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. 242 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. 243 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. 251 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(-). 253 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). 254 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. 255 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