J-
Cocommutative Hopf Algebras
with Antipode
by
Moss Eisenberg Sweedler
B.S.,
Massachusetts Institute of Technology
(1963)
SUBMITTED IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF
PHILOSOPHY
at the
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
August, 1965
Signature of Author .
.-.
. .
Department of Mathematics, August 31, 1965
Certified by ....-..
.
Thesis Supervisqr
Accepted by ..........
0...................................
Chairman, Departmental Committee on
Graduate Students
v/I
2.
Cocommutative Hopf Algebras
19
with Antipode
by
Moss Eisenberg Sweedler
Submitted to the Department of Mathematics on August 31,
1965, in partial fulfillment of the requirements for the
degree of Doctor of Philosophy.
Abstract
In the first chapter the preliminaries of the theory
of Hopf algebras are presented. The notion and properties
of the antipode are developed. An important filtration is
induced in the Hopf algebra by its dual when the Hopf algebra is split. It is shown conilpotence and an algebraically
closed field insure a Hopf algebra is split. The monoid of
grouplike elements is studied.
In the second chapter conditions for an algebra A -which is a comodule for a Hopf algebra H --to be of the
form A 'E B ® H (linear isomorphism) are given. The dual
situation is studied. The graded Hopf algebra associated
with a split Hopf algebra decomposes in the above manner.
Chapter III contains the cohomology theory of a
commutative algebra which is a module for a cocommutative
Hopf algebra. There is extension theory and specialization
to the situation the Hopf algebra is a group algebra.
Chapter IV is dual to chapter III.
Chapter V is devoted to coconnected cocommutative Hopf
algebras, mostly in characteristic p > 0 . There, the
notion of divided powers is developed and shown to characterize the coalgebra structure of a class of Hopf algebras.
The Hopf algebras are shown to be extensions of certain sub
Hopf algebras by their primitive elements.
Thesis Supervisor:
Title:
Bertram Kostant
Professor of Mathematics
3.
Contents
Introduction
4
............................................
7
Chapter I.
Preliminaries .......................... ..
Chapter II.
Decompositions ......................... . 36
Chapter III.
Cohomology ....................----.
. 47
Chapter IV.
Cohomology ..........................
. 87
Chapter V.
Cocommutative Coconnected Hopf Algebras. . 93
Bibliography . .........................
Index ........ .........................
Biography ....
... *0..........................-.....-
......
0......
000............
162
166
I
Introduction
A Hopf algebra as considered herein is simultaneously
an algebra and a coalgebra where the algebra structure
morphisms are morphisms of the coalgebra structure, (or
vice versa).
This differs from the graded Hopf algebras
of [2] Milnor and Moore, except in characteristic 2.
The
problem is to determine the structure of cocommutative Hopf
algebras.
Our first approach lies in a cohomology theory.
We
have constructed abelian cohomology groups
Hi(HA) ,
where
H
"C.H.A."
H(CH)i
is a Hopf algebra,
H-module,
H-comodule.
C
0 < i E Z
,
A
an algebra which is a left
a coalgebra which is a right "C.H.A."
We then determine the structures of algebras
(coalgebras) which are extensions of
H
(C) by
B
(H)
This theory applies to the algebra structure and the coalgebra structure--separately--of coconnected cocommutative
Hopf algebras.
We hope to develop an extension theory
where the extension is a Hopf algebra and is an extension
of one Hopf algebra by another.
Our cohomology theory gives the familiar group cohomology in case
G .
If
Ar
H =p-(G)
the group algebra of the group
is the group of regular (invertible) elements
5.
of
A
then
Hi(H,A)
Furthermore, if
underlying field
A
Hi(G,Ar)
=
is a finite Galois extension of the
k and
G
is the Galois group of
then the isomorphism classes of extensions of
by
A
A/k
H =F(G)
form a subgroup of the Brauer Group.
Kostant has shown--the results are unpublished--that
a split cocommutative Hopf algebra with antipode is a smash
product of a group algebra and a coconnected cocommutative
Hoof algebra; and that in characteristic zero a coconnected
cocommutative Hopf algebra is a universal enveloping algebra of the Lie algebra of primitive elements.
We present
proofs of these results and study coconnected cocommutative Hopf algebras in characteristic- p > 0
.
For a cer-
tain class (including all where the restricted Lie algebra
of primitive elements is finite dimensional) of coconnected
cocommutative Hopf algebras we are able to determine the
coalgebra structure.
The coalgebra structure is described
in a generalization of the Poincare-Birkhoff-Witt theorem.
We now outline the generalization.
In characteristic zero let
for the Lie algebra
algebra,
L ,
a Hopf algebra.
I=
LC U
(x
be an ordered basis
its universal enveloping
If
i = 0,1,2,...
6.
then
dx(
=
z
i=o
x2=0n
The Poincare-Birkhoff-Witt theorem is equivalent to:
a
i)
x
-
a <--m
x
=
mi O < e
forms a basis for
a
0,
E
Z
U
In any characteristic we say
xo, xlV x2,9..... n
.
is a sequence of divided powers if
n
dx xn=Zx.®
1=0
1&
n-i
We show how ordered products of divided powers--as in
i)--
form a basis for the certain class of coconnected cocommutative Hopf algebras.
For all coconnected cocommutative Hopf algebras in
characteristic
p > 0 , we show the Hopf algebra obtained
by factoring out the ideal generated by the primitive elements is isomorphic to a sub Hopf algebra of the original
Hopf algebra, when the vector space structure on the quotient is altered.
We also show the original Hopf algebra
is an extension--as an algebra and coalgebra--of the quotient
by the primitively generated sub Hopf algebra.
Chapter I
The study of Hopf algebras is a self-dual theory.
For this reason diagram notation is useful, as it makes
dual definitions and proofs evident.
For all time we fix the field
for all vector spaces.
over
k ,
n
If
k
which is the base
X1 ,...,Xn
are vector spaces
the permutation group on
a e
n-letters
we consider
a:
Often
a
Xe
-
X
->
x1@
-0 -@
x0
->x.
will be written
i ,...1
X
(1 ,...
...
ex
. (& x,
in)
where
1,2,...n) ; in this case
X@--
111n
OX
'''
0y
... e
X
1
If
X1
X2 =
n ,
@Xn
.
n
is a left G-
module.
If
X, Y
are vector spaces
a linear map from
X
to
"f: X -Y"
means
f
is
Y
Once we define a "right" object such as module or comodule, we consider the "left" object to be defined with
the mirror definition.
Similarly for "left" objects.
8.
Algebras
(A,m,-q)
space over
k
is
an algebra (over k) where
,
m: A @ A -+ A
T:
A
is a vector
k -+ A , if
the follow-
ing diagrams are commutative.
Igm
A®A A
3 A @A
-
1S
m®I
I)
AA
II)
-m
A
A (1k
A AA
is equivalent to associativity.
II)
k
) A
A®1@A
A
m
m
I)
Ti@I
k(DA~-
is equivalent to
TI(1),
a
is an algebra where
(rI)
is a unit.
is the identity,
m
is the usual
multiplication.
An algebra
A
AO A
is commutative if
m
(2,1)
A
is commutative.
