s Vis ...s VP
1
µj µl,..., µj ..... µq
µ1....,µg
it follows that the components of the contraction C; w are given by
n
µls sµq- 1 =
vls....Vp-1
rµls sµ1.- 1Aµ1.. sµq- 1
A=1
Vls....VJ-1AVjs....Vp-1
3.14. Tensorial Maps
Let E be an n-dimensional vector space. Then every linear automorphism a
of E determines for each pair (p, q) a linear automorphism 7 of ®(E*, E)
given by
T2(u* Q u) = (a®)-1u* Q a®u.
It is easily checked that
T02= T ° Ta
and
T,= i.
A linear map
D : ®(E*, E) --, ®(E*, E)
is called Censorial, if it satisfies
Fo Ta = Tao
for all linear automorphisms a of E.
As an example consider the contraction operator C. For the sake of
simplicity we let i = j = 1 and write C i = C. Then we have for z = x i p
OxpOx10...Ox9
7 (z) = X* -1 x i O ... p a* -1 xp p Xx 1 p ... p axq
whence
(Cl )(z) = <a*-1xr, ax1>a*-1x2 ®... Q a*-1xp Q axe Q ... Q axq
= <xl,xl>a*-1x2 O...O«*-1xp pocx2 p...Oaxq
= (Ta C) (z).
Thus C is a tensorial map.
3 Tensor Algebra
74
PROBLEMS
(v = 1, ... , n) be two bases of E and consider a tensor w E ®(E, E*).
1. Let
and the dual bases {e*v}, {e*v} are connected by
Assume that the bases
the relations
aye
e
a*v
A
A
Prove the following transformation formulas for the components of w:
V1, ... VP -
fJV1
VP
K1 . .
(A),(K)
2. Using the formula in problem 1 show explicitly that the components of a contracted
tensor satisfy the same transformation formula.
3. Show that <u*, u> = (C1)"(u Q u*) for all u e ®"E, u* E QpE*.
4. Assume that
c1: ®(E, E*) - ®(E, E*)
®(E, E*)
®5(E, E*)
is a pair of dual mappings and that 1 is tensorial. Prove that 1* is also tensorial.
5. Show that sum; composition, and tensor product of tensorial mappings is again
tensorial.
6. Let
: ®(E, E*) - ®(E, E*)
be a nonzero tensorial mapping. Prove that
r- p=s - q.
7. Consider the linear map
(a) - a O z,
where a e E Q E* is a fixed tensor. Show that qi(a) is tensorial if and only if a = ) t,
where t is the unit tensor.
8. Prove that every tensorial mapping c1: E Q E* + r is of the form
where C is the contraction operator.
9. Verify the relations
if
C;oC =
if
if
i<k,j<1,
i<k,j>1,
i>k,j<1.
75
Tensor Algebra Over an Inner Product Space
10. Consider the bilinear mapping
®(E, E*) x ®(F, F*) -+ ®(E O F, E* O F*)
defined by
(x1O...Oxp®x*l
0Y1)0...0(xp0
(x1
yp)0(x*1
0y*l)0...0(x*q®y*q).
(a) Show that this mapping is a tensor product
(b) Prove that
u e ®(E, E*), v e ®(F, F*)
CJ(u Q v) = CX u) ® CJ(v)
11. A tensor u e ®(E, E*) is called invariant if T (u) = u for every linear automorphism
(see Section 3.14).
(a) Show that if u 0 is an invariant tensor then p = q.
(b) If E has finite dimension show that a tensor u e E Q E* is invariant if and
only if u = ) t where t is the unit tensor.
(c) If E has infinite dimension show that every invariant tensor u e E Q E* must
be zero.
(d) Assume that E has finite dimension. Show that a tensor u is invariant if and
only if the components of u are the same with respect to all pairs of dual bases.
Tensor Algebra Over an Inner Product Space
3.15. The Induced Inner Product
Let E be an inner product space and consider the pth tensorial power
®pE. The inner product in E induces an inner product in ®pE such that
(xi, vi) ..
(x1 Ox ... Q xp, vi
(xp, Yp)
(see Section 1.25). In particular, if E is an n-dimensional Euclidean space
and if ev (v = 1, ..., n) is an orthonormal basis of E, then the products
p evp form an orthonormal basis of ®"E and so ®"E becomes a
ev1 p
Euclidean space as well.
The inner products in the spaces &J E determine an inner product in ®E
given by
(u, v) _
u, v e ®E,
(up, vp)
P
where
u=
up
P
and v=
vp
P
u p, v p e Q PE.
3 Tensor Algebra
76
3.16. The Isomorphism -r®
Let E be an n-dimensional inner product space with dual space E*. Then the
inner product in E determines a linear isomorphism -r: E - E* given by
<ix, y> = (x, y)
x, y e E.
This isomorphism -r in turn induces a linear isomorphism
i®: ®E - ®E*.
This isomorphism satisfies the relation
<i®u, v> = (u, v)
u, y e ®E.
(3.16)
In fact, let
and
u=x1O...Oxp
v=Y1O...OYp
be decomposable tensors. Then we have
<i®u, v> = <ix1
®... ®ixp, Y1
®... O Yp
= < x1, Y1 ... <ixp, Yp> = (x1, Y1) ... (xp, Yp)
= (x1 ®... O xp, Y1 ®... ® Yp) = (u, v)
Relation (3.16) shows that the restriction of -r® to ®"E coincides with the
isomorphism ppE ppE* induced by the inner product in ppE. Finally
note that
since r* = -r.
3.17. The Metric Tensors
Let E be an n-dimensional Euclidean space. Choose an orthonormal basis
{ev} (v = 1, ..., n) and set
g =
ev ® ev.
v
Since an orthonormal basis is self-dual (with respect to the inner product in
E), it follows that the tensor g is independent of the choice of the orthonormal
basis {ev}. It is called the contravariant metric tensor of E. Similarly, the
covariant metric tensor is defined by
g* =
e*v O e*v,
v
where {e*v} (v = 1, ..., n) is an orthonormal basis of E* with respect to the
induced inner product.
77
Tensor Algebra Over an Inner Product Space
The inner product of two vectors x and y can be expressed in the form
(x, y) = <g *, x O y> .
In fact, write
x =vey and
y=
vev .
v
v
Then we have
x O y> =
e*v x><e*v, y>
<e*v O e*v, x O y> =
v
=
V
v;7v = (x, y)
v
The same argument shows that
x*, y* E E*.
(x*, y*) = <x* O y*, g>
PROBLEMS
1. Given a Euclidean space E prove that
u1, u2 e ®E, v1, v2 e QX qE.
(u1 0 v1, u2 0 v2) = (u1, u2)(v1, v2)
2. Let E be a Euclidean space and E* be a dual space. Consider the metric tensors
g E E Q E and g* E E* ®E*.
(a) Show that
Ci(g OO g*) = t,
where t is the unit tensor.
(b) Prove that the metric tensor of ®"E is given by g O
Og
p
3. Verify the formula
(µ(t)u, µ(t)v) = n(u, v)
u, v e Q E,
where t is the unit tensor of E, n = dim E (see Section 1.26).
4. Let E be a Euclidean space and E* be a dual space. Consider the space L(E; E) =
E* Q E (see Section 1.27).
(a) Show that the adjoint of a linear transformation p = a* Q b is given by
P = ib Q i -1 a*.
(b) Show that the induced inner product in L(E; E) is given by
N
r
(`p,
/') = tr(q ° 1'i).
5. Given a Euclidean space E and an arbitrary basis
(u, v) =
9
µvpup
9V11
(v)(u)
(v = 1, ... , n) show that
V 1...., Up7M1...., NP
where
u=
U1, ..., vpevl 0 ... 0
(v)
and
ul ..., µpeul 0 ... 0 eµp .
v=
(u)
78
3 Tensor Algebra
6. Show that the components of the metric tensor g* with respect to an arbitrary
basis x (v = 1, ..., n) of E are the inner products (xv, xµ).
7. Let E be a Euclidean space and E* be a dual space. Given a pair of dual bases
{x*°} (v = 1, ... , n) show that
9* =
(xv, xµ)x*V O x
v, µ
and
g = >(x*v,x*M)xv®xM.
v, µ
8. Let E, E* be a pair of dual spaces and assume that E is a Euclidean space. Consider
the linear isomorphism i : E -+ E* defined by
<tx, y> = (x, y)
x, y e E.
Prove that the isomorphism i®: ®"E -+ ®PE* coincides with the linear isomorphism
Q pE + ®PE* which is induced by the inner product in ®"E. Show that i ® = i®
and that (i®)2 =z.
The Algebra of Multilinear Functions
In Sections 3.18-3.23 E denotes a finite-dimensional vector space.
3.18. The Algebra T(E)
Consider for each p >_ 1 the space Tp(E) of p-linear functions
p
In particular T 1(E) = E*. It will be convenient to extend the definition of
Tp(E) to p = 0 by setting T °(E) = F. The product of a p-linear function F
and a q-linear function 'P is the (p + q)-linear function 'F 'P given by
(F `P)(x1, ..., xp+q) = F(x1, ... , xp).'P(xp+ 1, ..., xp+q).
(3.17)
In the cases p = 0 or q = 0 we define the multiplication to be the ordinary
multiplication by scalars. This multiplication makes the direct sum
00
T(E) =
Tp(E)
p=0
into an associative (noncommutative) algebra with the scalar 1 as unit
element.
A linear map p: E -+ F induces a homomorphism p* : T '(E) E- T (F) given
by
((p*'P)(x 1, ..., xp) =((px ..., (pxp)
as follows directly from the definitions.
'P E T "(F)
The Algebra of Multilinear Functions
79
Moreover, a linear transformation co of E determines a derivation oT (cp)
in the algebra T(E) given by
P
(x1, ..., xv
1, ... , xP) =
v=1
3.19. The Substitution Operators
Fix a vector h e E and consider the operators iv(h) : T(E) -* T(E) given by
(iv(h)F) (x 1, ..., x,_1) = F(x 1, ..., h, ..., xP _ 1).
iv(h) is called the with substitution operator in T(E) corresponding to the
vector h. Clearly,
F, Y E T'(E).
iv(h) (iF + 'Y) = iv(h)F + iv(h)`J!
Moreover, it follows from the definition that for F E TP(E)
iv(h) F 'F
'F =
i h
v <_ p,
v _ p + 1.
This relation implies that for fv e E*
iv(h)(fi
...
... fP) =
fv
...
fP.
Now define operators iA(h) and is(h) on T(E) by
P
iA(h) =
(_ 1)v -1 iv(h)
v= 1
and
P
is(h) =
(p = 1, 2, ...).
iv(h)
v= 1
Then we obtain from (3.18) the formulas
iA(h)
((''F) =
iA(h)(''F + (-1)M ' iA(h)'
and
is(h) (i ' 'F) = is(h)( ' 'F + ( ' is(h)'
for all (J) E TP(E), 'F E T'(E).
(3.18)
3 Tensor Algebra
80
3.20. The Isomorphism ppE* - T"(E)
In this section we shall establish an isomorphism between the space T"(E)
and the pth tensorial power over E*. Consider the p-linear mapping
Tp(E)
n
given by
P(f1, ... , fp) = f1 ... fp
fV E E*.
It has to be shown that this map has the universal property. Fix a basis
..., in E and let {e*1, ..., e*"} be the dual basis.
We show that the products a*v1
e*vp (v, = 1, ..., n) form a basis of
Tp(E). In fact, every vector x e E can be written in the form
{e1,
x=
e*v e E*v.
e*v(x)ev
V
Now let F E T "(E). Then we have
e*V(xl)ev, ...,
xp) _
_
e*v(xp)eV
v
v
e*V1(x1) . . . e*VP(xp)F(eV1, . . . , eVP).
(v)
This equation can be written in the form
_
F(eyl,
. . .
(v)
, ev
P
)e*vl . .
. e*VP
and so it shows that the products a*v1
e*VP generate T"(E).
On the other hand, assume a relation
e*vl . . . e*VP = 0
(v)
Then
vP<e*V1,
x1>...
<e*VP, xp> = 0
xi E E.
(v)
Now fix a p-tuple (o,
..., ocp) and set x, = e21(i = 1, ..., p). Then the relation
above implies that dal, ..., aP = 0. Thus the products a*V1
e*Vp form a
basis of T"(E). In particular,
dim T "(E) = np
n = dim E.
Now consider the linear map a: ppE* - T"(E) induced by co. Then we
have the commutative diagram
E* x
x E* ( T(E)
81
The Algebra of Multilinear Functions
In particular
a(e*v1 O . . . O e*' ) = e*v1 .
e*vP-
. .
e*''P form a basis of Tp(E) it follows that a is an
isomorphism. Finally observe that, since oc(u* p v*) = oc(u*) oc(v*) a is
an algebra isomorphism
Since the products a*v 1
a: QE* + T'(E).
3.21. The Algebra T.(E)
Tp(E)(p >_ 1) denote the space of p-linear functions in the dual
space E* and set T0(E) = F. Observe that the space T1(E) is canonically
isomorphic to E under the correspondence a H fa given by
Let
fa(x*) _ <x*, a>
a e E.
Applying the results of Section 3.18 with E replaced by E* we obtain a
multiplication between the spaces T,(E)(p >_ 0) which makes the direct sum
T,(E) _
Tp(E)
p=o
into an associative algebra.
A linear map p: E -+ F induces a homomorphism co: * T,(E) -+ T, (F) given
by
t)(yi, ... , yp) _
p*yi,
..., p*yp)
Fe 7,(E)
and a linear transformation p of E determines a derivation in T,(E) given by
P
(x l , ..., xp) _
v=1
(xi, ..., (p*x*, ..., x).
Finally note that T.(E) is isomorphic to the tensor algebra over E (see
Section 3.20).
3.22. The Duality Between T"(E) and T(E)
Fix a pair of dual bases {e*v}, {ev} in E and E* and consider the bilinear
function <,>: Tp(E) x Tp(E) -+ F given by
<IF, 'I'> _
F(ev 1 , .. .,
evP)%Tt(e*v 1,
. . .
, e*vP).
(v)
Then we have in particular
<F, xl ' ... xp> _
F(eyl,
..., eyP)<e*v1, x1> ... <e*vp, xp>
(v)
_
<e*vl, xl>e vl, . . .,
vl
<e*vP, xp> evP
vp
.
3 Tensor Algebra
82
Since
><e*v,x>ev=x
x E E,
v
it follows that
<IF, x 1
...
F e T"(E), xv e E.
xp> _ iF(x 1, ... , x p)
Similarly
< fi
... fp Y> _ tY(fi,... , fp)
Y E Tp(E), fv e E*.
These relations imply that the bilinear function <, > does not depend on
the choice of the dual bases {ev}, {e*v} (v = 1, ..., n) and that it is nondegenerate. Thus it defines a scalar product in the spaces T"(E) and Tp(E).
On the other hand, we have a scalar product between the spaces ppE*
and ppE (see Section 3.8). A simple computation shows that the isomorphisms ®)E* T (E) and ®" E = Tp(E) preserve the scalar products.
3.23. The Algebra T(E)
Denote by T (E) the space of (p + q)-linear functions
f.
'i
n
In particular,
To (E) = T"(E)
and T q (E) = 7(E)
(see Section 3.18).
The product of two multilinear functions F E T (E) and Y E Ts(E) is
defined by
* ... , xq)(x
*
_ (x 1, ... , xp, x 1,
i, ... , xr, x i *, ... , xs*).
Form the direct sum T(E) of the spaces T (E) and identify each T (E)
with its image under the inclusion map. Then the multiplication above
makes T(E) into an associative algebra.
Now consider the bilinear mapping T "(E) x 7(E) - T (E) defined by
this multiplication. It follows from Section 1.20 that this bilinear mapping
has the universal property. Thus,
T (E)
T (E) Q 1(E)
T (E)
T (E) Q T.(E).
and
83
The Algebra of Multilinear Functions
Moreover, since for F1, F2 E T(E) and 'F1, 'F2 E T,(E)
(b 'F2) = (b
.
b2)
T(E) is the canonical tensor product of the algebras T(E) and T.(E).
Finally, let a : E 3 F be a linear isomorphism. Then an induced isomorphism 7: T(E) * T(F) is explicitly given by
for all F E T (E), y E F, y*' E F*.
4
Skew-Symmetry and Symmetry
in the Tensor Algebra
All vector spaces in this chapter are defined over a field of characteristic zero.
Skew-Symmetric Tensors
4.1. The Space Np(E)
Given a vector space E consider the pth tensorial power QpE and let Sp denote
the permutation group on p letters. Then every permutation 6 E Sp determines
a linear automorphism of QpE (also denoted by 6) defined by
6(x 1 O ... O xp) = xQ -1(1) Q ... O x y - 1 p
xy E E.
As an immediate consequence of the definition we have the formulas
(t6)u = ti(6u)
6,
E Sp, u e QpE
and
iu=u
(i being the identity permutation).
Now consider the subspace Np(E) of QpE generated by all products
O xp such that x, = x for at least one pair i j. Clearly Np(E) is
x1O
stable under 6 for each 6 E S.
It will be shown that for each u e QpE and each 6 E Sp
U - EQ 6u E Np(E)
For the proof we may assume that u is decomposable, u = x1 Q
84
Q x.
Skew-Symmetric Tensors
85
Consider first the case of a transposition i : i ± j. Then we have
u-;tu=x1Qx ...Qxi®...Qx;Q...Qxp
x;0...0xi0...0xp
+x1Q...QX
= x 1 QX ... QX (xi + x;) QX ... QX (xi + x;) QX ... QX xp
- x1 O ... O xi O ... O xi O ... O xp
- x 1 O ... 0 x; 0...0x; 0 ... QX x p E Np(E).
Now assume that u - EQ 6u e N"(E) for all permutations 6 which are products
of m transpositions and consider the permutation p = r6 where
is a
transposition and 6 is a product of m transpositions. By hypothesis we have
u - EQ 6u e N"(E).
Since N"(E) is stable under it follows that
iu - EQ i6u e N(E),
whence
; tu - EtQ t6u e N(E).
On the other hand we have
u - ; iu e N"(E).
Adding these relations we obtain
u - EtQ i6u e N(E).
Now (4.1) follows by induction.
4.2. The Alternator
A tensor u e QpE is called skew-symmetric if
for every 6 E Sp.
6u = EQ u
The skew-symmetric tensors of degree p form a subspace X"(E) of QpE.
The alternator in QpE is the linear map m: QpE -p QpE given by
1
P Q
It follows from the definition that for e Sp
1
TGA i =
=
1
PQ
EQ 6't =
P
EQt 6
Et
Q
1
p
Et
p P = Et TGA .
P
4
Skew-Symmetry and Symmetry in the Tensor Algebra
TGA o 'L = Et TGA
ieSp.
t 0 TGA = Et TGA
t E Sp .
Next we establish the relations
ker mA = N"(E)
and
(4.5)
Im ThA = X "(E).
In fact, let u = x 1 ®. p xp be a generator of N"(E). Then, for some
transposition -r, -ru = u and so Formula (4.2) implies that mA u = - ThA u
whence mA(u) = 0. This shows that N"(E) c ker
On the other hand, it follows from the definition of mA that for u e ®"E
mA.
ThA u - u =
1
PQ
(EQ 6u - u) E N"(E).
Hence, if ThA u = 0, then u e N"(E). Thus (4.4) is established.
To prove Formula (4.5) observe that, in view of (4.3), Im ThA c X"(E).
On the other hand, if u e X "(E) then JCA u = u and so X "(E) c Im ThA.
Next note that Formula (4.3) implies that
2
11A = A
and so ThA is a projection operator. Hence we obtain from Relations (4.4) and
(4.5) the direct decomposition
QpE = X "(E) Q N(E).
(4.6)
Thus every tensor u e QpE can be uniquely decomposed in the form
u= v+ w
v e X (E), w e N(E).
The tensor v = mA u is called the skew part of u.
4.3. Dual Spaces
Suppose now that E, E* is a pair of dual spaces, and let mA be the alternator
O xp are two decomposable
for QpE*, p > 2. If x* 1 Q
Q x*p and x1 O
tensors in QpE* and QpE respectively, we have, for any 6 E Si,,
O x*p, 6(x1 QX ... Q xp)> = <x*1, XQ- 1(1)> ... <x*p, X1 (p)>
x1>...
<x*p), x p>
_ <cr 1(x*1 O ... O x*p), x1 OX ... QX xp>
Skew-Symmetric Tensors
87
and hence we obtain the relation
<u*, 6u> = <6- 1u*, u>
U* E QPE*, U E QPE,
which shows that 6 and 6-1 are dual operators.
Since
=
1
E6 and
it A=
1
E6= 1
E-1
6
it follows that iA and i A are dual operators as well, i.e.,
u* E QPE*, u e QPE.
<U*, TGAU> = <TGAU*, u>
(4.7)
The duality of iA and iA implies that the restriction of the scalar product to
the subspaces Im iA = XP(E) and Im iA = XP(E*) is again nondegenerate
and hence a duality is induced between XP(E) and XP(E*).
Suppose now that
u=x*1x®...x®x*P
and
are decomposable tensors in QPE* and QPE respectively. Then we obtain
from (4.7) the formula
<JCA(x* 1 ®... XO x*P), JCA(xl
®... ® xp)>
= <x* 1 O ... 0 x*P, TGA(x1 0 ... ® xp)>
<x*1
=
1
pQ
®... 0 x*P, 1GA(x1 XQ ... ® xp)>
*
EQ<x1,
XQ- 1(1)> ...
<x*p,
XQ- 1(p)>,
whence
1
<mA(x*1 O ... O x*P), ThA(xl ®... O xp)> = i det(<x*`, x;>).
p
4.4. The Skew-Symmetric Part of a Product
Let QE =
QPE be a tensor algebra over E and consider the subspaces
NP(E) C QPE, p > 2. It will be convenient to extend the definition to the
cases p = 1 and p = 0 by setting N 1(E) = N°(E) = 0. Accordingly we define
iA to be the identity on ®'E and Q °E and then the previously established
formulas continue to hold in the cases p = 1 and p = 0.
It follows from the definition of NP(E) that
NP(E) O O9E C NP + 9(E)
O PE O N9(E) C N ° + 9(E)
p>Oq>0
-" -
(49)
88
4
Skew-Symmetry and Symmetry in the Tensor Algebra
Now let u e QpE and v e ®"E be two arbitrary tensors. Then we can
write
u= TGA u+ u 1
u 1 E N"(E)
V= TGA v+ v 1
v 1 E Nq(E),
and
whence
UQV=TGAUQTGAV+TGAUQV1 +U1QTGAV+U1Qv1.
Applying the projection mA to this equation and observing Relations (4.9)
and (4.4), we obtain the formula
TA(U Q v) = TtA(TCA U Q TGA v).
(4.10)
Since mA is a projection operator, it follows that
TGA(TGA U Q v) = mA(u Q v) = TGA(U Q mA V).
(4.11)
PROBLEM
Show that the mapping 6: QpE -+ QpE is tensorial (see Section 3.14), where QpE
is considered as a subspace of Q(E, E*). If the dimension of E is finite, prove that 6 is
generated by the operators µ(t) and C, t being the unit tensor for E and E*.
The Factor Algebra ®E/N(E)
4.5. The Ideal N(E)
Consider the direct sum
N(E) _
N"(E).
n
Formulas (4.9) imply that N(E) is a graded ideal in the graded algebra ®E.
Suppose now that u e QpE and v e ®"E are two arbitrary tensors. Then
we have
u Q v-(-1)pgv Q u e Np + q(E).
In fact, if 6 is the permutation given by
it follows that
6(u Q v) = v Q u and EQ = (- 1)I"
(4.12)
89
The Factor Algebra ®E/N(E)
and thus Formula (4.1) yields
u Q v-(-1)Pgv Q u= u Q v- EQ 6(u Q v) E NP + q(E).
Applying the operator ThA to (4.12) we obtain the formula
u e ®E, y e ®E.
ThA(u Q v) = (-1)P%(v Q u)
(4.13)
4.6. The Algebra ®E/N(E)
Consider the canonical projection
m: ®E -4 Q E/N(E).
(4.14)
Since N(E) is an ideal in QE, a multiplication is induced in QE/N(E) by
a, be QE.
ma mb = m(a Q b)
(4.15)
It follows from (4.15) that this multiplication is associative and that m(1) is a
unit element. Since the ideal N(E) is graded in the graded algebra QE, a
gradation is induced in the factor algebra ®E/N(E) by
®E/N(E) =
m(® "E)
P
and so ®E/N(E) becomes a graded algebra. Since N 1(E) = N°(E) = 0, we
have in particular that m(®1 E) and m(® °E) are isomorphic to Q 'E = E
and ®°E = IT respectively. Consequently, we shall identify m(®1 E) and
m(p°E) with E and IT respectively.
From (4.13) we obtain the commutation relation
uv = (-1)Pgvu,
(4.16)
for every two homogeneous elements of degree p and q in the algebra
Q E/N(E).
4.7. Skew-Symmetric Tensors
Define the subspace X(E) c QE by
X (E) =
X P(E)
P
and extend the projection operators mA: ®"E - ®"E (ThA = i in ®'E and
Q°E) to a linear map mA : QE -p QE. Then we have
ker ThA = N(E)
and
Im ThA = X (E).
90
4
Skew-Symmetry and Symmetry in the Tensor Algebra
Moreover, itA is a projection operator and
Q E = N(E) Q X (E).
If p denotes the restriction of the projection is to the subspace X(E) then
p : X (E) --> ®E/N(E)
is a homogeneous linear isomorphism of degree zero. Let mx : ®E -- X (E)
be the restriction of Tx to QE, X(E). Then we have the following commutative
diagram:
QE
r
X (E)
irl
(4.17)
4.8. The Induced Scalar Product
Let E, E* be a pair of dual vector spaces. Then a scalar product is induced in
QE, ®E* (see Section 3.8). It follows from (4.7) that the restriction of this
scalar product to the subspaces X (E) and X (E*) is again nondegenerate.
Since p : X(E) =- QE/N(E) is a linear isomorphism, a scalar product <, > in
the pair Q E/N(E), Q E*/N(E*) is induced by
<pu*, pu> = p !<u*, u>
u* E Xp(E*), u e XP(E).
(4.18)
Clearly the scalar product (4.18) respects the gradation. Moreover it follows
from (4.17) and (4.18) that
<mu*, mu> = <pm Xu*, pmx u>
= <pm a*, p7cA u> = p !<m u*, mA u>
u* E QX pE*, u e ®"E.
(4.19)
Now assume that u and u* are decomposable,
u = x1 QX ... QX xp,
u* = x*1 0X ... QX x*I.
Then formulas (4.19) and (4.8) yield
<m(x*1 O
... O x*p), m(x1 O ... Q xp)> = det(<x*`, xi>).
(4.20)
Suppose now that E is an inner product space. Then E is dual to itself
with respect to the inner product and hence we may set E* = E. It follows
from Section 3.15 that the induced scalar product in QE is again nondegenerate and, hence, so is its restriction to the subspace X(E). Hence an inner
91
Symmetric Tensors
product is determined in the factor space ®E/N(E) such that
(mu, itv) = p !(u, v)
u, v e X (E)
(cf. Formula 4.19). Formula (4.20) yields the relation
(m(x1 ®... Q xp), m(Y O ... O Yp)) = det(xi, y)
xi E E, y; E E.
PROBLEM
Define a multiplication in X (E) such that the linear map p : X (E) - ®E/N(E)
becomes an isomorphism. (This multiplication is necessarily uniquely determined.)
Prove that
?LA u
?LA U = ?LA (u Ox U)
u, v e Q E.
Symmetric Tensors
4.9. The Space Mp(E)
Consider the subspace M(E) of ppE generated (linearly) by the tensors
u - iu where u e QpE and i is a transposition. The space Mp(E) is stable
under every transposition. In fact, if v = u - iu is a generator of Mp(E) and
t' is a transposition we have
= (i'u - u) - (iu - u) + (iu - ftu) E Mp(E).
The same argument as in Section 4.1 shows that
u - 6u e M(E)
(4.21)
for every u e QpE and every permutation 6.
4.10. The Symmetrizer
A tensor u e QpE is called symmetric, if
6u=u
6ESp.
The symmetric tensors form a subspace Y"(E) of ppE.
Next consider the linear map Its: QpE -+ QpE given by
1
TGS=
(4.22)
pQ
An argument similar to the one given in Section 4.2 shows that
ker ms = MP(E)
(4.23)
92
4 Skew-Symmetry and Symmetry in the Tensor Algebra
and
Im 7s = YP(E).
(4.24)
Moreover, 7s is a projection operator,
(4.25)
7s = 7s
Thus we have the direct decomposition
Q pE = Y(E) 8 M"(E).
The operator 7s is called the symmetrizer in ppE and
(4.26)
7s u
is called the
symmetric part of u.
4.11. Dual Spaces
Suppose now that E, E* is a pair of dual spaces and let is be the symmetrizer
for ppE* (p >_ 2). The same argument as that used in Section 4.3 shows that
7s and is are dual operators,
u * E ®P*, u e ppE.
<u *, s u > = <Su *, u >
(4.27)
It follows from (4.27) that the restriction of the scalar product <, > to the
subspaces Yp(E*), YP(E) is again nondegenerate.
Now let
and
be decomposable tensors. Then we obtain from (4.27) and (4.25) that
<lrs(x * 1 O ... Ox x *p), TGs(xl
® ... ® xp)
= <x* 1 O ... O x*p, TGS (x 1 OX ... Q xp)>
= <x * 1 O ... O x *p, TGS(x 1 OX ... ® xp)
x* i x
=1
PQ
...
x*p x
(4.28)
Introducing the permanent of a p x p-matrix cx by
perm(x) _
aQc 1
...
Q
we can rewrite (4.28) in the form
1
<lrs(x*l
®... p x*p), 1s(xi O ... O xp)> = - perm(<x*`, x;>). (4.29)
P
4.12. The Symmetric Part of a Product
Let ®E be a tensor algebra over E, and consider the subspaces M"(E) c
ppE. As before we set M°(E) = M1(E) = 0, and define 1s to be the identity
on ®°E and p 'E. Then the formulas developed above continue to hold.
The Factor Algebra QE/M(E)
93
Now let v = u - -cu be any generator of MP(E), p > 2, and let w e ®"E
be an arbitrary tensor. Then we have
v®w=u®x w-iu®x w=upwwhere -r' E SP+q denotes the transposition given by
i '(v) =
i(v)
l< v< p,
v
p+
1<v<
_ _ p +q.
It follows that
MP(E) Q ®E c MP + q(E).
(4.30a)
®"E Ox M(E) C MP + q(E).
(4.30b)
Similarly we obtain
Now by the same argument as in Section 4.4 it is shown that
1s(u O v) _
u O 7s v)
_ 1S(u Q 1C v) _ lcs(lcsu Q v)
u E ®E, v e ®E.
(4.31)
PROBLEMS
1. Show that a bilinear mapping p: E x E -+ G can be uniquely written in the form
P = (p i +
where 1i is symmetric and p2 is skew-symmetric.
2. For each p > 2 prove that
X (E) n Y(E) = 0.
3. Let p: E x
.
x E -+ F be a symmetric p-linear mapping such that
p(x,... , x)=0
x E E.
Show that p = 0.
The Factor Algebra ®E/M(E)
4.13. The Ideal M(E)
Consider the direct sum
M(E) _
MP(E).
P
Formulas (4.30a) and (4.30b) imply that M(E) is a graded ideal in the graded
algebra ®E.
94
Skew-Symmetry and Symmetry in the Tensor Algebra
4
Suppose now that u e ®E and v e ®E are two arbitrary tensors. A
calculation similar to that in Section 4.5 shows that
u 0O v- v Qx u e M(E).
(4.32)
4.14. The Algebra QE/M(E)
Consider the canonical projection
it : Q E - Q E/M(E).
Since M(E) is an ideal in ®E, a multiplication is induced in ®E/M(E) such
that
it(a Q b) = ma itb
a, b e ®E.
(4.33)
It follows from (4.33) that this multiplication is associative, and that it(1) is
a unit element. (4.32) implies that the multiplication is commutative as well.
Since M(E) is graded, a gradation is induced in the factor algebra by
®E/M(E) =
ir(O PE)
P
and so ®E/M(E) becomes a graded algebra. Since M 1(E) = M°(E) = 0,
the restriction of it to Q 'E = E and Q °E = IT is an isomorphism. Consequently we identify it(Q 1 E) with E and it(Q °E) with F.
4.15. Symmetric Tensors
Let Y(E) c ®E be the space defined by
Y(E) =
YP(E).
P
Extend the projection operators its: QPE - QPE (its = l in Q 'E and ®°E)
to a linear map its : ®E - ®E. Then
ker is = M(E)
Im ms = Y(E)
and
Q E = M(E) Q Y(E).
The restriction 6 of it to the subspace Y(E) is a homogeneous linear isomorphism
6: Y(E) - QE/M(E)
95
The Factor Algebra QE/M(E)
of degree zero. If icY : ®E -- Y(E) denotes the restriction of its to ®E, Y(E) we
have the following commutative diagram :
Y(E)
OE
(4.34)
irl
4.16. The Induced Scalar Product
Let E and E* be a pair of dual vector spaces and consider the induced scalar
product in ®E and ®E*. According to Section 4.11 the restriction of this
scalar product to the subspaces Y(E) and Y(E*) is again nondegenerate.
Consequently, a scalar product is induced in the factor spaces QE/M(E),
QE*/M(E*) such that
u* E Yp(E*), u e Yp(E).
<6u*, 6u> = p ! <u*, u>
(4.35)
Clearly, the scalar product (4.35) respects the gradations. Moreover, it
follows from (4.35) and (4.34) that
<itu*, mu> = p !<msu*, msu>
u* E QE*, u e QE.
Now let u* and u be decomposable,
u = xi Q...Qxp,
u* = x*1 Q...Qx*p.
Then formulas (4.36) and (4.29) yield
O ... O x*p), m(xi O ... O xp)> = perm(<x*i, x;>).
(4.36)
5
Exterior Algebra
For this chapter E denotes a vector space over a field of characteristic zero.
Skew-Symmetric Mappings
5.1. Skew-Symmetric Mappings
Let E and F be two vector spaces and let
P
be a p-linear mapping. Then every permutation 6 E SP determines another
p-linear mapping hip given by
6(p(.7C 1,
... , xP) = P(aCQ(1) , ... , .7CQ(P)).
It follows immediately that
(#r6)q,
=
for every two permutations, and that
icp = co,
where i is the identity permutation. A p-linear mapping p is called skewsymmetric if
6(p = EQ (p
for every permutation 6 where EQ = + 1 or -1 if the permutation is, respectively, even or odd.
A p-linear mapping co is skew-symmetric if and only if
96
97
Skew-Symmetric Mappings
for every transposition -r. In fact, since a transposition is an odd permutation
it follows that every skew symmetric mapping satisfies (5.1). Conversely,
assume that co is a p-linear mapping which satisfies (5.1) and let 6 be an
arbitrary permutation. Now 6 is a product of m transpositions where m is
even (odd) if 6 is an even (odd) permutation. It follows that co satisfies
6cp = EQ co, and hence co is skew-symmetric.
This result implies that a p-linear mapping, gyp, is skew-symmetric if and
only if
p(x1,
..., xp) = 0,
(5.2)
whenever x, = x; for at least one pair i j. In fact, suppose that co is skewsymmetric and assume that x, = x; (i j). Let -r be the transposition i ± j.
Then
p(x 1, ..., xp) = - -rp(x 1, ..., x p) = - q (x 1, ..., x p),
whence p(x1, ..., xp) = 0. Conversely, assume that co satisfies (5.2). Then
if -r: i ± j is any transposition, it follows that
+ p(x1,...,x,..
xi
= p(x1,...,x, + x,...,x, + x;,...,xp)
cP+icp=0.
Hence p is skew-symmetric.
Since every transposition -r is a product of an odd number of transpositions of the form i ± i + 1, it follows that a p-linear mapping p is skewsymmetric if and only if
p(x 1,
..., xp) = 0,
whenever x, = x, + 1
1 <i <p - 1.
Formula (5.2) implies that a p-linear mapping, gyp, is skew-symmetric if
and only if
p(x 1,
... , xp) =0,
(5.3)
whenever the x, are linearly dependent. In fact, it is clear that this condition
implies that p is skew-symmetric. Conversely, let co be a skew-symmetric
map. Then if x 1, ... , xp are any linearly dependent vectors, we have
x; =
(for some j).
t1x,
Without loss of generality, we may assume j = p. It follows that
p-1
P(x l ... , xp) =
which proves (5.3).
) p(x 1, ... , xp _ 1, x,) = 0,
i
i=1
98
5 Exterior Algebra
From every p-linear mapping co we can obtain a skew-symmetric p-linear
mapping A by setting
1
A=
PQ
EQ 6cp.
To show that A is indeed skew-symmetric let p be an arbitrary permutation.
Then it follows that
p ! P(Aq,) =
EQ(P6)
EQ
Q
Q
= EP
Ep
t
Et P = Ep P !
whence p(Aq,) = Ep A.
The mapping A is called the skew-symmetric part of co, and the operator
A : p - Ac1, is called the antisymmetry operator. If co is skew-symmetric itself
we have 6cp = EQ co for every 6 and hence it follows that
-=cp.
PQ
1
This equation shows that a skew-symmetric mapping coincides with its
skew-symmetric part. Since Ac1, is skew-symmetric it follows that A 2 = A.
Proposition 5.1.1. Let
xE
cp:E x
p
be a p-linear mapping , where F is an arbitrary vector space, and let
f : ®"E - F be the linear map induced by co. Then p is skew-symmetric
if and only if N"(E) c ker f.
PROOF. co is skew-symmetric if and only if tp(x1, ... , xp) = 0, whenever
x = x for some pair (i , j), i j. But
p(x 1, ... , xp) = f (x 1 © ... ® xp)
and so co is skew-symmetric if and only if f is zero on the generators of
N"(E); i.e., if and only if N"(E) c ker f.
As an example of a p-linear skew-symmetric mapping, consider the plinear mapping
p
99
Exterior Algebra
defined by
°
where ThA is the alternator (see Section 4.2). Since ker mA = N"(E), it follows
from the proposition that /'A is skew-symmetric.
PROBLEMS
x E - F and by
1. Denote by Lp(E; F) the space of all p-linear mappings (p : E x
Ap(E; F) the subspace of skew-symmetric mappings. Assume that
T:Lp(E;F) - Lp(E;F)
is a linear map such that
T cp = (p
(p E A p(E ; F)
and
T(6p) = EQT((p)
6 E Si,, (p e Lp(E; F).
Show that T is the antisymmetry operator.
2. Let E be an n-dimensional vector space and suppose that A
0 is a determinant
function in E.
(a) Given an n-linear skew-symmetric mapping (p : E x
there is a unique vector b e F such that
x E -+ F show that
(P(x 1, ... , xp) = O(x 1, ... ,
(b) Show that every (n - 1)-linear skew-symmetric function b can be written in
the form
(x 1, ... , x,,_ 1) = O(x 1, ... , x,,_ 1, app),
where a E E is a fixed vector. Hint: Consider the n-linear mapping (p defined by
(-1)"-1b(xl,...,X.,...,x,,);.
=1
Exterior Algebra
5.2. The Universal Property
Let
x E- A
np:E x
p
be a skew-symmetric p-linear map from E to a vector space A. We shall say
that n p has the universal property (for skew-symmetric maps) if it satisfies
the following conditions:
A 1: The vectors A "(x1, ... , xp) (x, E E) generate A, or equivalently,
Im A p = A.
5 Exterior Algebra
100
A2:
If p is a skew-symmetric p-linear mapping from E into any vector
space H, then there exists a linear map f : A -p H such that the
diagram
'H
P
n°I
commutes.
A
Conditions A 1 and A 2 are equivalent to the following condition (the
proof being the same as in Section 1.4)
n : If
cp:E x
x
P
is a skew-symmetric p-linear mapping, there is a unique linear map
f : A - H such that the diagram above commutes.
The skew-symmetry of A p implies that
whenever the vectors x1, ..., xp are linearly dependent. On the other hand,
if the vectors x 1, ..., xp are linearly independent, then A "(x1, ... , xp) 0.
In fact, assume that A "(x1, ..., x,,) = 0. Then, by A2, /i(x 1, ..., x,,) = 0
for every skew-symmetric p-linear mapping i'/. In particular,
p= 0.
JCA(x 1
Thus, by Section 4.2, the vectors xv are linearly dependent.
5.3. Uniqueness and Existence
Suppose that
A
and
are two p-linear mappings with the universal property. Then there are linear
maps
f: A - A and g: A - A
such that
f o A"= A" and g o n p= A ".
Exterior Algebra
101
Now condition n 1 implies that
gof= i and f i g= i.
Thus f and g are inverse isomorphisms.
To prove existence, set
A "E = Qx "E/N"(E)
(see Section 4.1) and let A" denote the p-linear mapping
P
defined by
A "(x 1,
..., x p) = it(x 1 Q ... Q xp),
(5.4)
where it denotes the projection (see Section 4.6). In view of Proposition
5.1.1, A "is skew-symmetric. Property A 1 follows directly from the definition.
To verify A 2' let
q:E x
x
P
be any p-linear skew-symmetric mapping. Then p determines a linear map
h : p "E - H such that
h(x 1 Q ... Q xp) = p(x 1,
..., xp)
xy e E
(5.5)
(see Section 1.20). Since p is skew-symmetric, it reduces to zero in the
subspace N"(E) and so it induces a linear map f : A "E -p H such that
fom=h.
Combining this relation with (5.5) we obtain
cp(x 1, ... , xp) = f TG(x 1 Qx ... QX x p) = f A "(x1,..., x p)
whence co = f o A".
Definition. The pth exterior power of E is a pair (A, A p), where
A":E x
x
P
is a skew-symmetric p-linear mapping with the universal property. The
space A (which is uniquely determined up to an isomorphism) will also be
called the pth exterior algebra of E and is denoted by A "E.
The elements of A "E are called p-vectors.