Ae A
If
then
A
is an algebra,
-+
A
x
-+
l@ f(x)
If
where
A
and
B
vector spaces
f: X ->Y
1 E A
always denotes
'QA(l)
Y
kD
@f: X
where for an algebra
X, Y
A ,
are algebras
A& B
is
an algebra
9.
(A
B) q
(A @ B)
(>
A
A
BOB
m A mB
mA® B
and
nA e B = k @ 11B
f: A ->B
A(
B
AD k .
is a morphism of algebras if the following
diagrams are commutative:
A
A
A
ff
ff
f
k
B
B
B
4
B
-$A
k
A
If
space
A
M
will often be written
is an algebra a left
with a map
?P: A®D M -.
*M
k
k
A
A-module is a vector
satisfying:
k@I
A (A
M
M
)A
M
> A (M
M
A0 M
If
M, N
phism of left
'f M
are left
A-modules
A-modules if
f: M -+ N
is a mor-
I
10.
M
M-)
A
is commutative .
jf
SI@ f
N > N
A@N
&
An augmentation
eA : A -+ k
morphism
.
A
of an algebra
EA
is an algebra
An augmented algebra is
an algebra
with a fixed augmentation.
If
A
right)
is ar augmented algebra
k
is
a left (or
A-module by:
EI
N @A M
m
3 k
k
k
-+
N
A-module and
is a left
M
If
I
-
A0 k
a right
A-module
is a vector space such that
*N
NO A(
I
M
-
(9 ?M
) NO M
-
N (AM
_+ 0
is an exact sequence of vector spaces.
If
left
is an augmented algebra,
A
A-module, then
A+ = Ker e A,
k ®AM = Coker (A+O
) A0
M
i@I
or if
A+ - M denotes
Im(A+
) A
M
M
a
M) ,
?P
M - M)
i®I
k
A M = M/(A+ - M)
Coalgebras
(C,d,E)
space over
is a coalgebra over
k ,
k
where
C
is a vector
11.
d: C
-+
k
s: C -
CtC
if the following diagrams are commutative:
k
d
C
I
II)
II)
d C SC 0C
I
C (k
I)
C
d
dO I
d
C®C
C
C O----C
C
I)
C '
C@ C
E
is equivalent to coassociativity
E
is equivalent to
k
is an augmentation of a coalgebra.
d = k® I
is a coalgebra where
and
e = the
identity.
A coalgebra is cocommutative if
C@ C
C
(2,1)
C OC
If
C
is commutative.
d
X, Y
is a coalgebra and
are vector spaces
f: X - Y , then
E
C D X
If
C, D
f:
c
C®X
k
Y
are coalgebras
-
Y
®
C @) D
aYc
is a coalgebra where
12.
dC(
dD
>C(C(DeD
CQD
2D 4)
(1,3,
dC (
D
(CO D) 0
f: C -* D
C
eD
'C
ECOD
(CO D)
is a morphism of coalgebras if the diagrams,
CO C
C
f
C
f
dD
D'-
D
k
f
f
D
D
6D
are commutative.
If
space
C
M
is a coalgebra, a right
$: M -. M @ C
with a map
C-comodule is a vector
satisfying:
M
M
$C
MO C
>MP C
%
I
>MD C
()®d
C
I (
If
M, N
are right
morphism of right
C-comodules
C-comodules if
E
f: M -* N
is a
13.
M-)
m® C
If
N
,
fI
N
is commutative.
C
A unit of a coalgebra
C
is a coalgebra morphism
k - C.
A coalgebra with unit is a coalgebra with a fixed unit.
If
left)
C
is a coalgebra with unit
If
O
is a right (or
C-comodule where
k
M DN
k
N
N
-%
d
kk
is a left
I
TIC
C-comodule
M a right
C-comodule,
is a vector space such that
-+ M 0
N
-+ M(DN
)
M
A
N
is an exact sequence of vector spaces.
If
7: C -+ J
C
is a coalgebra with unit
then
M D
k = Ker (M
-
JC = Coker
M
C
iC
'
) M®J
)
We shall freely use facts such as kernels and cokernels of comodule morphisms are subcomodules or quotient
comodules, when the "dual " fact is well known.
If
X, Y
are vector spaces
Hom(X,Y) = ff: X -+Y.
14.
Hom(X,Y)
is a vector space.
Often
Hom(X,Y)
additional structure, for example if
and
A
an algebra then
Hom(C,A)
C
carries
is a coalgebra
has a natural algebra
structure as follows
Hom(C,A)
C,A)
) Hom(C D
Hom(C,A)
®
a
Hom(CA))
where
Hom(C,A)
g) = mA o f
a(f
g
d: C -+ C & C .
Multiplication in
denoted by
thus
in
*
Hom(C,A)
f*
-+
.
Hom(C,A)
X*
g = mA o f(& g o d .
(T)A 0
-C
'
X
is a coalgebra
-c. (X
0
X)*
a as above. a is injective for if
n
SAxi*
y * s
@
,We
can assume
i=1
linearly independent.
.
X*
Consider
X*@0 X*
> =
The unit
Hom(Xk) , as usual we consider
denote
We have just shown if
is an algebra.
<*,
will often be
Hom(C, A)
*A -+
X C X**
is induced by
is
k
We let
0
and
and
Then there exists
C X
Cy
where
x1 *
x
is
C X
<yi*,y >
=
1
where
.
Thus
15.
n
i-i i
® X*)
dim(X*
a
image under
(x@ X)*
=
k
-+
E =
T:
which shows in this
We identify
(X @
in
A
is finite dimensional
X
If
.
with its
X*@ X*
is a finite dimensional algebra; then
m: A @ A -+ A
A
is finite dimensional
X
X*@ X*
Suppose
TI:
If
Yj>
= (dim X) 2 = dim(X® X)*
is an isomorphism.
a
case
Xj0
is injective.
a
which shows
Y,*)
i*
induce transpose mappings:
A* -+ k* = k , d = tim: A* -+ (A
a finite dimensional coalgebra
A*
A)*
= A*@
A*
A
is
Similarly if
is a coalgebra.
A*
With these maps
@
is an algebra; however,
this is the same algebra structure as deduced above.
If
X, Y
are vector spaces
-+
a: X* ® Y*
Y)*
(x
x E X, y E Y
<a(x* @ y*),
x® y>
=
<x*,x><y*,7>
x* E X*, y* E Y*
a
is injective.
same proof works.)
a
in
(X @ Y)*
coalgebra then
(We proved this, above, for
Identify
If
M
X*
@
is a left
Y*
Y = X , the
with its image under
C-comodule,
C
a
16.
C*@ M* ,
) (C@ M)*
IPM*
defines
left
4'M* ,
and this gives
C*-module.
algebra,
M*
Similarly if,
M a left
A
the structure of a
is a finite dimensional
A-module by
AQ M
then
t M
-M
tPM: M* -+ (A
M
M)*
comodule structure on
A*Q M*
defines a left
M*
Hopf Algebras
(H,m,i3,d,E)
is
1)
(H,m,TI)
is an algebra with augmentation
2)
(H,d,e)
is a coalgebra with unit
3)
the diagram
H
a Hopf algebra when
m
-
H
d
H
)H
d)d
HOH
is commutative.