5 Exterior Algebra
102
Next we shall give a description of the pth exterior power of E in terms
of the subspace X "(E) c ®"E of skew-symmetric tensors. Consider the skewx E -+ ®"E given by
symmetric p-linear mapping p: E x
P(x 1
..., xp) _ iA(x 1 Q ... Q xp)
xy e E,
where itA denotes the alternator (see Section 4.2). By the universal property
it induces a linear map r: n pE -+ ®"E. We show that the diagram
QpE
AE -
,
opE
commutes. In fact,
rpc(x 1 Q ... Q xp) = i n "(x 1, ..., xp) = p(x 1, ... , xp)
_ iA(x 1 Qx ... ® xp)
and so
r O it _ TGA .
(5.6)
This relation implies that Im i = Im ltA (since it is surjective). But Im itA =
X "(E) and so we have
Im r = X "(E).
On the other hand, it is easy to check the relation
i
i
determines a linear isomorphism from
n pE onto X "(E).
r: n pE 3 X "(E).
Remark. Since i is injective, the formula i o 7 = iA implies that ker iA =
ker it. Thus we have the relation ker iA = N"(E) which was proved in
Section 4.2 in a different way.
5.4. Exterior Algebra
Extend the definition of n pE to the cases p = 1 and p = 0 by setting
n 1E = E and A °E = IT. Consider the direct sum, n E, of the spaces n pE
(p = 0, 1,...) and identify each n pE with its image under the inclusion
map. Then we can write
AE _
ARE.
p=o
103
Exterior Algebra
The projections it : QpE -+ A "E (with kernel N"(E)) determine a projection
ir:QE--, AE
with kernel N(E) = >J N"(E). Thus we have a linear isomorphism
f : Q E/N(E) =- AE.
Now recall from Section 4.6 that ®E/N(E) is an associative algebra. Hence
there is a unique multiplication in A E, denoted by A, such that f becomes
an algebra isomorphism. Thus we have
uE AE, vE AE,
u A v = m(u Qx v)
where u e ®E, v e ®E are elements such that mu = u and iv = v. This
multiplication makes A E into an associative algebra with the scalar 1 as
unit element. It is called the exterior algebra over E. It is generated (as an
algebra) by the vectors x e E and the scalar 1.
Formula (5.4) can now be written in the form
A "(x 1, ... , xp) = x 1 A ... A xp
x E E.
From (4.16) we obtain the relation
uAv=(-1)"vAu
ueARE, veAgE.
In particular,
u A v= v A u if p or q is even.
(5.7)
and
u AU =0 if p is odd.
The kth exterior power of an element u e A E is defined by
uk =
1
k.
k>1 ,
Au
uA
k
It follows that
Uk A U1 =
k+l)uk+1
k
u E A E.
Now let u e A "E and v e A qE be arbitrary and assume that p or q is
even. Then it follows from (5.7) that u A v = v A u. This yields the binomial
formula
(u + v)k =
u` A vj
i+ j=k
for u e A "E, v e A qE, p or q even.
5 Exterior Algebra
104
Now we shall describe the exterior algebra n E in terms of the subspaces
XP(E) ®"E (see Section 4.2). Consider the direct sum
X (E) =
X "(E)
P
and define a multiplication in X(E), denoted by r, by setting
a r b= TCA(a Q b)
a, b e X (E),
where mA denotes the alternator. Recall from Section 5.3 the linear isomorphisms ri : n "E 3 X (E). We shall show that ri preserves products,
ri(u n v) = ri(u) r i(v)
u e A "E, v e n qE.
In fact, write
u,vEQE.
v=m15
u=TCU,
Then we have, in view of the commutative triangle in Section 5.3 and Formula
(4.10)
ri(u A v) = ri(TC u A miS) =rim (u Q v)
= TCA(u OX v) = TCA(TCA u OX TCA v)
N
N
N
N
= TCA u n TCA v = riTCU r r7TCV = riu r riv.
Thus ri is an algebra isomorphism ri : n E
X (E).
5.5. The Universal Property of A E
Let A be an associative algebra with unit element a and let
h:AE - A
be a homomorphism. Then a linear map o : E - A is defined by
'p=h°i,
where i is the injection of E into n E. It follows from (5.7) that
(rpx)2 = 0
x E E.
(5.8)
Conversely, assume that 'p : E - A is a linear map satisfying (5.8).
Then there exists precisely one homomorphism h : n E - A such that
h(1) = e and
'p = ho i.
For the proof we note first that (5.8) implies that
cpx coy + coy cpx = 0
x, y e E.
105
Exterior Algebra
In fact, if x, y e E are arbitrary elements, we have
cpx coy + coy cpx = q (x + y). q (x + y) - cpx cpx - coy coy =
0.
To define h consider, for every p >_ 2, the p-linear mapping
x E-+A
oc:E x
p
defined by
a(x 1, ... , x p) = cpx, ... cpx p .
Then it follows from (5.9) that a is skew-symmetric and hence there exists a
linear map h": n pE -+ A such that
hP(x 1 n ... n xp) = (px, ... cpxp
p > 2.
Define h 1 and h° by h 1 = co and h°(1) = e and let h : n E -+ A be the linear
map whose restriction to n PE is equal to h", p >_ 0.
To prove that h is a homomorphism, let
u = xl n
n xp and
v = x,+, n ... n xp+q
be two decomposable elements. Then we have
h(u n v) = h(xl n ... A xp+q) = (x1 ... (pxp+q
= (px, ... (px p) (px p +, ... (px p + q)
= h(x, n ... A
xp) . h(xP + 1 n ... A x p + q) = hu hv.
The uniqueness of h follows from the fact that the algebra n E is generated by
the vectors of E and the scalar 1.
If A is a positively graded associative algebra, A = >P Ap, and co is a linear
map of E into A 1 it follows that the extending homomorphism h is of degree
zero.
Let U be an associative algebra with unit element 1 and let E : E -+ U
be a linear map with the following property : If o : E -+ A is a linear map
0,
into any associative algebra A with unit element a such that
h
:
U
-+
A
such
that
x e E, then there exists precisely one homomorphism
h(1)=e and
h o E _ cp.
Then we say that the pair (U, E) has the universal exterior algebra property.
An argument completely analogous to that found in Section 3.4 shows
that if the pairs (U, E) and (U', E') have the universal exterior algebra property
then there exists a unique isomorphism f : U -+ U' such that f o E = E'.
It follows from the results of this section that the pair (n E, i) has the
universal exterior algebra property, where i : E -+ n E is the inclusion map.
Now the above uniqueness theorem implies that for every universal pair
(U, E) there exists a unique isomorphism f : n E -+ U such that f o i = E.
5 Exterior Algebra
106
5.6. Exterior Algebra Over Dual Spaces
Let E, E* be a pair of dual vector spaces, and consider the exterior algebras
over E and E*. In view of the induced isomorphisms
f: ®E/N(E) n E and
g: Q E*/N(E*) 3 E*
it follows from Section 4.8 that a scalar product <, > may be defined between
n E and A E* such that
x*1
n ... n x*P, xi n ... A xP> = det(<x*`, xj>)
p
1
(5.10)
A,,1u E lT
if pq.
Condition A 1 implies that the scalar product <, > is uniquely determined
by (5.10). We also have that the restriction of < , > to the pair n PE*, A PE
is nondegenerate for each p, and so induces a duality between these spaces.
In particular, the restriction of < , > to n 1E* = E* and A 'E = E is just
the original scalar product.
Now expanding the determinant in (5.10) by the ith row we obtain the
formula
lx*
\ 1 n ... A x*P, x 1 n ... n XP> _
P
(- 1)i + j<x*i, xi>,
j=1
<x* 1 n ... n x*J n ... n x*P, x i n ... n xi n ... A XP>
p> 2.
(5.11)
5.7. Exterior Algebra Over a Vector Space of Finite Dimension
Suppose now that E is a vector space of dimension n and let {ev} (v = 1, ... , n)
be a basis of E. Then the products
ev 1 A ... n evp , (v1 < ... < vi,)
(5.12)
form a basis for n PE. In fact, it follows immediately from A 1 that the products (5.12) generate n PE. To prove the linear independence, let E* be a
dual space of E. If a*v (v = 1, ..., n) is the dual basis in E* we have, in view
of (5.10),
<e*''1 n
n e*vp, eµ1 n
.. A eµp> = det(<e*vi,
det(b).
(5.13)
This formula shows that the products ev n
A e*vp
n evp and e*v 1 A
(v1 <
< vP) form dual bases of n PE and A PE*. Hence, the dimension
of n PE is given by
,
1
dim AE _
n
P
(0 < p < n).
(5.14)
Exterior Algebra
1 07
For the dimension of the exterior algebra n E we obtain from (5.14)
it
dimAE=
n
=2",
p=o \PJ
while the Poincare polynomial of the graded vector space n E is given by
P(t) =
n
"
p=o P
t" = (1 + t)".
(5.15)
Every p-vector u can be uniquely represented in the form
cv1, ...,
u=
<
.Pev1 A ... A ev P
where the symbol < indicates that the indices (v 1, ... , vp) are subject to
VP
the condition v 1 <
< vp . The coefficients
are called the comV1 ° .
°
ponents of the p-vector u with respect to the basis {ev} of E. Formula (5.13)
implies that the scalar product of two p-vectors
U
=V1s ..., VPev1 A
.A
<
eyP
and
e*v1 A ... A e
*Vp
is given by
<u*,
u\/ =
V1, ..., vP'1
vt, ,vP
Inner product spaces. Suppose now that E is an inner product space and set
E* = E. Let A E be an exterior algebra over E. The isomorphism
f : QE/N(E) -- AE
(see Section 5.6) determines an inner product in n E such that
(x1 A ... A xp, Y1 A ... A yp) = det(x1, y).
If E is Euclidean and {eV} (v = 1, ... , n) is an orthonormal basis of E, it
follows that the products
eV 1 A ... A evP
v1 < ... < vp
form an orthonormal basis of n "E.
PROBLEMS
1. Let E, E* be a pair of n-dimensional dual vector spaces and consider the subspace
A c L(E*; E) consisting of the transformations satisfying p* = - 'p. Define a
bilinear mapping 'p : E x E - A by
cpa,bx* = <x*, a>b - <x*, b>a.
Prove that the pair (A, 'p) is a second exterior power of E.
5 Exterior Algebra
108
2. Show that a 2-vector z is decomposable if and only if z n z = 0.
3. Let E be a vector space of dimension n. Show that a 2-vector is decomposable if
and only if the matrix of its components (with respect to a basis of E) has rank 1.
4. Establish the general Lagrange identity
n
n
v1
v1
b11
=
1 ...
v1
n
v
...
niP
ZP1
...
'pP
yIpl
...
IpP
bvnp
p v
Hint: Employing a pair of dual bases {e*v},
x=
'11
)det(:
det
n
v1
...iP
Jev
and y*` =
(v = 1, ... , n), consider the vectors
(i = 1, ... , P)
'1 e*
v
Evaluate the scalar product
<x i n ... n xp , y* 1 n ... n
y*p>
in two different ways.
5. Assume that E has finite dimension, and consider a differentiable mapping
(J - n pE, t H u(t). Establish the formula
d
Uk = _U n Uk -
dt
1.
dt
6. Let E be any vector space and n E be an exterior algebra over E. Define a new
multiplication in the space n E by
unv=
+ q)
UAV
P!q!
u E n pE, v E n qE.
Prove that the resulting algebra n E is again an exterior algebra over E.
7. Consider the subspace _ o n 21'E of n E. Show that this subspace is a commutative
algebra which is algebraically generated by 1 together with a set of elements
satisfying w = 0. This algebra is called a Nolting algebra over E.
Homomorphisms, Derivations
and Antiderivations
5.8. The Induced Homomorphism
Let A E and A F be exterior algebras over the vector spaces E and F and
assume that a linear map co : E -p F is given. Then co can be extended in a
unique way to a homomorphism cP A , n E - n F such that PA(l) = 1. To
prove this consider the linear map rl : E - n F defined by ri = i o co where i
109
Homomorphisms, Derivations and Antiderivations
denotes the injection of F into A F. Then we have for every x e E
ri(x) A ri(x) = cpx A cpx = 0
(since
cpx e F).
Now it follows from the universal property of A E (see Section 5.5) that ri
(and hence co) can be extended in a unique way to a homomorphism
A E - A F. Clearly 'P is homogeneous of degree zero. Since 'P
A
preserves products we have
'A(u A v) = 'Au A (PAv
u,ve AE.
(5.16)
In terms of the multiplication operator (see Section 5.13) this can be written as
'PA
oµ(U)
=
n
u) ° 'P n
u e AE.
(5.17)
From (5.16) we obtain the formula
'Pn(x1 A...Axp)=(Px1 A...Acpxp
x,EE.
(5.18)
It follows immediately from (5.18) that Im 'P A = A (Im 'P). In particular,
if 'P is an onto map then so is 'P A The kernel of 'P A will be discussed in
Section 5.24.
For the identity map i : E - E we have obviously
(5.19)
=l
while the homomorphism (- z),, is given by (- i) u = (-1)"u, u e A "E;
hence (- z),, is the canonical involution of the algebra A E.
If ,Ii is a linear map of F into a third vector space G and A G is an exterior
algebra over G, it is clear that
(/' °'P) =
(5.20)
'/' A °'A
Formulas (5.20) and (5.19) imply that P A is injective whenever (p is injective.
In fact, if 'P is injective, there exists a linear map /i : E - F such that /' ° 'P = 1.
Now formulas (5.20) and (5.19) yield
/'A ° 'PA = (`Y ° 'P)
=ln=l
and hence 'P n is injective.
In particular, if E 1 is a subspace of E, then the injection i : E 1 - E induces
a monomorphism i A: A E 1 - A E. Hence, A E 1 can be identified with a
subalgebra of A E.
5.9. Dual Mappings
Suppose now that E*, E and F*, F are two pairs of dual spaces and that two
dual mappings
and 'P*:E*4 F*
5 Exterior Algebra
110
are given. Consider the induced mappings
(P A: A E - AF and ((p*) : A E* E- AF*.
The mapping (gyp*) A will be denoted by p ^ . Then co A and co ^ are again
dual,
(5.21)
SPA= (SPA)*
In fact let u e A E and v* E A E* be two arbitrary elements. Since co A and
P ^ are both homogeneous of degree zero we may assume that u and v*
are homogeneous of the same degree, say p. Furthermore, we may assume
that the p-vectors u and v* are both decomposable
U* = y* l A ... A
u= x 1 A ... A xp ,
y*P.
Then relations (5.18) and (5.10) yield
<v*, (p, u> = < y* 1 A ... A y*P, (px 1
A ... A Pxp
= det(<Y*`, px;>) = det(< p*Y*`, x;>)
=
A ... A (p*y*P, x1 A ... A xp> = <A v*, u>
whence
<v*, (PA u> = <rp^v*, u>
v* E AE*, u e A E.
If G, G* is a third pair of dual spaces with exterior algebras A G and A G*
and if l//: F -+ G, /i* : F* E- G* is another pair of dual mappings, we have
( ° P) = ((/I° q)*) = ((P* ° II*)A =
(I*)A = (P^ o//
whence
( ° P) = (PA ° ,A
(5.22)
5.10. The Induced Derivation
Let p: E -+ E be a linear transformation. Then p can be extended in a unique
way to a derivation, 0 A (gyp), in the algebra A E. The uniqueness of 0A(p)
follows immediately from the fact that the algebra A E is generated by the
vectors of E and the unit element 1.
To prove existence, consider the p-linear mapping
t/ip:E x
x E--> APE
n
defined by
P
Y'
(x1, ... , xp) =
x 1 A ... A (pxi A ... A xp
i= 1
I'1=(P and
/io=0.
p>2
Homomorphisms, Derivations and Antiderivations
111
If for some i <j, xi = x; = x, then
+x1 A...AXA".A(pXA
AXP
=(-1x1 A...AxAgxA...AxP
+(-1}'-1-`x1 A...AxAgxA...AxP
=0.
Hence P is skew-symmetric. It follows that there is a linear map,
8 A (gyp) : A E -+ A E such that
P
en
(x 1 A ... A xp) =
x 1 A ... A (pxi A
A XP
p>2
(5.23)
i= 1
and such that
8 A (cp)x = cpx
o
e
x E E,
a. e F.
Clearly OA(P) is a homogeneous (of degree 0) linear map extending 'p. To
prove that 8 A ('p) is a derivation let u = x1 A
A xP and v = Y1 A
A Yq
be arbitrary decomposable p- and q-vectors. Then
e n (co)
(u A v) = e n('p) (x 1 A ... A x p A Y 1 A ... A Yq)
_
P
(x1 A "A 'pXA . A xP AYlA...AYq
i= 1
q
+
(x 1 A ... A xp) A (Y 1 A ... A 'Yj A ... A Yq)
i= 1
= 8 A (cp)u A v+ u A 8 A (cp)v.
Now the linearity of 8 ('p) gives
8 A ('p) (u A v) = 8 A (cp)u
A v+ u A 8 A (cp)v
u, y e AE
and hence 8 A ('p) is a derivation. In terms of the multiplication operator this
formula reads
8 n ('p)
° z(u) = µ(e n ('p)u) + z(u) ° e n ('p).
(5.24)
For the identity map we obtain that
e A (l)u = pu
u E n PE.
If l//: E -+ E is a second linear transformation, then we have the relation
8 A (q // - /,'P) = 8 A ('p)e A (//)
- 8 (//)e
A
('P).
(5.25)
5 Exterior Algebra
112
For the proof, we notice first that the operation on each side of (5.25) is a
derivation in A E (see Section 5.6 of Linear Algebra) and consequently it
is sufficient to consider the restriction of these operators to E. But in this case
(5.25) is trivial.
Now suppose that cP : E - E, q : E* F-- E* is a dual pair of linear maps,
and consider the induced derivations OA(P) and 8 ^ (cp) = 8 ^ (gyp*). It will
be shown that 8 ^ (gyp) and 8 A (p) again form a dual pair
(5.26)
A x' e APE*. Then, in
Let u = x1 A
A x1 ,e APE and u* = x* 1 A
view of Formula (5.11), we have
P
<.x* 1 A ... A x*P, x 1 A ... A
<u*, 0A ()u> =
A ... A XP>
i= 1
p
(- 1)<x*j, (Pxi> <x* 1 A ... A x*1 A ... A x*p,
i, j = 1
x 1 A ... A 4xi A ... A XP>
/N
P
(
1)1+'<q*x, xi> <x*1 A
i, j = 1
_
Ax*jA...AX*P,
x1A...A4XIA...AX>
P
<x *1 n
n P*
j= 1
xA ... A x*P x 1 A ... A xP
= <e ^ (co)u*, u>,
whence
<u*, 8 ^ (cp)u> = <8 ^ (cp)u*, u>
u* e A E*, u E AE.
(5.27)
If /i : E - E and /i* : E* F- E* is a second pair of dual mappings, then
formulas (5.25) and (5.26) yield
e^
*/,*) = e ^ (//*)e ^ (p*) - e ^
e^(/,)e^(p) e^(p)e^(/i)
/,'p) = e ^ (//* p* -
-
whence
8
i/np) = 8 ^(/i)e
8
(/i).
(5.28)
5.11. Antiderivations
Let w be a homogeneous involution of degree zero in A E and let be a (not
necessarily homogeneous) antiderivation with respect to w. If co denotes the
restriction of S to E, we have
A x)=(px A x+wx A (px,
Homomorphisms, Derivations and Antiderivations
113
whence
(px A x+ wx A cpx = 0
x e E.
(5.29)
Conversely, every linear map co : E -+ A E which satisfies (5.29) can be
p) (with respect to w)
extended in a unique way into an antiderivation
of A E. Since A E is generated by E and the scalar 1, the uniqueness follows
the p-linear mapping
at once. To prove the existence of (co)
t/i p: E x x E -- A E
p
defined by
P
... , xp) _
. A wxv _ 1 A (pxv A xv + 1 A .. A x p
wx 1 A
p
2
v=1
1 = p and /io = 0.
It will be shown that
p
(5.30)
is skew-symmetric. In view of Section 5.1 it is
sufficient to verify that //(X1, ..., xi , xi + 1, ... , xp) = 0 whenever xi = xi + 1
Since wx A (Dx = 0 and x A x = 0, we obtain from (5.30) that
... , x, x, ..., xp) = wx 1 A .
. A wxi _ 1
whence, in view of (5.29),
l'/p(xl,...,X,X,...,xp) = 0.
The skew-symmetric mappings p induce a linear map cp) : A E -+ A E
such that
P
wx 1 A ... A wxv _ 1 A (pxv A xv + 1
A x p) _
SZ((p) (x 1 A
v=1
(5.31)
Now it will be shown that
involution co. Let u = x 1 A
(co)
an antiderivation with respect to the
A xp and v = xp + 1 A
A x p + q be two
decomposable elements. Then
(co) (u A U) =
(x 1 A ... A x p + q)
P
wx 1 A ... A wxv _ 1 A (pxv A xv + 1
A."AxP+q
v=1
q
+
wxl A ... A wxv_ l A (pxv A xv+ l A ... A xp+q
v=p+ 1
=
A v + (DU A
which completes the proof.
5 Exterior Algebra
114
It is clear that
p) is homogeneous of degree k if and only if Im o c
n k+ 'E. Assume now that
is in fact homogeneous of degree k.
The following two cases are of particular importance:
1. w = i A (derivations) : Condition (5.29) reduces to
(pxnx+xn(px=0
or, equivalently,
[1 + (- 1)k+ l]q x n x = 0.
(5.32)
If k is even (5.32) always holds: any linear map cp : E -+ n 2P + 'E can be
extended in a unique way to a homogeneous derivation in n E.
Now assume that k is odd. Then equation (5.32) reads
(px n x=0
xeE,
whence
(px n y=- spy n x
x, y E E.
Since k + 1 is even we have
(pynx=xn(py.
It follows that
(px n y = -x A py.
Now Formula (5.31) yields
n ... n xp) = i[1 + (-1)p+
n x2 n ... n xp.
(5.33)
It follows in particular that the restriction of cp) to every subspace n 2E
is zero.
2. w = (- i) (antiderivations) : Condition (5.29) becomes
A
(pxnx=xncpx,
i.e.,
[1 + (-1)k] cpx n x = 0.
If k is odd this condition is always fulfilled: any linear map p:E -+ n 2E
can be extended in a unique way to an antiderivation (with respect to the
canonical involution) in n E.
Now assume that k is even. Then the above equation implies that
px n x=0,
whence
ox^y= -coy nx
or equivalently
cpx n y = x n cp y.
115
Homomorphisms, Derivations and Antiderivations
Formula (5.31) now yields
n ... n xp) = 2 [1 + (- 1)P + i ] (px 1 n x2 n ... n xp.
(5.34)
to every subspace n 2E is zero.
It follows again that the restriction of
5.12. a-Antiderivations
Suppose now that F is a second vector space, WF is a homogeneous involution
of degree zero in AF and that (p : E -+ AE, /i : F -+ AF are homogeneous
mappings satisfying the conditions
(px n x+WEx n (px=0
xEE,
/iYAY+(DFYA/iY=0
yeF.
Assume further that a : E -+ F is a linear map such that
WF a A = a A WE and
/ia = a (p.
(5.35)
Then we have
(5.36)
= OC n ° E((P)
In fact, it is easy to verify that the operators
a A and a A °
are a-antiderivations (see Section 5.8 of Linear Algebra). Relation (5.35)
implies that the restrictions of these a-antiderivations to E coincide and
50 (5.36) follows.
PROBLEMS
1. Let E and F be two vector spaces, QE, QF tensor algebras and n E, A F exterior
algebras over E and F respectively. Consider the projections ltE : Q E -+ n E and
F: ®F - n F defined by
2E(x 1 0 ... QX x,,) = X1 A ... n x p
and
2F(Y 1 0 ... 0 Yq) = Y 1 n ... n Yq .
(a) If p: E - F is a linear map prove that
= 'PA °E .
(b) If p: E - E is a linear map show that
=
2. Let a e n kE be a fixed element, where k is odd. Define a linear map 9: n E -+ n E by
eu =
JaAU u E n "E, p odd,
0
u e n "E, p even.
Prove that 0 is a derivation in the algebra n E.
5 Exterior Algebra
116
3. Suppose w is an involution of degree zero in n E such that
WX n x=0
xEE.
Prove that w = l ^ or w = (- i) .
4. Show that there does not exist an involution w in n E such that (5.29) holds for every
linear map
p:E - E.
Hint: First show that for such an w the relation wx n x = 0 must hold.
5. Let E and F be vector spaces of finite dimension and consider a linear map p: E -+ F
of rank r.
Prove that
r(q) = 2'.
6. Let A E be the exterior algebra over E and consider an antiderivation S2 in A E
of degree 1. Define a new multiplication n in n E by setting
u n v= p+ q)UAV
uE npE, vE nyE.
p
Show that the operator C defined by S u = p S2u, u e A "E is an antiderivation with
respect to this multiplication.
7. Let d : ®E -+ ®E be a homogeneous linear map of degree 1 such that
d(u Q v) = du Q v + 6(u Q dv),
where 6 is a fixed permutation such that EQ = (-1)deg u. Assume that
ltA dltA = 1tA d,
where 7tA is the alternator (see Section 4.2). Define the operator S by S = 7rAd.
Prove that S is an antiderivation in the graded algebra X(E) (cf. the problem in
Section 4.8). Prove that S is a differential operator if and only if 7rA d 2 = 0.
8. Let l//: E -+ E be a linear transformation of an n-dimensional space E. Prove that
there exist uniquely determined transformations iii (i = 0, ... , n) of n E such that
n
('Y - Al)
=
tIiA"-`
AE F.
i=o
If cii") denotes the restriction of t ii to n" E, show that
tr
/4'
=ai
i=O,...,n,
where ai is the ith characteristic coefficient of cli (see Section 4.19 of Linear Algebra).
Prove that
o = (-1)nl,
li1 = (-1)"-
II' =
The Operator i(a)
117
The Operator i(a)
5.13.
Fix an element a e A E and consider the linear transformation µ(a) of n E
given by left multiplication with a,
µ(a)u = a n u
u E n E.
Since the algebra n E is associative, we have the relation
µ(a n b) = µ(a) ° µ(b)
a, b e A E.
(5.37)
Now consider the dual map
i(a) : n E* F- AE*.
It is determined by the equation
(i(a)u*, v> = <u*, a n v>
v e A E.
In particular,
i(t)u* = tu*
e F.
Now suppose that a is homogeneous of degree p. Then i(a) restricts to
linear maps
r> p
n rE* _ n r pE*,
and reduces to zero in n rE* if r <p. For u* E A PE* we have
i(a)u* = <u*, a>.
Dualizing formula (5.37) we obtain
i(a n b) = i(b) ° i(a)
a, b e A E.
(5.38)
In particular,
i(a n b) = (-1)' i(b n a)
a e n "E, b E ME.
(5.39)
Next, let F be a second vector space and let p: E -p F, p* : E* F- F* be a
pair of dual maps. Since co is a homomorphism,
'PA oµ(a) µ((PAa) °'PA
aE AE.
Dualizing this relation yields
i(a) ° 'P " = 'P " °
A
a)
a E A E.
(5.40)
Finally, let 'P and (p* be a pair of dual linear transformations of E, and consider the induced derivations H A ((P) and 8 A (p) Then Formula (5.24) implies
that
i(a) ° 8 " (rp) = i(8 A (sp)a) + 8 ^ (rp) ° i(a).
5 Exterior Algebra
118
5.14. The Operator i(h)
In this section we shall consider the operator i(h) for the special case h e E.
This operator is homogeneous of degree -1. Formula (5.39) implies that
i(h) o i(k) + i(k) o i(h) = 0
h, k e E.
In particular,
i(h)2 = 0.
Proposition 5.14.1. The operator i(h) is an antiderivation in the algebra n E*,
u* E n PE*, v* E A E*.
i(h)(u* A v*) = i(h)u* A v* +(-1)pu* A i(h)v*
PROOF. Consider the linear map tph : E* -+ IT given by
x* E E*.
(ph x* = <x*, h>,
It follows from Section 5.11 that c°h extends to an antiderivation h of degree
-1 in n E* (with respect to the canonical involution). We have to show that
i(h),
u* E A E*, v e A E.
v> = <u*, h n v>
(5.41)
n xp and
We may assume that u* and v are decomposable, u* = x* n
q
+
1,
both
sides
of
(5.41)
are
zero
and
so only
n xq . If p
v=x1n
the case p = q + 1 has to be considered. Then we have, in view of (5.11),
= <Qh(x* n ... A xp ), x 1 A ... A X,_1>
P
=
(v= 1
1)v-
n
1<x*, h> <x* n ... n x* A ... A
xp, x1 n ... A X_1>
and so Formula (5.41) follows.
Corollary I:
i(h) o 1u(h*) + µ(h*) o i(h) = <h*, h>l
he E, h* E E*.
PROOF. Apply the proposition with u* = h*.
Corollary II:
P
i(h)(x* n ... n xp) _
(_ 1)v -1
n
<x*
n ... n x* n ... n 4
v=1
Corollary III: Let F be a subspace of E and let Fl be its orthogonal complement.
Identify A F and A Fl with subalgebras of n E and A E* respectively. Then
i(a)(u* n v*) =(-1)pqu* n i(a)v*
a e A "F, u* E n qFl, v* E A E*.
119
The Operator i(a)
PROOF. We may assume that a = yl n
derivation, we have for y e F
n yp, yv e F. Since i(y) is an anti-
i(y)(u* n v*) = i(y)u* n v* + (=1)qu* n i(y)v*
= (_1)u* n i(y)v*.
It follows that
i(a) (u* n v*) = i(y1 n ... n yp) (u* n v*)
= i(yp) ... i(yi)(u* n v*)
= (_ 1)1 i * n i(yp) ... i(y i )v*
= (-1)P u* n i(a)v*.
Proposition 5.14.2. If an element u* E A PE* (p > 1) satisfies the equation
i(h)u* = O for every h e E, then u* = 0.
PROOF. Let v e A E be arbitrary. Since p >_ 1, we can write
v=
by n vy
by e E, vy e A
'E.
V
It follows that
=
<u*, by n vy> =
v
<l(hy)u*, vy> = 0
v
whence u* = 0.
PROBLEMS
1. Let u* e n PE*, p > 1 be an element such that i(a)u* = 0 for every a e n kE, where
k < p is a fixed integer. Prove that u* = 0.
Hint: Use the duality of the operators i(a) and µ(a).
2. Let E, E* be a pair of n-dimensional dual spaces and
bases. Given a linear transformation p: E -+ E show that
e ('P) =
{e*°} be a pair of dual
µ(coev)i(e*")
O\(p) =
3. Let E, E* be a pair of n-dimensional dual spaces. Prove the following relations:
(a)
(b)
(c)
µ(x)i(x*) + i(x*)u(x) = <x*, x>i
µ(e ji(e)u = pu
i(e*
(n - p)u
x* e E*, x e E
u e n "E
u e n "E.
5 Exterior Algebra
120
4. Let E, E* be a pair of dual spaces of dimension n = 2m. Prove the formula
h*Ai(h)UjAU1
J=
1
he
A2E*
(j = 1,...,m).
5. Show that the operator
1(a): n E* - n E*
a E A "E
is not an antiderivation unless p = 1.
6. Let a E A "E be arbitrary and assume that p < q. Prove the formula
i(a)(x*1 A
A x*q)
(-
=
1(v -1) la, x*V1 A
A x*Vp>x*Vp+ 1 A
A x*Vq,
v1<...<vp
where (vp+ 1, ... , vq) denotes the complementary ordered (q - p)-tuple.
Exterior Algebra Over a Direct Sum
5.15.
Let E and F be vector spaces and consider the anticommutative tensor
product of the graded algebras n E and A F (see Section 2.8.) On the other
hand, we have the exterior algebra over the direct sum E Q F.
Theorem 5.15.1. There is a canonical isomorphism between the graded algebras
AE Q n F and n (E Q F).
PROOF. Let
i1:E-+E +QF and
+QF
be the inclusion maps. They extend to homomorphisms
(i 1) A :
n E - n (E Q F)
and
(i2) A :
n F - n (E p F).
Hence a linear map
f:AEQ A F- A(E®F)
is given by
f (u Q v) = (i 1) A u n (2)A v.
(5.42)
Exterior Algebra Over a Direct Sum
121
We show that f is an algebra homomorphism. In fact, let u e A E, v e n qF,
u' E n rE, v' E A F. Then, if the multiplication in the algebra n E Q A F is
also denoted by A, we have
f [(u O v) n (u' O v')] _ (-1)qr f{(u n u') Q (v n v')]
_
(u n u') n (i2) (v n v')
A
n (il) n u' n (i2) v n (i2)
A
v'
= (ll)n U A (l2)A v A (ll)n U' A (l2)A v'
= f (u ® v) n f (u' ® v')
To show that f is an isomorphism we construct an inverse homomorphism.
Consider the linear map ri : E +O F -+ AE Q AF given by
ii(z) _ it 1 z 0 1+ 1 Q ic2 z
z e E +0 F,
(5.43)
where
it 1: E +Q F -- E and it2 : E +Q F -- F
are the canonical projections. Then we have
ii(z) n ii(z) _ (itlz O 1+ 1 O it2 z) n (itlz O 1+ 1 O it2 z)
_ (it1z n it1z) p 1+ it1z O it2 z- it1z Q it2 z+ 1 Q (it2 z n it2z)=0.
Hence r extends to a homomorphism
h: n (E Q F) -- AE Q AF.
Relations (5.42) and (5.43) imply that
hf(x®1)=h(i1x)=ri(ilx)=x
xeE,
hf(1Oy)=h(i2y)=rl(i2y)=y
yeF,
f h(z) = f (TG 1 Z xO 1 + 1 Q TG2 z) = i 1 T G 1 z + l2 TG2 z = z
z E E Q F.
Since the vectors x Q 1 and 1 Q y together with the scalar 1 generate the
algebra n E Q A F and the vectors x +0 y together with 1 generate the
algebra n (E Q F) it follows that
hof=l and foh=i.
Thus f is an isomorphism and h is the inverse isomorphism.
Corollary. Let E = E1 Q E2 be a direct decomposition of E into two subA E1 p n E2.
spaces. Then A E
5 Exterior Algebra
122
5.16. Direct Sums of Linear Maps
Suppose that E', F' is another pair of vector spaces and that linear maps
P : E - E', /i : F - F' are given. Then the linear map p Q /i : E Q F E' Q F' is given by
P +O = ii°(P°lt1 + i2°//°1t2,
(5.44)
where
and
denote the canonical injections. It will be shown that
+O/) =
®I1
(5.45)
From (5.44) we obtain
(' +O /,) ° it = ii °' and
//) ° i2 = i2 ° /i
(gyp
whence
('P +O') A ° (i 1) A =
('l) A °
A
and ('P +O') A °
(2)A =
('2)A °
Now let u e A E and v e A F be two arbitrary elements. Then we have
+O ) (u O v) =
=
+O Y') n ((i
O+
l) n u A (12)A v)
u] n
O+
v]
= ('I) A 'PA U A ('2)A `/' A 1,
=
('PA
OY1n)(UOv)
whence (5.45).
Now consider a linear map ' : E - F and suppose that two direct
decompositions
E=E1 f3E2 and F=F1 f3F2
are given such that 'Ei c F1 and 'E2 c F2. Then it follows from (5.45)
that
'PA = ('P1) O ('P2) '
(5.46)
where 'P1: E1 -p F1 and 'P2 : E2 -p F2 are the restrictions of 'P to E1 and E2.
Formula (5.46) yields, in view of (1.12) and (1.11), that
ker 'P
= (ker ('Pa) A) O A E2 + n E 1 O (ker ('P2) A)
and
Im 'P A = Im ('P 1) A O Im ('P2) A
Exterior Algebra Over a Direct Sum
123
5.17. Derivations
Suppose p: E -+ E and l// : F -+ F are two linear maps and consider the
induced derivations OA(P). A E -+ n E and 8 A (li) . A F -+ n F. Then
0A(Y' o /') = 0A(') o f+ Z o eA(/').
(5.47)
In fact, Proposition 2.9.1 implies that the mapping on the right hand side of
(5.47) is a derivation. Hence, it is sufficient to verify that the restrictions of
the operators to E Q F coincide.
Let z e E +O F be an arbitrary vector. Then
(e (gyp) ®i + l 0 e (/i))z = (e (q) O l+ l 0 e (/i)) (i z O 1+ 1 0 lt2 z)
_ (p7L1Z Q 1 + 1 Q ,/i7c2 z = il(p1L1Z + i2l'/TG2 z
_ (gyp +O i)z = e n ('p +O i) (z)
and so Formula (5.47) follows.
5.18. Direct Sums of Dual Spaces
Consider the dual pairs E, E* and F, F* of vector spaces. Then the induced
scalar product in E Q F and E* Q F* is given by
<z*, z> _ <x*, x> + <Y*, y>
z = (x, y), z* _ (x*, Y*)
The multiplication operator in the algebra n (E Q F) is given by
µ(a Ox b) = µ(a) o wq Q µ(b)
a E A E, b E n qF,
(5.48)
where wq denotes qth iterate of the canonical involution of the graded algebra
n E (see Section 6.6 of Linear Algebra).
Dualizing (5.48) we obtain the relation
i(a Q b) = wq o i(a) Q i(b).
(5.49)
In particular it follows that
i(h Q 1) = i(h) Q i
heE
i(1 Q k) = w Qx i(k)
keE
and
whence
i(h Q k) = i(h) Q i + w Q i(k).
(5.50)
5 Exterior Algebra
124
Since the vector spaces E Q F and E* Q F* are dual, a scalar product
(, ) is induced in n (E Q F) and A (E* Q F*). It will now be shown that
this scalar product coincides with the induced scalar product if n (E Q F)
and A (E* +O F*) are considered as the tensor products n E p n F and
A E* p n F*. In other words, it will be proved that
(5.51)
(u* O v*, u O v) = <u*, u> <v*, v>
for all u e A E, v e A F, u* E A E*, v* E A F*. Without loss of generality we
may assume that all the elements in (5.51) are homogeneous,
u e ARE,
If p + r
y e n rF,
u* E n qE*,
v* e A SF*.
q + s, both sides of (5.51) are zero and hence only the case
p + r = q + s remains to be considered. Then we have
(u* O v*, u 0 v) = l(u O v)(u* O v*) = wrl(u)u* O i(v)v*
If p = q and r = s, it follows that
i(u)u* = <u*, u>
and i(v)v* = <v*, v>,
q, it follows that either p > q or r > s, so that both sides of (5.51)
are again zero.
while if p
5.19. The Diagonal Mapping
Consider now the case F = E and let the diagonal mapping A: E - E Q E
be defined by
A=i1+i2.
Then the product u* n v* of two elements u* and v* of n E* can be written
in the form
u* n v* = A ^ (u* Q v*).
(5.52)
In fact, if j 1 and j2 denote the canonical injections of E* into E* Q E*,
Formula (5.42) yields
u* O v* = (J1) ^ u* n (J2) A v* = (T1) ^ u* n (Tt2) ^ v*.
Applying A ^ we obtain
A^(u* p v*) = A^(ic1)^u* A A^(ic2)^v*
=(itloA)^u* A
(i20A)^v*=
n l^v*
= u* n V.
Formula (5.52) shows that A" is the structure map of the algebra n E*
(see Section 2.1).
Exterior Algebra Over a Direct Sum
125
5.20. Direct Sums of Several Vector Spaces
Now consider the direct sum of r vector spaces
r
E= Q E,.
p=1
Then an r-linear mapping
/i: AEl x . x AEr - nE
is defined by
..., Ur) = (l)
A
U1 A ... A (r) A ur
up e Ep, i : Ep - E.
The r-linear mapping l'/ induces a linear map
f: AEl Q ... Q AEr - nE
such that
f (u 1 ®" O ur) = (i 1) A U 1 n ... n (ir) A Ur
(5.53)
The same argument as in the case r = 2 shows that f is a homomorphism
and, in fact, an isomorphism of the graded algebra n E 1 O
O A Er
onto n E. Formula (5.53) shows that f is homogeneous of degree zero.
Hence we may write
n (E 1 +0 ... 3 Er)
n E1
0 ... j n Er
and
U 1 O .. OX Ur = (l 1) n U 1
A
A (lr) n Ur.
Comparing the homogeneous subspaces of degree p in the relation we
obtain
n (E 1 ®... 0 Er)]p~
AP1E
P1++Pr=P
l 0 ... O
A PrEr
5.21. Exterior Algebra Over a Graded Vector Space
Let E = 7.
E. be a graded vector space, where the subspaces E. are homo1
geneous of degree k.. Then there exists precisely one gradation in the algebra
n E such that the injection i : E - n E is homogeneous of degree zero.
The uniqueness follows immediately from the fact that the algebra n E
is generated by the vectors of E and the unit element 1 (which is necessarily
of degree zero). To prove the existence of such a gradation consider first
the algebra n E,. By assigning the degree pk1 to the subspace n PEA we make
n E, into a graded algebra. Now writing
AE=
AElp...p AEr,
5 Exterior Algebra
126
we recall that the gradations of the n E, induce a gradation in the algebra
n E. Clearly the injection i : E -+ n E is homogeneous of degree zero and so
the proof is complete.
The algebra n E together with the above gradation is called the graded
tensor algebra over the graded vector space E. The subspace of homogeneous elements of degree k is given by
(n E)k =
O ' ` prEr
n p1E1 O
(p)
where the sum is extended over all r-tuples (Pi' ... , pr) subject to
r
= k.
i=1
Suppose now that the vector space E has finite dimension and that the
gradation is positive. Then the Poincare polynomial of the graded space
n E, is given by
n = dim E1, i = 1, ..., r.
Pi(t) = (1 + t")"i
Since the space n E is a tensor product of the spaces n E. we obtain for the
Poincare polynomial P(t) of A E (in view of Section 2.6) the expression
P(t) = (1 + tk 1)" 1 ... (1 + tkr)"r.
PROBLEMS
1. Let E be a vector space and A E be an exterior algebra over E. Show that
u,veAE
uAv=ltA(u®x v)
where n and
are the canonical projections of E Q+ E onto E and it = n + ?t2.
2. Let E = E 1 + E2 be a decomposition of E and set E 12 = E 1 n E2.
(a) Establish a natural isomorphism
/i:E1/E12 O+ E2/E12
E/E12
(b) Consider the canonical projections
p2: E2 - E2/E12
p1: E1 -+ E1/E12
p: E -+ E/E12
and let p : E 1 Q+ E2 - E be the linear map given by
p(x1, x2) = x1 + x2
x1eE1,x2eE2.
Show that the diagram
AE1 Q AE2
AE
cpl)
(E1/E12) Ox n (E2/E12) = n (E/E12)
is commutative.
Ideals in AE
127
Ideals in AE
5.22. Graded Ideals
Suppose that
1 =>
IP=1n APE,
P,
P
is a graded left ideal in the algebra A E. Then we have, for every p-vector
u e A PE and any element v= L vq E 1,
v A U=> vq A U=
q
(- 1)PgU A vq E I
q
and so 1 is a two-sided ideal. The same argument shows that every graded
right ideal is two-sided.
Now let a e A PE be an arbitrary homogeneous element, and consider
the graded subspace Ia of A E consisting of the elements u A a, u E A E.
Clearly Ia is a graded left ideal in A E, and hence it is a two-sided ideal.
Since a e Ia, it follows that Ia is the smallest (graded) ideal in A E containing
a, i.e.,
IQ =
fl1.
aEl
Ia is called the ideal generated by a. A homogeneous element a 0 is called
a divisor of an element u E A E if there exists an element v e A E such that
u= a A v or equivalently, if u E Ia.
More generally, every homogeneous subspace A c A PE generates a
graded ideal I A defined by
ui n a; ul E A E, a E A .
1A=
(5.54)
1 A is the intersection of all graded ideals containing A. If B is a subspace
of A it is clear that IB c IA. Now consider two homogeneous subspaces
A
A PE and B
A qE. It will be shown that 1 A = 1 B if and only if A = B
(and hence p = q). Clearly A = B implies that 1 A = 'B Conversely, assume
that 1 A = 'B Then every element b e B can be written in the form
b=
U1 n a1
U1 E A q- PE,
aEA
and hence it follows that q >_ p. The same argument shows that p q,
whence p = q. Consequently, the U1 are scalars, and so b e A, i.e., B c A.
Similarly we obtain that A c B, whence A = B.
As a special case of this result we have
Ia = Ib
if and only if a = ,%b, ,%
0.
a e A PE, b e A qE
5 Exterior Algebra
128
The ideal 'APE, p >_ 0, will be denoted by I". It follows from (5.54) that
ME.
1p =
j1
p
The ideals Ip form a filtration of the algebra A E, i.e.,
The ideal I' = I E is often denoted by A + E,
AE=
A'E.
j>o
If E is of dimension n we have
1n = A 'E
and
Ip=O
if p > n.
5.23. Direct Decompositions
Let
E=E1QE2
be a direct decomposition of E. Then we have (considering IE1 as an ideal
in AE)
(5.55)
I E1 = A + E1 Q A E2 .
In fact, since A+ E, c 1E 1, it follows that A+ E 1 Q A E2 c
'E1
Conversely,
let y A v (y e E1, v e A E) be a linear generator of IE1. Writing
v=
a1® b,
a1 e A E1, b,e A E2,
we obtain
yAveA+E1pAE2
and so
I E1 c A +E1 O A E2 .
Writing A E in the form
AE= AE1 Q AE2 = (r Q AE2) Q (A +E1 Q AE2)
= A E2 Q (n +E1 O A E2),
Ideals in AE
129
we obtain, in view of (5.55), the relation
AE = 'E1 +Q A E2 .
(5.56)
5.24. Linear Maps
Let p: E - F be a linear map and consider the induced homomorphism
co A: n E - n F. Generalizing the result of Section 5.8, we shall prove that
ker q
(5.57)
= 1 ker q .
Let E' c E be a subspace such that
E = ker p Q E'.
Then we can write co = 0 Q cp', where gyp' denotes the restriction of co to
E' c E. Since gyp' is injective so is (q' ) A and hence (see Section 5.16)
ker(p ^) = ker 0 A p n E' = n + ker co p n E'.
In view of (5.55), we have
n + ker p Q AE' = 1 ker p .
Combining these relations, we obtain (5.57).
5.25. Invertible Elements, Maximum and Minimum Ideals
zi, z E A `E, of A E is invertible if and
Proposition 5.25.1. An element z =
0.
z
is
nilpotent
if
and
only
if zo = 0.
only if zo
PROOF. Since nilpotency and invertibility are mutually exclusive properties,
it is sufficient to show that z is nilpotent (respectively invertible) if z e A + E
(respectively z n + E).
If z e A + E, then z e A + F, where F is a finite-dimensional subspace of E.
It follows that zm = 0 for m > dim F and so z is nilpotent. If z n + E, then,
for some
0,
,z = 1 - a
aE n+E.
Now consider the identity
(1 -a) n (1
1 -(k+ 1)!ak+i
Since a is nilpotent, it follows from this relation that 1 - a has an inverse
and hence z has an inverse as well.
5 Exterior Algebra
130
Corollary I. If z e A E is invertible, then z' is a polynomial in z.
Corollary II. Every proper ideal in A E is contained in A + E and so A E
has a maximum ideal, namely A +E.
A + E is an ideal in A E, then 1 contains invertible elements.
Hence, 1 E 1 and so 1= A E.
PROOF. If 1
Proposition 5.252. Let E be an n-dimensional vector space and let e be a
basis vector for n E. Then for every element u 0 of n E, there exists an
element v e A E such that
uAv=e and vAu=±e.
u,
e A `E and assume that ur 0 and u = 0 for i < r.
A e". Then
Choose a basis {ev} (v = 1, ... , n) of E such that e = el A
PROOF. Let u =
v 1..... vre
ur -
v1
A eyr.
A
Without loss of generality we may assume that l. r
r)_ ler+l
v=
A ... A en,
0. Multiplying u by
we obtain
A e"=e
u A v=ur A v=el A
and
vAu=vAur=er+l
A...Aer=(-1)r("-r)e,
which proves the proposition.
Corollary. If E has finite dimension, then every (two-sided) nontrivial ideal 1
in A E contains 1" = A" E and so A E has a minimum ideal, namely A "E.
Conversely, f E is a vector space such that A E has a minimum ideal, then E has
finite dimension.
0 be an ideal in A E and u 0 be an arbitrary element in I.
Then by the above proposition there is an element v e A E such that u A v = e
whence I" I. To prove the second part consider the ideals Iq = p,q A pE,
PROOF. Let 1
q >_ 0. If E has infinite dimension, it follows that flq Iq = 0 and so A E
has no minimum ideal.
5.26. The Annihilator
Let u e A E be a homogeneous element. Then a graded ideal N(u) in the
algebra A E is determined by
N(u) = ker µ(u).
(5.58)
Ideals in AE
131
N(u) is called the annihilator of u. It follows from the definition that
N(1)=O
N(O)= AE,
and that
N(u)
whenever u divides v.
N(v)
More generally, if U E A E is a homogeneous subspace of n E, the space
N(U) = n N(u)
(5.59)
UEU
is called the annihilator of U. As an intersection of graded ideals N(U) is
itself a graded ideal. For U = E we obtain that
N(E) _
if dim E _
E = n.
i' f dim
'
0
n "E
N(U) whenever V c U.
Now consider the special case U = n "F, where F is a subspace of E.
It follows from the definition that N(V)
N( A "F) consists of the elements u e A E satisfying
yIEF.
yi
It follows from the definition of N( A "F) that
N( A 2F) c ... c N( n pF)
N(F)
....
(5.60)
Proposition 5.26.1. Let F be an m-dimensional subspace of E. Then the
annihilator N( n m - p + 'F) coincides with the ideal generated by n "F,
N(nm-p+1F)=1APF
PROOF. If z1 n
0< p<m+ 1.
(5.61)
n zp, zi e F, is a generator of 1 A nF and y E F(j = p, ... , m)
are arbitrary vectors, then
µ(z i n ... A z p) (yp n ... A ym) = z 1 A ... A Zp n yp n ... A ym = 0,
whence 1 A pF c N(n m - p + 'F). To prove that
N(nm-p+iF)
1
APF
we show first that every element u e 1 APF can be written as
u = u' +
ai Q b.
U' E 1 A P+'F, a E A "F, b. E H,
(5.62)
where H is a complementary subspace to F in E. Without loss of generality
we may assume that u is of the form
u=anv
aen"F,veAE.
5 Exterior Algebra
132
Since E = F Q H, we have
v
eF,b;e AH,
= (a n v;)pb;.
u= a n
Let u' be the sum of all terms in this equation for which v; has positive degree.
Then we have
a E A "F, b; E A H, u' E I n p+ 1 F,
which proves (5.62).
Now we prove (5.61) by induction on p. For p = 0 we have
N( n m+ 'F) = N(0) = n E and I A of = n E
and so (5.61) is correct. Now assume that (5.61) holds for the integer p and
let u e N( n m- "F) be an arbitrary element. In view of (5.60) we have
u e N(n - P + 'F) and hence by the induction hypothesis, u e 1 A pF . In
view of (5.62), we can write
a p b,
u = u' +
u' E 1 n p+ 1F, a1 E A "F, b. E A H.
(5.63)
Now let {eµ} (µ = 1, ... , m) be a basis for F. Then (5.63) can be written in
the form
u = u' +
eµ1 n ... A eµp O cµ1, ..., µp
cµ1, ..., µp E / \ H.
(5.64)
Choose a fixed p-tuple (µ 1, ... , µ p) and let (µp + ,, ..., µm) be the complen eµm and observmentary (m - p)-tuple. Multiplying (5.64) by eµp +1 n
ing that u e N(n - pF) and u' E 1 A p + 1 F c N(n - "F), we obtain
0=ep
It follows that cµ 1.
µp
cµ1....°µp
eE AmF,e
0.
= 0 and so u = u' E 1 A p + 1F This completes the proof.
Applying Proposition 5.26.1 for the special cases p = 1 and p = m, we
obtain immediately the following
Corollary. Let u = xl n
n xm be a nonzero decomposable m-vector
and X c E be the subspace generated by the vectors x, (i = 1, ..., m). Then
N(u)=1X
(p= 1)
N(X)=I
(P=m).
and
In particular, u is divisible by a vector y
0 if and only f y n u = 0.
Ideals in A E
133
5.27.
As an application of the results of Section 5.26 we prove
Proposition 5.27.1. Let E be an n-dimensional vector space and cP : E -+ n E
a linear map. Then
x n (px = 0
xEE
(5.65)
(px=xnv
xEE
(5.66)
if and only if
for some fixed element v e A E. The element v is uniquely determined mod n "E.
PROOF. It is trivial that (5.66) implies (5.65). Conversely, suppose (5.65)
holds. Then it follows that
x n (py + y n (px = 0
x, y E E.
Now we proceed by induction with respect to n. For n = 0 the proposition
is trivial. Assume now that it is correct for dim E = n - 1. Choose an
arbitrary vector a 0, a e E. Since a n cpa = 0, it follows from (5.61) with
m = p = 1 that there is an element c e A E such that
spa = a n c.
Now define a mapping ti: E -+ n E by
6x=cpx-x n c.
(5.67)
6a=(pa-a n c=0,
(5.68)
x n 6x = x n cpx = 0,
(5.69)
Then we have
and
an6x=ancpx-anxnc=-xncpa+xnanc=0. (5.70)
Now consider the 1-dimensional subspace E 1 generated by a and write
E= E 1 Q F. Since
AE= AE1QAF,
it follows that
6x=1®i0x+a®Qi1x
xeE,
where i and 61 are well-defined linear maps. Multiplying by a we obtain
in view of (5.70)
0=a® i0x
xeE,
5 Exterior Algebra
134
whence t70 x = 0 and so
6x = a Qx 61 x
x E E.
(5.71)
Now let y e F be an arbitrary vector. Then relations (5.71) and (5.69)
yield
- (1 A a) OO (y A 61 y) _ (1 OO y) A (a OO 61 y) = y A 6y = 0
whence
y A 61y=0
yeF.
Now it follows from the induction hypothesis that
61 y= y A v1
y E F,
(5.72)
where v1 E A F is a fixed element. Since every vector x e E can be written
in the form x = ,a + y, . E IT, y e F, we obtain, in view of (5.68), (5.71)
and (5.72),
6x=6a+6y=6y=a®x 61y=a®x (y A v1)
= -y A (a A v1) = -X A (a A v1)
(5.73)
Combining (5.67) and (5.73) we find that
(px=6x+x A c=x A (c-a A v1)
and so the induction is closed.
To prove the uniqueness part of the proposition assume that u1 E A E
and u2 E A E are two elements such that
cpx=XAU1 and cpx=XAU2.
Then
xeE
XA(U2-U1)=0
and so u2 - u1 E N(E) = A'SE.
Corollary. Let p: E -p A k + 'E be a linear map such that
xA4x=O
xeE.
Then there exists an element u e A kE such that
cpx=xAU
If o
xeE.
0, then u is uniquely determined.
PROOF. In view of the above proposition, we can write
cpx = x A
v,
Ideals in AE
135
where v e A E. Applying the projection Pk +1: n E - n k + 'E to this relation,
we obtain
(Ox-Pk+1(x n v)=x n pkv=x n u
u= pkv
and so the first part of the corollary follows. Suppose now that
(px = x n u 1
and cpx = x n u2
u 1, u2 E n IE.
Then u2 - u 1 E ME. On the other hand we have that u2 - u 1 E n kE whence
u2 - u 1 E ME E n n kE. But, if p 0, it follows that k + 1 < n and so we
obtain u2 - u 1 = 0.
If co is a homogeneous linear map, Proposition 5.27.1 can be extended
to spaces of infinite dimension.
Proposition 5.27.2. Let (p: E - n E be a homogeneous mapping of degree k
such that x n cpx = 0. Then there exists an element u e n kE such that
cpx = x n u. If o 0 the element u is uniquely determined by co.
PROOF. We prove first the uniqueness of u. Assume that
cpx=x n u1 and (px=x n u2.
Then
xn(u2-u1)=0
xeE
and so
u2 - u 1 E N(E).
If the dimension of E is infinite we have N(E) = 0, whence u2 = u 1. If
dim E = n, we have N(E) = A" E. If o 0, it follows that k n and so
u2 -u1 =0.
To show the existence of u we state first the following
Lemma. Let p : E -p n k + 'E be a linear map such that for a subspace F of
finite dimension
y n (py = 0
yeF.
Then there exists an element v e n kE such that
(py = y n v
y E F.
The above lemma is proved by induction on dim F in the same way as
Proposition 5.27.1.
Now consider a linear map p: E - n k + 'E satisfying x n cpx = 0. We
may assume that o 0. Choose a E E such that cpa 0. Let H be a subspace of dimension k + 1 such that a e H. Then, by the above lemma, there
exists an element u e n kE such that
coy = y n u
y E H.
(5.74)
5 Exterior Algebra
136
Now we show that
cpx = x n u for every x e E.
Let x e E be an arbitrary vector and consider the subspace H 1 c E gener-
ated by x and H. In view of the above lemma there exists an element
v e n kE such that
cpz = z n v
zeH1.
(5.75)
From (5.74) and (5.75) we obtain
y n (u-v)=0
yeH,
whence u - v e N(H). In view of (5.61) we have N(H) = 1 n k+ 1H. On the
other hand u - v is of degree k and so u - v = 0.
Proposition 5.27.2 permits us to give explicitly the forms of a derivation
of odd degree and an antiderivation of even degree in n E.
Let 8 be a homogeneous derivation of degree k, where k is odd. Then
we have 8 =
p) where co denotes the restriction of 8 to E (see Section 5.11).
In view of (5.32), we have x n cpx = 0 and hence, by the Corollary to
Proposition 5.27.1, there exists an element a e n kE such that
(px =anx
x e E.
Now Formula (5.33) yields
Bu =
JaAU u e n "E, p odd,
0
u e n "E, p even.
If 8 is an antiderivation of even degree in n E, we have again x n cpx = 0,
where co denotes the restriction of 8 to E. Hence co can be written in the form
(px=anx
aen"E.
Now (5.34) gives
Hu =
anu uen"E,podd
un
E pE , p eve n.
o
PROBLEMS
1. Let 1 be a right ideal in AE. Assume that a n b e l and b
n E. Prove that a e l.
2. Show that the generator of a graded principal ideal is homogeneous. Conclude that
there exist nongraded principal ideals in n E.
3. Construct a left ideal in n E which is not a right ideal.
4. Given a subspace F
E prove that the algebra n (E/F) is isomorphic to the algebra
n E/1 F .
5. Let E = E 1 + E2 be a decomposition of E and consider the homomorphism
(P,,: nEl Q AE2 -+ AE,
137
Ideals and Duality
which is induced by the linear map
rp: E1 Q E2 - E1 + E2.
Determine the kernel of p .
6. Let (p be linear transformation of a finite-dimensional vector space E and denote
the Fitting null- and 1-components of (p by F0((p) and F1((p). Prove that
F0(q) = I F,.()
and
AF1((p).
Hint : See Problem 5, Section 2, Chapter XIII of Linear Algebra.
Ideals and Duality
5.28.
In this paragraph E, E* will denote a pair of dual finite-dimensional vector
spaces. Let 1 be a graded ideal in A E. Then the orthogonal complement Il
is stable under the operations i(h), h e E. In fact, if u* a Il is an arbitrary
element we have, for every u e 1,
<i(h)u*, u> = <u*, h n u> =0,
and so i(h)u* e I'. Conversely, if 1 is a graded subspace of A E such that Il
is stable under every i(h), then 1 is an ideal. Let u e 1 and h E be arbitrary.
Then we have, for every u* a I',
<u*, h Au> = <i(h)u*, u> =0
whence
h n u e (Il)l = I.
Now let F be a subspace of E and Fl be the orthogonal complement.
Then the subalgebra A (Fl) and the ideal 1 F are orthogonal complements
with respect to the scalar product in A E and A E*,
A (Fl) = (I
F)'.
(5.76)
In fact, consider the canonical projection
m : E --> E/F
and the injection
J: E*
F'.
In view of Section 2.23 of Linear Algebra, the spaces E/F and Fl are dual with
respect to the scalar product defined by
<Y*, mx>
=
<JY*, x>
Y* e Fl, x e E.
(5.77)
5 Exterior Algebra
138
This shows that the mappings it and j are dual. Consequently, the induced
homomorphisms
Th A: n E - n (E/F)
and
j A . n E* E- n (Fl)
are dual as well. This implies that
Im j A = (ker It )l.
A
Since ker it = F we have, in view of (5.57)
ker it A =
1F .
On the other hand, it follows from Section 5.8 and the definition off that
Im j = n Im j = n (F').
Combining the above relations we obtain (5.76).
As an immediate consequence of (5.76) we have the formula
(AF)' = IFl,
(5.78)
which is obtained by applying (5.76) to the subspace Fl c E* and taking
orthogonal complements on both sides.
Proposition 5.28.1. Let F c E, F* c E* be two dual subspaces. Then the
ideals IF and IF* are dual as well.
PROOF. Let F 1 = (F*)' and F i = Fl. Then we have the direct decompositions
E=F+QF1
and
E* = F* Q F i.
This yields, in view of (5.56)
nE = IF Q AF1
and
AE* = 1F* o AFT.
Since A Fi = n (Fl) = (IF)' and A F1 = n (F*1) = IF*, it follows that the
ideals 1 F and 1 F* are dual.
E
Proposition 5.28.2. Let F c E be a subspace. Then the subalgebra n F is
stable under the operations i(h*), h* E E*. Conversely, if A is a subalgebra
of n E stable under i(h*), h* E E*, then there exists a subspace F c E such
that A = AF.
139
Ideals and Duality
n yp, y E F, be any decomposable element of n F.
PROOF. Let Y1 n
Then, for every h* E E*,
i(h*)(Y1 n ... n yp) =
(-1y+l<h*, Y;>Y1 n ... n y; n ... n yp E AF.
Hence A F is stable under i(h*). Conversely, assume that A is a subalgebra
of n E which is stable under every i(h*). Then A is stable under every operator
i(u*), u* E A E. If A = 0 the proposition is trivial and so we may assume
that A 0. We show first that 1 E A. Let u 0 be an element of A. We write
u in the form
r
u=p=0u
u p E A "E, ur
p
0.
Since ur 0, there exists an r-vector u* E n rE* such that <u*, urn = 1.
Applying i(u*), we obtain
i(u*)u = <u*, ur> = 1.
Since i(u*) u e A and <u*, ur> = 1, it follows that 1 E A.
Now consider the subspace F = A n E. It will be shown that A = n F.
Clearly, A F c A. Let
n
u=
up
up e n pE, n = dim E,
p=0
be an arbitrary element of A and assume by induction that
uE n F for v > q.
Since n n + 'E = 0, this relation is correct for q = n. Define v by
n
v=u- v=q+ 1
(5.79)
Since u e A and uE A F c A, for v >- q + 1, we have v e A. Now let
u* E n q- 1E* be an arbitrary element. Applying i(u*) to (5.79) we obtain
l(u*)v = l(u*)uq _ 1 + l(u*)uq
<u*, uq- 1> + l(u*)uq
(5.80)
Since A is stable under i(u*), we have i(u*)v e A. Since 1 E A, it follows from
(5.80) that i(u*)uq e A. On the other hand i(u*)uq is of degree 1 and hence
i(u*)uq E A n E = F.
Now let y* E Fl be arbitrary. Then
Y* n u*, uq> = + <Y*, i(u*)uq> = 0.
Hence, uq is orthogonal to the ideal 'F'. Now it follows from (5.78) that
uq E A F and so the induction is closed.
5 Exterior Algebra
MO
PROBLEMS
1. Let 1 be a subspace of n E. Prove that 1 is a left ideal if and only if the orthogonal
complement 1 is stable under i(h) for every h e E.
2. Given a subspace F
E prove that
AF = n ker i(u*).
U* E IF1
3. Let h e E be a fixed vector. Define operators a : AE -+ AE and b : AE* + AE* by
au = µ(h)u
ueAE
and
bu* = i(h)u*
u* e AE*.
(a) Show that (n E, a) is a graded differential space and (n E*, (5) is a graded differential algebra.
(b) Prove that
H( A E) = 0 and
H( A E*) = 0.
A E.
4. Let E, E* be a pair of dual vector spaces and consider a graded subspace U
Show that U is an ideal in n E if and only if U1 is stable under the operations i(h),
he E.
The Algebra of Skew-Symmetric Functions
5.29. Skew-Symmetric Functions
Let E be an n-dimensional vector space and consider the space T "(E) of
p-linear functions in E (see Section 3.18). Then, if Fe T "(E) and 6 E S p,
an element 6F E T "(E) is defined by
(o) (x i , ... , xp) _ F(xa(1) , ... , xy(p)).
The function F is called skew-symmetric, if
o F = EQ .
The skew-symmetric functions form a subspace of T "(E) which will be denoted by A"(E).
Every p-linear function F in E determines a skew-symmetric p-linear
function A'F given by
A=1 E6.
PQ
AF is called the skew-symmetric part of F. In particular, if F is skew-symmetric, then A'F _ 'F.
141
The Algebra of Skew-Symmetric Functions
Thus the operator A : T "(E) -+ T "(E) is a projection operator. It is called
the antisymmetry operator.
Now let F e TP(E) and Y e T(E) and consider the (p + q)-linear function
1F Y (see Section 3.18). A simple calculation shows that
A(D kY) = A(AID kY) = A(D AtY).
(5.81)
This formula implies that
A(F kY) = A(A F AtY).
Thus the skew-symmetric part of a product depends only on the skewsymmetric parts of the factors.
5.30. The Algebra A (E)
The Grassmann product of two skew-symmetric functions F e AP(E) and
Y E A(E) is the (p + q)-linear skew-symmetric function F A Y given by
bn
()
tt
Explicitly,
A
x
=
... x
1
x
E
... x
'Y x
... x
Proposition 5.30.1. The Grassmann product has the following properties :
F A Y=(-1)P9' A F
F e AP(E), Y e A(E),
('F n kY) n X= F n (kY n X)
F, '-Y, X e A(E).
(A)
(B)
PROOF. (A) Observe that
where 6 is the permutation
(1,...,p,p + 1,..., p + q)-+(P + 1,..., p +q, 1,...,p)
Applying A to this equation and observing that E, = (-1)' we obtain (A).
(B) Let F e AP(E), Y e A(E), and X e Ar(E). Then
A
n X= p+ q+
r)!
.A
X
=(p p!q!r!
+q+r)!AA(D ) XJ
-(p+q+r)!
tr
Pq
A(D Y X ).
5 Exterior Algebra
142
Similarly,
=(p+q+r)!
tq. trt
p..
(
)
and so (B) follows.
The A -product makes the direct sum
n
A(E) =
AP(E)
p=o
into a graded associative algebra called the Grassmann algebra over E.
5.31. Homomorphisms and Derivations
A linear map p : E -+ F induces a homomorphism c0A : A(E) E- A(F) given
by
((PAkJ')(xi, ..., xp) =
'P E AP(F).
Next, let p be a linear transformation of E. Then p determines a derivation
in A(E) given by
P
1
(x 1, ..., (pxy , ..., xp).
... , xp) =
v=1
To show that this is indeed a derivation consider the derivation oT (cp)
induced by p in the algebra T(E) (see Section 3.18). It follows from the
definitions that
E A(E).
= eT (co)d
Moreover, the operator oT (cp) satisfies the relations
eT
o cr = 6 o eT (p)
whence
eA(co) ° A = A o oT (co)
It follows that
Op )(F n i
=(p+g)!OA
nY+Dn
OA(co)l A
Y+DA
OA(co)%Y.
The Algebra of Skew-Symmetric Functions
143
5.32. The Operator iA(h)
Recall from Section 3.19 the definition of the linear operator iA(h) : T(E) T'(E). It follows that if F E A(E), then
= i1(h)b.
iA(h)
Thus,
(iA(h) ) (x 1,
:
, x p - 1) = I (h, x 1, ..., x_ 1)
F E A( E).
iA(h) is called the substitution operator in the algebra A(E).
Lemma. Let A denote the antisymmetry operator. Then
iA(h) o A = A o iA(h).
PROOF. Let F E T"(E). It is sufficient to consider the case
D = f1 ... fp
f1 E E*.
Then we have
A=1
EQ
p
.fQ(p)
Q(1)
and thus,
1ACh) (4
) = i 1(h) (Ai:F)
1
=
p-
i 1(h)
EQ fa(1) ' ... fQ(p)
1
=
=
_
pQ
EQ fa(1)(h)fa(2)
'p
'p
p µ= 1 Q(1)=µ
...
fa(p)
EQ fµ(h)fQ(2) ' ... fQ(p)
(5.82)
fµ(h)Aµ
p µ=1
where
Aµ =
Q(i)=µ
EQ fa(2) ' ... fQ(p)
Now we show that
Aµ = (-1)"-1(p - 1)!A(fl O ... OO
fµ
...
fp).
In fact, fix µ and let 6µ E Sp be the permutation defined by
6µ(1)=2,..., 6µ0u- 1)=µ, 6µ0u)= 1, 6µ0u+
1,
...
6µ(p) = p.
144
5 Exterior Algebra
Then every permutation 6 E S, determines a permutation TQ E S, given by
iQ = 6 o 6µ .
Since
E(6µ) = (-1)µ-1,
E(#EQ) = (-1)µ - 1 E(6).
Thus we obtain
µ_
(-1)
Aµ =
,.
1
E(i)ft(1) ' ... ft(µ) ' ... ft(p)
t(µ) = µ
(-1)µ-1(p
=
- 1)!A(f1'...
fp).
(5.83)
Equations (5.82) and (5.83) yield
1)!
h A=
)
p
-1 µ-1fµ(h)A(fi
)
i A( ) (
_1
A
p
.. fµ
.. .
..
fp)
µ
(
1)µ - lfµ(h)fl ... ' fµ ' ... ' f p
(-
)µ-
µ
= 1A
p
µ
li µ(h) (fl ...
f p)
= A1A(h)1F
0
and so the lemma is established.
Proposition 5.32.1. The operator iA(h) is an antiderivation in the algebra A'(E),
lA(h) ( A 'F) = 1A(h)I) A 'F + (- 1)(F A LA(h)'F
(F E A(E), 'F E A(E).
PROOF. Since
we have to show that
1A(h)A( `F) =
p
p+q
A(iA(h) F 'F) + (_ 1)p
q A((F iA(h)`F).
p+q
By the lemma,
iA(h)A( 'F) =
1
p+q
p+qµ=1
( -1)µ -1 Aiµ(h)
(lF. 'F).
145
The Algebra of Skew-Symmetric Functions
Now,
i
h = iµ(h)
"()(
µ P
µ>-p+1
1(F' i_
µ (h)tY
p
)
(see Formula 3.18). Thus we obtain
i(h)A(F ) =
P
p+qµ=1
+
=
(-1)µ-1A(iµ(h)iF kY)
p+q
1
p + q µ=p+
P
1
p+q
+
(-1)µ - 1 A( iµ _ p(h)i)
(-1)µ - 1 lµ(h)D
A
1
p+q
P
p+q
ij
µ=1
A
(F.
p+q
µ=p+1
(-1)µ -1lµ - p(h)i
A(iA(h)l kY) + (-1)P
q A(F' iA(h)`Y),
p+q
which completes the proof.
5.33. The Isomorphism A E* 4 T' (E)
We show that the Grassmann algebra over E is isomorphic to the exterior
algebra over E*. In fact, consider the isomorphism a: ®PE* 3 T(E) (see
Section 3.20). It is easily checked that the diagram
TP(E)
jA
commutes, where icA denotes the alternator (see Section 4.2). Since
Im icA = XP(E*) and Im A = AP(E),
it follows that a restricts to a linear isomorphism
a : X P(E*)
AP(E).
Next observe that, in view of the commutative diagram above,
oc(u n v) = ajA(u Q v) = Aoc(u Q v)
=Aau av =
+q au A av
ueXPE* veXgE*.
146
5 Exterior Algebra
Thus the map
f3: X(E*) - A(E)
defined by
f3(u) = p !a(u)
u e X p(E*)
is an algebra isomorphism.
Composing this isomorphism with the algebra isomorphism
X (E*),
ri: AE*
obtained in Section 5.3, we obtain an algebra isomorphism
A E*
A(E).
X(E*)
Under this isomorphism,
(1) The homomorphism cP ^ : A E* - A F* corresponds to the homomorphism p": A(E) - A(F) (see Sections 5.9 and 5.31).
(2) The derivation 8 A ((p) corresponds to the derivation
5.10 and 5.31).
(see Sections
(3) The operator i(h) corresponds to the substitution operator iA(h)
(see Sections 5.14 and 5.32).
5.34. The Algebra A,(E)
Denote by A (E) (p > 1) the space of skew-symmetric p-linear mappings in
the dual space E* and set A0(E) = r. Then we have the n -multiplication
between A (E) and A9(E) (see Section 5.30). It makes the direct sum
n
A , (E)
=
A p(E)
p=0
into an associative algebra which is isomorphic to the algebra A E.
Now we show that the scalar product between T "(E) and T(E) defined
in Section 3.22 restricts to a scalar product between A"(E) and A (E).
In fact, it is easy to check the relation
<D, Q`P> = <o
D, 'P>
D E A(E), 'P e A(E).
It follows that
Now suppose that D 1 E A"(E) is an element such that
' 1 > = 0 for
every ' 1 E A (E). Then we have for every 'P ETp(E)
= <AD1, '> = <D1, A'> = 0
whence
= 0. In the same way it follows that if '1 E A (E) is an element
such that <D 1, ' 1 > = 0 for every 1 E A"(E), then 'P 1 = 0.
147
The Algebra of Skew-Symmetric Functions
Thus a scalar product <, >A between the spaces AP(E) and A (E) is defined
by
''
=
1
qi
HEAP E 'P EA(E).
P
D(x1, ..., xp)
xE E
<fl A" . A fps 'P>A - `I'(fl, ... , fp)
feE*.
x1 n ... n
<, x 1 ^ ... A x p>A =
=
1
P
x 1 A ... A xp>
2A(x1 OO ... OO xp)> = <A D, x1 OO ... OO xp>
= <(D, x 1 ® .. OX xp> = D(x 1, ..., xp).
The second formula is obtained in the same way.
Finally we show that
<fi n ... A fp, x1 n ... A xp>A = det(f (x;))
ff E E*, xi e E.
In fact,
<fl n ... ^ fps xl n ... n xP>A =
1
P
<fl n ... ^ fps xl n ... n xp>
= <f1 n ... n fp,A(xl p...Qxp)>
EQ f1(x) ... f(x)
Q
= det(f, (x;)).
Mixed Exterior Algebra
Throughout this chapter E*, E will denote a pair of dual vector spaces over a field of characteristic
zero.
The Algebra n (E*, E)
6.1. Skew-Symmetric Maps of Type (p, q)
A skew-symmetric map of type (p, q) from E*, E into a vector space H is a
(p + q)-linear mapping
xEH
x E* x E x
/i:E* x
p
9
which satisfies
YI (x (1), ... , x (p)' Xt(1),.. .
,
Y (x 1
... , x; X1,. . . , X9)
for all permutations o E Sp and i e S9.
Proposition 6.1.1. Every skew-symmetric map cui of type (p, q) determines a
unique linear map
f:APE*®
such that
... ^ xp) 0 (x 1 n ... n X9)] = Y' (aC i , ... , x; X 1, ..., X9).
PROOF. The uniqueness follows from the fact that the products
(x i
A ... n xp) O (x 1 A ... n x9)
span the space (A PE*) Q (A 9E).
148
149
The Algebra A (E*, E)
To prove existence define a linear map
g: (®PE*) ® (®'E) - H
by
g(x 1 ®... ® xp ® x 1
®... ® xq) = Y (x 1
..., xp ; x 1, ..., xq)
(see Section 1.20) and consider the bilinear mapping
f3:(®PE*) x (®9E)
H
given by
f3(u, u) = g(u ® u)
The skew-symmetry of c implies that f3(u, u) depends only on the vectors
7r1u and ire u where
7r1: ®PE* + APE*
and
2®E+ + A "E
are the canonical projections (see Section 5.3). Thus a bilinear mapping
y: A PE* x A qE -+H
is defined by
y(it1 u,
2 u)
= f3(u, u)
This map, in turn, induces a linear map
f:(ApE*)®(AlE)+H.
It follows that
.f [(x 1 n ... n xp) ® (x 1 A
A xq)]
y [ir1(x 1 ® ... ® xp ), 7r 2(x 1 A ... A xq)]
_ f3(xi
0."®xp,x1 ®."®xq)
= g(xi
0... ®xp ®x1 ®... ®xq)
= i/i(x i , ... , xp ; x 1, ... , xq)
and so the proof is complete.
6.2. The Algebra A (E*, E)
The mixed exterior algebra over the pair E*, E, denoted by A (E*, E) is
defined to be the canonical tensor product of the algebras A E* and A E
(see Section 2.2),
A (E*, E) = A E* ® AE.
6 Mixed Exterior Algebra
150
The multiplication in this algebra will be denoted by. and is determined by
the equation
(u* Q u) (v* p u) = (u* n u*) Q (u n u)
u*, u* E A E*, u; u e A E.
Thus A (E*, E) is a graded associative algebra with unit element 1 Q 1.
It is generated by the elements 1 Q 1, x* Q 1 and 1 Q x with x* E E* and
xeE.
If w e A (E*, E), we shall define wk (k >_ 0) by
wk=
1
w..... w
k>_1
k.
k
and
w° = 1.
Now consider the bigradation of A (E*, E) by the subspaces
A 9(E*, E) = A PE* Q A 9E.
Clearly,
w1 , w2 = (_ 1)P1P2+9192w2 . w1
w 1 E A 9i (E*, E), w2 E A 92 (E*, E).
(6.1)
Since
P1P2 + g1g2 = (Pi + g1)p2 + (p2 + g2)g1 (mod 2),
it follows that p 1 p2 + q 1 q2 is even whenever Pi + q1 and p2 + q2 are and
that
w1
w2 = w2 w 1
if p1 + q1 and p2 + q2 are even.
In this case we have the binomial formula
(w1 + w2)k =
w1
w2.
i+ j=k
The scalar product between A E* and A E defined by
<x 1 n ... n xp, x 1 ^ ... ^ xq > _
ifpq,
Jo
ldet(<x, x3>) if p = q,
(see Section 5.6) induces an inner product in A (E*, E) via
<u* Q u, u* Q u> = <u*, vXu*, u>
u*, u* E A E*, u, u e A E. (6.2)
(The symmetry and the nondegeneracy are easily checked.)
Now fix an element z e A (E*, E) and, denote by µ(z) the left multiplication
by z,
µ(z)w = z w
we A (E*, E);
let i(z) be the dual operator,
<i(z)w1, w2 > = <w1, µ(z)w2 >
w 1, w2 E A (E*, E).
The Algebra A (E*, E)
151
Then, if z e A 9(E*, E) and w e A s(E*, E),
i(z)w e A s- p(E*, E)
if r> p and s >_ q
and
i(z)w = 0 otherwise.
It follows from the definition that
µ(u* Q u) = µ(u*) Q µ(u)
u* E A E*, u e A E,
where the operators on the right-hand side are the left multiplications in
A E* and A E respectively. In particular,
µ(1 Q u) = 1 p µ(u)
u e A E
and
u* e AE*.
µ(u* O 1) = µ(u*) O l
Dualizing the relation
µCzl
z2) = µCzl) ° µ(z2)
z1, z2 E A (E*, E),
we obtain
i(zl . z2) = i(z2) o i(Z1).
(6.3)
Note that if z e A 9(E*, E) and w e A p(E*, E), then i(z)w is the element
in A g(E*, E) = r given by
i(z)w = <z, w>.
Next, consider the flip operators
QE: A (E*, E) -+ A (E, E*)
and QE* : A (E, E*) -+ n (E*, E)
given by
QE(u* Q u) = u Q u*
u* E A E*, u e A E
QE*(u O u*) = u* p u
u*E AE*,uE AE.
and
They are algebra isomorphisms as well as isometries with respect to the
inner products in A (E*, E) and in A (E, E*).
The subspace
of =
Op E,
where AE = A p(E*, E),
p>_0
of A (E*, E) is obviously a subalgebra. It is called the diagonal subalgebra.
Formula (6.1) implies that the diagonal subalgebra is commutative. Moreover, of is stable under the operators i(z) if z e AE. Finally, the restriction of
the inner product in A (E*, E) to of is nondegenerate.
6 Mixed Exterior Algebra
152
Next, let F, F* be a second pair of dual vector spaces and let
(p*:E*E-F*
and
,i:EE-F,
be a pair of dual maps. Then we have the induced algebra homomorphisms
/jA O (P : n (E*, E)
n (F*, F)
and
P ^ p /, A : A (E*, E) F-- A (F*, F).
It is easy to check that these maps are dual.
Since l/i ^ Q (p is an algebra homomorphism, we have the relation
(/jA O 4,
)(z w) _ (i/i O 4 n )z (/1 O 4 /, )W
z, w E A (E*, E)
or equivalently,
(/j ®4,)0/1(Z) = µ[/ O 4 )z] ° (I ®4A).
Dualizing, we obtain
i(z) ° ((p A O 'I' A) _ (`p A O 'I' A) ° i[(/i A O
`I'
)z].
(6.4)
Finally, note that a pair of dual isomorphisms P : E 3 F, 4* : E* + F*
induces an algebra isomorphism
cx, : A (E*, E)
A (F*, F)
given by
a=
l OO P
Next, consider the linear map
TE : A (E*, E) -+ L( A E; A E)
defined by
TE(a* p b)u = <a*, u>b
ueAE
(see Section 1.26). Since
= 0 if a* E A pE*, u E A 9E, p
TE restricts to linear maps
A 9(E*, E) - L( A pE; A 9E).
q,
The Algebra A (E*, E)
153
The dual of the linear transformation
TE(a* ®b) : n E -, AE
is the linear transformation TE(b ® a*). Finally, recall from Section 1.26 that
TE is a linear isomorphism if E has finite dimension.
6.3. The Box Product of Linear Transformations
Let (P1 (i = 1, ... , p) be linear transformations of E. Then a linear transfor-
mation
n"E
is given by
((p 1
...
n ... A xP) =
(P pX x 1
7P17(1) A ... A (pp
Q
It is called the box product of the (o. In particular,
((pl
co2Xx1 ^ x2) = (plx1 ^ (p2 x2 - (p1x2 ^ (p2x1.
The box product formula can be written in the form
((p 1
...
(p p)(x 1 A ... ^ x p) =
q) (1) x 1 n ... n
xP
Q
This show that the box product is symmetric,
In fact, let a SP be any permutation. Then we have
((pt(1)
...
(p(7 (1)x 1 A ... A (pQt(P) xP
41(P)Xx 1 A ... A xP) =
Q
Setting 6i = p, we obtain
(q, ...
4t(P))(x 1 A ... A x p) =
(Pp(1) x 1 A ... A P(P) xP
P
...
= (4 1
(p p)(x 1
n ... n xp).
It follows from the definition of the box product that
1
p
np(p.
P
Proposition 6.3.1. The operator TE satisfies the relation
TE(Z 1 ... zP) = TE(z l)
...
TE(z p)
; e E* ® E.
154
6 Mixed Exterior Algebra
PROOF. It is sufficient to consider the case z = y* ®
y e E (i = 1, ..., p). Then we have, for x, e E (i = 1, ... , p),
TE(z 1 ... Zp) (x 1
n ... n xp)
= TE(Y* n ... A yp ® Y t n ... n yP) (x 1 n ... n xp)
=<Yi n...Ayp,x1A."Axp>(Y1A."Ayp)
y p, xP)> (y1 n ... n yp)
=
Q
c <y, X(l)>Y1 n ... A <Yp, xe,(P)>Yp
=
=
c T (z 1)x (1) ^ ... A T
Q
(TE(z 1)
...
TE(z p)) (x 1 A ... ^ x p).
Corollary. Let z e E* ® E. Then
TE(z") = n P(TE z).
The composition Product
6.4
We now define a second multiplication in the space n E* ® n E which will
be denoted by o and called the composition product. Given u*, v* e A E* and
u,veAE,set
(u* ® u) o (v* ® v) = <u*, v>v* ® u.
(6.7)
(cf. Section 1.26).