L
H
Tj
H
mDm
H
(1,3,2,4)1 Hf
HO H@ H
e ,
17.
3)
and
"
is equivalent to
3)
and
(H,m,Tj)"
"E
is a unit for the coalgebra
"d:H -*H ® H
(H,d,E)"
is an algebra morphism".
is an augmentation for the algebra
is equivalent to
"m: H
DH
-+ H
is a coalgebra
morphism".
Since a Hopf algebra
Hom(H,H)
H
is a coalgebra and algebra
has the structure of an algebra.
algebra structure and if
H
H* has an
is finite dimensional
a coalgebra structure, as well.
In this case
H*
has
H* is a
Hopf algebra.
In
Hom(H,H)
consider
T
o s
k = k - 1 C H
is the unit.
so
e
We shall often
will denote the unit in
Hom(H,H) .
The identity
I: H -+H
Hom(H,H) .
S E Hom(H,H)
is an antipode for
I * S = E = S * I , that is
S
is an element of
H
if
is the inverse to
I.
Example:
Let
G
be a monoid (group without inverses).
the "monoid" algebra =
n
7(G)
L
Q) k - g
g E G
m
n,m
j=1
i,j=l
is an algebra where
where
F(G)
=
18.
-ri: k
F(G)
a Hopf algebra where
is
F(G)
d:
-+
g
s: F(G)
'(G) @ F(G)
-+
g@g
-
k
+1
g
r(G)
(G)
-+
has an antipode if
G
Suppose
.
and only if
G
S: F(G)
is a group define
is
F(G)
-
-1
g
then
S
is an antipode.
If
S
a group.
is an antipode,
n
e = 1 = I - S(g)
S(g) = g~
implies
gS(g)
=
,
,
hence
We now show that if
H
X g1 which
i=l
has inverses and is a group.
where
G
S(g)
d o S = (2,1) o S @ S o d
2)
S o m = m o S @ S o (2,1)
3)
E o S = S o
4)
If
=s
is commutative (as an algebra) or co-
commutative (as a coalgebra) then
We prove
1)
inverse to
52
=
S o S
=
as follows:
(2,1) o S 0 S o d e Hom(H, H@ H)
which is
d e Hom(H, HO H)
is a left
d o S ,
an algebra.
S , then
has an antipode,
1)
H
=
d
and
we show
(2,1) o S 0 S o d
d o S
is a right inverse
19.
to
d
This proves they are equal.
.
following:
f: C -+ D
If
C, D
are coalgebras,
a coalgebra morphism,
phism and
(h o g
We shall use the
g1 , g 2 E Hom(D,A)
A, B
h: A -+ B
are algebras,
an algebra mor-
then
o f) * (h o g 2 o f)
= h o (g1 * g 2 ) 0 f
The proof is clear from the definitions.
Consider
(d o S)* d = d o(S * I)= d o e = k
(d o e = d
We let
because
m's
.
d(l) = 1 0 1 ) .
mn =mo m@lo---om@
by associativity
mn
e
E
O ...
T
n -1
is independent of the order of the
Similarly we let
dn = d
T
o---o
o
d
T o d
n -1
dn
is independent of the order of the
tivity.
If
h e H
Z hi'@ h "
d's
by coassocia-
we often let
- --
hi(n+l) = dnh
Then
)
d * [(2,1) o S Q S o d](h) = Z h 'S(h
hi "S(h '")
I*s
=Z h
'S (h
'")
@
E)
=
20.
Z hi'S(hi")@ 1 = E(h)
i
1
I * S=
which verifies
For
2)
1) .
we show
m o S @ S o (2,1)
S o m
is left inverse to
m * [m a S® S o (2,1)](h&
Z h
For
and
4)
g) = Zh
") =
E(g)S(h
=
e
E = I *
For
Then
3)
a o I * S
=
'g'
H
m a S ® S
=
S(g")S(h ")
(e o I) * (E a
o E = (I o E) * (S o e)
suppose
E .
E(g)hi' S(h") =
=
S)
(g)E(h)
S
o 5
=
*E
=
o S
* S o E = S
a
is commutative so
m o S ( S o (2,1)
=
S o m
S * S2 (h) = Z S(h i)S2(h"') = z S(hi' S(hi"))
i
i
=
which shows
H
and
is right inverse.
(S o m)* M = (S * I) a m = E O m = 6
=
m
S2
is
is cocommutative
right inverse to
S
®S
o d
S * 32 (h) = Z S(hi')S 2 (hi")
i
= m o I
m o I®
S (Z [S(h)]
j
S ,
so
(2,1) o S
=
=
S o e(h) = e(h)
S2
®S
.
If
o d = d o S
S(h ')@ S(hi")
S(
i
[S(h)]J")
.
21.
I * S o S(h) = s o S(h) = s(h)
which shows
S2
is right inverse to
H
We have
of
H*
on
as an
S
H*-module
H** ; under this action
H
is a submodule.
We then concern ourselves with the action of
H**
a left
<a*, b*- h**>
h**= h e H,
Then if
1
H** D H , we define a left action
H , H*
Consider
S2
hence
on
H
H*-module by
<a*
=
H*
*
b*, h**>
dh =Z hi'
hi"
h =Z h ' <b*, hit > e H C H**
b*.
i
-
This shows
H
the action of
module.
is a submodule of
H*
on
H
.
H**
We consider
(We could have begun with
<b*, h** - a*> = <a*
module by
and explicitly gives
*
H**
H
as a left
as a right
b*, h**>
Again
H*H*H
would be a submodule where
h - a* = Z <a*, h '> h " s H
Then properties of the left representation of
H* on
H
have corresponding "mirror" properties to the right representation.)
H
is called split if as a left and right
all simple submodules are
1-dimensional.
H*-module
We give a
22.
H
criterion for
to be split.
@
g: H -+,H
Let
H
g = [(1,2) - (2,1)] o d
xi C H
Define subspaces
g
-+.
(H
g=
EZ
H
xi
Then
X
c X
E g
E®g
H
o
H3
Ker g
-
.
C -*
2
g
H-2
1
= Ker g .
x
-
--
>He H)
3
.
We call
H
U xi = H .
conilpotent if
i=1
Observe if
f = m o [(1,2) -
A
k
H*
= H
Let
(2,1)): H* &H* -+H*
kof
H*
fi =(
x
is cocommutative
H
H*
f
H*
.
1-1i
2
H*
f
-+ H*)
then
f
H*® H*
is
t
(H®(
H)*
commutative.
Im f
If
UX
= H
have proved:
Thus
C Im(tg ) c (Ker g1
then
H
Lie algebra under
f(x* ® y*)).
nXii
=[0j
0
I
so
Ohm f1
is conilpotent implies
[ ,
I .
We know if
(Where
k
=
H*
Io
.
We
is a nilpotent
[x*,y*] = x*y* - y*x*
is algebraically closed and
23.
H*
is
nilpotent as a Lie algebra then any finite dimen-
sional simple
H*-module (left or right) is
[2] , Corollary page 41.)
(Proved in Lie Algebras,
M C H
is a simple submodule then
M =H*
a*
h
=
04 h
h
where
n
Z
ni=l
h ' <a*, h
1-dimensional.