Then we have the relation
?'E(w 1 o w2) = TE(w 1) o TE(w2)
w 1, w2 e A (E*, E),
(6.8)
where TE is the operator defined by (6.5) and the right-hand side is the com-
position of the linear transformations TE(w 1) and TE(w2). Note that if
dim E < oo, then TE is a linear isomorphism (cf. Section 1.26) and so the
composition product is determined by relation (6.8) in this case.
It follows easily from the definition that the composition algebra is associative. In particular,
(ui ®u1)o(u2
-
<u1, u2/
®u2)o...o(up ®up)
<u2, u3/ ...
<u p_ 1, up/up ®u1.
The Composition Product
155
The kth power of an element w in the composition algebra will be denoted
by w®,
k
Note that the composition algebra has no unit element unless dim E < oo
(see Section 1.26).
It follows from the definition that
A 9(E*, E) o A s(E*, E) = 0
s
(6.10)
A;(E*, E) c A 9(E*, E).
(6.11)
if p
and
A 9(E*, E)
In particular,
A p(E*, E) o A p(E*, E)
A p - p(E*, E)
and so the subspaces AE = A p(E*, E) are subalgebras. Moreover, TE
restricts to homomorphisms
T, : 0 p(E) -+L(A "E ; ARE)
for each p.
Formula (6.11) shows that the composition product is not homogeneous
with respect to the usual gradation in A (E*, E). However, if we introduce a
new gradation (called the cross-gradation) by setting
Deg w= p- q
w e A 9(E*, E),
then we have, for any two homogeneous elements w 1 and w2 for which
w 1 o w2
0,
Deg(w 1 o w2) = Deg w1 + Deg W 2.
Thus, the composition product preserves the cross-gradation.
6.5
Lemma I. Let w 1, W 2 E A (E*, E). Then
w1 o w2 = [i O TE(w1)]w2 = [TE(w2)* O 1]w1
PROOF. We may assume that w1 = u* © u and W2 = u* Q u. Then formula
(6.7) yields
[i O TE(wl)]w2 = u* O TE(wl)V = <u*, U>(U* O u) = W1 o W2.
The second relation is established in the same way.
6 Mixed Exterior Algebra
156
Lemma II. Let w e A p(E*, E) and z e E* Q E. Then
(TE(z)^ Q a)w = w o zP
and
(i O 7E(z) A )w = z P o w.
PROOF. Applying Lemma I with w1 = w and w2 = zP, we obtain
w o ZP = [TE(zP)* 0 ljw.
By the corollary to Proposition 6.3.1, TE(z") = A P(TEz). Thus we obtain,
since w e A p(E*, E),
w o zP = [(A P(TE z))* Q i] w = [ TE(z) ^ Q i] w.
The second formula follows by a similar argument.
The following proposition states a relation between the multiplications in
the mixed exterior algebra and the composition algebra.
Proposition 6.5.1. Let zv e E* Q E (v = 1, ... , p), p > 2 and z e E* Q E.
Then,
l(z)(z 1 ... zP) - , <z, Z )Z 1 ... Zv ... ZP .
v
-
(ZµoZoZy + ZvoZoZµ).Z1
v<µ
PROOF. We may assume that zv = y* Q yv, where y* E E* and yv e E and that
z = x* ® x. Then we have (see Corollary II to Proposition 5.14.1)
i(z) (z1 ... zP) = [i(x) OO i(x*)J (Yi n ... A yp OX y1 n ... n YP)
=
(-1)``-v<Y*, x><x*, Yv>Yi A ... A yµ A ... A yp
O Y1 n ... n Yv n ... n yP.
Now write this sum in three parts with v = µ, v <u and v>
to obtain
i(z) (z 1 ... zP)
n
=<Y*,x><x*,Yv>Yin...n y*A...Aypoyl A...ApvA...A yP
V
-
<Yux><x*, Yv>(Y* OX Y, )[Yi n .
v
µ
P
v<µ
<Yu, x> <x* Yv> (Y* O Yµ) [Yi n ... n yµ n ... A y* n ... n Yp]
µ<v
0 y1 n...A5 A...Ayvn...A yr].
157
Poincare Duality
Since, in view of (6.9)
<Ya, x><x*, Yv>(Y* ® Yµ) = zµ ° Z ° zv
it follows that
i(Z)(Z 1 ... Zr,) =
<z, Z>Z 1 ... Zv ... ZP
V
- (zµ°Z°Zv).Z1 ... 2 ...Zµ ...z,
V<µ
-
(Zµ°Z°Zv).Z1 ...Zµ...ZV...Zp.
µ<V
Interchanging the roles of µ and v in the last term and combining it with the
preceding term now completes the proof.
Corollary. Let z e E* ® E and w e E* ® E. Then
i(z)w" = <z, w>wp-1 - (w o z o w). wp- 2
p > 2,
Poincare Duality
In Sections 6.6-6.16 E denotes an n-dimensional vector space over a field of
characteristic zero.
6.6. The Isomorphism TE
Consider the linear isomorphism TE from A (E*, E) to L( A E; A E) given
by
TE(a* ® b)u = <a*, u>b
ueAE
(see Section 6.5). It restricts for each pair (p, q) to an isomorphism
TE : A 9 (E*, E) 3 L(A "E ; ME).
This isomorphism satisfies the relation
tr(TEw1 ° TEw2) = <w1, w2>
w1 e T9(E*, E), w2 e T (E*, E). (6.12)
In fact, let w 1 = a* ®b, a* e APE*, be ME, w2 a b* ®a, b* e
a e A "E. Then
(w1, w2> = <a*, a><b*, b>.
On the other hand, Formulas (6.7) and (6.8) yield
tr(TE(a* ® b) ° TE(b* ® a)) = <a*, aXb*, b>
and so Formula (6.12) is established.
and
6 Mixed Exterior Algebra
158
6.7. The Unit Tensors
Define tensors 1E e A (E*, E) and & e A (E*, E) by
DE = TE 1''AE
and
p=TE1(l,1)
(P=0,...,n).
Then E is the unit element of the composition algebra whilep is the unit
element of the subalgebra A(E). The tensorp will be called the unit tensor of
degree p. We shall denote1 simply by . Thus,
= TE 1(iE).
The tensor " can be written in the form
" = e Q e,
where a*, a is a pair of dual basis vectors of A "E* and A "E. Clearly,
"
E=
(6.13)
gyp.
p=0
The unit tensor & coincides with the pth power of in the algebra A (E*, E),
p
= fp
(p = 0, ..., n).
(6.14)
In fact, the corollary to Proposition 6.3.1 yields
TE(MP) = A "TE(S) = A plE = 'APE
and so (6.14) follows.
Next observe that Formula (6.12) applied with w 1 =
yields
<p , w> =
tr
(6.15)
p
and w2 e A p(E*, E)
TE(w).
Thus,
<
p, w> = tr TE(w)
w e A p(E*, E).
(6.16)
u* e A PE*, u e A "E.
(6.17)
In particular,
<, u* Q u> = <u*, u>
Poincare Duality
159
Proposition 6.7.1. The unit tensors satisfy the relation
P
9
)p+q
q
(p+q\
p
q
(6.18)
and
i( ) =
n
p+q
q
)p-q
p >_ q
()
6.19
PROOF. The first part follows directly from Formula (6.14). To prove the
second part consider first the case q = 1. Then the corollary of Proposition
6.5.1 (applied with z = w = ) yields
l( P= <,4 P-1 -( a o )
P-2.
= , we obtain
i(P = nP - 1 - (p - 1)t3' -1
Since, by (6.16), <, > = n and
o
=(n-p+1)tP-1.
Thus (6.19) is correct for q = 1. Now the general formula follows via induction on q.
.
i.
<pp, p) _
n
(p = 0,.
.
.,
n)
P
II.
i( q) n = n - q
(q = 0, ..., n).
6.8. The Poincare Isomorphism
Choose a basis vector e of A "E and let e* be the unique basis vector of A "E*
such that <e*, e> = 1. Then, as noted previously,
e*©e=
Now define linear maps
De: AE -- AE*
and
De:AE*--,AE
by setting
De u = i(u)e*
uEAE
and
Deu* = i(u*)e
u* e AE*.
160
6 Mixed Exterior Algebra
In particular, De(1) = e*, De(e) = 1 and De(1) = e, De(e*) = 1. Note that,
for any nonzero scalar ,,
De = a 'De
and D =
As immediate consequences of the definitions we have
De(u n v) = i(v)De(u)
u, v E A E
and
De(u* n v*) = i(v*)De(u*)
u*, v* E A E*.
Moreover, the operators De and De restrict to linear maps
p=0,...,n
^E*
Dp:APE-_
and
D": APE* -- ^ " PE
p = 0, ..., n.
Theorem 6.8.1
1. De and De preserve the scalar products
<Deu,
Deu*>
_ <u*, u>
2. The duals Dp and (DP)* are given by
Dp = (-1)1
P
p = 0, ..., n
and
(DP)* _ (-1)Pt" - PAD" - P
3.
DPoD"_p=(-1"-P)l
p=0,...,n.
p=0,...,n
and
DP o D" - P = (-1)P(" - P)l
p = 0, ... , n.
In particular, the maps De, De, D, and DP are all linear isomorphisms.
PROOF
1. It is sufficient to consider the case u e A PE, u* E A PE*. Then we have,
in view of (6.17),
<De u,
Deu*>
_ <i(u)e*, l(u*)e>
<in- p, i(u)e* ® l(u*)e> = /in- p, i(u* ® u)
Since the diagonal subalgebra is commutative,
<in- p, i(u* ® u) n> = <u* ® u, lC n- p
n
Thus we obtain, in view of the Corollary to Proposition 6.7.1,
<De u,
Deu*>
_
<u* ® u, ip> _ <u*, u>.
>.
Poincare Duality
161
2. Let u e A PE and v e A" - E. Then
Dell, v> = <i(u)e*, v> = i(v)i(u)e*
= (- 1)I' - P)i(u)i(v)e* = (- 1)'
- P)l D" _ P v, u>.
This proves the first relation. The second relation is obtained in the same
way.
3. Let
Then, by (2) and (1),
vE
(DP ° D" -,,)u, v> = (- 1)"" - P)<D" _ P u, D" - Pv> = (- 1)Pc" - P)<u, v>
which gives the first part of (3). The second part is established in the same
way
6.9. Naturality
Let (p : E
F be a linear isomorphism from E to a second vector space F.
Choose a basis vector e of n "E and set f = (p A e. Note that the vector
f * = (qp ^) - le* satisfies < f '*, f> = 1.
We show that the diagrams
AE
"
' AF
"' DI
ae
AE*
c^)
AF*
AE*
car
1+
AE -
AF*
"' DI
AF
commute.
In fact, Formula 5.40 yields
((p A ° D f ° (p A )U = i(u)gp ^ f * = De u
and so the first diagram commutes. The commutativity of the second diagram
follows in the same way.
6.10. The Isomorphism DE
In this section we introduce a canonical linear isomorphism
DE : n (E*, E)
A (E*, E)
(not depending on the choice of a basis vector of n "E). In fact, let
DEW = i(w)w e A (E*, E).
If a*, a is a pair of dual basis vectors of ME and n "E*, we have,
DE(u* ® u) = i(u* ® u)(e* ® e) = De u ® Deu*.
(6.20)
6 Mixed Exterior Algebra
162
Thus, the operators De, De, and DE are related by
(6.21)
DE = (De e De) ° QED
where QE denotes the flip map defined in Section 6.2.
The corollary to Proposition 6.7.1 shows that
DE 1P =n - P
(p = 0, ... , n).
In particular,
0
and DE(1) = 1.
DE(1) =
Moreover, it is immediate from (6.3) that
DE(wl w2) = i(w2)DE wI
w1, w2 e A (E*, E).
(6.22)
To state the analogue of Theorem 6.8.1 we introduce the involution E of
A (E*, E) given by
E(Z) = (- 1)P(n - P) + 9(n - 9)Z
z e A 9(E*, E).
Then we have
Theorem 6.10.1
1. DE is an isometry.
2. The dual operator is given by
DE = SZE ° DE .
3. DE-czE.
PROOF
1. By Theorem 6.8.1, Part (1) and Formula (6.20) we have
(DE(u* +® u), DE(v* +® v)> = <1)eu ® IYu*, De v ® Dev*>
= <De u, Dev*>(De v, DeU*>
= <u* ® u, v* ® v>.
2. This follows from Theorem 6.8.1, Part (2) and (6.20x).
3. This is a consequence -of (1) and (2).
CJ
Finally, observe that the isomorphism DE restricts to a linear automarphism DA of the space AE.
Theorem &10.1 implies that
DA = Dit
and
Da = i.
163
Poincare Duality
6.11. Naturality
Let q : E - F be a linear isomorphism and consider the induced isomorphism
A (F*, F) (see Section 6.2).
; : A (E*, E)
Then the diagram
A (E*, E) ;) A (F*, F)
DE
DF
A (F*, F)
A (E*, E)
commutes. This follows from the naturality of De and De, Formula (6.21) and
the relation
QFo; =
where aq = QP A Q (q, ^) -1.
6.12. The Intersection Product
We introduce a second product structure, the intersection product, in n E by
setting
u rn v= De[(De) -1 u n (De) - 1 v]
u, v e A E.
Thus if u E A PE and v E A 9E, then u r v E A P+ 9-"E. In particular, u ( v= 0
ifp+q>2norp+q<n.
It is immediate from the definition that
ur
P)(" - 9)v
rn u
u E APE, v E A 9E.
Moreover, if e e A "E is the element used to define De, then
u r a=er u=u
u e A E.
The intersection product makes A E into an algebra called the intersection
algebra.
Now we show that De is an isomorphism from the intersection algebra to
the exterior algebra,
De(u r v) = De u n De v
u, v e A E.
In fact, let u e A PE and v e A 9E. Then Theorem 6.8.1, Part (3) gives
De(u rn v) = (- 1)P(P)+( 9" - 9)(D" - P) - 1 u ^ (D" - 9) - 1 v
= DP u A D9 v = De u n De v.
6 Mixed Exterior Algebra
164
Then u r v is a scalar. It will be denoted by
Now let u e A PE and v e
J(u, v). Theorem 6.8.1, Part (3) implies that
J(u, v) _ (u r v)<e*, e> _ <De(U r v), e>
_ <De U A De v, e> _ <De v, i(De U)e>
_ <De v, DeDe U> _ (-1)P(" - P)<De v, U
_ (-1)P("
P'<e*, v n u> _ <e*, U n v>.
Thus,
J(u, v) _ <e*, u n v>.
To obtain a geometric interpretation of J(u, v) let E be a real vector space.
Orient E via the determinant function 0 given by
0(x1, ... , x") _ <e*, xi n ... n x">
x E E.
Now consider two oriented subspaces E 1 and E2 of dimensions p and
n - p. Choose positive bases
a 1, ..., aP
and aP + 1, ... , a"
of E 1 and E2 respectively and set
U=a1
v=aP+1
Then
J(u, v) = 0(a 1, ..., a").
This shows that
i. J(u, v)
0 if and only if E 1 + E2 = E ; that is, if and only if E = E 1 p E2.
ii. Suppose that E = E1 p E2 holds. Then the orientation of E coincides
with the orientations induced by E1 and E2 (see Section 4.29 of Linear
Algebra) if and only if J(u, v) > 0.
6.13. The Duals of the Basis Elements
Now let {e}, {e*v} be a pair of dual bases. Then the dual images of the basis
< vP) of A PE are given by
(v1 <
n
vectors eel n
De (eel ^
^
1)Ep=1(Vz - Oe*vp + 1 ^
... A e*V I,
where (v+ 1, ..., v") is the complementary (n - p)-tuple.
To prove (6.23) we write e* in the form
e* = E, a*v1 n ... n e*Vn,
where o denotes the permutation (1, ..., n) - (v1, ... , v").
(6.23)
Poincare Duality
165
Then we have
De(evl ^ ... A evp) = E i(ev1 A ... A evpxe*v1 A ... A e*v")
= E,ri(evp) ... i(ev1xe*v1 ^ ... n e*' )
= E a*vp + 1 n ... n e*v".
The relations v 1 <
< vp and vp + 1 <
< vn imply that
E = (-1)''_°
and so Formula (6.23) is proved. In the same way it follows that
De(e*v1 ^ ... ^ e*') = (-
evn.
In view of Theorem 6.8.1, Part (3) and (6.23) we obtain
De(e*p + 1 ^ ... n e*vn) = DeDe(_ 1)Ep=1(v1- i)ev l ^ ... A evp
= (_ 1)P(n-P)(_
n ... A evp
i.e.,
1)"(-
De(e*Vp + 1 ^ ... n
n ... A evp , (6.24)
(6.23) yields, in the -case p = n - 1,
De(ei A ... A ei n ... A en) = (-1)n- `e*`.
(6.23)
6.14. The External Product
x E into
Consider the (n - 1)-linear skew-symmetric mapping of E x
n-1
E* defined by
[x 1, ... , xn _ 1] = De(x 1 A ...
^ xn - 1).
The vector [xi, ... , x_ 1] is called the external product of the vectors
x, (i = 1, ..., n - 1). Clearly the definition of the external product depends
on the choice of the basis vector e* e A nE*. It follows that
<[x1, ...,n1x-], xi>=<e*, x1 n...nxn1 nx1>=0 (i=1, ..., n-1)
showing that the external product is orthogonal to all factors. Now consider
the external product of n- 1 vectors x*v e E*,
[x*1,
..., x*n-1] = De(x*1 n ... n x*n-1).
From Theorem 6.8.1, Part (1) we obtain
<[x1,...,
xn1], [x*1, ..., x*"- 1]> = {x*l A ... ^ x*n- 1, x A ... A x
= det(<x*`, x;>).
6 Mixed Exterior Algebra
166
This yields the Lagrange identity
... , x,,_], [x*l, ... , x*"-1]> = det(<x*`, x;>)
0 < i, j < n - 1.
For the external product of (n - 1) basis vectors e we obtain, from (6.25)
i=
, ..., n.
This formula gives, in the case n = 3,
[e1, e2] = e*3,
[e3, e1] = e#2
[e2, e3] = e* 1
6.15. Euclidean Spaces
Suppose now that E is an n-dimensional oriented Euclidean space. Then all
the spaces A "E (1 <p <- n) are Euclidean and there exists precisely one unit
vector e e A "E which represents the given orientation. Since E is self-dual we
may set E* = E. Then the Poincare isomorphisms coincide and are given by
De u = i(u)e
u e AE.
De maps A "E onto A "- pE (0 < p < n). Theorem 6.8.1, Part (1) implies
that
(Deu, DeV) _ (u, u)
u, u e A E
and so De is an isometric mapping. From Theorem 6.8.1, Part (2) we obtain
Dp =
(6.26)
where Dp denotes the adjoint of D.
Assume now that n is even, n = 2m. Then the operator Dm defines a linear
transformation of AmE,
Dm : A mE
A mE.
The above formula yields
Dm = (-1)mDm.
It follows that the transformation Dm is selfadjoint or skew depending on
whether m is even or odd.
PROBLEMS
1. Define the operation of GL(E) on A (E*, E) by
az = [(a ^ )-1 p «^ ]z
Prove the following relations :
a. a(z 1 + z2) = az 1 + az2
b.
c. (ad)z = a(flz)
d. iz = z (i being the identity map).
x e GL(E).
Poincare Duality
167
Show that
(Ocp<n),
app=gyp
where
is the unit tensor.
2. Verify the relations
i(1Qu)'=u
uEApE
i(u* Q 1)p = u* Q 1
n+q,p n w
=
n
9
u* E A PE*
<q, w
w E A (E*, E)
P
3. Verify the formulas
De i(v*)u = (-1)q(n _ )v* n Deu
D%(v)u*
v n Deu*
DE i(z)w = (-1)P(S - n)+ qc' -
DE w
U E ARE, V* E /\*
y e A I E, u* E
qE*
z e A (E*, E), we A (E*, E).
4. Prove that
Deµ(z) = i(z)DA
z e 0(E*, E).
5. Let E and F be vector spaces of dimension n and let a : E - F be a linear isomorphism.
a. Denote by DQ and Df the Poincare isomorphisms
Deu* = i(u*)e,
Df v* = i(v*) f,
where a is a basis vector of A' E and f = a e. Show that
Df o (a ") - i= a A 0D.
b. If F = E prove that
DE(ocz) = aDE(z)
z e A (E*, E)
(for the definition of az see Problem 1).
6. Let E be an oriented Euclidean plane. Show that the linear map
De : E 4 E
is a rotation with rotation angle + ir/2.
7. Let E be an oriented 3-dimensional Euclidean space. Prove that
x x y= De(x n y)
x, y c E
(see Section 7.16 of Linear Algebra). Use the above relation to prove the formulas
(x1 X YX2 X Y2) = (x1, x2)(Y1, Y2) - (x1, Y2xx2, Yi)
and
(x x y) x z = y(x, z) - x(y, z).
6 Mixed Exterior Algebra
168
8. Let {e}, {e*"} be a pair of dual bases of E and E*. Prove that
DL, (x 1 n ... n xp)
=
v1<...<VP
p!(- 1 Z =1(vi - i) /e*V1, x \ ... <e*vn, x >e*vP + 1 ^ ... A a*v".
1
In the above sum, (vp+ 1, ... , vn) denotes the ordered complement (v1, ..., vp) in
the natural order.
9. Show that the restriction of the isomorphism DE to the subspaces A (E*, E) and
A °(E*, E) is the identity.
10. Show that the components of DE Z are given by (see Problem 8)
(DEZ)p+ 1..... Vn = (- 1)' 1
(v1 -
1
1 < ... < Vn,
1
where (v1, ... , vp) is the complementary p-tuple of (vp+ 1, ... , vn) and (/2i,. . . , pq)
is the complementary q-tuple of (pq + 1, ... , pn) in the natural order.
11. Let E* and E be two dual 3-dimensional spaces and a e E, a & 0 and b* E E* be two
given vectors. Prove that there exists a vector x e E such that [a, x] = b* if and only
if <b*, a> = 0, and prove that, if xo is a particular solution of [a, x] = b*, then the
general solution is given by x = xo + Aa, e f.
12. Consider an (n - 1)-linear skew-symmetric map q of E into a vector space F. Prove
that there exists exactly one linear map x : E* - F such that
(p(x1, ... , x,,- 1) =
x[xl,...
x,, E E (V = 1, ... , n - 1).
, xn_ 1]
13. If a is a linear automorphism of E, prove that
det a((x-1)*[x1, ... , x,,_1]
[wc1, ... ,
x,, E E (V = 1, ... , n - 1).
14. Let e,, (v = 1, ..., n) be a basis of E such that e1 n
n e n = e. Given n - 1
vectors
xi =
(i = 1, ... , n - 1),
4 c eV
v
show that the vector y* = [x1, ..., xn
1
1
5
,Iv = (1)_V
] has components
A
1
(v= 1, ... , n).
n-1 ' n-1 " n-1
Av
1
n
15. Using the formula in Problem 14 derive the relation
n
H
V1
v=1
nl
n
=
V= 1
n
n
v
V1
n - 1 nv ' '
v
1
n-1
V1
from the Lagrange identity.
1
zAv
yn-1
n
sn-1
n ... n
A
169
Applications to Linear Transformations
16. Find the minimum polynomial, the characteristic polynomial, the trace, and the
determinant of D. Hint: In case dim E = 2m consider the restriction of Ds to the
subspace n m(E*, E).
17. Suppose E is a Euclidean space of dimension n = 2m and set E* = E. Prove that
n m(E, E) = X 2(A mE) 8 Y2(A mE)
is a decomposition of n m(E, E) into orthogonal spaces stable under Ds (see Sections
4.2 and 4.10 with E replaced by n mE).
Applications to Linear Transformations
In Section 6.16 E and E* denote a dual pair of n-dimensional vector spaces.
If p : E - E is a linear transformation, then the induced homomorphism
P A : AE - AE will also be denoted by A (P and the restriction of (p to AE
will be denoted by n p(p.
6.16. The Isomorphism T
Let t e E* ® E be the unit tensor for E and E* (see Section 1.26). Then t
determines a linear map
T:L(E;E)_E*®E
defined by
q e L(E ; E).
T (q,) = (i
(Notice that this isomorphism T is the inverse of the T in Section 1.26.) In
particular, we have
T(i) = t.
Now let a e E and b* a E* be arbitrary vectors. Then it follows that
<T(co), b* ® a> =
®:)t, b* ® a> = <t, P*b* ® a>
= <(p*b*, a> = <b*, cpa>,
whence
(p e L(E; E), a e E, b* e E*.
<T(cp), b* ® a> = <b*, cpa>
(6.27)
Using this relation, we obtain
<T (q,), t> =
<e,
T(cp), t> = tr p
tr (p,
p e L(E; E).
(6.28)
170
6 Mixed Exterior Algebra
Similarly, a linear map
T : L(E*; E*) _ E* ® E
is defined by
?(x) = (x ® c )t
x e L(E* ; E*).
An argument similar to that given above shows that
x e L(E* ; E*), a e E, b* E E*.
< T (x), b* O a = <xb*, a> = <b*, x*a)
Comparing this relation with (6.27) we find that
T(qp) = T(rp*)
(p e L(E; E).
(6.29)
Now let l/i e L(E; E) be a second linear transformation. Then we have
= (:® Y') [(l ® p)t] = (l ® Y') T ((p),
T (Y' p) = (l ®
whence
T(fi ° q,) = (i/i ®
(6.30)
l)T(co).
With the aid of (6.28) and (6.30) we shall prove that T and T are linear
isomorphisms. In fact, for any p e L(E; E), i/i* e L(E*; E*) we have
T(/,*)) =
l)t) = <(l ®
t)
= <T(i/i ° qi), t> = tr(i/i ° qi),
< T (q,), T (,fr*)> = tr(/i ° q,)
rp, l/i e L(E; E).
Since the bilinear function (i//, rp) - tr(i/i o q,) is nondegenerate, it follows that
T and T are linear isomorphisms.
The Skew Tensor Product of n E* and A E
6.17. The Algebra A E* p n E
Recall that so far we have defined two multiplications in the space A E* ®
A E, the canonical tensor product of the algebras A E* and A E (see Section
6.1) and the composition product (see Sections 6.4 and 6.5). We shall now
introduce a third algebra structure in this space, namely the skew tensor
product of the graded algebras A E* and A E. The corresponding multiplication will be denoted by A.
171
The Skew Tensor Product of A E* and A E
Thus,
(u* ® u) n (U* ® U) = (-1)9r(u* n U*) ® (u n U)
U* E A E*, u E A 9E, U* E A rE, U E A E.
A E* © A E is an associative algebra with unit element 1 ® 1. It is generated
by the elements 1 ® 1, x* ® 1 and 1 ® x where x* a E* and x e E. The
formula above implies that
W1
1 A W2
-
(- 1)(P+ 9)(r+s)W2 A W 1
w 1 E A PE* ® A 9E, W 2 E A rE* ® A E.
(6.31)
as is easily verified. In particular, if p + q or r + s is even, then
W1AW2=W2AW1.
(6.32)
For w e A E* ® A E we shall set
wk=
1
k>1
k.
k
and
Then (6.32) yields the binomial formula
wi n ww1, w2 e A PE* ® ^ 9E, pq even.
(w1 + w2)k =
i+j=k
Note that the products in the algebras A E* Q A E and A E* ® A E are
connected by the relation
(u* ® u) A (U* ® U) = (-1)9r(u* ® u) (U* ® U)
u E A 9E, U* e A rE*.
In particular,
z1 n ... A zk = (_ 1)k(k- 1)/2z1 ... zk
zi e E* ® E,
(6.33)
and so
1)k(k - 1)/ 2 zk
zeE* ® E.
6.18. The Inner Product in E* © E
Consider the bilinear function (, >> in the space A E* ® A E defined by
*
*
®U,V ®U>> =
((-1y <u*, U)<U*, u>
0
if p = s and r = q,
otherwise,
where u* e A PE*, u e A 9E, U* e A rE*, v e A E. It is easily checked that this
bilinear function is an inner product.
172
6 Mixed Exterior Algebra
Next, recall from Section 5.15 the isomorphism f : A E* Q A E -+
A (E* Q E) given by f(u* ® v) = 'A u * n JA v where i : E* -+ E* Q E and
j : E - E* Q E denote the inclusion maps.
Proposition 6.18.1. Define an inner product in E* Q E by
(x* © x, y* © Y) = <x*, y> + <Y*, x>.
Then f is an isometr y,
«.f (u* ® u), f(v* ® v)) = (u* ® u, v* ® v>>.
PROOF. We may assume that
u=xln...nxq
and
v*=y*A...A y*,
v= y1 A... Ay.
Then we have
f (u* ® u) = (ix t A ... A ix) A (jx 1 A ... ^ Jxq)
f(v* ® v) = (iy* n ... n iy*) ^ (jy 1 n ... A JYj
whence
(f (u * ® u), .f (v* ® v)) = (<(ix* n ... n i4) n (jx 1 n ... A Jxq),
(iy n ... A iy*) n (j y1 n ... A
Thus,
(f(u* ® u), f(v* ® v)) =
0 ifp+r q+s.
On the other hand, if p + r = q + s, then
(f(u*
u
v*
v
= det «iy*>> «ixQ,JYo>
(1 x p ,
iy
) (1 x , jya)
But, since
(ix, iy) = 0
a = 1, ... , P,
(ixa , JYa) = (x, Ya>>
P = 1, ... , q,
(Jxp, iy*) = «Y*, xp>>
Y = 1, ..., r,
(Jxp, JYa>> = 0,
b = 1, ... , s,
it follows that
( f (u* ® u), f (v* ® v)) = 0 unless 5 = p and r = q.
173
The Skew Tensor Product of A E* and A E
Ifs = p and r = q, we obtain
«.f (u* O u),.f (U* O U)) = (-1)P9 det(<x« , ys>)det(<yy, , xp>)
-(
1)' <u*, U><U*, u>
= ru* ®
which completes the proof.
,
®
a
Finally note that the inner product defined above and the inner product
defined in Section 6.2 are obtained from each other by the formula
(W1, wz) = (-1)' <w1, w2>
w1 E A 9(E*, E), w2 E A p(E*, E). (6.34)
In particular,
«Z1 A ... ^ ZP, W1 ^ ... n WP>> = (- 1)P<Z1 n ... ^ ZP, W1 ^ ... n WP>
zi EE* Q E, wi EE* Q E. (6.35)
7
Applications to Linear
Transformations
In this chapter E continues to denote an n-dimensional vector space over a field of characteristic
zero.
All the problems concerning this chapter are collected at the end of the chapter.
The Isomorphism DL
7.1. Definition
Recall from Section 6.2 the isomorphism
TE: A (E*, E) -> L(n E; A E).
Let
DL:L(AE; AE)
L(AE; AE)
denote the isomorphism defined by the commutative diagram
n (E*, E) -
L( A E ; A E)
A (E*, E)
L( A E; A E)
where DE is the Poincare map. If u* E A E* and u e A E, the linear transformation DL ° TE(u* ® u) is explicitly given by
[DL o TE(u* ® u)]u = <Deu, u>Deu*.
From the results of Section 6.10 and Formula (6.12) we obtain the following relations:
DL = SZL ° DL
(7.1)
DL = SZL
(7.2)
DL ('APE) = lnn-PE
tr(DL X Dj) = tr(x ° f3)
a, fi e L( A E; A E),
where cZL denotes the involution in L( A E; A E) given by cZL = TE ° E °
174
The Isomorphism DL
175
The last relation implies that
a e L(A E; AE).
tr(DL a) = tr a
Proposition 7.1.1. Let e be a basis vector of n E. Then we have for a, fJE
L(nE; AE)
DL a = D° a* ° (De}-1
DL(x ° fi) = DL(f3) ° DL(a)
PROOF. Let a = TE(u* ® u). Then we have, by Theorem 6.8.1,
CDe ° TE(u* ® u)* ° De J(U) = De(<De u, u>u*) _ <Deu, V>Deu*
= DL TE(u* ® u)(U)
and so (1) follows. (2) is an immediate consequence of (1).
0
Next observe that the image of the diagonal subalgebra 0(E) ander TE is
L( A "E; A "E). Thus DL restricts to a linear automorphism
the space
DL of p= o L( A "E; A "E). Moreover, since cZL reduces to the identity in this
space, Formulas (7.1) and (7.2) show that DL is a selfadjoint involution.
Let (P be a linear transformation of E and consider the induced transformation i ® cP A of A (E*, E). To simplify notation we shall set
Then
Now Formula (6.4) yields the relations
i(Z) o 4j* = * o i(4Z)
i(z) o
Moreover, if
=
o i(4" z).
(7.3)
= TE(z), Lemma II of Section 6.3 implies that
w = cD(w)
w° p= 4* w
w e A (E*, E)
we /1 E*
(7.4)
Setting w = gyp, we obtain
zp = gy(p)
(P = 4, ... , n).
(7.5)
7.2. The Determinant
Recall from Section 4.4 of Linear Algebra that the determinant of a linear
transformation ( is defined by the equation
i
px 1, ... ,
det P A(x 1, ..., x).
176
7 Applications to Linear Transformations
This can be written in the form
(A ncp*)e* = det p e*
where e* is a basis vector of A "E*. Thus,
det (p = tr( A nip*) = tr( A "gyp).
Similarly,
( ^ e* = (A
det p e,
where a is a basis vector of A E.
Since " = e* ® e, where a*, a is a pair of dual basis vectors of n "E* and
A nE, the relations above imply that
fi(n) =
det (p -" .
Now Formula (7.5) shows that
z" = det ( &
z e E* ® E.
Proposition 7.2.1. Let z e E* ® E and set TE(z) = (p. Then for 0 < p < n,
l(DE zP) = det ( "- P
( DE zP) = det
n _p
1.
and
Zp o DE(z" - P) = det (p ' Lp
2.
DE(z" - P) o zP = det (p .P .
PROOF. Using (7.5), the second Formula (7.3) and Part (ii) of the Corollary to
Proposition 6.7.1 we obtain
z'(DE 9) =
n
= (1( p) }tn = det pi(p) n = det
which proves the first Relation (1). The second Relation (1) is obtained in the
same way.
(2) In fact, the first Relation (7.4) and Part (1) of the proposition yield
ZP DE Z" P = I'(DE Z" P) = det (p.
P.
The second Formula (2) is established in a similar way.
Corollary (Laplace formula). Let p be a linear transformation of E. Then
Pip ° DL( ^ "- P(p) = det (p 1^PE
1.
and
2.
DL(^ Pip) ° A n P( = det (p i
pL .
The Isomorphism DL
177
PROOF. Apply TE to Formula (2) in the proposition and observe that TE(zp)
= A pq,
Proposition 7.2.2. Let z e E* ® E and set T (z) =gyp. Then
i(DEZp)Zq=
1.
p-q
(2n
p
n
det , `p+q-n
and
2.
DEZp DEZq =
2n
p
n-
pq
det q, . DE zp+q-" .
PROOF. In fact, applying (7.5) the second Relation (7.3) and Part (1) of
Proposition 7.2.1 we obtain
i(DE ZP)z = 1(DE Zp)D( q)
= I i(D*DE
det
2n-p-q)det
n-p
(2n
q
cDi(4n _
P _ `gyp + q - n)
- p - 91cp
)det
n-p
z
(2) Since the restriction of DE to the diagonal algebra A(E) is an involution, we have, in view of (6.22),
DEL1(DE ZP)Zq] - DEL1CDE Zp)DE(DE zq)
= DE[DE(DE Zq DE ?)] = DE(zq) ' DE(zp).
Now (2) follows from (1), since AE is commutative.
7.3. The Adjoint Tensor
Consider the symmetric (n - 1)-linear mapping
Ad:(E*®E) x ... x (E*®E)_.,E*®E
given by
Ad(Z 1, ..., Zn _ 1) = DE(Z l ... Zn _ )
; e E* ® E,
and set
ad(z) =
1
(n-i).
Ad(z,
... , z).
178
7 Applications to Linear Transformations
Thus,
ad(z) = DE(Zn - 1) ;
ad(z) is called the adjoins tensor of z. Observe that ad(z) does not depend
linearly on z except in the case n = 1.
Since DE is an involution in the diagonal subalgebra AE, we have the
relation
z e E* Q E.
DE ad(z) = zn -1
Moreover, Proposition 7.2.1, (2) implies that
ad(z) o z = z o ad(z) = det TE(z)&
Next, let z e E* Q E and define tensors B p(z) E E* Q E (p = 0, ..., n - 1)
by the expansion
p- 1
_
ad(z +
B p(Z)a," p - 1.
p=0
Proposition 7.3.1. The tensors B p(z) are given by
B p(z) =
(P = 0, ..., n - 1).
1
PROOF. Since, by the binomial formula (see Section 6.2)
n-1
1- pp n - 1 p
(Z + ), )n - 1 =
p=0
we have
n-1
ad(z + A) _
DE(zp n
_ pP," - 1-p.
p=0
But, since A(E) is commutative,
DE(Z"
' n-p-
1) = 1(Zp 4n - p - 1) n = 1(Zp)
p+1
and so we obtain
n- 1
ad(z +
)=
1(Zp), + 1 an p - 1.
p=0
It follows that
B p(z) =
1
(P = 0, ..., n - 1).
Next we prove the Jacobi identity
DE(ad z)" = (det TE(z))p -1 z" p
1 <p < n.
Proposition 7.2.2, (1) yields
i(ad z)z9 = (n + 1 - q)det TE(z)z4-1
(7.6)
The Isomorphism DL
179
whence, by induction,
i ad z p z9 = n+ p
q det T
Z pz9 - p.
P
In particular,
i(ad Z)p - 1)Z" - 1 = p(det TE Z)p - 1z " - P
1 < p < n.
On the other hand,
i [(ad Z)p - 1] Z" - 1 = i [(ad Z)p - 1 ] DE ad z
and thus, in view of Formula (6.22)
i((ad Z)p - 1)Z" - 1 = DE [(ad z). (ad z)" - 1]
= pDE(ad z)p.
Combining these relations we obtain the Jacobi identity.
Setting p = n - 1 in this identity yields the formula
ad ad z = (det TE(z))"- 2z
z e E* ® E, n > 2.
7.4. The Classical Adjoint Transformation
4' -1 be linear transformations of E. Fix a nonzero determinant
Let
function 0 in E. Then, for every n-tuple of vectors (x 1, ..., x"), a linear transformation i (x 1, ..., x") of E is defined by
SZe(x 1, ... , x")h =
EQ 0(qp 1
4n - 1 XQ(n - 1) '
It is easy to check that SZs(x 1, ... , x") is skew-symmetric in the x. Thus the
correspondence
(x1, ..., x") H
1, ..., xn)
defines a skew-symmetric n-linear mapping
Ex
x E -+L(E ; E).
n
Hence, there is a unique element in L(E; E), denoted by Ad(p1, ..., c°?
such that
SZs(x 1,
... , x") = 0(x 1, ..., xn)Ad(q,1, ...,
c_ 1).
Now the above definition reads
Ad(cp 1,
... ,
c_ 1)h
0(x 1, ..., x") =
E, 0((p 1 X(1), ... , 4_ 1 x_ 1) ' h) x7(n) .
Q
180
7 Applications to Linear Transformations
In particular, if {e1,
we have, for h e E,
..., en} is a basis of E such that 0(e 1, ..., en) =
1, then
c, 0((p1 eQ(1), ... , (pn - 1e(_ _ 1),
Ad((p 1, ... , q_ 1)h =
Q
We shall show that the function Ad is symmetric. In fact, let
tion of (1, ..., n - 1) and set
t(aC 1, ... , xn)h =
c
be a permuta-
(pt(n - 1)X7(_ 1), h)x0(n)
(1)
Q
It has to be shown that SZt = SZo . Extend i to a permutation of (1, ..., n) by
setting i(n) = n and let a = o r'. Then we have
t(aC 1, ... , xn)h =
Ea 0((p 1xa(1), ... , (pn - 1 Xa(n - 1), h)xa(n)
a
= SZA(x 1, ..., xn)h,
whence SZt = SZo .
For a single linear transformation (p we set
1
ad(q,) = (n _ 1)
Ad((P> ... > (p).
n-1
Since
n
0(cox 1, ... , h, ... , (pxn)xQ(n) _ (n - 1) !
Q
o(rpx 1,
..., h, ... , (pxn)xi ,
i=1
Q(n)
it follows that ad((p) coincides with the classical adjoint of (p as defined in
Section 4.6 of Linear Algebra.
Proposition 7.4.1. The operators DL and Ad are connected by the relation
1.
DL(q l D ... D (Pn - 1) = Ad((p 1, ... , (pn - 1)
(p e L(E; E), v = 1, ... , n - 1.
In particular
2.
DL(^n-1(p) = ad(p)
P e L(E; E).
PROOF. Without loss of generality we may assume that the q are of the form
(p= TE(a* Q a)
Fix a basis {e1,
a* a E*, aa E.
..., en} of E such that 0(e 1, ..., en) = 1 and set
e 1 A ... n en = e
e* 1 n ... n e*n = e*.
The Isomorphism DL
181
Then we have for h e E
=
E, 0(<a i ,
1, ... , <an - 1, eQ(n - 1)>an - 1, h)e (n)
Q
=
1)> . 0(a 1, ...,
<a_1,
i,
It follows that for h* E E*
<h*, Ad((p1,... , (Pn-1)h>
=
0(a1, ... , an- 1, h)
<an- 1, eQ(n- 1)><h*,
= <ai A
A an_1 n h*, e1 A
A
n aq_1 n h*,e> <e*,a1 n
<ai n
n
n h>
= <h*, a><a*, h>,
where
a = i(a i n ... n a_ 1)e
and a* = i(a 1 n ... n a_ 1)e*.
Thus,
<h*, Ad((p1, ...,
<h*, a><a*, h>.
(7.7)
On the other hand,
DL((p1
...
p, - 1) = DL TE[(a i ® a 1) ... (a_ 1 ® a_ 1)]
= D L TE [(a 1 ^ ... A a_ 1) ® (a1 A ... A a_ 1)]
TEDE[(al ^ ... ^ an_ 1) ® (a1 A ... A an- 1)]
= TE[i(a 1 n ... ^ a_ 1)e* ® i(a i n ... n a_ 1)e]
= TE(a* ® a).
It follows that
<h*, DL((P1
...
(Pn- 1)h> = <h*, TE(a* ® a)h>
= <h*, a><a*, h>.
Now the proposition follows from Relations (7.7) and (7.8).
Corollary I. Let p be a linear transformation of E. Then
ad(p) = det (p
i
and
ad(p) o p
= det p
i
In particular, ad(p) is a linear isomorphism if and only if p is.
7 Applications to Linear Transformations
182
PROOF. In fact, Proposition 7.4.1 and the Laplace formula (see Section 7.2)
yield
ad((p) ° (p = (p o ad((p) = (p ° DL( ^ "- 1(p) = det (p
1.
Corollary II. If p and cui are linear transformations, then
ad(/i o (p) = ad((p) o ad(/i).
PROOF. In view of Proposition 7.4.1 and Proposition 7.1.1, (2),
1(p)
ad(/i ° (p) = DL ^ "- 1(Y, ° (p) = DL(^ "- 1,1/ ° ^ "= DL( A "- 1(p) o DL( A1 ,/i) = ad(/i) o ad((p).
Corollary III. Let (p be a linear transformation of E. Then
DL(A p ad (p) = (det (p)p - 1 ^ "
(P = 1, ..., n).
p(p
In particular,
n > 2.
ad ad((p) = (det (p)"-2(p
PROOF. Apply the proposition and the Jacobi identity (see Section 7.3).
Corollary IV. Let a* a E*, av e E (v = 1, ..., n - 1) and set
(Pv = TE(a* ® a).
Assume that either {a*} or {av} is linearly dependent. Then
Ad((p1,
..., (p"-1) = O.
Characteristic Coefficients
7.5. Definition
Consider for each p > 1 the p-linear function C, in L(E ; E) given by
Cp((p 1, ..., p1,) = tr((p 1
...
p1,)
and set
Co = 1.
Since the box product is commutative, the functions Cp are symmetric. Note
that
C 1((p) = tr (p
(p e L(E ; E).
Characteristic Coefficients
183
The function C will be denoted by Det,
...
Det(p1, ..., cn) = tr((p 1
cn)
Now consider the tensors z e E* a E given by
v= 1,...,p.
TE1(4))
Then Proposition 6.3.1 and Formula (6.16) imply that
...
Cp((p 1, ..., p p} = tr(TE(z 1)
TE(zp)) = <z 1 ...
(7.9)
In particular,
Cp=O ifp>n.
For a single linear transformation P we set
=
C
1C
P
...
_
Pop)
(p = 0, ..., n).
Thus, in view of (6.6),
Cp((p) = tr( A
Clearly,
A e F.
C(AP) = APCp(q,)
Relation (7.9) implies that
(P = 0, ..., n).
Cp(4) = <(T E 1(p)P, p>
Recall from Section 7.2 that
det cP = tr( A p).
Thus
Cn(4) = det gyp.
Equivalently,
1
-!
n
Det(cp,... , 4)) = det (p.
Replacing cp by Q + /i in this relation we obtain the formula
1
det(gp +) = n.p+q=nDet(gp, ..., (p,
P
..., i/).
9
Proposition 7.5.1. The CP(cp) satisfy the relation
n
det(gp + )i) =
C_
A e F.
p=0
Thus
p(cp) is the pth characteristic coefficient of cp.
(7.10)
7 Applications to Linear Transformations
184
PROOF. Let z e E* ® E be the unique tensor such that TE(z) = gyp. Then we have
(p+)1= TE(z+'U)
whence
det(gp + ti) = <(z +
)", i">.
Expanding by the binomial formula we obtain
"
<p z" -
det(gp + t1) _
p
p
(= 0, ..., n).
p = C"- p(co)
It follows that
"
det(gp + i) =
p=o
C" - p(co) V'.
Corollary. If q, and i/i are linear transformations, then the characteristic
polynomials of i/i o q, and q, o ,/i coincide.
PROOF. In fact, since the trace of a composition does not depend on the
order,
Cp(11/ ° (p) = tr n p(/, ° q,) = tr( n p,/, ° n pep)
= tr( n pip o n " 1i) = Cp(co ° i/i)
(p = 0, ..., n).
7.6. The Linear Transformations Ap(p)
Let q be a linear transformation and write
n-1
ad(gp + 1)
- p=o
where A p(ip) E L(E ; E).
In particular,
A_ (q,)
(q,) = ad(gp).
Proposition 7.6.1. The transformation A p(co) is given by
P
=0
(-1)vCp-
0, ..., n - 1).
Characteristic Coefficients
185
PROOF. From the formula (see Corollary I to Proposition 7.4.1)
((p + i) o ad((p + i) = det((p + ti).
we obtain
n- 1
n
L,
A,_ 1((P)An P +
p=0
p=1
Comparing the coefficients of
recursion formulas
n
A p(co)1
P
=
C p((p)),n
P
1.
p=0
on both sides of this equation yields the
(p = 0, ..., n),
(P ° Ap- i(P) + A(p) = Cp((P) l
where A _ ((P) = 0 and An((p) = 0. From these relations we obtain
P
A p((P) =
v=0
(P = 0, ..., n).
(-1)VCp -
(7.11)
In particular, for p = n - 1,
n- 1
ad((p) =
v=0
(- 1)VCn - v - 1((P)(Pv
Cayley-Hamilton theorem. Every linear transformation cp satisfies its characteristic equation; that is,
n
L, (- 1)VCn - v((p)(pv = 0.
v=0
PROOF. Apply (7.11) for p = n observing that An((p) = 0.
Proposition 7.62. The trace of the linear transformation A p((p) is given by
(p = 0, ..., n).
tr Ap((P) _ (n - p)Cp((P)
PROOF. Let z e E* Q E be the unique tensor such that TE(z) _ (p. Consider the
tensors B p(z) determined by the expansion
n- 1
ad(z + i) _
B
p(Z)a,n- p- 1
p=0
(see Section 7.3). Since
(P = 0, ..., n - 1),
TEBP(z) = A(p)
(p = 0, ...,
tr A (p) _ <, B p(z)>
n-
Next observe that, by Proposition 7.3.1,
Bp(z) = i(zP}p +
(P = 0, ..., n - 1).
1).
(7.12)
186
7 Applications to Linear Transformations
It follows from (7.10) that
<, B ,,(z)> = l( )B p(Z) = l(
1
= i(z")i()ip+ 1 = (n - p)<z", p> = (n - p)Cp(co).
(7.13)
Relations (7.12) and (7.13) yield
tr A p((p) = (n - p)Cp(co)
(p = 0, ... , n).
Finally, we show that the characteristic coefficients of the classical adjoint
transformation are given by
Cp(ad (p) = (det (p)P 1C_(p) i(p = 1, ..., n).
(7.14)
In particular,
and tr ad((p) = C" _ 1((p).
det ad((p) = (det (p)" -1
In fact, taking the trace on both sides of the Jacobi identity (7.6) and
observing that for a E L( A E; A E)tr DL X = tr a (see Section 7.1) we obtain
tr A" ad((p) = (det
(p)P- 1
tr n ""p
(p = 1, ... , n)
whence (7.14).
7.7. The Trace Coefficients
Let Trp denote the symmetric p-linear function in L(E ; E) given by
Tr
...
=
1
tr
p
>1
and set
Tro = n.
The pth trace coefficient of a linear transformation p is defined by
Trp((p, ... , gyp)
p >_ 1
and
Tro((p) = n.
Thus
Tr p((p) = tr (pn
p > 1.
In particular,
Trp(i) = n.
Note that, in contrast to the characteristic coefficients, the trace coefficients
do not vanish in general for p> n.
Characteristic Coefficients
187
Proposition 7.7.1. The trace coefficients and the characteristic coefficients are
connected by the relation
1 p-1
Cp(co) = -
_
(- 1)P
v
P v=0
- 1 Cv((p)Trp -
-
p ? 1.
(p)
PROOF. Taking the trace in the formula in Proposition 7.6.1 and using Proposition 7.6.2 we obtain
P
(n -
v=0
(-1)VCp P
= nCp(co) +
v=i
(- 1)VCp - v((p)tr (pv
It follows that
P
Cp(p) = - -
P v=1
_
1
(-1)vCp - v((P)tr pv
(- 1)P
q9 v.
P v=o
7.8. Application to the Elementary Symmetric Functions
Fix a basis {e1,
..., en} of E and consider the linear transformation cP given by
cpev = ,vev (v = 1, ..., n).
A simple calculation shows that
..., n)
C(p) =
while
where Qp and sp denote the symmetric polynomials given by
... , ,n) =
,v1 ...
v1<...cvp
P
P>1
and
sp(a,1,
...,
P
=
vi
Now Proposition 7.7.1 yields the classical recursion formulas for the Qp in
terms of the sp,
1
op= P- v=0 (- 1)P - 1 ov sp - v
v
(p= 1, ... , n).
188
7 Applications to Linear Transformations
7.9. Complex Vector Spaces
Let E be an n-dimensional complex vector space and let E denote the 2ndimensional real vector space. Let DE be a nonzero determinant function in
E. Regard DE as a C-valued n-linear function in E and set
0 = (- l)n0E n DE ,
where DE is defined by
DE(x 1,
Then 0 is linear over
... , xn) =
DE(X 1,
xv a E.
... , xn)
and skew-symmetric. To show that 0 is real-valued
and nonzero (and hence a determinant function in
choose a basis
{al, ... , an} of E. Since 0(zl, ... , zn) = 0 whenever the vectors {z} are
linearly dependent (over C) it follows that
0(al, ... , an, ial, .
lan)
(- i)n
(n!)2
.
,
a7(n))DE laid l),
... ,
= DE(a 1, ... , an)AE(a 1, ... , an)
= I DE(a 1, ... , an) 12.
Thus,
and so 0 is a nonzero determinant function in E. If DE is replaced by ),DE,
where ), is a nonzero complex number, then 0 changes into I , U 20. This
shows that the orientation of E determined by DE is independent of DE
and so E carries a natural orientation.
Proposition 7.9.1. Let cp be a linear transformation of E and let cps denote the
corresponding linear transformation of E. Let f and f denote the characteristic polynomials of p and
Then
li(t) _ .f(t) f(t).
PROOF. We show first that
det (p = det (p det (p.
(7.15)
In fact, let DE be a nonzero determinant function in E and set 0 = inAE n
DE . Then
q A = (- i)n(p*DE A (p*DE
But
(p*DE = det (p DE,
(p*DE = det cP DE
Characteristic Coefficients
189
and so we obtain
0 = (-1)" det (p det P 0E 0E _ det (p 20
whence (7.15).
Replacing
by co - ti in this relation yields
det(q, - ti) = det(gp - ti)det(cp - ti)
whence
Je(t) =.f(t) I(t).
Corollary. The characteristic coefficients of p are given by
Cr((P) =
(r = 0, ..., n).
p+q=r
PROBLEMS
In the problems below T is the map defined in Section 6.16.
1. Find all linear transformations p of E such that T(cp) is decomposable.
2. Let
r
T(p) _ > a O a*`
a E E, a*` E E*
be a representation of the tensor T(p) such that the vectors a1 and a*` (i = 1, ... , r)
are linearly independent. Show that r =
3. Let E, E* be a pair of finite-dimensional dual vector spaces and consider the linear
map
f:E*QE_ L(E;E)
defined by f (b* O a)x = <b*, x>a. Prove that
a. f= T*
b. 1= T'.
4. Verify the relation
<zp, (ad z)p> =()(det
n
T -'(z))"
p
1 < p < n, z e E* O E.
5. Show that
det A nip det A "- nip = (det 'p)0 < p < n.
6. Show that the coefficient of )J - n in the characteristic polynomial of an n x nmatrix is (-1)" _ p times the sum of all principal minors of order p.
7 Applications to Linear Transformations
190
7. Let E be a Euclidean space and set E* = E. Write
L(E; E) = S(E; E) Q+ A(E; E)
where S(E; E) denotes the space of selfadjoint transformations and A(E; E) denotes
the space of skew transformations (see Linear Algebra Problem 3, Chapter IX,
Section 2). Prove that
T(S(E; E)) = Y2(E)
and
T(A(E; E)) = X 2(E).
(see Sections 4.2 and 4.10).
8. Let E be a real vector space and consider the bilinear function b in 0(E, E*) defined
by
I(u, v) = <Dsu, v>
u, v e 0(E, E*).
a. Prove that D is symmetric and has signature zero.
Hint: Make E into a Euclidean space and set E* = E; then consider first A m(E, E)
in the case dim E = 2m. See problems 7 above, Problems 3 and 4 after Section 6.14,
Problem 4, Section 5, Chapter IX of Linear Algebra, and Problem 3, Section
2, Chapter IX of Linear Algebra.
b. Given two linear transformations 'p, t// e H( A E; A E) (see problem 13) prove that
D(T(co), T(/i)) =
and conclude that the signature of the bilinear function
F(cp, /,) = b(T(p), T(/i))
is zero.
c. If dim E = 2m and gyp, cli are any two linear transformations of E such that p is
regular, prove that
T(i)m) = det p tr A "(p' ° cli)
and
tr A m = det p tr A m -1.
Conclude that
= det p
where am and &m are the mth characteristic coefficients of 'p and P - 1 respectively.
d. If n = 2 show that,
T(cli)) = tr p tr cl' - tr(' ° cli).
Characteristic Coefficients
191
9. Use one of the formulas (2) in Proposition 7.2.2 to derive the classical Laplace
expansion formula for a determinant:
Let A = (a;) be an n x p matrix and fix a p-tuple (2i, ... , 2p) such that 21 <
< 2p. Let (2p+ 1, ... , 2n) denote the complementary (n - p)-tuple in increasing
order. Then
det A =
V1<...<Vp
(-1)E =1det
det
where (vp+ 1, ... , vn) is the complementary (n - p)-tuple of (vl, ... , vn). Here
denotes the q x q matrix consisting of the rows 111..... /39 and the columns
OC 1, ... , OC9 .
10. Show that the rank of the adjoint transformation is given by
r(ad gyp) = n
if r(p) = n
r(ad gyp) = 1
if r(cp) = n - 1
r(ad gyp) = 0
if r(cp) < n - 2.
Use these relations to prove the formula
ad ad p = (det p)n - 2(p.
11. Given an n x n matrix A =
define the matrix
by
(v, u = 1, ... , n).
1 ... an
µ ...
«n
n
Using the Jacobi identity in Section 7.3 prove the classical Jacobian identities:
«µP + 1
Vp + l
«µP + 1
Vn
(det A)'1
«µn
vr + 1
. .. «µn
vn
(v1 <...<vp,µl <...<µp)
where (µp+ 1, ... , µn) and (vp+ 1, ... , vn) are the complementary (n - p)-tuples of
(µl, ... , l p) and (vl, ... , vp) respectively.
12. Use the formula
<De u, v> = <e*, u n v>
u e A "E, v E A n- pE
to derive the classical Laplace expansion formula (see Problem 9).
define the vectors x by
Hint: Given an n x n matrix
x1 = >Ja4 e
i = 1, ... , n.
v
Then apply the above formula with
u = x1 n
n xp and v = xp +1 n
n x.
7 Applications to Linear Transformations
192
13. Let E be a Euclidean space and consider the space H( A E; A E) of homogeneous
linear transformations A E -+ A E of degree zero. Then the Poincare isomorphism
induces a linear automorphism DH of H( A E; A E). Given an isometry cp : E -+ E,
prove that A p is an eigenvector of DH with eigenvalue + 1.
14. Consider the isomorphism
*:H(AE; AE) + H(AE*; AE*)
defined by
Prove that * commutes with DL,
DL
(DLcO)*
P E H(A E; AE).
Skew and Skew-Hermitian
Transformations
Throughout this chapter E denotes an n-dimensional vector space over a field of characteristic
zero.
The Characteristic Coefficients of a Skew
Linear Transformation
8.1. Definition
Let E be an n-dimensional inner product space see Section 1.25 and denote
the inner product by (, ). It determines a linear isomorphism o
Q:E-=>E*
via
x,yEE.
<0x, y> = (x, y)
Thus E can be regarded as self-dual. Hence every subspace F
E determines
a second subspace Fl (the orthogonal complement of F). Its dimension
is given by
dim Fl = n - dim F.
In particular, if the restriction of the inner product to F is nondegenerate,
then F r Fl = 0 and so we have the direct decomposition
E=FQ+F'.
A basis {e1,
...,
of E is called orthogonal, if
eµ) = 0
Note that (e', ev)
v
µ.
0 (v = 1, ..., n), since the inner product is nondegenerate.
193
194
8 Skew and Skew-Hermitian Transformations
Every inner product space admits an orthogonal basis. In fact, consider
the function
Q(x) = (x, x)
xEE.
Since
(x, y) = i (Q(x + y) - Q(x) - Q(y)),
it follows that Q
0. Thus there is a vector e1 such that Q(e 1)
0. Let
E 1 denote the subspace spanned by e 1. Then
E=E1Q+Ei
and the restriction of the inner product in E to E i is again nondegenerate.
Now assume by induction that Ei admits an orthogonal basis {e2, ...,
Then {e1, e2, ...,
is an orthogonal basis of E.
Every linear transformation ( of E determines an adjoins transformation
P by the equation
x, y e E.
((Px, y) = (x, 5y)
If P = - cp, then cP is called a skew linear transformation. The skew linear
transformations form a subspace of L(E; E) which will be denoted by Sk(E).
Its dimension is given by
l
dim Sk(E) _ (21.
8.2. The Isomorphisms 'E and lI'E
Consider the linear operator D : A 2E -+ Sk(E) given by
E(a n b)x = (a, x)b - (b, x)a
x e E.
We show that 'DE is an isomorphism. Since
dim Sk(E) _
(2) =dim n ZE,
it is sufficient to prove that 'E is injective.
Choose an orthogonal basis {e1, ..., en} of E and set
E(e n e;) =
(i <J).
Then
Bpi,{x) = (e,, x)e; - (e;, x)ei
x e E.
195
The Characteristic Coefficients of a Skew Linear Transformation
Thus the relation
0
i <j
implies that p1!' = 0 and so J?E is injective.
Thus
E: A 2E - Sk(E)
is a linear isomorphism. The inverse isomorphism will be denoted by 'FE.
Remark. In Section 8.3 we shall derive a formula which expresses the
characteristic coefficients of a skew transformation p in terms of the tensor
"E(P) (see Theorem 8.3.1).
Next consider the projections
E*:E* O E* -+ ^ 2E*
7LE:E Q E -+ A 2E,
and
7C: LE -+ Sk(E),
where
E*(a* O b*) = a* n b*
2E(a p b) = a n b,
and
1P E LE.
p) = P q
We shall show that the isomorphisms 'DE and TE : E* Q E - LE (see Section
(6.2) are connected by the following commutative diagrams :
E* Q E
3
E* Q E
LE
' LE
W
EQE
E* Q E*
n
n
IrL*
nE
W
A 2E - Sk(E)
A 2E*
` - A 2E
E
In fact, let a* E E* and b e E. Then
O l)(a* O b) = T la* n b
and so
O l)(a* O b)x = (a-
la*, x)b
- (b, x) r la*
8 Skew and Skew-Hermitian Transformations
196
On the other hand, set TE(a* Q b) = (p. Then
cpx = <a*, x>b = (Q- la*, x)b.
Hence
px = (b, x)Q la*
and so
((p - P)x = (Q- la*, x)b - (b, x)Q- la*.
It follows that
O l)
7Lo TE = E°
and so the first diagram commutes. The commutativity of the second diagram
follows from i o E (Q -1 Q i) = 7LE* o (i Q a).
8.3. The Isomorphism i
Let E* be the dual space of E. Then the bilinear function (, ) defined by
Kx* © x, y* © y>> = <x*, y> + <y*, x>
(cf. Section 6.18) determines an inner product in the direct sum E* p E, as is
easily checked. Observe that this inner product does not depend on the inner
product in E.
Now consider the isomorphism Q : E - E* induced by the inner product
in E (see Section 8.1) and define a linear transformation i of E* Q E by
setting
1x* + x).
(x* p x) = (x* + ax) +O (_ox*
A simple calculation shows that
2(x* p x) = 2(ox p
(-o-1x*))
and so i is a linear automorphism of E* Q E.
Moreover, i is selfadjoint with respect to the inner product (, ). In fact,
i(x* +O x), y* +O y» = <x*, y> + <Qx, y> - <y*, Q-1x*> +
= <x*, y> +
<y*, x>
<y*, x>
+ (x, y) - (y*, x*)
= (x* +O x, i(y* © y)>>.
Now extend i to an algebra automorphism
i: A (E* Q E) - A (E* Q E).
Then i A is selfadjoint with respect to the induced inner product in A (E* Q E).
197
The Characteristic Coefficients of a Skew Linear Transformation
The automorphism i A determines via the canonical isomorphism
f:AE*Q A E 4 A(E*QE)
(see Section 5.15) an algebra automorphism of n E* 2 n E which will also
be denoted by i A A straightforward computation shows that
i A (x* p 1) = x* Q 1- 1 Q 7
1x*
x* E E*
and
tA(1Qx)= JXQ1+1®x
xeE.
These relations yield
i A (x* p x) = (x* n Qx) O 1- 1 O (Q- lx* n x)
+ x* Q x+ ox O o
1x*
x* E E*, x e E.
(8.2)
Let QE be the linear automorphism of E* Q E given by
QE(x* O x)
= Qx 0 Q- ix*.
Then Formula (8.2) yields
X A QE(x* Q x) = (Qx A x*) Q 1- 1 Q (x A cr
x*)
+ ox ® Q- 1x* + x* O x.
(8.3)
Adding (8.2) and (8.3) we obtain
n
x* E E*, x e E.
[x* 0 x + QE(x* OX x)] = 2(x* OO x + QE(x* OX x))
Thus,
X n (U + QE U) = 2(u + QE U)
u E E* Q E.
In particular, if QEu = u, then
i A u - 2u.
(8.4)
On the other hand, subtracting (8.3) from (8.2) yields
iA [x* O x - QE(x* Q x)] = 2[(x* n ox) O 1 - 1 O (o - tx* n x)]
= 2[7LE*(x* p ox) 0 1- 1 0 E(° 1x* 0 x)],
where
7LE : E Q E + A 2E
and 7LE* o E* Q E* -+ A 2E*
are the projections defined in Section 8.2. Thus we have the relation
'L
(u - QEu) = 2[7CE*(l O Q)u O 1 - 1 O
E E* p E. (8.5)
In particular, if QEu = - u, this reduces to
Xn
u- 7LE*(l Q Q)u Q 1- 1 Q 7LE(° t QX l)u.
(8.6)
198
8 Skew and Skew-Hermitian Transformations
Next, consider the linear isomorphism TE: E* Q E -+ L(E; E) (see
Section 6.2). It is easily checked that the adjoint transformation of TE(u)
is given by
u e E* Q E.
TE(u) = TE(QE u)
Thus TE(u) is skew if and only if u satisfies
QE(u) = -u.
Lemma. Let (P be a skew linear transformation and set
u = TE 1C(P),
Z = E((P)
Then
i A u- 2(Q A Z Q 1- 1®z).
PROOF. Since (P is skew, u satisfies
QEU = -U.
Thus Formula (8.6) yields
t, u= 7LE*(l Q Q)u Q 1- 1 Q 7LE(Q- 1 Q l)u.
In view of the second diagram (8.1)
7LE*(l 0 Q)u = 7LE*(l 0 Q) TE
1(4))
= Q n 'E 7L((P)
Since (P is skew, 7r((p) = 2(p, and so we obtain
E*(l Q Q)u = 20 `I'E((P) = 20 z.
(8.7)
On the other hand, the first diagram (8.1) yields
7LE(Q- 1 0 l)u = 7LE(Q- 1 0 l)TE 1((p) = PE 1ir((P) = 2'PE 1((p) = 2Z.
(8.8)
Relations (8.7), and (8.8) yield
i u= 2(QA z Qx 1- 1 ®z)
and so the lemma is proved.
Theorem 8.3.1. The characteristic coefficients of a skew linear transformation
are given by
C2 p(4)) = ((' E (P)p, ('PE 4)))
and
C2p+1C4)=0
p>_0.
PROOF. Set (p = TE(u), u e E* Q E. Then, according to (7.10),
Ck((P) = <Uk, k> = <uk, k>
(k = 0, ... , n).
199
The Characteristic Coefficients of a Skew Linear Transformation
Next, observe that (see Section 6.17)
uk =
(- 1)k(k - 1)/2uk
k=
(- 1)k(k- 1)/2tk
and
(see Formula (6.33) of Section 6.17). Using Formula (6.34) in Section 6.18
we obtain
Ck(p) = <uk, tk> = (
(8.9)
1)k<<uk, tk>>.
Now consider the automorphism i. Since p is skew, the lemma shows
that
TA u= 2(z* Q 1- 1 Q z),
(8.10)
where
and
z = `PE cp
On the other hand, since QED _ , Formula (8.4) yields
= 2&
Thus, since TA preserves products in the algebra n E* Q A E,
TA() = 2kk.
Inserting this into (8.9) and observing that TA is selfadjoint with respect to
the inner product (, >> we obtain
k(uk
Ck(co) = (
(
i n (k)»
1)k2 -
1)k2 - k«i (uk)
k>>
(- 1)k2 - k«(i u)k,k»
(8.11)
.
Raising (8.10) to the kth power (in the algebra n E* Q A E) and using
the binomial formula yields
n
u)k = 2k(z* QX 1 - 1 QX z)k = 2k
i+j=k
(- 1)'zQ
(8.12)
From (8.11) and (8.12) we obtain
Ck(co) = (
1)k
i+j=k
(
1)J
\\z*i O z,,
Now observe that
E n 2jE* O n 2jE
while
k E n kE* QX n kE.
k>>.
(8.13)
200
8 Skew and Skew-Hermitian Transformations
Thus Formula (8.13) shows that
if k is odd
Ck((o) = 0
(see the definition of the inner product (,) in Section 6.18).
On the other hand, if k = 2p, then, (8.13) yields
C2 ((p) _ (_ 1)P«z#n OX
P
Since
P=(_1)Pc2P-1)2P=(_1)P2P
(see (6.33), Section 6.18) it follows that
C2p((P) = <<z#P ® zP, 4'>>
and so, in view of (6.33)
C2p((p) _ <z#P ® z,
2P>.
Finally, applying (6.17), we obtain
C2 p((p) _ <z#P, zP> = <(C A z)", z"> = (z", z") = (('11E (P)PS ('PE (P)P)
This completes the proof of the theorem.
The Pfaffian of a Skew Linear Transformation
8.4. The Pfaffian
Let E be an inner product space of dimension n = 2m and let p,,
, p,
be skew linear transformations. Every cp determines an element
E A ZE.
Thus,
"E((P 1) A ... A "E((Pm) E A "E.
Now fix a basis vector a of A "E and set
Pfa((P 1,
... , (Pm) _ (`I E((P 1) n ... A E((Pm), a)
Equivalently,
E(o1) n...n
E(cm)
= a1 aPfa(o1> ..., co)
m a.
The scalar Pfa((P 1, ... , Corn) is called the Pfaffian of (P 1, ..., co. Since every
two elements q'E(rpµ) commute, it follows that Pfa is a symmetric m-linear
function in Sk(E).
The Pfaffian of a Skew Linear Transformation
201
The Pfaffian of a single skew transformation (p is defined by
f=
a((P)
1
m
Pf...
a
((P,
> (P) .
m
This can be written in the form
Pfa((p) = (`I'E((P)m, a)
Clearly,
EV
mPfa((p)
Proposition 8.4.1. If i is an isometr y of E, then
Pfa(t ° (p j ° 'L - 1, ... , 'L ° (p m ° - 1) = det -r Pfa((P 1 , ... , (pm)
PROOF. Since i preserves the inner product we have
E(ix n x y) = i° E(x n y) o f -1
x, y e E
whence
i n `1'E(p) =
° 'L -1)
(P E
Sk(E).
It follows that
E(t° (1 ° - 1) n ... n LIJE( ° (m °
= 'L A
- 1)
E((p 1) ^ ... ^ 'L A E((Pm) =
n (`I'E((p
1) ^ ... A qJ ((pm))
whence
Pfa(t°p1or 1,...,'L°(p m°'L-1)
(t n (`I'E((p1) ^ ... A E(Pm)), a) = (1 E((p1) ^ ... A E((pm), x n
But since i is an isometry
i l a= det i- 1 a= det i a
and so we obtain
Pfa(t ° (p1 ° 'L - 1, ... , 'L ° (p m ° 'L - 1) = det i(LI'E((p 1) ^ ... A "E((pm), a)
= det
Pfa((p 1, ... , (pm)
Proposition 8.4.2. Let a and b be basis vectors of A E. Then
Pfa((P) Pfh((p) = (a, b)det (p.
In particular,
Pfa((p)2 = (a, a)det (p.
1 a)
202
8 Skew and Skew-Hermitian Transformations
PROOF. In fact, since
'
m=
E P)
((a, a)) Pf((P)a= (,(b, b) Pfbb
1
1
a
)
it follows that
((q'E
P)"`,( 'E
P)"`)
=
Pfa((p)Pfb((p)
asbb
(a, b) = a1b Pf
Pf
b = Aa, t e r). Now applying Theorem 8.3.1 with p = m and observing
that
det cp, we obtain
det
=
1
(a, b))
(,
PfP)
a( Pf
.
8.5. Direct Sums
Let E and F be inner product spaces. Then an inner product is induced in
the direct sum E p F by
(x1 +O Yi, x2 +O Y2) = (x1, x2) + (Y1, Y2)
x1, x2 E E, Yi, Y2 E F.
Let
iE:E-E +QF,
iF:FEE E)F
and
denote the natural inclusions and projections. Then the isomorphisms
t'E,
and t'E®F (see Section 8.2) are connected by the relations
E®F(1E) n (x n Y) = 'E° E(x A Y) ° E
x e E, y e E
E®F(1F) n (x A Y) = 1F ° DF(x A Y) ° F
xeF, yeF,
and
as is easily checked. The relations above imply that
° E)
P E Sk(E)
(F) A '11F(k) = '11E©F('F ° Y ° F)
'I' E Sk(F).
(iE) n "E(SP) = '11E®F('E °
and
Since
(P E 3i// = lE0(P07LE + lF0l//07LF,
we obtain
(E) A "E(SP) + (iF) n F(Y') = '11E©F('P
4')
E Sk(E), c e Sk(F).
(8.14)
203
The Pfaflian of a Skew Linear Transformation
Proposition 8.5.1. Let E and F be inner product spaces of dimensions 2p and 2q
respectively. Choose basis vectors a and b of A 2E and A 2"F. Then
PfQ®b((P Q i) = PfQ(q,) Pfh((p)
cP e Sk(E), ,,Ii e Sk(F).
PROOF. Set "E('P) = u, LI'F(l//) = v and `I'E©F((P O+ i//) = w, where
w = (lE) n u +
(1F) n v.
It follows that
wp + q =
k+1= p+q
((lE) n u)k A ((1F) n
v)1
=
k+1= p+q
uk 0 v'.
Since u k = 0 for k > p and v 1 = 0 for 1> q, this formula reduces to
Wp+q = up OX Uq.
It follows that (see Section 5.15)
PfQ®b((P Q i//) = (wp+q, a O b)
= (up Q uq, a Q b) = (up, a) . (vq, b)
= Pf p)
8.6. Euclidean Spaces
Let E be an oriented Euclidean space of dimension n = 2m and let a be the
unique unit vector in A "E which represents the orientation. Then we set
PfE(co1.... , om) = ("E('P 1) n ... A E((Pm), e)
pv e Sk(E)
and
PfE (SP)
=
P e Sk(E).
1
m
Observe that if the orientation of E is reversed then PfE is changed to -PfE .
Proposition 8.4.1 shows that the Pfaffian is invariant under proper
rotations. Proposition 8.4.2 implies that
PfE((P)2 = det cP
cP e Sk(E).
Finally, by Proposition 8.5.1,
PfE©F( +O i/i) = PfE((p) PfE('')
cP, cli e Sk(E).
EXAMPLE 1. Choose an orthonormal basis {e1,
coe2µ _ 1 =)e2 µ
cpe2µ = - i%µ e2µ - 1
...,
of E and define cP by
)µ E r
(µ = 1, ..., m).
204
8 Skew and Skew-Hermitian Transformations
Then, when n = 2, ' E(cp) = ,1(el n e2), and so
PfE((p) =
as follows from Proposition 8.5.1.
EXAMPLE 2. Let E be an n-dimensional complex vector space with a Hermitian
inner product (, )H and let E denote the underlying real vector space. Give
E the natural orientation (see Section 7.9) and the positive definite inner
product
x, y e E.
(x, y) = Re(x, y)H
Let i be the skew transformation of E given by
Then
1
as follows from Example 1.
Skew-Hermitian Transformations
In Section 8.7 we shall derive an analogue of Theorem 8.3.1 for skewHermitian maps.
8.7. The Isomorphisms 6 and i
Let E be an n-dimensional complex vector space and let E* be the complex
dual space. Define an inner product in E* Q E by setting
(x* Q x, y* O+ y) = <x*, y> + <y*, x>.
Extend this inner product to the (complex) exterior algebra A (E*
This inner product determines via the canonical isomorphism
A (E* Q E)
(8.15)
E).
AE* Q AE
(see Section 5.15) an inner product in A E* Q A E. A simple calculation
shows that (see Formula (6.34))
<<u* OO u, v* OO v>> = (-1)p<u*, v> <v*, u>
u*, v* e A PE*, u, v e ARE.
Next, assume that E is equipped with a Hermitian inner product (, )H .
Then an inner product is defined in the underlying real vector space E by
(x, y) = Re(x, y)H
x, y e E.
205
Skew-Hermitian Transformations
The multiplication by i in E determines a skew linear transformation in
E which will be denoted by i (see Example 2, Section 8.6)
i(x) = i
x.
Recall from Section 8.2 the inverse isomorphisms
SkE and L:
IE : n 2E -
n 2E
and set
J =E(lpg)
Then J E n 2 E.
Next define a (real) linear isomorphism Q : E E* by
Qx, y> = (x, y) - i(x, i Y)
X E E, y e E.
Since i is skew and an isometry it follows that
QiR(x) = - iQ(x)
x e E.
The map o- determines a (real) linear map i : ER
i(x) = kr(x) Q x
(8.16)
E* Q E by
x E E.
Formula (8.15) implies that
(ix, iy) = <iQx, y> + <icJY, x>
= i(x, y) + (x, i y) + i(y, x) + (y, ix)
= 2i(x, y) + (x, i y) + (y, ix) = 2i(x, y).
Thus i satisfies
«ix, iy) = 2i(x, y)
x, y e E.
(8.17)
Next observe that ix n ix = 0, x E E, where the symbol n c indicates
that the multiplication is taken in the complex algebra n (E* Q E), and so i
extends to a homomorphism (see Section 5.5)
i A: n E + n (E* Q E).
Relation (8.17) implies that
<<X A
u, i u) = (2i)"(u, u)
u, y e
pE .
Lemma 1. Consider the direct decomposition
n 2(E* +E
E) = [n 2E* O 1] +O [E* O E] +E [1 O n 2E]
and write
iAU=iLUQ1+XMU+10XRu
uEn2E,
(8.18)
206
8 Skew and Skew-Hermitian Transformations
where iL u E A 2E*, iM u E E* O E, i R u E A 2E. Then the following relations
hold:
(p e Sk(E)
XL "E[l08 4] _ - 2it L '11E(4)
(p E
XR 1-E[l , (p] = 2lxR '11E(4)
Here [, ] denotes the commutator and q'E : Sk(E) - A 2E.
PROOF. It follows from the definition of i that
XL(x n y) = krx n c iO y = - Qx A
Qy.
This equation yields, in view of (8.16),
t L(l 08 x n y+ x n i y) _ -o i (x) n c 0Y - Qx n c Qi y
= iQx n Qy + Qx n c 10y
= 2krx n Qy = - 2ixL(x n y).
(8.19)
Next observe that the isomorphism t'E satisfies
[(p E(x n y)] _ IE((px n y + x n (py)
(p E
Applying this relation with (p = i we obtain
[ l 08
,
E(x A y)] = E(' 08 x A ) + x A l y)
whence
i y.
'E[i08 E(x n y)] =
(8.20)
Formulas (8.19) and (8.20) imply that
XL '1'E[, E(x A y)] _ - 2ixL(x A y)
whence
XL'E
[lpg, "E u] _ - 2lxL(u)
ue
Since 'DE: A 2E - Sk(E) is an isomorphism, the first formula in the lemma
follows. The second formula is proved in the same way.
Corollary. If cp is a skew-Hermitian transformation of E, then
iLY'E(4') = 0
(8.21)
XR '11E(4') = 0.
(8.22)
and
PROOF. Apply the lemma observing that
(p] = 0.
Lemma II. Let (p be a skew linear transformation of E and consider the complex
linear transformation i ° (p + (p ° i . Then
XM `I'E(4) = T E 1(i ° (p + (p ° i08)
Skew-Hermitian Transformations
207
PROOF. It follows from the definition of i that
XM(x A y) = icrx ®c y
- icry ® x.
This yields, in view of (8.16),
TE XM(x n y)(z) _ - <icry, z>x + <icrx, z>y = - i<Qy, z>x + i<Qx, z>y
_ -i{(y, z) - i(y, ijz)}x + i{(x, z) - i(x,
i
i{(x,z)y - (y, z) x} + {(x,
E(x n y)z + DE(x n y)i z.
Thus we have the relation
TE XM(x n y) = i FE(x n y) + E(x n y) ° i
whence
TE XM(u) _X08
cDE(u) + E(U) o lpg
u e n 2Epg .
Setting DE(u) _ cP we obtain
TE iM'1'
p) = i$8 ° cP + cP o i
cP
Sk(ER)
and so the lemma is proved.
Corollary. A skew-Hermitian transformation (p of E satisfies the relation
(8.23)
TE XM `PE((p) = 2up.
Lemma III. A skew-Hermitian transformation of E satisfies
A''E cP = 2l T E 1(co)
In particular,
X A J = -2k.
PROOF. Apply formulas (8.21)-(8.23) observing that
xAu=xLu®l+xMu+1®iRu
Theorem 8.7.1. The characteristic coefficients of a skew-H ermitian transformation are given by
Cp(p) = i"(CI E p)p, JP)
(p = 0, ..., n).
Thus, if p is even, then Cp(cp) is real and if p is odd, then C,((p) is imaginary.
PROOF. Set T E' (gyp) = u. Then we have, in view of formulas (7.10), (6.33),
and (6.34)
<U", i"> = <ub, fb> = (
1)p«up, »
8 Skew and Skew-Hermitian Transformations
208
Now set LI'E (P = z. Then, by Lemma III,
X A Z = 2iu
and
X A J = -2-t.
It follows that
1
1
Cp((P) =
2P(2i)
n Zp, 'L A
Jp
Now using relation (8.18) and (6.33), we obtain
C(p) =
1p
1
)
(2,
2
p
(2l)2p(zp, JP) = i (zp, JP) = i"((tPE (P)p, JP).
Corollary. Let E have the natural orientation (see Section 7.9). Then the
Pfaffian of p is given by
Pfq = 1 det
.
PROOF. Applying Theorem 8.7.1 with p = n we obtain
det cP = i"(J", (`I'E P)")
Now write
J" _ , e,
where a is the positive unit vector in A 2nEpg . Since J =
that
=
1
(see Example 2, Section 8.6). Thus,
det cP = l"(e, (LE q)") = j"Pf((p)
it follows
Symmetric Tensor Algebra
In this chapter E denotes a vector space over a field of characteristic zero.
Symmetric Tensor Algebra
The results of this chapter are, in most cases, isomorphic to analogous
results in Chapter 5. Consequently, most proofs are omitted or presented
in a highly abbreviated form. Modulo occasional changes in terminology
the reader should be able to read the proofs in Chapter 5 as substitutes for
the proofs omitted here.
9.1. Symmetric Mappings
Let E and F be vector spaces and let
P
be a p-linear mapping. Then cP is called symmetric if cP = Qcp for every permu-
tation o (see Section 5.1). Since every permutation is a product of transpositions, it follows, that a mapping p is symmetric if and only if p = icp
for every transposition i. Every p-linear mapping p determines a symmetric
p-linear mapping Scp by
Scp is called the symmetric part of cP and S is called the symmetry operator.
If p is symmetric, we have Scp = cp.
209
9 Symmetric Tensor Algebra
210
x E into F and
Proposition 9.1.1. Let ( be a p-linear mapping of E x
f : ®"E -+ F be the induced linear map. Then (p is symmetric if and only if
M(E) c ker f (cf. Section 4.9).
PROOF. If (p is symmetric, we have, for each transposition i,
f(x1
...O x, - i-1(x1 ®... O x,))
_ (p(x 1, ..., xp) - i(p(x 1, ... , xp) = 0.
Since the products x1 Q
® x, generate ®-"E, it follows that f reduces
to zero in M"(E). Conversely, if M"(E) c ker f, it follows that
((p - x p) (x 1, ..., xp) = f[(x1 ®... O x p) - (r - 1(x1
for every transposition i and hence (p is symmetric.
©... O x,)] = 0
9.2. The Universal Property
Let
P
be a symmetric p-linear mapping from E into a vector space S. We shall say
that V p has the universal property (with respect to symmetric maps) if it
satisfies the following conditions:
v 1: The vectors V p(x 1, ... , xp), x a E, generate S.
v2: If
P
is any symmetric p-linear mapping, then there exists a linear map
f:S - H such that the diagram
H
V 1'
S
commutes.
Conditions v 1 and v 2 are equivalent to the following condition
v : If
x E-->H
cp:E x
P
is a symmetric p-linear mapping, then there is a unique linear map
H which makes the diagram above commute.
f:S
Symmetric Tensor Algebra
211
Suppose now that
x E-+S and Vp:E x
W:E x
p
x E-+S
p
are symmetric maps with the universal property. Then there is a linear
isomorphism f : S
S such that
fo Vp= Vp.
This is shown in exactly the same way as in the skew-symmetric case (cf.
Section 5.3).