M.
">
If
M
is cyclic; i.e.,
n
dh = Z h ' h "
If
which shows
M C the space spanned by
i=l
h
n
and any cyclic--hence simple--submodule of H
1=1
is finite dimensional. Thus if k is algebraically closed
and
H
is conilpotent
Suppose
H
H
is split.
is conilpotent, we do not assume
algebraically closed, then since
H
k
is
is the union of its
cyclic submodules--hence of finite dimensional submodules-it follows, from
theorem 5 page 40,
[2]
as a left
H*-
module
H=
Ha
Ha
,
submodules;
a e 01
a* e H*
and for
as a transformation on any finite dimen-
sional submodule of
prime polynomial
Ha
,
P(a,a*)
the minimal polynomial is the
to some power.
If
H
is also
split any of the prime polynomials must be of the form
P(a,a*) = X - A(a,a*)
that within
Ha
a
A(a,a*)
E k
.
We shall show
there is precisely one simple submodule,
and a unique element
dga = ga 0
where
and
g
in this simple submodule where
X(aa*) = <a*,
g > .
24.
Assume only that
y E H
dy = ye y
where
simple submodule.
E(y)y
H
H* - y = k - y
then
E(y) = 1
Note
MC H
Suppose
is a split Hopf algebra.
y =
because
If
is a
E *
is a simple submodule, 0
1
I(y)
=
M.
y
n
dy
y'1
=
i=1
pendent;
yi" we can assume
M
is one dimensional and
dy = y'@ y" ; I
=
E
dy =
implies
s
y-
Y
if
dg
®
and
E(y')
and
:
0
f
0
.
Thus
g
E
H
is grouplike
and
E(y)
y
g
y
implies
e (y") = y
y.
E(y')y" = y
@
E(y')
.
An element
E(y)
E(y)
=
Cy
c M
E(y) = E(y')E(y")
=
E(y)
implies
0-=
e-(Y")
d
I * E
=
Similarly
E(y") 4 0 .
E *
is linearly inde-
g
We have just shown a simple submodule
.
M
of a split Hopf algebra contains a unique grouplike element
which is
grouplike
Let
y
E(y)
for any
E(g) = 1
and
G(H) = G =
0
4
y e N
H* . g
Conversely if
.
g
is
is simple.
grouplike elements of
H
.
Then
we have shown there is a bijective correspondence between
G
and the simple submodules of
logical interpretation of
morphisms from
7 (1)
where
k
to
G
H
.
We observe the homo-
is as the set of coalgebra
H , since
g s G
corresponds to
:
'y:k
g
-.
H
is a coalgebra morphism.
25.
Note for
g e G
the minimal polynomial for
H* - g
an operator on
X - <a*, g> .
is
is conilpotent as well as split (so
Ma C Ha
is simple and
dg
in
Ma
a*
restricted to
then
= ga
Ma
a*
H*
E
as
Thus when
H =
Ha)
H
if
the unique grouplike element
ga
and the minimal polynomial of
X - <a*, ga>
is
A(a,a*) = <a*, ga>
This shows
polynomial of
ga
a*
because the minimal
is also X - x(a,a*)
a_
If ~a were a second simple submodule of
H
and g
the grouplike element of Ma then similarly:
<a
which implies
ga>
restricted to
=
?(a,a*)
a
ga
H
(=<a*,
g>)
Ma = M3.
and
potent split Hopf algebra
Ma
Thus for a conil-
H
H
=
g E G
and the minimal polynomial of
sional submodule of
Hg
Again we just assume
linearly independent set.
a*
acting on a finite dimen-
is a power of
H
is split.
If
X - <a*,g>
We show
G
is a
G
is not linearly indepenn
dent there is a minimal relation Z X g = 0 , (n > 2)
1=1
where g E G are distinct
ki . By the minimality
of
n
a* E H*
g 2 'g3 ' 'n
where
are linearly independent.
<a*, g2 > = 1
,
<a*, gi> = 0
Choose
i=3,.. .n
26.
n
Then
X1 <a*, g > -
-
iA <a*,
g > =-2
1=2
n
n
0 = a* - 0 = a* -(
%g)
X
g <a*, g1 >
=
1=1i=
= -A2 g1 + XA
2 92 =
This implies
X2 = 0
minimality of
n
X2 (g2
g1 = g2
or
d(gh) = (dg)(dh) = (g ) g)(h
which contradicts the
g, h e G
Observe if
.
g1 )
-
QDh)
= gh(® gh
Thus
.
G
a monoid and since it is a linearly independent set in
the space spanned by
G
T(G)
as an algebra.
,
is
H
We consider
T(G) C H .
If
H
dS(g) = (2,1) o S 0 S o dg
gS(g)
so
S(g)
S , g e G
has an antipode
=
I
=
S(g)(D S(g)
*
S(g) = E(g) = 1
is the inverse to
g
in
Filtration of
G
H
is split.
so that
and
G
S(g) E G
;
is a group.
H
We exhibit a natural filtration on
exhaustive when
then
H
which is
The multiplication and dia-
gonal map respect the filtration as does an antipode when
present.
The filtration leads to a graded Hopf algebra.
It is also an invaluable basis for induction.
Within
H*
let
J = F(G)
.
It is a straight
forward verification that for any subcoalgebra
C C H
27.
Thus
C4
C C H @H)
(do CC®
J
is a
We assume
il=
(J1+1L
1)
H = UHi
2)
H H
3)
dH
let
c
Z
D:
then
.
.
J i+l
a**b* E J +
b*,
x>
b*
x = 0
let
x
x> = 0
<a*, b* -
=
-h = Ot
a
XEHi,b*EJi+l
<a* * b*,
implies
Hm@ H
Hi = 7h E HI
a* E H*
<a*
i=o,1,...
C H1+j
We show
for any
Define
is split and wish to verify:
H
m+n=1
C:
H* .
2-sided ideal in
H
H* .
2-sided ideal in
is a
2-sided ideal, thus
and
But
.
x>
(rheHI Ji+l - h = 0]
<b*, x> = <E * b*,
b* E Ji+1
x> = <E, b* - x> = 0
;
and we are done.
H* 0
H*
O+ z E H 0 H
is
"dense " in
there is
In fact there is
<a*® b*,
0
L
z> + 0
prcve
(H
z* E H*
a*@ b* E H*
.
H)*
in
H*
that if
where
<z*,
H* , a*, b* e H*
z> $ 0
where
Thus to
d(a*
.
h) = Z h i'
a* - h "
we need
28.
only show
d(a* - h)> = <b*
<b*@ c*,
c*, Z h'@ a* - h If.
The left hand side is
<b* * c*, a*
.
h> = <(b*
c*) * a*, h>
= <b* * (c* * a*),h> = z <b*, hi '><c* * a*,
i
= Z <b*, hi><c*, a* . h i>
i
h1 ">
which is the right hand side.
Thus if
H -+ H
?P: H*
is the left action we have
proved
H*0 H
H
-
SIQd
H
H*(
H
4I
d
(2,1,3)
H @ H*@ H
-
H @H
commutative.
To prove
n = dim(h - H*)
h = Ag
g E G
dim(h - H*)
lows:
x
= g
H
H
=
UH
showing
we proceed by induction on
h E Hn-l
? e k , whence
.
If
h e T(G) = H0
= n > 1 , choose a basis for
is split so there is
n = 1
.
h - H*
then
Suppose
as fol-
g E G , g E h - H* , let
extend to a basis x ,...Xn
n
dh = Z hi'
x
i=l
for
h - H* .
Then
~~1
29.
for suitable {h '
n
da* - h =
C H
h '0 a*
.
h) - H*) < n
By
this shows
C
H*
H*
-*
H=
(Ji+l.L
m(
,
and by induction
.
Thus
a* - h e Hn-2
h e Hn-l
1
dH
To prove
1
-
a* e J
n
=I2 h '( a* - x
i=2
x
i=l
dim((a*
If
IZ Hn
n+m=1
Hm
we deduce from:
m = tdl H* ® H*
H*
Jn
I
n+m=i+l
jm) c Ji+1
(where
J 0 = H*)
d(H )C
i+l n2
( z J
that
ilni
n=0
1+1
n
i+1-n)--
n=0
= (H
Hi)
(H@
Hi 1
+ HO0
H)
n
H) n --
(HO Hi-2 + H,
H) n H 1OH
*- (HQ H 0 + Hi
=
Hn
Hi-n
n=0
Thus
d: H
-
H@ H
is a morphism of filtered algebras
since the natural filtration on
i
(H @ H)i=
Z Hn )Hi-n
n=0
H(@ H
induced by
H
is
30.
We show
i + j
0
=
tion.
H H
C H
HO = I"(G)
,
1 + j .
by induction on
which is closed under multiplica-
Suppose the result is true for smaller
dHi C HO
Hi + H,
dH
H
d
C HO
When
Hi
+ --
- - - - - - -
i + j
+ Hi@ HO
+ H
@
H
is an algebra morphism
thus
dHi H
C H
HiH
H + H ® H0 + H1+j-1@
J - H H c Hi+j-1
This implies
2
0
and by
m: H @ H - H
Thus
C H1+j
+j-1
H
respects the
filtration.
If
S
S(H0 ) C HO
is an antipode for
by
C Hi OHO + Hi
dS(H i) C HO
H
S (Hi)
e: H -+ k
k
C
+ ---
H
so
n < i
1
+ HO
H
and induction
H + H OHO + H
J - S(H i) C Hi
This implies
c G
CHn
(2,1) o S ® S o d = d o S
If
S(G)
By induction suppose for
.
S(H n)
dH
H
) Hi
.
and by
.
is filtered by
k = kO , then
respect filtration.
TI:
k -*H
Thus we can pass to
31.
Co
(
S=gr(H) =
H0
where
i=0
H i=
H /Hi
= k
= k.
When
H
i=1,2,.
1
If
H
H
HO
which is a Hopf algebra over
has an antipode then so does
is conilpotent as well as split
submodules of
H
H
a* s H*
ensional submodule of
is a power of
Hg
H is a submodule implies
dH
J: H*® H -+ H
as a left
is
C Hg (S H
X - <a*, g>
We wish to show
.
the left action consider
H OH
He H*
H (H=
decomposes as
)
H
H
@
Hg
ing on a finite dimensional submodule of
X - <a*, g>
a*
b*
=
x
Hg
Hg
is
acta
it suffices
the minimal polynomial
is a power of
H*x
H
a* E H*
He Hg
dHg C H 0
To show
H* - dx
acting on
H* - dx = Z
If
.
x E Hg , a* e H*
to show for any
of
H(E H
H.
where
g sG
is a submodule and the minimal polynomial of
power of
Hg
T
(1,3,2)
H® H
dHg C H
H*-module by
H*
= dH*
x
X - <a*, g>
by
a* - <a*, g>
b*n
H* - dx = b*ndH*
for suitably large
n ; since
,
acting on a finite dim-
dHg C Hg @ Hg , this follows from when we show
If
H9
=
g e G
H*
The
under the left action of
minimal polynomial of any
H.
x = db*
x e H
H*
implies
x = 0
.
I
32.
b*n
x=
H*
Thus
0
for large
dHgCH9
@H
n HQ)H
For any subset
X C H
Xn=
Hg
is
a
let
n + m , HnHm h c Hgh .
H 0-= k - g , H0 h = k - h
because
c H00
dh
h
dHm c HO
+
Hn
h
@Hm
+
Ng
0+
Ho
h
h
HO
+ Hm
Clear for
Suppose the
n + m
true for smaller values of
dHn
and
X n Hn
We show by induction on
result is
= Hg@ Hg
H .
subcoalgebra of
n + m = 0
n
by induction
dH n
n
h C H gh
m
0
@H
+ H(
H gh + H gh
O
n+m-1
@H
gh
n+m-l
x e HgHmh , a* E H*
Thus if
(a* - <a*, gh>)
x e Hgh
and the minimal polynomial of
X - <a*,
a power of
ular
H1 = He
thus
He
is
gh>
.
a*
Hence
acting on
H* - x
HgHh C Hgh .
is
In partic-
is a subalgebra as well as subcoalgebra,
a sub Hopf algebra.
If
H
S , a similar induction to that showing
has an antipode
H nHmh c Hgh
-1
shows
S(HO) c H
is the antipode for
= Hg
,
thus
S(He) C He
and
SIHe
He .
As another application of the filtration we characterize
~~~~~1
33.
when elements of the algebra
(invertible); where
algebra.
If
are regular
is a split Hopf algebra and
B
an
is the group of regular elements of
f e Hom(H,B)
then
/
Br
H
Hom(H,B)
is regular if and only if
B
f(G) C Br
U4
1
=
= f(g) f-1 (g)
E(g)
hence
f(G) C Br
struct
f
1
thus
f~1
=
f * f -1()
For
H0
g e H0 = 7(G)
H0 .
is defined on
is defined on
for any
Conversely suppose
.
by defining it on
by induction.
f~
is regular with inverse
f
Suppose
Hn
a
,
then
g
E
G
f(G) C Br
and
we con-
and extending it to
f 1 (g) = (f
let
)
f~
Suppose by induction
; we wish to extend it to
En
H
.
H
Let
n
da
m
Z gio'@ aiO
=
+2
a
ai
't
+
-+
Z a, '0 a
in n
n
10 1
E,
E
H,? H
HOg Hn
where
g
I(a) = a
E*
'EG.
E
Hn
@
implies
0
m
"
= a
10=1
thus it suffices to define
If { a*
C H*
where
the right action of
a - a*
(mod Hn- 1) ;
f~ (ai O
<a*j, gi O>
on
H*
a
=1,...m
then under
=6
il0
H
O
.
(mod Hn- 1 )
H0
34.
and it suffices to define
- a* )
m
a* ) = 0i
1
d(a
0
01
= gj'Q
E *
a* jp
aj0
+ ---
X E Hn 0
aj
I(a - a* ) = a - a*
a "
=
a
* a*
+ Za i
n n
Hn- 1
For such
f~1
a + X
,
- a*j
it
ai
n
*
implies
(mod Hn-) -
Thus we have shown it suffices to define
da = go
I
g e G , X e Hn (
f
1
where
(a)
Hn-1
define
a
(a)
1
(g)[E(a) - m o f(@ f -I
=
f
=
f1 (g)[e(a)
o (d (a)
- g c a)]
- f (Xi)f 1(x
i
where
X = Z x '
i
D
x I.
f * f~1 (a) = f(g)[f~ (g)[E-(a)
- f x
I)f~ (x l?