To establish existence, set
V pE = ®E/M(E)
(where M"(E), the subspace of ®"E, is defined as in Section 4.9) and define
Vpby
V "(x1, ... , xp) = ?L(x 1 ®... ® xp),
where ?r denotes the projection.
Definition. The pth symmetric power of E is a pair (S, V p), where
p
is a symmetric p-linear mapping with the universal property. The space S is
also called the pth symmetric power of E and is denoted by V "E.
9.3. Symmetric Algebra
Consider the direct sum
VE= Q VpE
p=0
(where V °E = f and V 'E = E) and identify each V pE with its image under
the canonical injection ip : V pE -+ V E. We thus obtain
VE=
V pE.
p=0
Assigning to the elements of V pE the degree p, we make V E into a graded
vector space.
As in Section 5.4 on exterior algebra we construct the homogeneous
linear isomorphism
f:®E/M(E)-+ V E
9 Symmetric Tensor Algebra
212
such that
fit(x1 0 ... O xp) = x 1 v ... V xp
and use it to induce a multiplication in V E. Then V E becomes an associative
commutative graded algebra with the scalar 1 as unit element. It follows
from the definition of the product that
(x1 V ... V xp)(xp+1 V ... V xp+9) = x1 V ... V xp+9.
Hence we shall denote the product of two elements u and v by u v v. Then
the above formula reads
(x1 V ... V xp) V (xp+1 V ... V xp+9) = x1 V ... V xp+9.
(9.1)
The graded algebra V E is called the symmetric algebra over E. It is clear that
the vectors of E together with the scalar 1 form a system of generators for
the algebra V E.
As in Section 5.4 on exterior algebra, we define the kth power of an element
ue VE by
uk=
1
k.
k> 1
Vu
uV
k
u° = 1.
Then we have the binomial formula
(u + v)k =
u` V vj
u, v e V E.
(9.3)
i+j=k
The algebra V E has the following universal symmetric algebra property:
Let A be an associative algebra with unit element a and cP : E - A be a
linear map such that
(px (p y = (p y px.
Then there exists a unique homomorphism h : V E -+ A such that h(1) = e
and h o i = cp, where i denotes the injection E -+ V E.
Moreover, if U is any associative algebra with unit element and E : E -+ U
is a linear map such that the pair (U, E) satisfies the universal property
above, then U is the symmetric algebra over E.
9.4. Symmetric Algebras Over Dual Spaces
Let V E, V E* be symmetric algebras over a pair of dual spaces E, E*,
and consider the induced isomorphisms
f : QE/M(E) - V E,
g: QE*/M(E*) -= V E*.
Symmetric Tensor Algebra
213
It follows from Section 4.16 that f and g induce a scalar product <, > in V E
and V E* such that
/x*1 v ... v x*P, x1 v ... v xp> = perm(<x*i, xj>)
p>_1
= ,1µ
<V PE*, V IE> = 0
if p
q.
From v1 we obtain that <, > is uniquely determined by (9.4). It follows
from (9.4) that the restriction of <, > to the pair V PE*, V pE is nondegenerate
for each p, and so induces duality between these spaces. In particular, the
restriction of <, > to V 1E* = E* and V 'E = E is just the original scalar
product.
9.5. Homomorphisms and Derivations
Suppose that cP : E -+ F is a linear map. Then cP can be extended in a unique
way to a homomorphism cP : V E -+ V F such that cP (1) = 1. The homomorphism cP is given by
q
v...vcpxp
(x1
x,EE
and is homogeneous of degree zero. If l/i : F -+ G is a linear map into a third
vector space G, then we have
and the identity map of E induces the identity in V E,
'V = 1.
It follows from (9.5) and (9.6) that if cP is injective (surjective) then cP is also
injective (surjective). The fact that p preserves products can be expressed
by the relation (see Section 9.6 for µ(a))
P V ° µ(a) =
a) ° cP ,
a e V E.
(9.7)
Suppose now that p* : E* -- F* is a linear map dual to cp. Then the
induced homomorphism (gyp*) : V E* * V F* is dual to cP
('P*) _ ('P v)*.
The homomorphism (gyp*) will be denoted by (p V .
If lfi* : F* - G* is a linear map dual to cli we have the composition formula
(f,°q,)v=(Pv°JV.
9 Symmetric Tensor Algebra
214
Now let p be a linear map of E into itself. Then p extends in a unique
way to a derivation B (cp) in the algebra V E. The derivation 8 (') is
given by
P
x 1 V ... V (px j V ... V XP
(x 1 V ... V xP) =
ev
j=1
and is homogeneous of degree zero. The derivation property of O ((p)
can be expressed by the formula
8
V
µ(a) = µ(e (4)a) + u(a) ° 8 (q)
a e V E.
(9.8)
If tji : E - E is a second linear map, we have the composition formula
O ,
'I' - /i °i') = 0 (p)0O OI') - e AcIi) ° e
If (p* : E* * E* is dual to cp, then the induced derivation 0 (p*) of the
algebra V E* will be denoted by 8 " (gyp). The linear maps 8 (gyp) and 0 V ((p)
are dual,
ev(e) =
If /j* : E* - E* is dual to i//, we have the composition formula
(9 V (p 0-,/J 4') = e V(/,) 0 O"() - e v (`I,) o O v (cI,)
9.6. The Operator i(a)
Fix a E V E and consider the linear map µ(a): V E -+ V E given by
µ(a)u = a v u
u E V E.
Clearly,
µ(a v b) = µ(a) o µ(b)
a, b E y E.
Now let
i(a) : V E* 4
V E*
be the dual map. It is determined by the equation
<i(a)u*, v> = <u*, a A U>
u* E V E*, u e V E.
If a E V PE, then i(a) is homogeneous of degree p. In particular,
i(a)u* = <u*, a>
u* E V E
and
i(a)u * = 0
u * E V rE, r <p.
(9.9)
215
Symmetric Tensor Algebra
Dualizing Formulas (9.9) and (9.7), we obtain
i(a v b) = i(b) o i(u)
a, b e V E,
and
i(a)°cp" _ cP" oi((p,a)
a e V E, cpe L(E; F).
Finally, if (p : E - E, (p* : E* - E* is a pair of dual linear transformations,
then, dualizing (9.8), we have
a e V E.
i(a) o 0 V (gyp) = i(V V ((p)a) + 0 V (p) ° i(a)
Now consider the operator i(h), where h e E. Observe that i(h) is homogeneous of degree -1. As in Proporition 5.14.1 one may show that i(h) is a
derivation in the algebra V E*,
i(h)(u* v v*) = i(h)u* v u* + u* v i(h)u*.
This yields the Leibniz formula
(r) i(h)Pu* v i(h) v*.
i(h)r(u* v u*) _
p+q=r p
-
Finally, observe that if an element u* E V PE (p 1) satisfies the equation
i(h)u* = 0, for every h e E, then u* = 0 (cf. Proposition 5.14.2).
9.7. Zero Divisors
In this section it will be shown that the algebra V E* has no zero divisors
(of course, the same holds for V E).
Consider, for each p > 0, the subspace I P c V PE* given by
IP =
V µE*.
Ii-P
Clearly, Ip is an ideal in the algebra V E* and we have the sequence
VE*=JoD11D'2D....
The ideal I 1 is also denoted by V +E* . Every two ideals
J, =
V µE*
and F' =
L>p
V µE
µ>_P
are dual. If a e V PE, then i(a) restricts to an operator from Iq to Iq _ P for
q<p.
Lemma. Let u* a IP be an element such that
i(h)Pu* = 0 for every h e E.
Then u* = 0.
9 Symmetric Tensor Algebra
216
PROOF. If p = 1, the lemma follows from the remark at the end of Section
9.6. Now assume that the lemma is true for p - 1 (p > 2) and let u* E IP,
be an element satisfying
i(h)Pu* = 0
h E E.
Replacing h by h + k (. E I,) we obtain
P
p
i(h)i(k)P - u * = 0.
a=o µ
Since . is arbitrary, this yields
(µ = 0, ... , p).
i(h)µi(k)P - µu* = 0
In particular,
i(h)P-1 i(k)u* = 0
h, k E E.
Thus, by induction,
i(k)u* = 0
k E E.
Now applying the lemma for p = 1 we obtain u* = 0 and so the induction
is closed.
Proposition 9.7.1. The algebra V E* has no zero divisors.
PROOF. Let u* and U* be two nonzero elements in V E*. Assume first that
u* and U* are homogeneous of degree p and q respectively and that p > 1
and q > 1. In view of the lemma, there exists h e E such that i(h)Pu* 0.
Now consider the elements i(h)Pu* (µ = 0, 1, ...). Since i(h)°U = U 0
and i(h)4+ 1 U* = 0, there is an integer r > U such that
0
i(h)rv*
while
i(h)r + 1 U* = 0.
Now the Leibniz formula yields
ihP+ru*
V U* =
P+r
ihPu* V ihrv*.
P
Since i(h)Pu* is a nonzero scalar and i(h)rv*
i(h)P+r(U*
whence u* V U*
V U*)
0, it follows that
0
0.
In the general case
P
u* _
u
u E V aE*, up
0
UK E VK E*, U9
0.
a.=o
and
9
U* _
Un
K=0
217
Symmetric Tensor Algebra
Then
u* v U* _
ua*. v UK + uP v U9.
0. This completes the proof
Since up v u9 0, it follows that u* v u*
of the proposition.
9.8. Symmetric Algebra Over a Direct Sum
Consider two vector spaces E and F and the direct sum E Q F. In this section
we shall establish an isomorphism between V (E Q F) and the canonical
tensor product V E O V F. Define a linear map
f:VE® VF-+ V(EQF)
by
f(u OO u) _ (i1)
u
v (i2) u,
where it and i2 are the inclusions. A straightforward calculation shows that
f is an algebra homomorphism (cf. Section 5.15).
To show that f is an isomorphism, consider the linear map
ri:EQF-+ VEQ VF
given by
ri(z) = 7riz Q 1+ 1 O 2 z
z e E Q F,
where 7r1 and r2 denote the projections. Since
1(z1) . rI(z2) _ (Z2) ' j(z)
ri extends to an algebra homomorphism
z1, z2 E E Q F,
ri:V(EQF)-+ VEQ VF
(see Section 9.3).
It is easy to check that
j f (x O 1) = x 0 1,
?1.f (1 O y) = 1 0 y
and
flj(x®y)=x®y
x e E, y e F.
These relations imply that
and fotj=i.
Thus f is an isomorphism.
Next, let E* and F* be spaces dual to E and F respectively. Define a
scalar product between E* Q F* and E Q F in the usual way and consider
9 Symmetric Tensor Algebra
218
the induced scalar product between V (E* p F*) and V (E p F). Then we
have the relation (cf. Formula (5.51))
<f (u* ® v*), f(u ® v)> = <u* ® v*, u ® v> = <u*, uXv*, v>.
Finally, assume that F = E and let A: E -+ E p E be the diagonal map.
Then we have the relation (cf. Formula (5.52))
0" f(u* ® v*) = u* v v*
u*, v* e V E*.
9.9. Symmetric Tensor Algebras Over a Graded Space
Let E =
. i E1 be a graded vector space and let the vectors of E, be
homogeneous of degree k,. Then there exists precisely one gradation in the
algebra V E such that the injection i : E - V E is homogeneous of degree
zero. V E together with this gradation is called the graded symmetric algebra
over the graded vector space E. The subspace of homogeneous elements of
degree k is given by
(V E)k =
(V p' E 1)
®... ® (V E,),
(P)
where the sum is extended over all r-tuples (pi, ... , pr) subject to
r
p1k1 = k.
9.10. Symmetric Algebra Over a Vector Space of Finite
Dimension
Suppose now that E is a vector space of dimension n and that ea (a = 1,
is a basis of E. Then the products
ea 1 v ... v ea
P
a 1 < a2 < ... < ap
..., n)
(9.10)
form a basis of V "E. In fact, it follows immediately from v 1 and the commutativity of V E that the products (9.10) generate V E. To prove linear
independence let E* be a dual space of E and a*(a = 1, ..., n) be a dual
basis. Then Formula (9.4) yields
<e* v ... v e*RP, ea1 v ... v eaP> = perm(<e*f;, e21>) = perm(5 )
and thus the products (9.10) are linearly independent.
The above result shows in particular that
dimVPE=(n+p-11
n
1
p>0.
(9.11)
219
Symmetric Tensor Algebra
The basis vectors of V PE can be written in the form
n
n
JJ k! e1 V ... V en"
kv>=p.
i= 1
v= 1
9.11. Poincare Series
For the Poincare series of the graded algebra V E we obtain, from (9.11)
(t)
t
P=o
Y.(
t
1)
p
p=o
P
(1
whence
P(t) = (1 - t) - n.
(Here E has dimension n.)
Now suppose that E =
Ei is a positively graded vector space of finite
dimension and that the vectors of E. are homogeneous of degree ki . Then
the Poincare series of V Ei is given by
Pi(t) = (1 - t"9 - ni
ni = dim E1.
Hence, the Poincare series of E is
P(t) = (1 - tkl)-nl .. . (1 - tkr) - nr.
9.12. Homogeneous Functions
A homogeneous function of degree p in E is a map h : E -- F which satisfies
h(i%x) = il!h(x)
,% E f.
The homogeneous functions of degree p form a vector space HP(E). The
product of two homogeneous functions h and k of degree p and q respectively
is the homogeneous function h k of degree p + q given by
(h k) (x) = h(x)k(x)
x e E.
This multiplication makes the direct sum
H(E) _
Hp(E)
P
into a commutative associative algebra. Its unit element is the homogeneous
function ho of degree zero given by
h0(x)=1
x E E.
9 Symmetric Tensor Algebra
220
Now let D be a symmetric p-linear function in E. Then a homogeneous
function h, of degree p is given by
he(x) =
... , x).
In terms of the substitution operator we can write
h,(x) =
It follows from the definition (see Section 9.15 for S"(E)) that
= 1,h + µh,,
D, 'P E S"(E), , µ E r
and that
h , = h h,
1D, 'I' E S"(E).
Thus, the correspondence D - h determines an algebra homomorphism
i : S(E) -+ H(E).
The map i is injective. In fact, assume that iD = 0 for some D E S"(E). Then
i(x)M = 0
for every x e E and so the lemma in Section 9.7 implies that D = 0.
On the other hand, i is not in general surjective. As an example let E be a
Euclidean space and define h e H 1(E) by
h(x) = (x) (x, x)112,
where E is a function in E satisfying
E(x)
. > 0,
Then h is homogeneous of degree 1 but it is not additive and hence not a
linear function. Thus it is not contained in Im i.
PROBLEMS
1. Consider the problems of Chapter 5, and carry them over to symmetric algebra
whenever possible.
2. Let F c E be a subspace, and define IF to be the ideal in V E generated by the vectors
of F. If F1 is a complementary subspace, prove that
VE=IFp VF1
and
IF =V F® V F1.
3. If p : E - F is a linear map, prove that
Im ep
= V Im p and ker (p
= Iker p.
Polynomial Algebras
221
4. If
P
is a simultaneously symmetric and skew- symmetric p-linear mapping (p >_ 2),
prove that p = 0.
Polynomial Algebras
9.13. Polynomial Algebras
A monomial of degree p in n variables in a field f is a function
P
which satisfies
P(t 1,
... , tn) = P(1, ... , 1)til ... tnn
+ kn = p.
Thus every monomial of degree p can be written in the form
where k 1 +
P(t1,... , tn) = atil ... rn"
p
k= p.
In particular, a monomial of degree zero is an element of r.
The monomial of degree 1 given by
Pi(t 1,
(i= 1, ... , n)
... , tn) = ti
will be denoted by t.
The monomials of degree p generate a vector space with respect to the
usual operators. It will be denoted by rn.
The product of a monomial P of degree p and a monomial Q of degree q
is the monomial of degree p + q defined by
(P ' Q) (t 1, ... , tn) = P(t 1, ... , tn)Q(t 1, ... , tn).
This multiplication makes the direct sum
rn -
rnP
P
into a commutative associative algebra called the polynomial algebra in n
variables over r with the scalar 1 as unit element. It is generated by 1 and the
monomials ti (i = 1, ..., n). The elements of rn are called homogeneous
polynomials of degree p.
9 Symmetric Tensor Algebra
222
9.14.
Now we shall establish an isomorphism between V E (dim E = n) and the
polynomial algebra in n variables. In fact, fix a basis {e1, ...,
of E. Then
every element u e V "E can be uniquely written in the form
u=
en"
vi
(V)
= p, c 1 ...VP e f
ii
(cf. Section 9.10). Thus it determines a homogeneous polynomial Pu of
degree p given by
Pu(t 1,
t:"
..., tn) -
Vi = p
Since,
= I,Pu + 4uP,
A, p e f
u, v E V "E,
this defines a linear map
q:V"E-urn.
Conversely, every polynomial
P(t1,
..., tIt) =
cvt,
(v)
homogeneous of degree p, determines an element Up e V "E given by
and so we have a linear map
/i : f - V E.
It follows from the definition that
(p01/1=1
l/i o p = i,
and so q and i/i are inverse isomorphisms. Finally, observe that
uVV=
Pu Pv
U, v e V E
and thus q is an algebra isomorphism,
(p : V E
F~n .
In particular, we have
(p(e)=t i
(1= 1, ... , n).
223
The Algebra of Symmetric Functions
It follows that the polynomial- algebra has no zero divisors since V E has
no zero divisors (see Proposition 9.7.1).
PROBLEMS
1. Prove that V E is a principal ideal domain if and only if dim E = 1. Hint: See Chapter
XII of Linear Algebra.
2. Let E be a pseudo-Euclidean space of dimension n (see Section 9.17 of Linear Algebra
for the definition) and index r. Consider the symmetric tensor algebra V E and choose
for each p >_ I a subspace T" V "E of maximal dimension such that the restriction
of the scalar product to T" is negative definite. Prove that the Poincare polynomial
T" is given by
of the graded space T =
iii
1
PT(t)=2(lt)[(lt)s_(l+t)s]
s=n-r.
Prove an analogous formula for the exterior algebra :
PT(t) = i (I - tf [(I + t)5 - (1 - to
s = n - r.
The Algebra of Symmetric Functions
9.15. Symmetric Functions
A p-linear function I in E is called symmetric, if
Q E Sp.
The symmetric functions form a subspace of TP(E) (cf. Section 3.18) denoted
by S(E).
Every p-linear function fi determines a symmetric function Sit' by
Pa
called the symmetric part of I.
If t e S(E), then ScI = ct and so S is a projection operator. Moreover, S
satisfies (cf. Formula (5.81))
S(I a '1') = S(S o '1') = S(I o S'V)
, P E T(E)
whence
S(cb Q '1') = S(S Q S'1').
The symmetric product of
e S' (E) and P E 5(E) is defined by
Vq'=(p+q)!SI xlP.
pq!
9 Symmetric Tensor Algebra
224
Explicitly,
(v
) (x 1, ..., xp + q) _
1
(1)' ... ,
1) ' ... ,
q)).
The symmetric product satisfies the relations
and
(bv'1')v X= bv('l'vX).
This multiplication makes the direct sum
S(E) = >2SP(E)
p=0
into a commutative associative algebra, called the algebra of symmetric
functions in E.
A linear map gyp: E - F induces a homomorphism p* : S(E) - S(F)
given by
((p* qI) (x 1, ..., xp) = '4'((px 1, ... , (px p)
'I' E SP(F)
and a linear transformation q of E determines a derivation 0S(p) in the
algebra S(E). It is defined by
(())(x1, ... , xp) =
P
v=1
(x 1, ..., rpx,, , ... , xp).
This is shown in the same way as for the algebra A(E) in Section 5.31.
9.16. The Operator is(h)
Let is(h) : S'(E) - S'(E) also denote the restriction of the operator is(h) defined
in Section 3.19 to S'(E). Thus,
(is(h)(b) (x 1, ... , x,_) = b(h, x 1, ... , x,_)
b e S(E).
is(h) is called the substitution operator in the algebra S'(E).
In exactly the same way as in Section 5.32 it is shown that
is(h) o S = 50 is(h)
(see the lemma in Section 5.32) and consequently,
is(h) (st' v 'F) = is(h)1 v 'F + 'b v is(h)'F
(see Proposition 5.32.1). Thus, is(h) is a derivation in the algebra S(E).
Note that the operator is(h) is dual to the multiplication operator ,us(h)
in S(E).
225
The Algebra of Symmetric Functions
Finally, we have the commutative diagram (see Section 3.20)
OpE*
)
T"(E)
ns
1°
O pE * ) T "(E)
where its denotes the symmetrizer (see Section 4.15). It implies that a restricts
to a linear isomorphism V PE* = S"(E) for each p and in fact to an algebra
isomorphism Y(E*) 3 S(E).
Y(E) (see Section 4.15), it follows that V PE* S"(E).
Since V PE*
In particular,
dimSpE =
n+p-
1
(p=0,1,...).
p
9.17. The Algebra S,(E)
In exactly the same way we obtain the algebra
S(E),
S,(E) =
p=0
where S(E) denotes the space of symmetric p-linear functions in E*. The
scalar product between T"(E) and Tp(E) determines a scalar product between
S"(E) and S(E) given by
tb, W = 1
p
tb'b ESpE q' eS E.
It follows from the definition that (cf. Section 5.34)
<tb, x1 V ... V
b(x1, ... , xp)
xE E
and
<f1 V ... v fps'P>s - 'P(f1, ... ,
fp)
fEE*.
In particular,
<fi V
V fp, x1 V
V xp> = perm(f (x;)).
9 Symmetric Tensor Algebra
226
9.18. Homogeneous Functions and Homogeneous
Polynomials
Let
P(t 1, ... , tn) =
Ck l
, ..., k t
k= pp
!' ... tnn
(k)
v
be a polynomial of degree p. It determines a homogeneous function hp of
degree p by
h
C
{ 1)k ... (
n)kn
(k)
(here the kare exponents!), where x = L vev . This yields a homomorphism
6: rn - H(E).
On the other hand, the polynomial P determines an element Up E S"(E)
given by (cf. Section 9.14)
C 1,...,kf(f i
Up =
i ... (f1)
where {f 1,
kv = p
v
(k)
..., f'} is the dual basis of {e1, ... , end (the kv are again exponents).
Moreover, the correspondence P H u defines an isomorphism
i: rn - S'(E)
(see Section 9.16).
We shall show that the diagram (see Section 9.12 for z).
S(E)
IT,,
(9.12)
H(E)
commutes. Since all maps are algebra homomorphisms, it is sufficient to
show that
= 6(ti)
(i = 1, ..., n).
But, since l/Jti = f', we have ii/i(ti) = f On the other hand, 6f i = ` = f `
whence (9.12).
Clifford Algebras
10
In this chapter E denotes a vector space over a field with characteristic zero and ( , ) denotes a
(possibly degenerate) symmetric bilinear function in E.
Basic Properties
10.1. The Universal Property
Let A be an associative algebra with unit element eA. A Clifford map from E
to A is a linear map p which satisfies
(gpx)2 = (x, x)eA
xeE
or equivalently,
(P(x)(P(y) + (P(y)(P(x) = 2(x, y)eA
x, y e E.
A Clifford algebra over E is an associative algebra CE with unit element e
together with a Clifford map iE : E -- * CE subject to the following conditions:
Cl : The subspace Im iE generates the algebra CE.
C2 : To every Clifford map gyp : E -- * A there exists a homomorphism
f : CE -- A which makes the diagram
E
A
(10.1)
iE
CE
commutative.
Conditions Cl and C2 are equivalent to the following condition
C : To every Clifford map p : E -- * A there exists a unique homomorphism
f : CE - A such that Diagram (10.1) commutes.
227
228
10 Clifford Algebras
In fact, it is easy to check that Ci and C2 imply C. Conversely, assume that
iE satisfies C. Then C2 follows immediately. To establish Ci denote by A
the subalgebra of CE generated by Im iE and by iE the map iE considered as a
map into A. Then, clearly, iE is a Clifford map. Hence there is a unique
homomorphism f : CE - A such that
fo 'E = lE.
On the other hand, if j : A - CE is the inclusion map, we have
J°lE = lE
Now consider the map j of: CE - CE. Then the relations above imply that
r,
(J ° f) ° lE = J ° (f ° 1E) = J ° lE = lE
.
On the other hand,
t ° lE = lE .
Thus the uniqueness part of Condition C implies that
j°f= 1.
Hence j is surjective and so Ci follows.
10.2. Examples
(1) Let
be the real axis with the negative definite inner product given by
(x, y) _ - xy
x, y E
.
We show that C is a Clifford algebra over . Let iE :
given by
iE(x) - i x
- C be the linear map
x E {l.
Then iE satisfies C1. To establish C2, let
be any Clifford map. Since p is linear,
q(x) =
xE[Q
Set p(1) = a. Since p is a Clifford map, it follows that a2 = - eA. Thus qP is
of the form
p(x) = x a,
a2 = - eA .
Now define f : C - A by setting
f(x+iy)=xeA+ya
x,ye.
Basic Properties
229
Then f is an algebra homomorphism as is easily checked. Moreover,
f lE(Y) = f (iy) = y' a= (p(y)
ye
and so iE satisfies C2.
(2) Let 2 be the plane with a negative definite inner product. We show that
the algebra of quaternions (see Linear Algebra Section 7.23) is a Clifford
algebra over 2.
Choose an orthonormal basis {x 1, x2} of 2 and let {e, a 1, e2, e3} be an
orthonormal basis of 0-0. Define a linear map iE : 2 - 0-{1 by setting
iE(xl) = e1,
jE(x2) = e2.
Then
(iE(x))2 = - (x, x). e = (x, x)- e
xe
and so iE is a Clifford map. It is easily checked that iE satisfies C 1. To establish
C2, let p: 2 -+ A be a Clifford map and set q (x1) = a1, q (x2) = a2. Then
(p is of the form
(px = 21a1 + 22a2,
where
x = 21x1 + 22x2
and the vectors ai (i = 1, 2) satisfy the relations
a= a2
a== -e
ala2 + a2 a1 = 0.
Now define f : D-LI - A by
f(2e+21e1 +22e2 +23e3)=2eA+21a1 +22a2 +23a1a2.
Then f is a homomorphism and satisfies
fiE(x) = f'E(21x1 + 22x2) = 21a1 + 22a2 = cpx
xE
2
fo'E=(p.
10.3. Uniqueness and Existence
In this section we shall show that there is a Clifford algebra over every inner
product space, unique up to an isomorphism.
First suppose that CE, CE are Clifford algebras over E and let iE : E - CE
and iE : E -+ CE denote the corresponding Clifford maps. Then, by C2, there
are homomorphisms f : CE -p CE and g : CE - CE such that
lE = f o lE an
lE = g o lE .
10 Clifford Algebras
230
It follows that
iE = (f ° g) ° iE
and
lE = `g ° f) ° iE .
Now Condition C 1 implies that
fog=i and
g°
= i.
Thus f and g are inverse isomorphisms. This shows that a Clifford algebra
over E, if it exists, is uniquely determined up to an isomorphism.
To prove existence consider the tensor algebra ®E (see Section 3.2) and
let J denote the two-sided ideal in ®E generated by the elements
xeE
where 1 is the 1-element of Q °E = r. Define CE to be the quotient algebra
CE = Q E/J
and let
7L : O E - CE
be the canonical projection. Let iE : E - CE be the linear map
lE = 7L ° JE ,
where JE : E --* ®E denotes the inclusion map. We shall show that (CE, 'E)
is a Clifford algebra over E.
First observe that, for x e E,
(iE x)2 = (7tJE x)2 = lt(JE x)2 = 7L(x QX x) = (x, x). 1
and so iE is a Clifford map. Since the algebra ®E is generated by E and since
it is surjective, it follows that iE satisfies C1. To establish C2, let p : E - A be
any Clifford map. In view of Section 3.3, qP extends to a homomorphism
h:®E-+A.
This homomorphism satisfies
h(x O x - (x, x). 1) = (q x)2 - (x, x)eA = 0.
It follows that J c ker h and so h factors over it to induce a homomorphism
f:CE - A.
Clearly,
flE(x) - firjE(x) =,r1C(JC) _ h(x) = lp(JC)
JC E E
and so C2 follows.
Thus to every inner product space E there exists a unique Clifford algebra
CE (up to an isomorphism). It is called the Clifford algebra over E.
231
Basic Properties
, ) = 0, then J is generated by the products x Q x and so
CE = n E. Thus the Clifford algebra is a generalization of the exterior
EXAMPLE. If (
algebra.
10.4. The Injectivity of iE
Proposition 10.4.1. The map iE:E - CE is injective.
PROOF. If ( , ) = 0, then CE = A E and the injectivity of iE follows from
Section 5.8. Now assume that (
,
) is nondegenerate. Then, if xo E ker iE,
we have for y e E
2(x0, y)e = 1E(x0) ' IE.Y + 1EV 1E(x0) = 0
whence x0 = 0.
In general write E = E0 Q E1, where E0 is the nullspace of ( , ) and E1
is a subspace such that the restriction ( , ) of( , ) to E 1 is nondegenerate.
Then the Clifford map i, : E1 - CEI is injective.
Let it0 : E -+ E0 and n, : E -+ E1 be the projections and consider the map
(pl E
E1
CEI.
Then
((p1x)2 = (llitlx)2 = (itlx, it1x)
where e1 denotes the unit element of CEI. Now,
(?t, x, ?t, x) = (x, x) - 2(x, 7t0 x) + (7t0 x, it0 x) = (x, x)
xeE
and so we obtain
((p, x)2 = (x, x)e,
x e E.
(10.2)
Next consider the linear map p: E - n E0 Q CEI given by
(px = tax a e, + 1 Q p1 x
X E E.
Then we have, in view of (10.2),
((px)2 = (7c0
x)2
O e 1 + ic0 x OO (p 1 x - 7C0 X O (p 1 x + 1 O ((p 1x)2
= (x,x).(1 Oe1)
and so (p is a Clifford map. Thus, by C2, there is a homomorphism f : CE -p
n E0 Q CE such that (p = f o 'E. Since (p is injective, it follows that iE is
1
invective.
Henceforth we shall identify E with its image under iE . Then E becomes a
subspace of CE and we have the relation
x,yeE.
10 Clifford Algebras
232
In particular,
xeE.
x2 =
Now properties C1 and C2 can be rephrased as follows:
C1: The algebra CE is generated by the subspace E.
C2 : Every Clifford map p: E - A extends to a homomorphism f : CE - A.
10.5. Homomorphisms
Let F be a second vector space and let ( , ) be a symmetric bilinear function
in F. A linear map p: E - F is called an isometry, if
(qx, coy) = (x, y)
x, y e E.
Let p: E - F be an isometry. Then there is a unique homomorphism
cPc : CE -+ CF which makes the diagram
E
'P )F
(10.3)
CE
p
CF
commutative. In fact, consider the linear map
Then
(cPF x)2 = (lF (px)2 = ((ox, (px)eF = (x, x)eF
x e E,
and so cPF is a Clifford map. Thus it extends to a homomorphism cPc : CE - CF.
To prove uniqueness, suppose that CPC : CE - CF is a second homomorphism making Diagram (10.3) commute. Then
Pc 1E(x) = cc 1E(x)
xeE
and so C 1 implies that Pc = cPc
If i/i : F - H is a second isometry, then clearly,
( ° P)C - C ° CPC
Moreover, the identity map of E induces the identity in CE,
These properties imply that if p is an isometric isomorphism, then cPc is an
algebra isomorphism.
233
Basic Properties
10.6. The 712-Gradation of CE
Consider the linear automorphism w of E given by
w(x) = - x
x E E.
Since w is an isometry, it induces a homomorphism
WE : CE -k CE.
Moreover, since w2 = i, it follows that
wE = 1.
Thus WE is an involution of the algebra CE.
Next, consider the subspaces CE and CE of CE consisting of the elements
CE = ker(WE - 1)
and
CE = ker(WE + 1).
Since WE is an involution, it follows that
CE = CE $ C.
Moreover,
CE CE c CE
C. CE c CE,
C. CE c CE
C. CE
Ccc C.
In particular, CE is a subalgebra of CE.
The elements of CE (respectively CE) are called homogeneous of even
(respectively odd) degree. This defines a 12-gradation of CE. The map WE is
called the degree involution of CE.
It follows from the definitions that the subspaces CE and CE are linearly
generated by the products
C: x
1
xp
x1 E E, p even
C: x 1
xp
x1 E E, p odd.
and
A subspace U of CE which is stable under the degree involution is a
graded subspace. To prove this, set
Uo =Urn CE,
U1 = UnCE.
Then
V = U0 E1 U 1.
10 Clifford Algebras
234
In fact, let u e U and set
u0 = 2(u + (J)E u),
u1=2u
(J)E u)
Then u0 e U0, u1 e U1, and u = u0 + u1.
10.7. Direct Decompositions
Let E and F be inner product spaces. Define an inner product in the direct
sum E Q F by setting
(x1 0 Y1' x2 0 Y2) = (x1, x2) + (Y1' Y2)
x1, x2 E E, Y1' Y2 E F.
Now consider the Clifford algebras CE, CF, and CE®F .
Theorem 10.7.1. The 12-graded algebra CE®F is isomorphic to the skew tensor
product of the 712-graded algebras CE and CF,
CE®F" CE QX CF
PROOF. To simplify notation we write E Q F = H. Let i : E - H and j : F - H
denote the inclusion maps. Since these maps are isometries they induce
homomorphisms
IC: CE - CH
and Jc : CF -+ CH.
Now define a linear map f : CE Q CF -+ CH by
.f (a O b) = ic(a) 'jc(b)
a E CE, b E CF .
We show that f is an algebra homomorphism. Since iC and jc are homomorphisms it is sufficient to show that
ic(a) ' jc(b) = (-1)jc(b) ' ic(a)
a = deg a, f = deg b.
Moreover, in view of C 1, we may assume that
a = x1 xp
x1EE
b=Y1 ... Yq
yjeF.
and
Then
ic(a) ' jc(b) = ic(x 1) ... iC(x p) ' ic(Y 1) .. ic(Yq)
Since every two vectors iC(xl) and JC(yq) are orthogonal with respect to the
inner product in H, it follows that
jc(YJ) + jc(Yj) '
0.
Basic Properties
235
Thus we obtain
iC(a) JC(b) _ (- 1)'j(y1) ... Jc(yq)lc(x 1) ... ic(x p)
_ (-1)' jc(b) ic(a) = (- 1)j(b) ' ic(a).
To show that f is an isomorphism we construct an inverse homomorphism.
Consider the linear map
given by
(x O+ y) = x Q eF + eE Q y
x E E, y E F.
It satisfies
y))2=x2QeF+x®y-x®y+eEQy2
(rl(xEJ
= x 2 Q eF + eE Q y2 = [(x, x) + (y, y)] eE O eF
_ (x O+ y, x O+ y)eH
Thus ri is a Clifford map and so it extends to a homomorphism
g : CH - CE Q CF.
It follows from the definitions off and g that
gf (x O eF) = gic(x) = g(x O+ 0) = ri(x Q 0) = x Q eF
x e E.
Similarly,
gf (eE O+ y) = eE Q y
y e F.
Finally,
gf (eE O eF) = eE O eF
Since the algebra CE Q CF is generated by the elements x Q eF, eE Q y, and
eE Q eF ; this implies that
gof=i.
On the other hand,
fg(x O+ y) = f(x O+ y) = f(x O eF) + f(eE O y)
x e E, y e F
= i(x) + j(y) = x Q y
and
fg(eH) = eH.
These relations imply that
fig=i.
Thus f is an isomorphism.
10 Clifford Algebras
236
EXAMPLE. Suppose that the inner product in E is degenerate and let Eo denote
the null-space. Choose a second subspace F of E such that
E=E0O+F.
Then we have, in view of Theorem 10.7.1 and the example in Section 10.3,
CE
A Eo ©CF .
Proposition 10.7.2. Let p : E - F be an isometry. Then
1. If p is injective, so is Pc;
2. If co is surjective, so is Pc.
PROOF OF 1. Set Im p = F 1. Then p determines an isometric isomorphism
QP i :E 4 F 1. Now write F = F 1 $ F2 where F2 is orthogonal to F 1. Then
the diagram
CE
(
CF
1
CF
commutes as is easily checked (i 1 denotes the obvious inclusion map). The
diagram shows that c°c is injective.
PROOF OF 2. Set ker p = E1 and write E = E1 $ E2 where E2 is orthogonal
to E1. Then q induces an isometric isomorphism P2: E2 -- F and the diagram
CE f CE QX CE2 2 CE2
1
CF
commutes, where it2 is the obvious projection. Since it2 is surjective, it follows
that c°c is surjective.
10.8. The Involution SE
Given a Clifford algebra CE consider the opposite algebra CEP (cf. Linear
Algebra Section 5.1) and let denote the multiplication in CE P. Then the inclusion
J
--* C oPP
E
is a Clifford map and so it extends to a homomorphism
SE : CE - CE P
For x1 E E we have
SE(x 1
... xp) = x i x2 ... xp = xp ... x 1.
237
Clifford Algebras Over a Finite-Dimensional Space
Clearly,
S2E
-t
and so SE is an involution. Moreover,
SEX=x
xeE.
Finally, SE commutes with the degree involution,
SE°0E _ 0E°SE.
Next, let u e CE and set
u - SE WE(u).
In particular,
x = -x
xeE.
The correspondence u H u defines a linear involution of CE. It satisfies
u ' 3 = SE WE(u U) - SE(W E(u) ' WE(V ))
- SE(wE(U)) ' SE(WE(u)) = U u.
PROBLEM
Show that the map a - a (see Section 10.8) in the cases CE = C and CE = H coincides
with the usual conjugation.
Clifford Algebras Over a
Finite-Dimensional Space
In this paragraph E denotes an n-dimensional vector space.
10.9
Proposition 10.9.1. Let dim E = n. Then
dim CE = 2.
PROOF. First consider the case n = 1. Fix a nonzero vector a in E and let A
denote the vector space generated by a and a. Then
e a = a e = a and a2 = (a, a)e.
10 Clifford Algebras
238
Thus A is an algebra. It is easy to check that the inclusion map E - A
extends to an isomorphism CE 3 A. Thus, dim A = 2.
In the general case choose an orthogonal basis {ei} (i = 1, ... , n) of E
and denote by Ei the 1-dimensional subspace of E generated bye, (i = 1 , , n).
Then we have the orthogonal decomposition
E=E1Q+...D+En.
Thus, by Theorem 10.7.1 (cf. Section 5.20)
CE = CE 1 O
O CEn
and so
dim CE = 2.
Corollary. If {xi} (i = 1, ... , n) is any basis of E, then the 2n vectors
<J),...,xl ...xn
form a basis of CE.
PROOF. It follows from the relation xix; + x;xi = 2(x1, x;)e that the vectors
above generate the space CE. Since, by the proposition, dim CE = 2, they
must form a basis of CE.
10.10. The Canonical Element eo
Consider the linear map
E:AE-4CE
given by
E(x 1 A ... A xp) =
1
P
E x(11 ... xp)
(1 <- P <- n).
We show that E is a linear isomorphism. In fact, choose an orthogonal basis
an so
{e1, ... , en} of E. T en ei e; = - e; ei (i
E(eil A ... A eip) = ei1 ...
eip
(11 < i2 < ... < ip).
Thus E takes a basis of A E into a basis of CE and so it is a linear isomorphism.
Now choose a nonzero determinant function A in E. Then E determines
an element ee e CE by the equation
E(x 1 A
A xn) = 0(x 1, ... , xn) ee
xv e E.
(10.4)
eo is called the canonical element in CE (with respect to the determinant
function A).
239
Clifford Algebras Over a Finite-Dimensional Space
Now choose a basis {e1, ..., en} of E such that (e1, e;) = 0 (i
0(e 1,
..., en) = 1. Then
ee = e1
j) and
(10.5)
en.
Next observe that the determinant function 0 determines a scalar 2e such
that the Lagrange identity
det((xi, yj)) = 2e 0(x1, ... , xn)0(y1, ... , yn)
xv e E, yv e E,
(10.6)
holds. Setting xv = yv = ey we obtain
(e1, e 1) ... (en , en) = 2e .
(10.7)
Relations (10.5) and (10.7) imply that
ee = (_ 1)ncn - 112e2 .. en
= (_ 1)n(n - 112(e1, e 1) ... (en , en) ' e
= (_ 1)n(n - 1)/22 e e.
Thus the square of the canonical element is given by
ee = (_ 1)ncn - 1)/22e e.
Since 2e
(10.8)
0 if and only if the inner product in E is nondegenerate it follows
that
1. If the inner product is nondegenerate, then ee is invertible in CE,
2. If the inner product is degenerate, then ee = 0.
Proposition 10.10.1. The canonical element ee satisfies the relation
xeE.
ee x =
Thus,
uECE.
In particular, fn is odd,
uECE
and if n is even,
ee u = COE(u) ee
uECE.
PROOF. Choose an orthogonal basis {e1, ..., en} of E and write
ee =
A. e 1 ... en
A E r.
(see Formula (10.5)). Then we have
ee ei = 2e 1 ... e e, =
1)n
`2(e1, e.)e 1 ...
e1... E
and
i
=2ei .e1 ...en =(-1)`-12(ei., e.)e1
...e....e
i
n
i
10 Clifford Algebras
240
whence
eo ei = (-1)n -' et eo .
Thus, by linearity,
xeE.
10.11. Center and Anticenter
The center of a Clifford algebra CE, denoted by ZE, consists of the elements a
which satisfy
a subalgebra of CE. Since CE is generated by E, it follows
that an element a e CE is in the center if and only if
xeE.
Next observe that ZE is stable under the degree involution. In fact, if
a e ZE, then
WE(a) ' x = - WE(a) ' WE(x) _ - WE(a ' x) = - WE(x ' a)
= - WE(x) ' wE(a) = x ' wE(a)
and so WE(a) E ZE. Thus ZE is a graded subspace of CE and hence a graded
subalgebra,
ZE = ZE O+ ZE,
ZE = ZE r C,
ZE = CE r ZE.
The anticenter of CE, denoted by AZE, consists of the elements a which
satisfy
uECE.
Since CE is generated by E, an element a e CE is in AZE if and only if
xeE.
As above it follows that the anticenter is stable under the degree involution
and hence it is a graded subspace of CE.