")]
+ Z f(x ')f~l(x ")
=
Thus
f
has a right inverse.
E(a)
Similarly it
has a left
inverse and is regular.
We know that
pode.
kt
G(H)
is
a group when
The above result shows that when
H
H
has an antiis split it has
35.
an antipode if and only if
if
G(H) =1l
then
H
G(H)
is a group.
has an antipode.
In particular
36.
Chapter II
In this chapter we prove two decomposition theorems
for modules and comodules of a Hopf algebra with antipode.
The theorems are dual.
We apply the theorems to a split
Hopf algebra with antipode by means of the grading.
For a
split conilpotent Hopf algebra with antipode we find
m
H
He ®r(G) '
is a linear isomorphism.
comodule,
is a vector space which is a right
X
Suppose
(H
Let
H-comodule.
H
AX,
~
X
to make
@
X
a right
: X -+ X @ H be the comodule structure,
then the comodule structure on
#2:4,2X®
X
We use the
considered as a coalgebra).
multiplicative structure of
H-
X
0
X
is denoted
H(1,3,2,4)~XX® 0
XXO H0
HO X @.H
>130j),X0 X (DH (DH
I@ I~m
)
For an algebra
algebra we require
comodules.
A
to be a right
m: A
That means
@
X@X®H
H-comodule as an
A -+ A , k I A
are morphisms of
37.
A @ AO H
A®A
m
Im
A
I9
A@H
-
I@ k
k
k
H
A-
A
H
are commutative.
The diagrams can be rewritten:
1, 3,2, 4
A0
A
syA
H
A
H
),A
A@H
A
@H
A QH
k
A
Hence an equivalent condition for
comodule as an algebra is that
morphism.
right
@HA,
A
to be a right
$: A -+ A @ H
H-
is an algebra
Unless otherwise specified when an algebra is a
H-comodule it is as an algebra.
H
is a right
H-comodule
38.
d: H -+ H @ H
under
B = A 0H k
Let
H-comodule.
and
B
where the algebra
Since we consider
is a subalgebra of
Theorem 1:
A
(as an algebra).
H
i)
a
a: H -* A
then
In
fact
P = mo I
a~1
(
o
A
A
4:
A -+B
a,P
a-1
>A
@g
other set theoretically.
*
a;
m
A --- > A) C B.
a .
B-modules;
H-comodules.
a
By
o E = a l
(
are morphisms of left
We shall show
Tj
then
POI
are morphisms of right
Proof.
=
is a linear isomorphism.
H I
A 0 H ----
is the inverse isomorphism to
aP,P
H-comodules
a * a~1
where
0
Observe
k
is regular, i.e., there is
A
Im(.A
0H
where
E Hom(H,A)
®H
S ;
is the comodule structure),
a e Hom(H,A)
a: B
B = A
H-comodule;
is a morphism of right
a~
B = $-'(A & k)
,
is a Hopf algebra with antipode
(d: H -+ H@ H
ii)
is a right
A
an algebra which is a right
Suppose
If
k C H
A
and
i)
0
are inverse to each
~~~~~~1
39.
H
A
d
H (H
A(
-
H
is commutative.
We wish to show
rd
H
$
H 0H ---H
2,1
is commutative.
verse
$ o a~
H
7A (
$ o a
Hom(H, A(@H)
In
a@ I o d
and
H
-1
has right in-
has left inverse
o~ @ S o (2,1) o d , which establishes the commutative
P%0
al o.
diagram. Define P: A -+ A
p
== MolI
Im(.) C B
is equivalent to commutivity of
A
A
I
I
A
A@ H
A
=
$
A-3)A@ H-)
-
al
A
-+
I
k
,
A@H
m
4
A~ A---A-~~-A®H
1<
A --
yA
A@ A
H ->A
(1,3,2,4)
=
A
A
m@ m
)
AOA(@H@H
I
I@
) A(@ H ® H
AH
->
Ho@ A@ H
$®a~1
____->a s
A
0
H
---
AQ H
o (2,1)
-
A @H(@H
A OH
(1, 3, 2,4)
--------m
A --
+A
H
;> A
lI1a.
A
H
I
-
1
d2
->
A
H
H
A
H
H
H.@ H
I(@d
-1
H
-
m
I
I *
A0H
s 0 a-1
)
me I
A QH
-
)A
A0 H@HOA
m
->
A
A
H
H
s
0
> A
A®A0H®H
(1,3,2)
A
®S > A 0 H 0A 0H
(1, 4,2,3)
=A -
mA
@R H (D H (D H
(1, 3, 4,2) )A
=A
A @ A @HOH
(1, 2,4,3)
$@d
=
+
H
A ® H ® A
41.
=A --
mO k
I @ a~1
H ---->A @ A ----> A(DH
A
P
k
=A
AQH.
A
Thus we can define
image restricted to
is
P
A@H @H ~~~~~~
- B@ H
I
A® A(
---
A
I~m
H HA
A
A -
A
d
=A--A@ H>
Ia
a
A(
A
m
A
>-;
---A
)A@
4
m
A@ H
A@ A -9A
I
=
Consider
A
Pa:
In the diagram
m oi
B
H
I
B
A
-I &
Im
A
H
m
i@a
m@I
4
with its
B
A-A)H
=A-
P
as =
Now consider
A
P: A -+B
k
B
I
B SH
B
)6B (D H
I
42.
the square commutes because
phism.
4
is a left
The triangle commutes because
@o a
=
m@ I o I®@ a~I
I o $@ I o
M
I 0 I@
I=
1I
Io
~1
= a * a 1 g
= k® I
5) o a
d o $ o a
= m(@1
1 g a~O
0 I o I (d
-me01(I)a-0
Hence
B-module mor-
o a E I o d
I o d
.
Pa = IB @H
Q.E.D.
Now we consider the dual situation.
H-module
(H
considered as an algebra) where
is the structure morphism then
under
*2
2
2 : HOX@X
If
X0 X
d01II
-----------
If the coalgebra
ip: H @ X -+ X
is a left
H-module
>H®H@X@X
A
H&
is a left
X@ H @D X>X@X.
X
H-module the conditions:
d: A -+A
A
e: A -+ k
is a morphism of
is a morphism of
H-modules
H-modules
are equivalent to:
L
is a left
where
(1,3,2,4)
*:
X
H@ A - A
is
a coalgebra morphism.
43.
They both imply the same commutative diagrams.
coalgebra
A
is a left
H-module as a coalgebra if either
of the conditions is satisfied.
Within a coalgebra
subspace where
r: C -* C/I
coideal
-+
C/I
IT: C
-+
D
C
ICKerE
a
I
2-sided coideal
dICI®C + COI .
,
is the canonical map then if
C/I
r: C
m: H @ H -+ H .
H-module under
is a left
Whenever we speak of a
H-module it is always as a coalgebra.
coalgebra as a left
H
The
I
is a
If
is a
2-sided
has a natural coalgebra structure where
is a morphism of coalgebras.
Conversely if
is a surjective morphism of coalgebras then
I = Ker II
is a
2-sided coideal and the factor map
C
p
D
IT
C/I
is a morphism of coalgebras.
Within
morphism so
a
let
H
H+
H+
is a
A
since
4(H+
4
B = k OH A = A/H+ - A
If
7r: A -*B
A
E
Then
is a coalgebra
H+
H 0 A , where the coalgebra
H-module; and
coideal in
Ker E .
2-sided coideal.
2-sided coideal in
a left
denote
@
A) = H+ - A
is a
is a coalgebra morphism.
A
is
A
is
2-sided
Thus
has a quotient coalgebra structure.
A
is naturally a right
d
A(@ A
I
) A (B
B-comodule where
.
Theorem 2.
A
H
is
a coalgebra which is a left
P: A -> H
Suppose
is a morphism of left
ii)
P
is a regular element of
P: A
a
H®
A
d
Ker(A--
H-modules
Hom(A,H)
B
is a linear isomorphism.
P~0 I
*
H@ A ~~)A)~)H+ -A .
A @0A
is the factoring
d
A
A
w
a: H 0
phism to
k OH A
=
PG r
) A
In fact
So if
B
S ;
where
P
d
then
H-module;
i)
then
B
P~10 I
'o H
QA
~
-
N
A
.,"I
)A
a
B '
H@ A
.A
is
the inverse isomor-
P .
Observe
P,a
a Hopf algebra with antipode
0,a,a
are morphisms of right
are morphisms of left
B-comodules,
H-modules.
The proof is dual to that of theorem 1.
We now present applications of theorem 1.
H
is a graded Hopf algebra with antipode
S .
00
i=O
i
HiHiC
Hi+j
dHi
C
$
I=O
Hn(
Hi-n
Suppose
That is,
45.
H
S(H )C
Tk
-+H
e: H
,
are morphisms of graded vector spaces where
Then
H0
k = k0
is a sub Hopf algebra with antipode and the
Tr: H -+ H0
natural projection
is a morphism of Hopf algeH
bras, namely an algebra and coalgebra morphism.
right
H 0 -comodule under
d
<: H '-en Hn
ar: HO0 -+ H
10 T
H )
H
HO
the natural imbedding is an
morphism with inverse
S o a .
If
is a
H0- comodule
H DHk
B C H , B
then by theorem 1
i@
B
a
m
) H
H
H--
H
is a linear isomorphism.
H
In case
is a split Hopf algebra with antipode we
obtained a graded Hopf algebra with antipode,
H
=
H
= 1(G).
H HO
the subalgebra
k
Let
.
R .
B
denote
The n
m
is a linear isomorphism.
In case
antipode
H
is a split conilpotent Hopf algebra with
46.
(
=
H
g E
Hg
G
We put a F(G)-comodule structure on
b: H
-+
H-
h
H
E
H
by
He T(G)
g ,
H
H9,I $(h)
=
h@ g.
is a T~(G)-comodule as an algebra since
CHgh
H
HO
F-(G)
a
-+ H
S o a .
= He
is
a
.
The natural injection
of
~(G)-comodule morphism with inverse
Hence by theorem 1
isomorphism.
a
He @ F(G) -+ H is a linear
47.
Chapter III
Cohomology
We define abelian cohomology groups
i=0,1,2...
where
H
is a cocommutative Hopf algebra,
is a commutative algebra which is an
certain conditions.
Hi(H,B)
In case
H
H-module satisfying
is a group Hopf algebra
Hi (HB) is the familiar group cohomology
Br
H (G,Br)
is the subgroup of regular elements of
B
where
We define
.
the notion of an algebra which is an extension of
B
B
H
by
and show the isomorphism classes of extensions corres-
pond to
of
over
k
H2(H,B)
and
k
,
H
.
When
B
is a Galois field extension
the group algebra of the Galois group of
then the extensions of
elements of the Brauer Group.
H
by
B
B
correspond to
We show how to multiply ex-
tensions so that the correspondence between extensions and
H 2 (H,B)
is a group isomorphism.
In the next chapter we define cohomology groups
H(B,H)i
when
B
is a cocommutative coalgebra
mutative Hopf algebra and
B
is an
H
a com-
H-comodule satisfying
certain conditions, we outline a dual extension theory to
H(B,H) 2
the previous case and outline how
corresponds to
isomorphism classes of extensions.
With
H
a Hopf algebra
B
an algebra which is a
48.
left
H-module under
left Hopf algebra
?P: H
@
B -+ B
then
B
is called a
H-module if the following diagrams are
commutative:
H
I (Dm
B 0B
HOB
I10
He k
He B
2
?
J®I
k
m
B
B
>B
B
T1
The left diagram means
B -+ B
m: B
morphism; the right diagram means
H-module morphism.
B
H
A
Regn (C,A)
Regn(C,A)
B
0 < n E Z
For a coalgebra
C
n
n
n
Hom (C,A) = Hom(C ® '-.-C,A)
B -.B
Regn (H,B)
Homo(C,A) = Hom(k,A) = A
Now we assume
@
let
,
H-module
regular elements of
s: H
is a commutative algebra
is a multiplicative subgroup of
We let
is a left
regular (invertible) elements of
=
is a left C.H.A.
group.
B
a cocommutative Hopf algebra.
and algebra
Let
k -+ B
I:
H-module
is called commutative Hopf algebra
H-module (C.H.A. H-module) if
and
is a left
A
B
is a left C.H.A.
.
If
is an abelian
and
Rego(C,A) =
H-module where
Regn(H,B)
Define
,n: Regn(HB)
Homn(C, A)
= Ar
is the module action.
is an abelian group.
Homn(C,A).
-+ Homn+1(HB)
0 < n E Z
49.
n
f(-)
on7
o (I)
---
I 0I -.m
)
~1=1
i - 1
n - i
(-1)n+l
where
n > 0 ; for
f E Regn(HB)
h
b E Br
H
E
We shall let
6 0 b(h) = (h-b)b~
n = 0
F[a]
denote
a
F
Q)
---
unit.
In
Q)F
n > 0 ,
Regn(H,B)
e .
This is often denoted
and given
to
morphism.
6 01 (h) = (h - 1)1
h - 1 = e(h)
H.A.
Regn+1(H,B)
6n
6n
so
E(h)
is a group homo601 = E .
is one of the conditions for
B
Note
to be an
H-module, we now use the other:
6 0 (bc) (h) = (h - (bc))c~Ab~A
=
Z (hi 'I - b)b~
(hi
=Z (hi
i
For the case
,n(E)
- c)(c
1b 1
i
= [6 0 (b) * 60(c)](h)
.
n > 0
n
T
?P 0
E ()i
n
E *
E *
i=1
k
as a map from
and that
=
on
6 n()
= 6 n(f) * ,n(g)
* g)
These imply that we can consider
Regn(H, B)
the
We shall show
6 n (f
f,g e Regn(H,B)
is
E[n]
E =
E
o
[i
[n-i
E
)
50.