Proposition 10.10.1 shows that
1. If n is odd, then eo E ZE,
2. If n is even, then eo E AZE .
Next, assume that the inner product in E is nondegenerate and consider
the linear map'PE CE - CE given by
coE(u) = eo u
u E CE .
Since eo is invertible, APE is a linear isomorphism.
Clifford Algebras Over a Finite-Dimensional Space
241
Now let u e ZE. Then we have, in view of Proposition 10.10.1,
coE(u)x = eoux = eoxu = (-1)"-'xeou
x e E.
= (-1)" 'x cPE(u)
Similarly, if a e AZE, then
(PE(a) x = (-1)"x (PE(a)
These relations show that
x e E.
1. If n is odd, then APE restricts to linear automorphisms of ZE and AZE,
2. If n is even, then cPE interchanges ZE and AZE.
Lemma I. Assume the inner product in E is nondegenerate. Then (AZE)' = 0.
PROOF. Let a E (AZE)'. Then
ax=-xa
xeE
and so
aeo=(-1)"eoa.
On the other hand, Proposition 10.10.1 yields, since a e CE,
eo a = wr' (a) eo = (-1)" - 'a eo .
These relations imply that
aeo=0.
Since the inner product is nondegenerate, eo is invertible and we obtain
a=0.
Lemma II. Assume that the inner product in E is nondegenerate. Then ZE = (e).
PROOF. The lemma is trivial for n = 1. Assume that it holds for n - 1 and let
E be an n-dimensional inner product space. Choose a vector a e E such that
(a, a) 0 and write
E = (a) $ F,
where F denotes the orthogonal complement of a. Thus we can write
u = 1 ®v + a ®w
ueCE
veCF, weCF
and
x= 1®y+a®eF
xeE,yeF.
These relations imply that
ux=a®v-a2®w+ 1®vy+a®wy
=a®v-(a,a)1®w+ 1®vy+a®wy.
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10 Clifford Algebras
Similarly,
1Qyv-a®x yw.
Now assume that u is in the center of E. Then u x = x u and we obtain
a 0 (wy + yw) = 0
yeF
(10.9)
and
1 Q (vy - yv) - 2(a, a)1 Q w = 0.
(10.10)
Thus, w e AZE and so w e (AZE)'. Hence Lemma I implies that w = 0. Now
Formula (10.10) yields
vy=yv
yeF.
This shows that v e ZF and so v e ZF. Hence, by induction, v = 2. eF. It
follows that
u=
and so the induction is closed.
Proposition 10.11.1. Assume that the inner product in E is nondegenerate. Then
1. If n is odd, ZE = (e) + (ee), AZE = 0,
2. If n is even, ZE = (e), AZE _ (eo).
PROOF OF 1. Let n be odd. Assume that a e AZE. Then ax = - xa, x e E,
and so
It follows from Proposition 10.10.1 that a ee = 0 whence a = 0.
Next observe that, by Lemma II, ZE = (e). Since n is odd, the map APE
restricts to an isomorphism ZE 4 Z. Since (pE(e) = ee we obtain ZE = (ee)
whence ZE = (e) + (ee).
PROOF OF 2. Let n be even. Then we have, for a e ZE, ax = xa, x e E, and so
On the other hand, by Proposition 10.10.1,
a ee = WE(a) ee.
It follows that WE(a) = a whence a e Z. Now Lemma II implies that
a = 2. e whence ZE = (e).
Since the map c°E for even n interchanges center and anticenter, it follows
that AZE = (ee). This completes the proof of the proposition.
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Clifford Algebras Over a Finite-Dimensional Space
10.12. The Algebra C _ E
Given an inner product space E denote by - E the space E with the inner
product
(x, y) _
_ - (x, y)
x, y E E
and let C _ E be the Clifford algebra over - E. Denote the multiplication in
C_E by
Thus,
xeE.
Now fix a nonzero determinant function 0 in E. Then it is easy to check that
' A - (-1)"/LA
(10.11)
(cf. Formula (10.6)).
Now define a linear map p: E -+ C _ E by
px = ee ° x
x E E,
where ee denotes the canonical element of C_ E Then we have, in view of
Proposition 10.10.1,
cpx°cpx = ee °x°ee °x = (-1)"-1ee gee °x°x.
Since, by (10.8),
122e = (- 1)"
ee o ee =
1)12( -1)"AA e
and
x°x = (x,x)_ e=
it follows that
px o cpx =
(-1)"
1)12(x, x)AA e.
Now assume that 0 can be chosen such that
A = (-1)" - 1)12
(10.12)
Then the relation above reads
cpx o cpx = (x, x) e
x e E.
Thus p extends to a homomorphism
(p: CE -+ CE.
We show that
+ 1.
en
p(eA) _ (-1)" - '2(e)
(10.13)
244
10 Clifford Algebras
In fact, choose an orthogonal basis {e1, ..., en} of E such that A(e1, ..., en)
= 1. Then eA = e 1
en and thus
(P(eA) _ (p(e 1) o ... o (p(en)
oe1)o...o(eA oen)
(eA
= (_ 1)n(n - 1'2(e
A )n
+1
Similarly, consider the linear map il/ : - E - CE given by
xE-E.
Suppose that A can be chosen such that
A = (_ 1)n(n + 1)/ 2
Then
e
and hence, by the result above (applied to - E), il/ extends to a homomorphism
- CE.
Proposition 10.12.1. Assume that n is even, n = 2m, and that A can be chosen
such that 2A = (-1)m. Then the algebras CE and C_E are isomorphic.
PRooF. Since n = 2m and 2e = (-1)m,
A = (_ 1)n(n - 1)/2 _ (_ 1)n(n + 1)/2
Thus we have the homomorphisms
and
(p: CE - C_E
Ii: C_E - CE.
To show that (p and il/ are isomorphisms we establish the relations
and
il/ o q = w
(poll/ _ (,gym E .
In fact, let x e E. Then
i/i(ee o x) = i/i(ee) il/x
_ (_ 1)mee+ 1
_ (_
2
= (_ 1)menA+ 1
, eA x
. x = (- 1)m(eo)m +1 . x.
Since eo = e, (by (10.8)), it follows that
i/np(x) = (- 1)mx = w(x)
x e E.
But (p and i// are homomorphisms and so the equation above implies that
il/
(p = w.
In the same way it is shown that (p o i// = o/
E.
Clifford Algebras Over a Finite-Dimensional Space
245
10.13. The Canonical Tensor Product of Clifford Algebras
If E is an inner product space and E = + 1 we shall set E = E if E = 1 and
E be an even-dimensional inner product space which
admits a determinant function 0 such that ee = E e (E = + 1). Then, for any
inner product space F,
CE®eF ti CE ® CF
PROOF. Write E Q EF = H and let i : E - H and j : F - H denote the inclusion
maps. Since E has even dimension, Proposition 10.10.1 implies that
xeE.
(10.14)
Moreover, if is : CE - CH and Jc : CAF - CH are the induced homomorphisms,
then (cf. the proof of Theorem 10.7.1)
ic(eA) ' Jc(b) =
b e CAF .
Jc(b). ic(eA)
(10.15)
Now let p : F - CH be the linear map given by
YEF
Sp(y) = ic(eA) ' Jc(Y)
Then, in view of (10.15),
P(Y)2 = ic(ee)Jc(YZ) = E(Y, Y)E ic(eE) ' Jc(eF)
= E2(Y, y). eH = (Y, y) ' eH.
Thus, qP extends to a homomorphism
p: CF - CH.
Next note that, in view of (10.14),
ic(x) ' SP(Y) = ic(x)ic(eA)Jc(Y) = - ic(eA)ic(x)Jc(Y)
= sp(y).c(x)
x e E, y e F
whence
ic(a) Wp(b) = Wp(b) ic(a)
a e CE, be CF.
Next, define
D : CE O CF -+ CH
by setting
(a O b) = ic(a) Wp(b)
a e CE, b e CF.
Then b is a homomorphism. To show that it is an isomorphism we construct
an inverse map.
10 Clifford Algebras
246
First define a linear map i/i : H - CE Q CF by
(xO+Y)=xOeF+C Qy.
Then, in view of 10.14,
(ifr(x O+ Y))2 = x2 Q eF + E[(xeA + eA x) Q y] + E2(ee O y2)
[(x, x) + E(Y, Y)]eE O eF = (x O+ Y, X O+ Y)H eE Q eF.
Thus i/i extends to a homomorphism
kY:CH- CEOCF
Note that
'T'ic(a) = a Q eF
a e CE.
It follows that
('1' ° 1) (x O eF + eE O y) _ '[i(x) + ic(eA)J(Y)]
= x Q eF + E(eA O eF)(eA O Y)
= x Q eF + E2(eE O Y)
=x®eF+eEQy
xeE,yeF.
This shows that q'0 i _ t. On the other hand,
(I ° ') (x O Y) _ l [x O eF + E(eA O Y)]
= i(x) + Eic(eA)ic(eA)Jc(Y)
= i(x) + E2ic(eE)J(Y)
= i(x) + j(y) = x Q y
x e E, y e F
and so I o ' = i. It follows that I is an isomorphism.
10.14. The Direct Sum of Dual Spaces
Let E 1 and E2 be dual n-dimensional spaces and consider the direct sum
E=E1QE2.
Then a nondegenerate inner product is defined in E by
(x1 0 x2, Y1 0 Y2) = 2[<x1, Y2> + <Y1, x2>]
x1, Y1 E E1, x2, Y2 E E2
(note that this is not the usual inner product in the direct sum!). In particular,
the restriction of twice the inner product in E to E1 x E2 coincides with the
scalar product between E 1 and E2.
Proposition 10.14.1. There is an isomorphism CE 4 L( A E 1).
Clifford Algebras Over a Finite-Dimensional Space
247
PROOF. First recall the definition of the multiplication and the substitution
operators in A E 1. Identifying E 1 with E2 we have the relations
µ(x1)2 = 0
x1 EE1
i (x2)2 = 0
x2 E E2
and (see Corollary I to Proposition 5.14.1)
x1 E E1, x2 E E2.
1(x2) °µ(x1) + µ(x1) ° 1(x2) = <x1, x2>1
Now define a linear map p: E - L( A E 1) by setting
(p(x) = µ(x1) + 1(x2)
x E E,
where x = x 1 $ x2, x 1 E E 1, x2 E E2. Then the relations above yield for all
xeE
p(x)2 = µ(x 1) ° µ(x 1) + µ(x 1) ° 1(x2) + 1(x2) ° µ(x 1) + 1(x2) ° 1(x2)
=
Thus p extends to a homomorphism
q . CH - L(A E 1).
We show that p is an isomorphism. Since
dim L( A E1) = (2n)2 = 22n = dim CE,
it is sufficient to show that p is surjective. This is a consequence of the
following.
Lemma. Let E*, E be a pair of finite-dimensional dual spaces. Then the algebra
L( A E) is generated by the operators µ(x) and i(x*), x e E, x* E E*.
PROOF. Recall from Section 6.2 the linear isomorphism
TE: APE*Q AE4L(A'E; AE)
given by
TE(a* Q b)u = <a*, u>b.
This can be written in the form
TE(a* Q b) = µ(b) ° i(a*).
Since every vector a* E A PE* is generated by products x; A
x* E E* and every vector b e A qE is generated by products Y1 A
y E E, and since
µ(Y 1 A ... A Yq) = µ(Y 1) ° ... ° µ(Yq)
and
1(x i A
... A x) = i(x) ° ... ° i(x 1)
the lemma follows because L( A E) = Qp L( A pE; A E).
A xp,
A yq,
248
10 Clifford Algebras
Proposition 10.14.2. Let E be a 2n-dimensional vector space with a nondegenerate inner product. Assume that there exists an involution w of E such
that th = - w. Then the Clifford algebra CE is isomorphic to the algebra of
linear transformations of A E 1, where E 1 = ker(w - i).
PROOF. Consider the subspaces E 1 and E2 consisting of the vectors x which
satisfy, respectively, wx = x and wx = -x. Then E = E1 $ E2. In fact, let
x e E and set
x l - 2(x + wx)
x2 - 2(x - wx).
For x1 E E1, yl E E1, we have
(x1,
yl) = (wx1, wy1) = -(w2x1, yl) = -(x1, yl)
whence (x1, yl) = 0. Similarly,
(x2 , y2) = 0
x2 , y2 E E2 .
Thus the restrictions of the inner product to E1 x E1 and E2 x E2 are zero.
On the other hand, the restriction of the inner product to E1 x E2 is nondegenerate. In fact, fix x1 E E 1 and assume that (x 1, y2) = 0 for every y2 E E2.
Then we have for y e E
(x1, y) - (x1, yl) + (x1, y2) = 0
whence x1 = 0. Thus a scalar product between E 1 and E2 is defined by
<x1, x2> = 2(x1, x2)
x1 E E1, x2 E E2.
It satisfies the relation
(x1 O+ x2, yl e y2) = (x1, y2) + (yl, x2)
= 2[<xl, y2> + <y1, x2>]
Thus Proposition 10.14.1 implies that CE ^L( A E 1).
PROBLEMS
1. Let CE be the Clifford algebra over an n-dimensional inner product space and denote
the left multiplication by an element a e CE by µ(a). Show that
det µ(x) = (x, x)2" -
1
xEE
and
tr µ(a) = 0
if a E CE .
2. The isomorphisms bE and 11E. Let E be an n-dimensional vector space with a nondegenerate inner product. Identify E with the dual space and denote by i(x) (x e E) the
substitution operator in n E.
249
Clifford Algebras Over a Complex Vector Space
i. Show that the isomorphism bE defined in Section 10.10 satisfies the relations for
all x e E and all a e A "E
b E(x A a) = x b E(a) - b E 1(x)a
and
bE(a A x) = ,(a) x + (-1 )'Eb1(x)a.
ii. Let 11E denote the inverse isomorphism. Show that for all x e E and all u e CE
'1E( x ' u) = x A ii ,(u) + I(x)rl E(U)
and
11E(U ' x) _ 11E(U) A x -
iii. Let 7rE. CE - r be the linear map given by
nE u = no r1 E(u)
u E CE
where no : A E -+ r is the obvious projection. Prove the following relations for all
xeEandallu,veCE:
iE(x ' u) = n0 I(X)E(U)
iE(U ' x) _ - n0
E(wE u)
nE ° WE = nE
iE(U v) = 7rE(v. U).
3. Use the linear map nE (see Problem 2 (iii),) to define a bilinear function QE in CE by
setting
QE(u, v) = nE(u v).
i. Show that QE is symmetric and nondegenerate.
ii. Show that QE(x, y) _ (x, y), x, y e E.
iii. Prove the relation
QE(U ' W, v ' W) = QE(W U, w v)
U, V, W E CE.
Clifford Algebras over a Complex Vector Space
10.15. Clifford Algebras Over Complex Vector Spaces
Let C" be an n-dimensional complex vector space and let ( , ) be a nondegenerate symmetric bilinear function in C". Note that if ( , )1 and ( , )2
are two such bilinear functions, then there is a linear isomorphism p of C"
such that
(qx, (Py) i = (x, y)2
x, y e C".
Thus the corresponding Clifford algebras are isomorphic. We shall denote
the Clifford algebra over C" by C,,.
10 Clifford Algebras
250
EXAMPLE : n = 1. The Clifford algebra C 1 is isomorphic to the algebra
C Q C. In fact, choose a vector e 1 E C such that (e 1, a 1) = 1. Then the vectors
{e, e 1 } form a basis of C 1. Now define a linear map sb: C 1 - C Q C by
setting
b(ae + f e 1) = (a + fl, a - f3)
a, f e C.
Then 'b is an algebra isomorphism as is easily checked. Thus,
C1
CQC.
The Element ee. A normed determinant function in C" is a determinant
function A which satisfies
0(x1, ... , xn) . 0(y1, ... , yn) = det((xi, yj))
xi, yj e C".
It is easy to see that a normed determinant function always exists and that it
is determined up to sign. Thus the canonical element ee corresponding to a
normed determinant function satisfies (see (10.8))
Hence Theorem 10.13.1 yields an isomorphism
(10.16)
C2m+k ti C2m ® Ck.
Proposition 10.15.11. Let n be even, n = 2m, and assume that the inner product
is nondegenerate. Then
(10.17)
L( A Cm).
C2m
PROOF. Write C" as an orthogonal sum
C"= AQB
where dim A = m and dim B = m. Choose orthonormal bases {aµ} and
{bµ}, µ = 1, ..., m in A and B respectively and define an involution w of
C" by setting
u= 1, ... , m
w(aµ) = ibµ
iaµ
µ = 1, ... , m.
Then for µ, v = 1, ..., m we have the relations
0,
(waµ ,
(wbµ ,
0
and
(waµ ,
(aµ,
i(bµ ,
w is a skew transformation. Now Proposition 10.14.2 shows that
C"
L(A E1),
where E 1 is the subspace of E determined by the equation wx = x.
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Clifford Algebras Over a Real Vector Space
Combining Formulas (10.16) and (10.17) we obtain for all m > 1 the
relations
C2m
L( gym)
and (using the example above)
C2m+ 1
L( A Cm) Q+ L( A Cm).
10.16. Complexification of Real Clifford Algebras
Let F be a real inner product space (not necessarily of finite dimension) and
consider the complexification E _ C Q F. Define an inner product in E by
(A O x, µ O Y)E _ 2µ(x, Y)
2, u E C, x, y E F.
Then the inclusion map j : F - E is an isometry and so it extends to a (real)
homomorphism jc : CF -p CE. Now consider the complex linear map
(p.CQCF - SCE
given by
(p(A O a) _ 2 ' Jc(a)
2 E C, a E CF.
To show that (p is an isomorphism we construct an inverse homomorphism. Consider the linear map i/i : E - C Q CF given by
i/4x+iy)= 1Qx x+i®y.
Then
(li(x + iy))2 = 1 O x2 - 1 O Y2 + i(x y + y x)
_ [(x, x) - (Y, Y)](1 O e) + 2i(x, Y)(1 O e)
e)
x,yeF.
Thus i/i extends to a homomorphism i/i : CE -+ C Q CF It follows from the
= i. Thus p is an isomorphism,
definitions that ll/ o co = l and co o
p:COCF4CE
Clifford Algebras Over a Real Vector Space
10.17
Let E be a real n-dimensional vector space with a nondegenerate inner
product. Recall that E can be decomposed in the form
E=E+QE-,
10 Clifford Algebras
252
where the restriction of the inner product to E + (respectively E) is positive
(respectively negative) definite. If dim E + = p and dim E - = q, we shall say
that the inner product is of type (p, q). Clearly, p + q = n. The difference
s = p - q is called the signature of the inner product.
The Clifford algebra over an inner product space of type (p, q) will be
denoted by C(p, q). We shall write
C(n, 0) = C(+)
and
C(0, n) = Cn( - ).
Thus, by Examples 1 and 2 of Section 10.2,
C1(-) ^C and C2(-) ^Q-i.
Next recall from Section 9.19 of Linear Algebra that a normed determinant
function in E is a determinant function A which satisfies
det((x1, y;)) =
1, ... , xn) 0(y 1, ... , yn)
xv e E, yv e E.
Every inner product space admits a normed determinant function A and A
is uniquely determined up to sign. Thus the scalar 2n (see Section 10.10)
corresponding to a normed determinant function is given by
o=
This implies that the corresponding canonical element ee of the Clifford
algebra C(p, q) satisfies
eo = (-1)(1/2)n(n-1)+'?e.
In particular,
1. If (
,
) is positive definite, eo = (-1)(1 /2)n(n - 1)e
2. If (
,
) is negative definite, eo = (-1)(h/2)(t+ 1)e
3. Ifn=2mandp=q=m,eo=e.
Theorem 10.17.1. There are algebra isomor phisms
C(p,q)©C2(+)=C(q+2, p)
p,q?0
C(p, q) © C2(-)
p, q ? 0.
and
C(q, p + 2)
In particular, for n > 0,
Cn(-) © C2(+) ~ Cn+2(+)
and
C(+)® C2(-) ti Cn+2(-).
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Clifford Algebras Over a Real Vector Space
PROOF. Observe that the canonical elements of C2(+) and C2(-) satisfy
ee = -e and apply Theorem 10.13.1.
Theorem 10.17.2. Assume that s - 0 (mod 4). Then
C(q, p)
C(p, q)
In particular,
if n - 0 (mod 4).
PROOF. Set s = 4k. Then
n=2q+s=2q+4k=2m,
wherem=q+2k
and
m-q=2k.
Thus, if A is the normed determinant function, then
=(-1)q=(-1)m.
Now apply Proposition 10.12.1.
10.18. Inner Product Spaces With Signature Zero
Suppose that E is an inner product space with signature zero. Then p = q and
so dim E = 2p. We shall show that
CE
L(n E 1),
where dim E 1 = p.
Choose an orthogonal decomposition E = E + Q E - such that the restriction of the inner product to E + (respectively E) is positive (respectively
(v =
negative) definite. Thus dim E+ = dim E- = p. Let
and
1, ... , p) be orthonormal bases of E + and E - respectively,
(ar, aµ) = 5vµ
V, U = 1, ... , p
(br, bµ) =
v, µ = 1, ... , p.
Define an involution w of E by setting
wa= b and wb - av = 1, ... , p.
Then w is a skew transformation as is easily checked. Thus by Proposition
10.14.2,
CE
L(n E ),
where E 1 is the kernel of w - i.
EXAMPLE. Consider the algebra C(2, 2). Then, by the result above,
C(2, 2) ^'
L(4).
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10 Clifford Algebras
On the other hand, the second formula in Theorem 10.17.1 applied with
p = 0 and q = 2 yields
C(2, 2)
C2(-) O C2(-) "' fNl Q u-fl.
Thus,
fl--fl Q fl--fl
The following proposition establishes an explicit isomorphism between these
algebras.
Proposition 10.18.1. There is a canonical algebra isomorphism
fb: fl--l Q u--fl -
PROOF. Identify
with 0-fl and define a linear map
f : fl--l Q fl--fl by setting
fb(pOq)x =
where q denotes the conjugate of q. It is easily checked that fb is an algebra
homomorphism. To show that it is an isomorphism define positive definite
inner products in 0-fl ® 0-fl and L(4) by
(p O q, p' O q') = (p, p')(q, q')
and
(q,, /,) = 4 tr(rp ° /,)
P, /i e
respectively. Observe that the adjoint transformation of fb(p Q q) is given by
(p O q) = DO O q)
Thus we have
(fb(p O q), fb(p' O q')) = 4 tr(fb(p O q) ° fb(p' O q'))
= 4 tr fb(pp' O qq')
Now it is easy to check that for a e 0-fl and b e 0--fl
tr fb(a Q b) = 4(a, e)(b, e).
It follows that
(fb(p O q), fb(p' O q')) = (pp', e)(qq', e) = (p', p)(q', q)
=(p®q,p'Qq').
Thus fb is an isometry and hence a linear isomorphism.
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Clifford Algebras Over a Real Vector Space
10.19. Clifford Algebras of Low Dimensions
In this and the following section we shall determine the structure of
for 1 < n < 8.
and
(1) n = 1. We show that
C 1(+)
In fact, let e 1 be a unit vector in . Then the vectors e, e 1 form a basis of
C 1( +). Now define a linear map t': C 1( +) -+ [ Q [ by setting
J(ae+fe1)=(a+f3,a-f3)
a,fe.
Then 1 is a homomorphism as is easily checked. Moreover, 1 is a linear
isomorphism and so an isomorphism of algebras. This proves that C 1( +)
On the other hand, it has been shown in Example 1, Section 10.2, that
C1(-)
C.
(2) n = 2. We show that
C2(+)
Fix an orthonormal basis {e1, e2 } of 2 and define a linear map p: 2 L(2) by setting
gp(x)e 1 = ae 1 + f3e2
gp(x)e2 = fe 1 - ae2 ,
where
x=ael+le2.
Then it is easy to check that
p(x)2 = (a2 + f32)1 = (x, x). 1.
To show that 1 is
Thus q extends to a homomorphism t : C2(+) -+
an isomorphism note that the transformations t(x), x e E, are selfadjoint and
have trace zero. On the other hand, t(e) = 1. Thus, if A denotes the subspace
of CE spanned by e, e 1, and e2, 1 determines an isomorphism from A to the
space of selfadjoint transformations of 2 Finally, it is easy to check that
(e2 e1) is the transformation e1 -+ e2, e2 -+ -e1 and so it is skew. Since
every transformation of 2 is the sum of a selfadjoint and a skew transformation, it follows that 1 is an isomorphism.
10.20
Using the results of Section 10.19 and Theorem 10.17.1 we shall now determine the Clifford algebras Ck( +) and Ck( -) for k < 8 explicitly. Recall that
the direct sum of two algebras A and B is the algebra A Q B with multiplication defined by
(a1 Q b1) (a2 Q b2) = (a1a2 Q b1b2).
256
10 Clifford Algebras
We have the following isomorphisms :
n = 3:
C3(+)
C1(-) ®C2(+)
C O L(I2)
$l)®D'
C3(-)C1(+)®C2(-)(l
I1-OeI1-O
n =4: C4(±)
C2(-) O C2(+)
U-0 O L(11R2)
n = 5: C5(+)
C3(-) O C2(+)
(U-0 e U-0) O L(l 2)
I1-0 O L(l 2) e Il-0 O L(l 2 )
C5(-)
C3(+) O C2(-)
C O L([1R2) p U-O
n =6: C6(+)
C4(-) O C2(+)
U-0 O L(l 2) O L(l 2)
U-0 ® L(l 4)
C6(-) ti C4(+) ®C2(-) = DO®L(E2) ®I1-O ~ D-O ®I1-O ®L(E2)
L(l 4) Ox L(l 2) L(l 8)
n = 7:
C7(+)
C5(-) OO C2(+)
C Ox L(1R2) O U-0 O L(1R2)
C O U-0 O L({R4)
C7(-)
C5(+) Q C2(-)
[U-0 O L(2) $ U-0 O L(2)] O U-0
D-0 O D--0 O L(R 2) e D--0 O D--0 O L(2 )
O
n = 8:
C8(+)
O
O+
C4(-) O C4(+)
O+
O U-0 O
U-0 O
O
O
L(W6).
These results are combined in the following table:
C(+)
C(-)
C
L(B 2)
CO
0-U
U-U Q+ U-U
0-U® L(B2)
(0-il O L(B2) O (0-ll O L(B2))
0-U ®L(B4)
C® L(B2) O H
C® 0-U ®L(B4)
L(B16)
L(B 8) O L(B 8)
L(B16)
U-() O L(B2)
L(B 8)
and
10.21. The Algebras
Since the canonical element of C8( +) ^C8( -) satisfies eo = 1, Theorem
10.13.1 yields isomorphisms
Cn+8(+) ti Cn(+) ® L(W6)
and
Cn + 8(-) ti Cn(-) ® L(W 6).
257
Clifford Algebras Over a Real Vector Space
These isomorphisms, together with the isomorphisms in the table, determine
for all n.
and
the structure of the Clifford algebras
10.22. The Algebras C(p, q)
Finally, we determine the algebra C(p, q) for an indefinite inner product of
type (p, q)
Case 1: p = q. Then it was shown in Section 10.18 that
(10.18)
C(p, p) ^L( A DU').
Case 2. p > q. Write p = q + s. Then we have an orthogonal decomposition
= + 0 (f
= (f 0 f)
n=
has even dimension and since the canonical element of
Since
C(q, q) satisfies ee = e, Theorem 10.13.1 gives
C(p, q)
C(q, q) O CS( +).
Now using Formula (10.18) we obtain
C(p,q)L(A )QCS(+)
p>q,s= p - q.
Case 3: p <q. Set q - p = r. Then we obtain
C(p, q) = C(p, p + r)
C(p, p) O Cr(- )
Thus
C(p, q)
p <q, r = q - p
L( A fiP) © Cr(-)
Now the results of Section 10.21 give the structure of C(p, q) for all p, q.
EXAMPLE. The Clifford algebra over the Minkowski space (p = 3, q = 1) is
given by
C(3, 1) = C2(+) Q L(A W)
L(E2) O
PROBLEMS
1. Establish explicit isomorphisms
C3(+)
L(C2)
C3(-)
H$H
C7(+)
L(Cs)
C7(-) "' L(Ps) O+ L(B8).
L(4).
258
10 Clifford Algebras
2. Let C2 be a complex vector space of dimension 2 with a positive definite Hermitian
inner product
i. Consider the linear transformations (p which satisfy
q) +rp=trq).i.
Show that these transformations form an algebra (over B) and that this algebra is
isomorphic to 0-fl.
ii. Use this algebra to establish a canonical isomorphism
COH
L(C2).
3. Let E be an n-dimensional Euclidean space and consider the symmetric bilinear
function QE in CE defined in Problem 3 after Section 10.14.
i. Show that the signature s of QE is given by
cos
nn
4
+sin
nn
4
n> 1.
ii. Conclude that
5n+4 = -
5n+8 = 16s
n > 1.
iii. Show that QE is positive definite only if n = 1.
4 Derivations. Let E be a vector space over a field F.
i. Let 9 be a derivation in CE which restricts to a linear transformation ( of E. Show
that (p is skew.
ii. Show that every skew transformation 'p of E extends uniquely to a derivation in CE.
5. Antiderivations. Recall that an antiderivation in CE is a linear transformation Q
which satisfies
Sl(ab) = Sl(a)b + WE(a)b
a, b E CE
(cf. Section 5.8 of Linear Algebra).
i. Show that the map Q = i - wE is an antiderivation in CE.
ii. Assume that the inner product in E is nondegenerate. Show that every antiderivation Q in CE which restricts to a linear transformation of E is of the form
Q = , (1- wE)
, E r.
6. Clifford algebras over even dimensional spaces. Let CE be the Clifford algebra over an
even-dimensional real vector space with a nondegenerate inner product. Show that CE
is simple (that is, the only two-sided ideals in CE are CE and (0)). Conclude that a
homomorphism p from CE to any associative algebra A which satisfies ap(e) = eA is
injective. Hint: Consider the complexification of CE and apply Proposition 10.15.1.
7. Clifford algebras over spaces of odd dimension. Let CE be the Clifford algebra over an
odd-dimensional space and assume that the canonical element satisfies eo = e.
Consider the linear transformations
spa = 2(e + eo)a
a E CE
259
Clifford Algebras Over a Real Vector Space
and
via = 2(e - ee)a
Set J" + = Im p and % - = Im /i. Show that
and that
a E CE.
and % - are two-sided ideals in CE
CE=T
Apply the result to complex and real Clifford algebras.
8. Show that the center of a Clifford algebra over an infinite-dimensional vector space
with a nondegenerate inner product is the subspace spanned by e.
11
Representations of
Clifford Algebras
Basic Concepts
11.1. Representations of an Algebra
Let A be an associative algebra with unit element e. A representation of A
in a vector space V is a homomorphism R :A - L(V) where L(V) denotes
the algebra of linear transformations of V such that R(e) = i. A representation is called faithful, if the map R is injective.
A subspace W c V is called stable under R, if it is stable under every
transformation R(a), a E A. A representation is called irreducible, if the only
stable subspaces are W = V and W = 0. In particular, if R is surjective, then
the representation is irreducible. In fact, assume that W is a stable subspace.
Then R is stable under every linear transformation of V. This is only possible
ifW=0orW=V.
Two representations R 1 and R2 of A in V1 and V2 respectively are called
equivalent, if there is a linear isomorphism b: V1
V2 such that
o R 1(a) = R 2 (a) o
a E A.
In this case we shall write R1 ' R2.
Next let P and Q be representations of A in U and V respectively. Then a
representation of A in U $ V, denoted by P $ Q, is given by
(P O+ Q) (a) = P(a) $ Q(a)
a E A.
It is called the direct sum of P and Q. Similarly, the tensor product of P and
Q, denoted by P O Q, is the representation in U O V defined by
(P O Q) (a) = P(a) O Q(a)
260
a E A.
261
Basic Concepts
Clearly, if P,
P2 and Q1
Q2, then
P, e Q, --P2 e Q2 and P, Q Q, "'P2 Q Q2.
11.2. Representations of a Clifford Algebra
Let CE be the Clifford algebra over an inner product space E and let R
be a representation of CE in an n-dimensional vector space V. Then R
restricts to a linear map RE : E - L(V). We show that this map is injective
if the inner product in V is nondegenerate. In fact, assume that RE(xo) = 0
for some xo E E. Then
R(xo y + yxo) = R(xo) ° R(y) + R(y) ° R(xo) = 0
for every y e E.
Since
xo y + yxo = 2(x0, y)e,
we obtain
(x o, y) = 0
yEE
whence xo = 0. Thus RE is injective.
11.3. Orthogonal Representations
A representation of a Clifford algebra CE in a Euclidean space V is called
orthogonal, if
(R(x)u, R(x)v) = E(x, x) (u, v)
x e E, u, v e V,
where c = + 1. It is called positive orthogonal, if c _ + 1 and negative
orthogonal, if c = -1.
Thus, if R is positive orthogonal, then
(R(x)u, R(x)v) = (x, x). (u, v).
It follows from this equation that i(x) o R(x) = (x, x). ,,x e E.
On the other hand, R(x) o R(x) = R(x2) = (x, x). i, x e E. These relations
imply that i(x) = R(x). Similarly, if R is negative orthogonal, then the transformations R(x) are skew.
Proposition 11.3.1. Assume that the inner product in E is positive (respectively
negative) definite. Then every representation of CE in a Euclidean space is
equivalent to a positive (respectively negative) orthogonal representation.
PROOF. Let dim E = k and choose a basis {e,, ..., ek} of E such that
a;) = E
b1;
(i, j = 1, ... , k).
11 Representations of Clifford Algebras
262
Then
(i, j = 1, ..., k).
e1 e; + e; e1 = 2c51. e
In particular, e? = E e, and so the elements e1 are invertible. Thus, if CE
denotes the multiplicative group of invertible elements in CE, then e1 E C.
Let G denote the subgroup of CE generated by the elements e (i = 1, ..., k)
and e. The relations above imply that G is a finite group.
Now introduce a new definite inner product in V by setting
(u, v)0 =
(R(a)u, R(a)v).
aeG
Then we have for g e G
(R(a)R(g)u, R(a)R(g)v)
(R(g)u, R(g)v)0 =
aeG
=
(R(ag)u, R(ag)v) =
(R(a)u, R(a)v)
aeG
aeG
= (u, v)0
u, v E V.
Thus,
(R (g)u, R(g)v)0 = (u, v)0
(11.1)
g e G.
Next observe that, since both inner products of V are definite of the same
type, there is a linear automorphism'b of V such that
(b(u), b(v)) = (u, v)0
u, v e V.
Now set
P(a) =
o R(a) o b -'
a e CE.
Then P is a representation of CE equivalent to R. Moreover, Relation (11.1)
yields
(P(g)u, P(g)v) = (bR(g) b 1(u), bR(g) b 1(v))
= (R(g) F 1(u), R(g) F 1(v)) o
= (F 1(u), F-'(v))0 = (u, v)
Thus,
(P(g)u, P(g)v) = (u, v)
In particular, if we set P(e1) = o (i = 1, ..., k), then
(6i u, 6i v) = (u, v)
u, v E V,
and so
6 o a. = i
(i = 1, ..., k).
g e G, u, v e V.
The Twisted Adjoint Representation
263
On the other hand, we have
6i o 6i = 6i = P(e) _ (e,, e.) l = c. l
(1 = 1, ... , k).
These relations yield
61 = E
a1
(1= 1, ... , k)
and so, by linearity,
P(x) = E P(x)
x E E.
It follows that
(P(x)u, P(x)v) _ (P(x)P(x)u, v) = E (P(x)2u, v)
= E (P(x2)u, v) = c. (x, x) (u, v)
x e E.
This relation shows that P is an orthogonal representation. In particular, if
the inner product in E is positive (respectively negative) definite, P is a
positive (respectively negative) orthogonal representation.
The Twisted Adjoint Representation
11.4
Definition. Let E be an n-dimensional vector space with a nondegenerate
inner product. Denote by CE the multiplicative group of invertible elements
in CE. Then a representation of the group CE in CE is defined by
ad(a)u = WE(a)ua -'
a e CE, u e CE,
where WE denotes the degree involution, ad is called the twisted adjoint
representation of C.
It follows from the definition that
ad WE(a) _ WE o ad(a) o
(11.2)
We show that the kernel K of ad consists of the elements e, t
ad() e)u =
=u
0. Clearly,
u e CE
and so ) e e K. Conversely, assume that ad(a) = i. Then
WE(a)u = ua
u E CE .
(11.3)
Setting u = e we obtain WE(a) = a. Now Equation (11.3) yields
au = ua
u E CE
whence a e ZE and so a e Z. Now Lemma II, Section 10.11, implies that
a = , e,
t e T. Since a E CE, it follows that t
0.
264
11 Representations of Clifford Algebras
11.5. The Clifford Group
Let rE denote the subgroup of CE consisting of those elements a for which E
is stable under ad(a). rE is called the Cli/ ord group of E. Every element
0 is contained in FE. In fact, since for x e E
h e E which satisfies (h, h)
ad(h)x =
-hxh-1
= x - 2(h'x)h
h
it follows that ad(h)x e E, x e E, and so h e E
Proposition 11.5.1. The Cli/ ord group is stable under the degree involution
and under the antiautomorphism SE (see Section 10.8).
PROOF. (1) Let a E E Then Formula (11.2) yields
ad(wE a)x = WE ad(a)wE 1(x) _ - WE ad(a)x = ad(a)x e E
Thus, WE(a) E
x E E.
E
(2) Let a E E Then a' E rE and so we have
x E E.
WE(a -1)xa E E
Applying SE yields
SE(a)x SE WE(a -1) E E
xeE
whence, since WE commutes with SE
WE(SE(a))x(SE(a)) -1 E E
x e E.
Thus, SE(a) E rE .
11.6. The Map 2E
Recall from Section 10.8 the antiautomorphism a - a defined by a =
SE WE(a). Proposition 11.5.1 implies that the Clifford group is stable under
this map.
Now consider the (nonlinear) map o: CE - CE given by
o(a) = as
a E CE .
Proposition 11.6.1. If a E rE, then
o(a) _ Aa e
Aa E r*
Moreover, the map 2E rE - r* given by AE(a) _ 2a is a homomorphism from
rE to r* (the multiplicative group of r).
255
The Twisted Adjoint Representation
PROOF. Since rE is stable under conjugation, a e fE. Now let x e E and set
y = wE(a)xa-1. Then y e E and so SE(y) = y. It follows that
SE(a) - 1 xSE(wE a) = WE(a)xa - 1
whence
xSE(wE a)a - SE(a)W E(a)x
x e E.
Since
SE(wE a) = a
and
SE(a) = W E(a),
we obtain
x(aa) = WE(aa)x
x e E.
Thus, setting as = b, we have
xb = WE(b)x
x e E.
Now write b = bo + b 1 with bo C CE and b 1 C C. Then the equation above
yields
xbo = box and xb 1 = - b 1 x
x C E.
Thus, bo C ZE and b 1 C (AZE)1. Now Lemmas I and II, Section 10.11, show
that bo = Aa a and b 1 = 0 whence b = ,a e and so 'bE(a) =
e. Finally,
since a is invertible, it follows that Aa 0 and so C r*.
Now consider the map AE : rE - r* defined by
E(a) = AE(a)e
a C rE .
To show that 2E is a homomorphism, let a e rE and b e E Then
E(ab) = ab ab = abba.
By the first part of the proposition,
bb = AE(b) e.
It follows that
E(a b) = AE(b)aa = AE(b)AE(a)e = AE(a)AE(b)e
whence
AE(ab) = AE(a)AE(b)
Corollary I. The map 9 satisfies
8(WE(a)) = 9(a)
a e 1E.
PROOF. In fact,
e(W E(a)) = WE(a)WE(a) = WE(a)WE(a) = WE(aa) = ,E(a) - e = 0(a).
266
11 Representations of Clifford Algebras
Corollary II. The homomorphism 2E satisfies the following two relations:
1. AE(WE(a)) _ AE(a)
aErE.
2. AE(ad(b)a) _ AE(a)
a, b e
E.
PROOF. (1) follows from Corollary I.
(2) Let a E rE and b e E Then, since 2E is a homomorphism, the first
relation yields
AE(ad(b)a) _ AE(w(b)ab -1) _ AE(w(b))AE(a)AE(b) - 1
_ AE(b)AE(a)AE(b) - 1 = AE(a)
Proposition 11.6.2. Fix a E rE and let to denote the restriction of ad(a) to E.
Then to is an isometry.
PROOF. Since for x e E
O(x)= - (x, x) e,
it follows that
AE(x) _ - (x, x)
x e E.
Now Part (2) of Corollary II to Proposition 11.6.1 yields
(ad(a)x, ad(a)x) _ - AE(ad(a)x) _ - AE(x) _ (x, x)
whence
(tax, tax) _ (x, x)
x E E.
In particular, consider a vector h e E with (h, h)
h e fE). Since
hx + xh = 2(x, h)e,
we obtain
(h, x)
ih(x) = x- 2 h h x
x E E.
This equation shows that
ih(h) _ - h,
while
ih(Y) = Y
if (h, y) = 0.
Thus th is the reflection in the plane orthogonal to h.
0 (recall that then
267
The Twisted Adjoint Representation
11.7. The Homomorphism E: rE -- O(E)
Let O(E) denote the group of isometries of E. Then a homomorphism
E : rE - O(E)
is defined by
(FE(a) = to
a E rE .
Proposition 11.7.1. The homomorphism (FE is surjective.
PROOF. Let h e E be a vector such that (h, h) 0 and let ph denote the reflec'tion in the plane perpendicular to h. Then (FE(h) = ph.
By Lemma I below, every isometry of E is generated by reflections and
so the proposition follows.
Lemma I. Let E be an n-dimensional vector space with a nondegenerate inner
product. Then every isometry t of E is the product of at most n + 1 reflections.
PROOF. We show first that if a and b are any two vectors such that (a, a) =
0, then there is a reflection p such that p(a) = ± b.
(b, b)
In fact, since
(a + b, a + b) + (a - b, a - b) = 4(a, a)
0,
it follows that (a + b, a + b) 0 or (a - b, a - b) 0. Replacing b by -b
if necessary, we may assume that (a - b, a - b) 0. Now set h = a - b
and consider the reflection
P()
((h, h))
Then p(a) = a - (a - b) = b.
Now we prove the lemma by induction on n. If n = 1, then ix = E x,
E = ±1, and so t is a product of at most two reflections. Suppose now that
the lemma holds for a space with dimension n - 1 and let t be an isometry of
E (dim E = n). Choose a vector a e E such that (a, a) 0 and set b = i(a).
Then (b, b) = (a, a) and so, by the remark above, there is a reflection p such
that p(a) = ± b. Now set
ti =p-lot.
Then t i (a) = + a and so t i restricts to an isometry of the orthogonal complement, E1, of a. Thus, by induction, t i is the product of at most n reflections
of E1. Every such reflection extends to a reflection of E. Thus, t is the product
of at most n + 1 reflections and the induction is closed.
Corollary I. The Clifford group rE is generated by the elements h e E which
satisfy (h, h)
0.
268
11 Representations of Clifford Algebras
PROOF. Let a E rE and set (FE(a) = t. By Lemma I, t is of the form
=
where (h1, h.)
Phi°".°Phr
h1EE
0. Since (FE(h1) = ph;, it follows that
FE(a - t h l ... hr) = t 't = 1.
Thus,
a-1h1 ...hr =
2*Ef*
and so
h, _ (2 tht)h2 ... h,..
a
Corollary II. The homomorphism (FE satisfies
det (FE(a) a = WE(a)
a E rE .
(11.4)
PROOF. Consider the homomorphism p: rE - I~E given by
(p(a) = det (FE(a) a.
Then we have, for x e E, (x, x)
0,
(p(x) _ - x = WE(x)
and so (11.4) holds in this case. Now apply Corollary I.
11.8. The Group Pin
From now on E will be a real vector space with a (nondegenerate) inner
product of type (p, q). We shall write CE = C(p, q) and rE = r(p, q). Recall
from Section 11.6, the homomorphism AE : r(p, q) The group Pin(p, q)
is the subgroup of r(p, q) consisting of the elements a which satisfy
AE(a) = ±1.
Since, for x e E, AE(x) _ - (x, x), it follows that all the vectors x e E
for which (x, x) = ±1 are contained in Pin(p, q). In particular, - e e Pin(p, q).
Moreover, Corollary I to Proposition 11.7.1 shows that Pin(p, q) is generated
by these vectors.
Next, denote by O(p, q) the group of isometries of U r and consider the
homomorphism (FE: r(p, q) - O(p, q) defined in Section 11.7. We shall show
that the restriction (F of E to Pin(p, q) is still surjective. In fact, let a e O(p, q).
Then, by Proposition 11.7.1, there is an element b e r(p, q) such that
(FE(b) = a. Now set
a=
b
tr2 .
12E(b) I
269
The Twisted Adjoint Representation
Then
, E()
a=
2E(b)
E( b)I
-
=+1
and so a e Pin(p, q). Moreover, tE(a) = 'E(b) = a. Thus tE restricts to a
surjective homomorphism
: Pin(p, q) - O(p, q)
Now we show that ker I = S°, where S° is the subgroup of Pin(p, q)
consisting of the elements a and - e. In fact, if a e ker I, then a e ker ad,
and so a = 2. e (see Section 11.4). It follows that AE(a) = 22. On the other
hand, since a e Pin(p, q), ( 2E(a) ( = 1. These relations yield 22 = 1 whence
2= + 1 and so a = ±e.
In view of the results above we have the exact sequence of groups
1 - S° ` Pin(p, q)
O(p, q) - 1
where i denotes the inclusion map.
11.9. The Group Spin
Let Spin(p, q) denote the subgroup of Pin(p, q) consisting of the elements
which satisfy WE(a) = a. Thus,
Spin(p, q) = Pin(p, q) n C°(p, q).
Formula (11.4) shows that an element a e Pin(p, q) is contained in Spin(p, q)
if and only if det t(a) = + 1; that is, if and only if P(a) is a proper isometry.
Thus the homomorphism I restricts to a surjective homomorphism k
from Spin(p, q) to the group SO(p, q) of proper isometries. Since
S° c Spin(p, q) it follows that the kernel of this homomorphism is again S°.
Thus we have the exact sequence
1 - S° ` Spin(p, q) Z SO(p, q) - 1.
PROBLEMS
1. Show that the homomorphism E and the bilinear function QE (see Section 10.14,
Problem 3) are connected by the relation
)E(a) = QE(a, a)
a E FE.
2. Let B" be an n-dimensional Euclidean space and write Pin(n, 0) = Pin(n) and
Spin(n, 0) = Spin(n). Determine the groups Pin(n) and Spin(n) (n = 1, 2, 3) explicitly.
In particular, show that the group Spin(3) is isomorphic to the group of unit
quaternions.
270
11 Representations of Clifford Algebras
The Spin Representation
11.10
Let F be a 2n-dimensional Euclidean space. Recall that a complex structure
in F is a linear transformation J which satisfies
J 2 = - l and (Jx, J y) = (x, y)
x, y e E.
These relations imply that
j= -J
and so J is a skew transformation.
Next, consider the 2n-dimensional complex vector space E = C Q F
and define an inner product in E by
(2 O x, µ O y) = 2µ(x, y)
2, µ E C, x, y e F.
Let w denote the linear transformation of E given by
w(2 Q x) = i2 Q Jx
2 E C, x e F.
Then we have
w2(2Qx) _ (-2)Q(-x) _ 2®x
whence
cc,2=l.
Thus w is an involution. Moreover,
(w(2 O x), 2 Q x) = (i2 Q Jx, 2 Q x) = i22(Jx, x) = 0
2 E C, x e E
and so w is skew.
Let E1 and E2 denote the following subspaces of E:
E1 = {x (wx = x}
and E2 = {x (wx = -x}.
Then Proposition 10.14.2 shows that
CE
Moreover, the isomorphism CE
L(A E1).
L( A E 1) is obtained as follows: Let
(p: E - L( A E 1) be the linear map given by
gp(x)u = x 1 A u+ i(x2)u
x = x1 Q x2
xeE
x1 E E1, x2 E E2
and extend it to a homomorphism RE : CE - L( A E 1). Then this homomorphism is an isomorphism,
RE : CE 3 L(A E 1).
In particular, the representation RE is irreducible.
The Spin Representation
271
11.11. The Spin Representation
Recall that the inclusion map j : F - E induces a homomorphism jc : CF -p CE.
Thus the representation RE determines a representation RF of CF by complex
linear transformations of n E 1 given by
RF(a) = RE(1 Q a)
a e CF.
RF is called the spin representation of CF. A representation of a real algebra
in a complex vector space V is called irreducible if the only stable (complex)
subspaces are W = V and W = (0). We show that the spin representation is
irreducible. In fact, assume that W is a stable subspace of n E 1. Let b e CE and
write
b = 2Qa
AEC,aECF
(see Section 10.16). Then we have for w e W
RE(b)w = RE(2 O a)w = ARE(1 Q a)w = ),RF(a).
Since W is a complex subspace of n E 1, it follows that RE(b)w E Wand so W is
stable under RE. But RE is irreducible and so it follows that W = n E 1, or
W = (0).
11.12. The Hermitian Inner Product in A E 1
Since E = C Q F, we have the complex conjugation z H z in E given by
2 Q x H Z Q x. Now introduce a positive definite Hermitian inner product
in E by setting
(z 1, Z2)H = (z 1, z2)
z 1, z2 E E.
Then we have an induced Hermitian inner product in n E. It is given by
(z 1 n ... n Z, W 1 n ... n Wp)H
= (z1 n
n Z ,w1 n
n w,)
Proposition 11.12.1. Let x e F. Then the transformation RF(x) is Hermitian
symmetric.
PRooF. Let iH(z) denote the substitution operator in n E 1 corresponding to
the Hermitian inner product. We show that
iH(z) = 1(2)
z E E.
(11.5)
11 Representations of Clifford Algebras
272
In fact, let z, z2, ... , z p E E 1 and w1,..., w, E E 1. Then
(iH(Z)(w 1 n ... n Wp), Z2 n ... A Zp)H
= (W 1 A "A WP, Z A Z2 A ... A Zp)H
= (W 1 n ... A W p, Z A Z 2 A ... A Z p)
= (i(Z)(W 1 n ... n w y), z2 A ... A Zp)
= (i(Z) (W 1 n ... A Wp), Z2 A ... A Zp)H,
and so Formula (11.5) follows.
Next, let x e F and set
x1 =2(x+wx)=2(1ax+i®Jx)
and
x2=2(x-wx)=2(1Qx-IQJx).
These relations show that x2 = xl.
Now consider the linear transformation RF(x) of A E 1. Then, since
l(x2) = i(x 1) = lH(x 1)
(RF(x)u, V)H = (x 1 A U, V)H +
(l(x2)u, V)H
= (u, 1H(x 1)V)H + (IH(x 1)u, V)H
= (u, l(x2)v)H + (u, x 1 A V)H = (u, RF(x)v)H
and so the proposition is proved.
Corollary. If x e F, then
RF(x)v)H = (x, x)(u, V)H
(RF(x)u,
u, v e A E 1.
In particular, f x is a unit vector, then RF(x) is a unitary transformation.
PROOF. In fact, by the proposition,
(RF(x)u, RF(x)v)H = (RF(x)2u, v)H = (x, x `u, v)H
11.13. The Half Spin Representations
The spin representation restricts to a representation RF of the subalgebra
CF in A E 1. Now write
ME1 and (A E1)- =
(A E1)+ =
ME1.
p odd
p even
Then the spaces (A E 1) + and (A E 1) - are stable under the transformations
RF(a), a E CF °. Thus we have induced representations
RF : CF - L(A E 1) +
273
The Wedderburn Theorems
and
RF :C F - L(n E1)called the half spin representations.
Proposition 11.13.1. The homomorphisms RF and RF are isomorphisms. In
particular, the half spin representations are irreducible.
PROOF. First observe that the maps RF and RF are injective. Since
E 1) + = 2" -1, we have dims L(n E 1) + = 2" - 2 and so dimR L(n E 1) +
= 22n -1. On the other hand, dim CF = 22" - 1. Thus RF is an isomorphism.
The same argument applies to R.
The Wedderburn Theorems
In this paragraph we establish an algebraic structure theorem which will be used later
to study the representations of C8( - ).
11.14. Invariant Linear Maps
Let E and F be finite-dimensional vector spaces and let R be a representation
of the algebra L(E) in F. Thus R is a homomorphism
R : L(E) - L(F).
A linear map a : E - F will be called R-invariant, if it satisfies
a o P = R(qp) o a
qP E L(E).
(11.6)
The R-invariant linear maps form a subspace of L(E; F) denoted by LR(E; F).
A linear transformation ll/ of F is called R-invariant, if
R(ip) ° il/ = i/i ° R(ip)
cP E L(E).
These transformations form a subspace of L(F) denoted by LR(F).
11.15. The Isomorphism R
Let
R:LR(E;F)OE-+F
be the linear map given by
R(a Q x) = a(x)
a e LR(E; F), x e E.
(11.7)
274
11 Representations of Clifford Algebras
We show that the diagram
LR(E;F) Q E
LR(E;F)QE
R
R
F
F
commutes. In fact, let p e L(E) and a E LR(E;F). Then we have, in view of
(11.6),
(FRL(1 O P)(a O x)] =
R(co)FR(a O x)
Thus,
(Z O 4,) = R(co) ° (FR
(p E L(E).
(11.8)
11.16
Theorem 11.16.1 (Wedderburn). R is a linear isomorphism,
(FR : LR(E ; F) O E 4 F.
PROOF. We shall construct an inverse map
- LR(E; F) O E.
Choose a pair of dual bases {e*v}, {ev} (v = 1, ..., n) of E* and E and set
Pv = TE(e*µ O ev),
where TE : E* Q E 4 L(E) is the canonical isomorphism defined by (6.5).
Observe that TE satisfies
(p0 TE(x* Q x) = TE(x* Q cpx)
(p e L(E).
(11.9)
x e E, y e F.
(11.10)
Next, define linear maps Tµ:F - L(E; F) by
Tµ(y)x =
<e*v, x>R(cpv)y
v
Lemma. The maps Tµ have the following properties:
1. Tµ(y) E LR(E; F), y E F.
2. Tµ(ax) = <e*µ, x>a, a E LR(E; F).
T '(y)eµ = y, y e F.
3.
µ
PROOF. (1) It has to be shown that
T µ(y) 0 ( = R(4,) 0 T µ(y)
i/i e L(E).
The Wedderburn Theorems
275
Fix y E F. Then, by (11.10)
R(p)Tµ(y)x =
<e*v, x>R(p °
On the other hand,
T µ(y)cox =
)y
<e*v,
v
Hence we have to establish the relation
<e*v, x>R(co °
(11.11)
<e*v,
v
v
Formula (11.9) yields, for x* = e*µ and x = ev,
= TE(e*µ O coev)
P°
It follows that
<e*v, x> TE(e*µ O Peti,)
_
<e*v, x>cP °
v
v
= TE(e*µ O px) = TE (e*tL Q
<e*v, t x>ev
v
_
<e*v, (px> q
.
v
Thus,
<e*v, (Px>qv
<e*v,
v
v
Applying R to this formula we obtain (11.11).
(2) Let a E E. Then
Tµ(ocx)a =
<e*v, a>R(q )(ax)
v
_
<e*v, a>a(q,vx)
v
Since
(pox = <e*µ, x>ev,
it follows that
Tµ(ocx)a =
<e*v, a><e*µ, x>a(ev)
_ <e*µ, x> ' a(a)
whence
Tµ(ax) _ <e*µ, x>a.
276
11 Representations of Clifford Algebras
(3) In fact, since
(e*% ®eV) = 1,
TE
it follows that
T L(y)e,
Y E F.
=Y
L
This completes the proof of the lemma.
Now let
be the linear map given by
TL(y) © eµ .
11'(Y) _
L
Then we have for a e LR(E; F) and x e E
kIIbR(a Q x) = '(ax) _
TL(ax) Q eµ
L
_
<e*L, x>a Q eL = a Q x
L
whence
On the other hand, for y e F,
R '41(Y) = R TL(y) O eL =
T L(Y)(eL) = Y
whence FR qi = i. It follows that 'b is an isomorphism.
Corollary.
dim F = dim LR(E; F) dim E.
Thus dim E divides dim F.
11.17. The Isomorphism BR
Observe that the composition map
L(F) x L(E ; F) - L(E ; F)
restricts to a bilinear mapping
LR(F) x LR(E; F) - LR(E; F).
The Wedderburn Theorems
277
To simplify notation we set LR(E; F) = U. Then a linear map
8R : LR(F) - L(U)
is defined by
0R(Ii) =
°a
/ E LR(F), a e U.
Clearly,
OR(i/i2 ° 'i'1) = OR(i/i2) ° OR(i/i1)
and so 9 is an algebra homomorphism.
11.18
Theorem 11.18.1. (Wedderburn). eR is an isomorphism.
PROOF. We construct an inverse map
L(U) - LR(F)
Let y e L(U) and set
R(Y) = R ° (Y 0 1)0 1
1.
Then SPR(Y) E L(F). Since, by (11.8), R(p) ° cZR(Y) =CR(Y) ° R(ip), p e L(E),
it follows that SZR(Y) E LR(F).
Now we show that
eR ° R = l and R ° eR = 1.
In fact, let y e L(UR) and a e U. Then
[(OR ° R)(Y)]a = [OR(tR ° (Y © l) ° I ' )] a
Now fix x e E. Then; by definition of tR,
It follows that
i)(a ® x) = tR(Y(a) © x) = y(a)x.
[OR
Thus,
(eR °
Y
Y E L(U)
and so the first relation (11.12) follows.
To establish the second relation, let i/i e LR(F). Then
= R ° (0R(fr) ©l) ° R ' .
(11.12)
278
11 Representations of Clifford Algebras
From the proof of Theorem 11.16.1 we have for y e F
R 1(Y) =
TL(y) O eµ
µ
whence
(OR(l//) OX l)tR 1(Y) = L, OR(lfr)T µ(Y) O eµ .
µ
It follows that
OO
l) I
1(Y) = L,
µ(Y)eµ] =
L, T(y)e] = (Y)
µ
µ
(see the lemma in Section 11.16). We thus obtain
R ° (OR(S) ® 1)0 R 1
i.e.,
0R) = l/J
i/i e LR(F)
This completes the proof of Theorem 11.18.1.
11.19
Theorem 11.19.1. Let A be an associative algebra with unit element a and let
R be a representation of the algebra A Q L(E) in a vector space F. Then there
exists both a representation R of A in a vector space U and an isomorphism
I: U Q E 3 F which makes the diagrams
UQE
R (a)®q
UQE
1i
a E A, cP E L(E)
(11.13)
F
commute. Thus R is equivalent to R Q i where i denotes the standard representation of L(E) in E.
PROOF. Define representations P and Q of A and L(E) in F by setting
P(a) = R(a Q i)
aEA
and
Q((P) = R(e O q,)
(P E L(E).
The Wedderburn Theorems
279
Let LQ(F) denote the subspace of F which is invariant under Q. We show that
P(a) E LQ(F)
a E A.
In fact, let p e L(E). Then
P(a) ° Q(q,) = R(a O l) ° R(e O P)
= R(a O q,) = R[(e O q,) ° (a O l)]
= R(e O q,) ° R(a O l)
= Q(q,) ° P(a).
Now set U = LR(E; F). Then, by Theorem 11.16.1, there is an isomorphism
UxE +F
such that
e L(E).
1 ° (l O co) =
(11.14)
By Theorem 11.18.1 there is an algebra isomorphism
Q : L(U) 3 LQ(F).
It is defined by
Q(y) =
-'
° (y O l) °
(11.15)
y e L(U).
Thus a representation Rv of A in U is given by
Ru(a) = SZ - 'P(a)
a E A.
Relations (11.16) and (11.15) imply that
P(a) = QRu(a) =
° (Ru(a) Q t) o I
1
Thus,
° (Rv(a) Q l) = P(a) °
(11.17)
a e A.
Relations (11.14) and (11.17) yield
° (Ru(a) O 4') =
° (Ru(a) O l) ° (l O
4:))
= P(a)°t 6(i O co)
= P(a) ° Q((p) ° Y
= R(a O 4,) ° (1'.
Thus Diagram (11.13) commutes and the proof of Theorem 11.19.1 is complete.
280
11 Representations of Clifford Algebras
-
Representations of Ck( )
11.20. The Radon-Hurwitz Number
In this section we shall denote Ck( -) simply by Ck. Let R be a representation
of Ck in a Euclidean n-space IJn. We show that then k - n - 1. In fact, by
Proposition 11.3.1 we may assume that R is a (negative) orthogonal representation. Let {e1, ..., ek} be an orthonormal basis of fik and set R(e1) =
(i = 1, ..., k). Then we have the relations
6io6 + 6;o6 = - 25,
t
in particular, 6? = - i (i = 1, ..., k). Moreover, the remark preceding
Proposition 11.3.1 shows that o is skew (i = 1, ... , k). Now fix a unit vector
a e n and set a,(a) = a,(i = 1, ... , k). Then
(i= 1, ... , k)
(a, a3= (a, o 1 a) = 0
and
(at, a) = (a' a, 6; a) = (6; 6 a, o a).
It follows that
2(ai ,
a) = - ((6; o + o o )a, a) =
a) = 2
whence
(a1, a) = &,
(i, J = 1, ..., k).
Thus the vectors a, a1, ..., ak form an orthonormal (k + 1)-frame in Rn.
This implies that k + 1 - n. Thus to every n > 1 there is a largest k > 0
such that Ck represents in pn. This number is called the Radon-Hurwitz
number of n and will be denoted by K(n). It follows from the above that
K(n)
-n-
(11.18)
1.
Proposition 11.20.1. The Radon-Hurwitz number satisfies the functional
equation
n > 1.
K(16n) = K(n) + 8
PROOF. In view of Theorem 10.17.1, Section 10.21, and the table in Section
10.20, we have an isomorphism
'V: C,,8
Ck O L(U 16).
Thus, if P is a representation of Ck in lin, then
Q=(PO1)0k
is a representation of Ck + 8 in [jn a [ 16
16n It follows that
K(16n) > K(n) + 8.
Representations of Ck( -)
281
Conversely, let Q be a representation of C,, + 8 in g 16n and set R = Q o I11.
Then R is a representation of C,, Q C8 in 16n By Theorem 11.19.1 (applied
with A = C,, and E = p 16) there is a representation of C,, in a vector space
U, where
g16n
U0
It follows from this relation that dim U = n. Thus,
K(n) > K(16n) - 8.
Theorem 11.20.2. Write
a>0,0-b-3,godd.
n=
Then the Random-Hurwitz number of
Din is given by
K(n)=8a+26- 1
n> 1.
In particular, fn is odd, then K(n) = 0.
PROOF. In view of Proposition 11.20.1 we have only to show that
K(2". q) = 2b - 1
0
- b < 3, q odd.
This will be proved in Lemma II of the next section.
11.21
Lemma I. Let q be odd and 0
- b < 3. Then
K(26 . q) <7
PROOF. Assume that C,, represents in n and k > 8. Write k = l + 8, l > 0.
Then C, ^C1 Q C8 ^C1 Q L('6) represents in Rn. Thus, by Theorem
11.19.1, 16 divides n and so n cannot be of the form 2b q(0
Lemma II. Let q be odd and 0
- b - 3, q odd).
- b - 3. Then
K(26 q) = 2 b - 1.
PROOF. It has to be shown that, for odd q,
1. K(q) = 0.
2. K(2q) = 1.
3. K(4q) = 3.
4. K(8q) = 7.
(1) K(q) = 0: By Lemma I, K(q) < 7. Thus it has to be shown that if C,,
represents in D and 0
k < 7, then k = 0. By the table in Section 10.20
-
11 Representations of Clifford Algebras
282
every Ck (1 <_ k < 7) contains C as a subalgebra. Thus a representation
of Ck (1 < k < 7) in
determines a representation of C in . This is
impossible, since q is odd. It follows that q = 0.
(2) K(2q) = 1: We show that
K(2q) <_ 1.
(11.19)
Lemma I implies that K(2q) <_ 7. Now the isomorphisms in Section 10.20
show that Ck contains 0-0 as a subalgebra if 2 < k < 7 and so a representation
of Ck (2 <_ k < 7) in 2j induces a representation of 0-0 in 2j This is impossible since 2q is not divisible by 4 (see Lemma III below). Thus, k = 1
and so (11.19) is proved.
On the other hand, C1
C represents in 2j and so
K(2q) >_ 1.
It follows that K(2q) = 1.
(3) K(4q) = 3: We show first that
(11.20)
K(4q) _< 3.
By Lemma I, K(4q) <_ 7. Thus we have to show that for 4 < k < 7 the algebra
Ck does not represent in
By the isomorphisms in Section 10.20 every
such algebra is of the form
Ck = Bk O
where Bk contains 0-0 as a subalgebra. In fact,
B5 = 0-0 Q C,
B4 = 0-0,
B6 = 0-0 Q 0-0,
B7 = 0-0 Q 0-0 Q 0-0 Q 0-0.
Now let R be a representation of Ck in
Then Theorem 11.19.1 (applied
with A = Bk and E = 2) shows that there is a representation R of Bk in
a vector space U where
X44
U O 2.
Since 0-0 c Bk, R determines a representation of 0-0 in U. Thus dim U is
divisible by 4 (see Lemma III below) and hence dim
is divisible by 8.
This is impossible since q is odd, and so (11.20) follows.
On the other hand,
K(4q) _> 3.
(11.21)
In fact, write
C3 = 0-0 Q 0-0
(see Section 10.20). Then a representation R of C3 in
R(p Q+ q)x =
xE 0-0.
(0-0) is given by
283
Representations of Ck( -)
Thus
RQ...QR
4
is a representation of C3 in D" and (11.21) follows.
(4) K(8q) = 7: By Lemma I,
K(8q) < 7.
To show that
K(8q) > 7,
we construct a representation of C-, in
8q. Write
Q
C-, =
(see Section 10.20) and set
a,1 E L(8).
R(a Q 1) = a
and so
Then R is a representation of C-, in
RQ...QR
4
is a representation of C-, in 8q. Thus, K(8q) > 7.
This completes the proof of Lemma II and hence the proof of the theorem.
EXAMPLES
K(2)=21-1=1, K(4)=22-1=3
K(6)=21-1=1, K(8)=23-1=7
K(10)=2'-1=1, K(16)=8+2°-1=8.
Lemma III. Let R be a representation of 0-II
Then n is a multiple of 4.
an n-dimensional vector space E.
PROOF. Write
R(p) (x) = p. x
p e 0-0, x e E.
We shall say that a family of vectors x,, ..., xk E E generates E over 0-[l, if
every x e E can be written in the form
k
x=
i=1
pi.xi
piE 0-0.
284
11 Representations of Clifford Algebras
Let m be the least number such that E is generated by m vectors and let
X1 , ... , xm be such a family. Then it is easy to show that the relation
m
i= 1
implies that pt = 0 (i = 1, ... , m). Now choose a basis {e, a 1, e2, e3} of 0-fl.
Then it follows that the 4m vectors
(l = 1, .
xi , e 1 xi , e2 xi , e3 xi
m)
form a basis of E over B. Thus, n = 4m.
11.22. Orthogonal Multiplications
Let E and F be Euclidean spaces. An orthogonal multiplication between E
and F is a bilinear mapping E x F - F, denoted by (x, y) - x y with the
following properties :
i. Ix'YI = Ixllyl, xeE, yeF.
ii. There is an element e E E such that e y = y, y E F.
Property (i) implies, in view of the symmetry of the inner products, that
' Y x2 .y) _ (x 1, x2) I Y I2
x1, x2 EE, yeF
(x'Y1, x'Y2) = IxI2(Y1, Y2)
xeE,Y1, Y2 EF.
(x 1
and
As an example consider the algebra of quaternions (E = F ^0-[l) (see Section
7.23 of Linear Algebra).
Given an orthogonal multiplication denote the orthogonal complement
of e by E 1. Every vector a e E 1 determine a linear transformation 6a of F
given by
6a(Y) = a ' Y
Y E F.
Relations (i) and (ii) imply that this transformation satisfies
(6a Y, 6a Y) =1 a l 2 '
I Y 12
and
(6a Y' Y) = 0
a E E1, y e F.
In particular, if I aI = 1, then 6a is a rotation of F.
Representations of Ck( -)
285
11.23. Orthogonal Systems of Skew Transformations
Let F be an n-dimensional Euclidean space and let {o, ..., 6k} be a family
of skew linear transformations of F. The family {o, ..., 6k} will be called
orthogonal, if
y e F.
(ay, 6; Y) = (Y, y)
(11.22)
This relation is equivalent to the relation
0,0cr + 6jo6 =
(11.23)
In fact, (11.22) implies that
y, z e F.
(0 Y 6 j z) + (6i z, 6; Y) = 2(y, z) 5,
Since the Q are skew, it follows that
(6 j 6i Y, z) + (z, 6i 6; Y) = - 2(Y, z)
whence
(6jai -+ 6iaj)y - -2(Sijy
y e F.
Conversely, assume that (11.23) holds. Then we have
(6i Y 6 j Y) _ - (6; 6i Y, y) = 2i5r,(Y, y) + (a16; Y, y)
y e F.
= 2511(y, y) - (6; Y, o 1 Y)
It follows that
(61Y, 6; Y) = 51,(Y, y).
Every orthogonal family of k skew transformations of F determines an
orthogonal multiplication between +1 and F. In fact, choose an orthonormal basis {e, a 1, ..., ek } in
x ' Y = AY +
k+ 1 and set
A'a1(Y)
xe
1, y
e F,
where
Then, clearly, e y = y, y e F. Moreover,
xy2 =
22
1y12 + 22
21(Y, o 1 Y) +
A`A'(a. Y, ay)
= 221Y12 + aAiVtyJ2
= (22 +
2121). FyI2 =
and so we have an orthogonal multiplication k +1 x F - F.
286
11 Representations of Clifford Algebras
Conversely, let k +1 x F - F be an orthogonal multiplication. Denote
the orthogonal complement of e by k and choose an orthonormal basis
{el, ... , ek} of R. Define 6 by
a.(y) = e, y
y E F.
Then the 6 form an orthogonal system of skew transformations.
11.24. Orthogonal Multiplications and Representations of Ck
Let E x F - F be an orthogonal multiplication and denote the orthogonal
complement of e by E 1. Consider the linear map P : E 1 - L(F) given by
P(a)y = a y
a E E, y E F.
Then
(y, P(a)y) _ (e, a) (y, y) = 0
and so the transformations P(a) are skew. This implies that
(P(a)2y, z) _ - (P(a)y, P(a)z) _ - (a ' y, a ' z) _ - (a, a) (y, z)
y, z E F,
whence
P(a)2 = -(a, a). i.
Now introduce a negative definite inner product in E 1 by setting
(a, b) - _ -(a, b)
a, b e E 1.
Then the equation above reads
P(a)2 = (a, a) -
l
aEE1.
Thus P extends to a homomorphism from the Clifford algebra Ck,
(dim E 1 = k) into L(F) and so it determines a representation of Ck in F.
Conversely, let R be a representation of Ck in F. By Proposition 11.3.1
R is equivalent to a (negative) orthogonal representation P of Ck in F. P
satisfies the relations
P(a)2 = P(a2) _ (a, a) -
l = -(a, a). i
aeE1
and
(P(a)y, P(a)y) _ -(a, a) ' (y, y) _ (a, a) ' (y, y)
a e E 1, y e F.
In particular, the transformations P(a) are skew.
Now set E _ (e) Q k and define a bilinear mapping E x F - F by
setting
x'y=Ay+P(a)y,
where x e E, y e F, x= Ae + a, a e k.
Representations of Ck( -)
287
Then, clearly, e y = y. Moreover,
I x y 12 = 221 y 12 + 22(y, P(a)y) + I P(a)y 12 = 22 I y 12 + (a, a) '
=IxI21yI2
I y 12
xeE,yeF,
and so this bilinear mapping is an orthogonal multiplication between E
and F.
Thus there is a one-to-one correspondence between orthogonal multiplications E x F -* F and representations of Ck in F (dim E = k + 1).
Theorem 11.24.1. Assume that there exists an orthogonal multiplication
" x " -* Rj". Then n = 1, 2, 4, or 8.
PRo0F The orthogonal multiplication in " determines a representation of
C,_1 in . Thus, n - 1 <_ K(n). On the other hand, in view of Relation
(11.18) K(n) < n - 1. Thus,
K(n) = n - 1.
(11.24)
Now write
a>0,0<b<-3,godd
Then, by Theorem 11.20.2,
K(n) = 8a + 2b - 1
and so Equation (11.24) implies that
8a + 2b = 16a 2b q;
(11.25)
that is,
1)
0<b<-3.
It follows that
8a>16a-1.
Since 8x < 16X - 1 for x> 1, x e , we obtain a = 0. Now Formula (11.25)
shows that q = 1. Thus, n = 2b (0 -< b < 3); i.e., n = 1, 2, 4, 8.
Remark. If n = 1, 2, 4, or 8, there are indeed orthogonal multiplications
in p". For n = 1 and n = 2 we have the ordinary multiplication of real
(respectively complex) numbers, for n = 4 the algebra of quaternions and
for n = 8 the algebra of Cayley numbers (see Problem 5).
11.25. Orthonormal k-Frames on S"-1
Let S"' be the unit sphere in p". A (continuous) orthonormal k -frame on
S' is a system of k continuous maps X1: S" -1 -p satisfying
(x, X.(x)) = 0 and (X 1(x), X;(x)) =
x e S'.
288
11 Representations of Clifford Algebras
Now let 61, ..., 6k be an orthonormal system of k skew transformations
of " and set
X.(x) = 61(x)
x e S" - 1, (i = 1, ..., k).
Then we have for x e
(x, X.(x)) _ (x, ajj(x)) = 0
and
(X 1(x), X,{x)) =
Thus every orthogonal system of k skew transformations of " determines
an orthonormal k-frame on S" -' . Now, combining the results of Sections
11.23 and 11.24 and Theorem 11.20.1, we see that the (n - 1)-sphere admits
an orthogonal k-frame with k = K(n).
1-frame on S 1,
3-frame on S3,
1-frame on Ss,
7-frame on S',
Hence there is an orthonormal
1-frame on S9,
8-frame on S 15,
9-frame on 531
(see the examples in Section 11.21).
Remark. F. Adams has shown that this is in fact the best possible result;
that is, there are no orthonormal k-frames on S" - ' if k> K(n), (cf. Ann. of
Math. 75 (1962) p. 603.).
PROBLEMS
1. Suppose that an orthogonal multiplication is defined in a Euclidean space E. Denote
the orthogonal complement of e by E1. Show that the following conditions are
equivalent:
-(x, y)
1.
2. x2 =
x,yEE1.
xEE1.
2. Cross-products in Euclidean spaces. A cross-product in a Euclidean space F is a
bilinear mapping F x F -+ F, denoted by x, which satisfies the conditions:
1. (x, x x y) = 0 and (y, x x y) = 0 for all x, y E F.
2. lx x yl2 =
(x,y)2 forallx,yEF.
i. Show that a cross-product is skew-symmetric.
ii. Let E be a Euclidean space with an orthogonal multiplication which satisfies
the relation
(x y, e) _ (x, y)
x, y E E.
289
Representations of Ck( -)
Let F be the orthogonal complement of a and denote by n the projection nx =
x- (x, e)e, x e E. Define a bilinear mapping F x F-+ F by
x x y= n(x y)
x, y e F.
Show that this map is a cross-product in F.
3. The complex cross-product. Let C3 be a 3-dimensional complex vector space with a
positive Hermitian inner product (, ). Choose a normed determinant function D.
Then the complex cross-product of two vectors a and b is defined by the equation
(a x b, x) = D(a, b, x)
x e C 3.
Show that the complex cross-product has the following properties :
1. ( l a l + 2a2) x b = .Z1(a 1 x b) + Z2(a2 x b) for all ) 2 E C .
(a x b, a) = 0 and (a x b, b) = 0.
2.
3. a x b = -bxa.
4. (al x b1, a2 x b2)
_ (al, a2). (b1, b2) - (a1, b2)(b1 , a,).
5.
la x
6.
a x (b x c) _ ()b - (a, b)c.
bl2
= 1a121b12 - I(a b)12.
4. Orthogonal multi plications in C4. Let C4 be a complex 4-dimensional vector space with
a positive definite Hermitian inner product. Choose a normed determinant function
0 and a unit vector e. Let C3 denote the orthogonal complement of e. Then a normed
determinant function D is defined in C3 by the equation
.yi E C3.
D(y1, .y2, .y3) = O(e, .y1, .y2, .y3)
Define a multiplication in C4 by the equations
xl .yl = -(x1, y1)e + x1 x yl
/1,e
.y 1 - /2y 1
x1
(/1,e) _ ) x 1
()).(e)= )µ e
#
)EC
u E C.
i. Prove the formula
Ix yl2 = Ixl2. lyl2
x, y E C4.
ii. The conjugate of an element x e C4 is defined by x = a,e - x1 , where x = Ae + x 1,
E C, x 1 E C3. Show that
=x =
xeC4.
iii. Verify the formulas x y2 = (x y) y and x2 y = x (x y).
5. The algebra of Cayley numbers. Let E be an 8-dimensional Euclidean space. Choose
a complex structure J in E (that is, a skew transformation J which satisfies J2 = - i)
and use it to make E into a 4-dimensional complex space H. Define a Hermitian
inner product in H by setting
(x, y)H = (x, y) + i(x, J y)
x, y E H.
Choose a normed determinant function 0 in H.
i. Show that the multiplication defined in Problem 4 makes E into a real (nonassociative) division algebra with a as unit element. It is called the algebra of Cayley
numbers.
290
11 Representations of Clifford Algebras
ii. Verify the relations
(ax, ay) = a 2(x, y)
x, y E E
and
(xa, ya) = (x, y) I a f 2.
6. The cross-product in B. With the notation and hypotheses of Problem 5, let F
denote the (7-dimensional) orthogonal complement of a in E.
i. Prove the relation
(xy, e) = - (x, y)
x, y E F.
Conclude that the Cayley multiplication in E determines a cross-product in F.
ii. Let x E E and write x = ae + y where a E R, y E F. Show that the conjugate
element of x (see Problem 4, (ii)) is given by x = ae - y. Conclude that if x is a
non-zero Cayley number, then
x
--
x
(x, x)
iii. Show that the relation x x y = 0 holds if and only if y = fix, ,. E fly.
7. Recall from Section 10.20 that C6(-) L(B ). Construct an explicit isomorphism
b : C6(-) 3 L(B ) in the following way: Regard 6 as the underlying real vector
space of C 3 and define a negative inner product in B6 by setting (x, y) _ = -(x, y),
L(B 8) given by coa (x) = ax
where (x, y) = Re(x, y)H. Show that the map gyp:
is a Clifford map and that the homomorphism
(Cayley multiplication) a E 6, x E
extending p is an isomorphism.
b: C6 Consider the linear transformations of B8 given by
(x, e)H a
WA(x) = (e, x)H e,
w(x) = (x, e)e and
xHx
and describe the corresponding elements in C.
8. Use Problem 7 to construct explicit isomorphisms
C7(-) 3 L(P8) O L(B8)
and
C8(-)
L(B ' 6).
Index
A
C
adjoint tensor
A(E)
A.(E)
canonical element eo 238
Cayley numbers 289
178
142
146
C_ E
243
T(E)
78
characteristic coefficients 182
classical adjoint transformation 179
classical Jacobian identities 191
Clifford algebras 227f f
anticenter of 240
center of 240
complexif ication of real Clifford
algebras 251
existence of 229
uniqueness of 229
Clifford group 264
T.(E)
81
Clifford map 227
algebras
A(E)
142
146
A.(E)
C_ E
243
Cn(+) 256
C(-) 256
C (P, q)
257
S(E) 224
S.(E) 225
T(E) 82
C(+) 256
®E/M(E) 94
®E/N(E) 89
Cn(-)
alternator 85
annihilators 130, 131
anticommutative flip operator 46
anticommutative tensor products of
graded algebras 46, 120, 170
antiderivations 43, 112, 144, 258
a-antiderivations 115
of graded algebras 48
antisymmetry operator 98, 141
C (P, q)
bilinear mappings
box product 153
1, 18
257
cross-products
288
D
DE
B
256
composition algebra 33
composition product 154
contraction operator 72
161
degree involution 233
derivations 66, 70, 110, 123, 142, 258
diagonal mapping 124
diagonal subalgebra 151
divisors 127
291
Index
292
DL
174
dual spaces 31, 71, 92, 106, 123
dual differential spaces 54
dual G-graded spaces 46
dual graded differential spaces 56
is (h) 224
isomorphisms
DE
DL
161
174
248
rlE
E 194
R 273
E
194
'T'E
a 204
T 36, 169
238
rlE 248
eo
exterior algebras 103
over a direct sum 120
over a graded vector space 125
over dual spaces 106
over inner product spaces 107
exterior power
of an element 103
of a vector space 101
external product 165
F
157
TE
196, 204
T
T®
09R
SE
76
276
248
J
Jacobi identity
178
K
filtrations 128
flip-operator 42
Kiinneth formula for graded differential
spaces 55
Kunneth theorem 53
G
graded differential spaces 55
graded ideals 127
Grassmann algebra 142
Grassmann products 141
L
Lagrange identity 108, 166
Laplace formula 176
AE
H
264
M
half spin representations 272
homogeneous functions 219
M(E) 93
I
mixed exterior algebras 149
mixed tensors 71
MP (E) 91
multilinear mappings 3
i (a)
117, 214
iA (h)
i (h)
143
118
i(h)
79
intersection algebra 163
intersection product 163
invariant linear maps 273
N
N (E) 88
Nolting algebras
N( E)
84
108
Index
293
0
Spin representation
operators
1(a) 117,214
of the homology algebra 56
substitution operators 79, 143, 224
symmetric algebras 212
graded symmetric algebra over a
graded vector space 218
270
structure map 41
i (h)
iA (h)
118
143
(h)
is (h)
79
224
opposite algebra 236
orthogonal multiplications
over a direct sum 217
284, 286,
289
orthogonal systems of skew transformations 285
over dual spaces 212
symmetric functions 223
symmetric power 211
symmetric product 223
symmetrizer 92
orthogonal k-frames on S' 287
T
P
E 194, 267
P 169
196, 204
R 273
T
Pin 268
Poincare duality 157
Poincare isomorphism 159
Poincare series 44, 219
polynomial algebras 221
7'(E) 82
7''(E) 78
'T'E
194
R
Randon-Hurwitz number 280
representations 260
equivalent 260
faithful 260
irreducible 260
orthogonal 261
S
S(E) 224
S.(E) 225
SE
236
a 204
skew-Hermitian transformations 204
skew linear transformations 194
Pfaffian of 200
skew-symmetric functions 140
skew-symmetric mappings 96, 148
skew tensor products (see anticommutative tensor products)
Spin
269
7'E
157
7'.(E) 81
tensor algebras 62
graded tensor algebra over a graded
vector space 126
mixed 72
over a G-graded vector space 67
over an inner product space 75
tensor products 10, 26
existence of 9
intersections of 19
of algebra homomorphisms 42
of algebras 41
of basis vectors 17
of Clifford algebras 245
of differential algebras 57
of differential spaces 50
of direct sums 14
of dual differential spaces 54
of factor spaces 13
of G-graded vector spaces 44
of inner product spaces 33
of linear maps 21
of representations 260
of subspaces 13
uniqueness of 8
tensors 60
adjoint 178
decomposable 60, 71
invariant 75
metric 76
skew-symmetric 85, 89
Index
294
symmetric algebras 212
symmetric p-linear mappings
tensor algebras 63
symmetric 91
®E/M(E) 94
®E/N(E) 89
9R
®E 62
276
trace coefficients 186
trace form 37
twisted adjoint representation
263
w
U
Wedderburn theorems 273
universal property for
bilinear mappings 5, 6
exterior algebras 105
multilinear mappings. 5, 6
skew-symmetric maps 99
x
SE
248
210
9
780387
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