Note
o I
&
e = e
because
-
and
(f
(1)~
* g
o
bc(-1)us
[i-]
1
because
I
is an H.A.
H-module.
?P o I()(f * g)= (7P o I
This also guarantees
(f * g)(~l)
B
g(-l)i
*
m
I
is clear
~
(f * g)(l)
and thus
6n(f *
g) =
coogb
(-)
[1-1
p
=
*
(f(l)n+ll
I n- ]
m
is a coalgebra morphism.
E
6 n(f)
I[n-i]
[n-i] I
mi1
M0in-il
I)
1
It
(-1)n+1
n(g)
We have the sequence of abelian groups and homomorphisms
0
Reg (H,1B)
60
) Reg (H,B)
1
2
-' Reg (H,B) -
62
)
We shall now show
6 n+1 6 n
= E: Regn (H,B) -+Regn± (HB)
We do this by expanding
-0
* term *
---
41
*
term *
6 n+1 6 nf
.
In our notation
expands to
42
term * term *
The explanations of the expansions follow the computation
and a typical note is of the form:
"texplanation".
g)
because the group is abelian
m
I
f) * (p o I
4 -+ 41 * 42 * 43:
51.
6 n+1 6 n
=
(where
n > 0)
1
o 1O(p
0
f-
r
1 @f
0(1[i-
®m
[1-i])
* f (_)
*
2
/"--,
n+1
-----------------
i
n
(
o I e f
*
o(I[i-e
f
m
1=1
j=1
I[n-l)
(-1) J
f (-1)n+1
0
( 1 i]
0
m
0 1 [n+1-j])
3
"K
* (*
o I
rf
0([1-110 Me
f
1=1
(-+1) _n2
*
11
=*0 I
1 n-i3)
12
(* o
) f)
o (11-ilomer
i=1
13
*
4
I
f
-
))
52.
21
n+-1
j1:1
'i)
(*, oIOf(i
o -1 (I 0 M®@ I[n-1j])
22
-------------[1-]
*(-1)
[ -1
o (
1[J1]l(Dm®
I[n+1-j
*
I [n+l-i])
1<i<n
1<J <n+1
23
*
n+1
ir
J=1
( (_,)nl++j
o (Jl]
31
(* o Iq
11111
lli
,,,, ,,
''I'll'',
................
n
f ,n+2
r f _,i+n+20
1-=1
32
t.'-'h,~
*
I-
(-1)
q)Eo e
..
I[~l0m~~~]
.............
(
53.
111
So
121
SI
me f
T[i
o
- 1=1
(I
*
I
-L-
131
n+
211
*
212
+1
w
~1*
p o m@
o I1@ (f
~ j
o
[J-2]
®
m
[n+1-j]
*
j=2
221
it
(-
1
)
_________
o (il i-1J (t
m
j-1-2jg0 m 0 Itn+1-j])
1<i<j-2<n-1
222
f
r-l
me I
(g me I [iJ-
o (I[1
l<j<isn
223
(
1<i=J -1<n
r
j-
2]
2
0 ®m e)
[
ILIn
1
-4j)
*
)-
*
*
54.
224
-1)I
Q[j-1]
m2
y[n-j]1
l<i=j<n
231
n
(
n+1+j
j
Iin-j]))
F]
*
j=l
232
ni-1 -- n-1-
(-1)
) E0m
*
311
312
1~
( 1 )n+2
o (I
[1l]m
I[n- i))E
i=1
321
*
which cancels to
L
E
f
EO
m
as follows:
111
211
221
222
121
212
223
224
131
311
231-
312
232-
321
.
55.
In the expansion:
1-)
* o I@&f * g = (P o I
11*12*13:
f ) * (7p 0 I
g)
2--- 21*22*23:
f * g o (I[a]® mQ
@I[b])
=(
0 1[a] (
[b]) ,
(g o I [a]
(f
*
g)
1
(P o I @
(f
= f 1
* g~
= ?P o I (@ f~ 1
)~
=f-1 0 (I[a] (gm
a (I[a]0 m@ I[b]))-1
3
31*32:
m@ I[b])
(f * g)
0
E =
E))9 *
(f
(g
6)
E
0p I[b]
,
(f
= f
(p a I .f) -1 = 7 o I9
f
(f
-
o I[a]g m @ I[b]))-1
)
* g)
1
g -1
1
f-1 a (1[a] me I[b])
The rest of the expansion is clear.
In the cancellation
221)
221--
222
7brc~n2 f(_)a-c+n+2
a
because
(1[a]@m
a+b+c=n -2
0I[b]
mO I[c])
(-)a-c+n+1
222) =
o (I[a]
a+b+c=n-2
m
I[b]@mO
@I(c])
56.
The rest of the cancellation is clear.
culation we assumed
In the above cal-
n > 0 ; however, it is true
6 60
We omit the straightforward verification.
We have a complex
6)
)Reg
61
(HB)-
H, B
and
Reg(HB
which depends upon
2
62
Reg (H,B)-.
*: H(& B -+ B .
We define
the cohomology
Hn(H,B)
=
Ker 6n/Im
H (HB)
=
Ker 60
6 n-1
n > 0 .
Example
B
is a commutative algebra and
automorphisms of
extension of
B
k
B
.
and
is a left C.H.A.
tion.
Recall
G
B
is a group of
may be a Galois
the Galois group.
Let
H = F(G)
H-module under the induced representa-
Thus we have the cohomology groups
Br = fregular elements of
group and is a
sense.
For example
G
B
.
H (H,B) .
Br
is an abelian
G-module in the usual group theoretic
We show
Hn(G,Br)
=
Hn(H,B)
where the left term refers to cohomology of groups and the
right term to Hopf algebra cohomology.
We do this by show-
ing the "standard complex" in computing the group
,
57.
cohomology is isomorphic to
Reg (H,B)-)
Reg (HB).)
In the standard complex for group cohomology:
Cn =
f,# g
E
Cn
f| If: G X--X G -+Br
fg(91 ,... gn
E G X..-X
(g1 ,...gn)
=
f(g,...gn)g(g
. . .9)
G .
n: Cn -+ Cn+l
6n
l(g,'n+
n
i
'
(g2 ''.' gn+1
(-1)
T [f('1 , '''9i1l
g181+1'
1=1
gl'-''n)
]( , n+1
n
We consider
f . Cn
GX. - X G
as a basis for
extends uniquely to an element
so that
f Homn(HB)
Clearly the map
Cn
is injective.
by
G(H®@-*-OH)
Since
T
-+ Homn(H,B)
,
Regn(H,B)
= G X - XG
and
58.
C"
-+
f
-+T
Regn(HB)
is surjective hence bijective.
verification that
It is a straightforward
Cn -+ Regn(H,B)
is a group homomorphism,
hence, isomorphism and
A2
Al
>
CO
Reg 0 (HB)
2
C1C
4
) Reg (H,B)
60
Reg 2 (HB)
61
is an isomorphism of complexes.
)62
Hn(H,B) = Hn(G,Br)
Hence
A Normal Complex
Within
Regn(H,B)
we define for
n > 0
f
H@Reg (H,B) =
@H-
)B
f e Regn(HB)I
k
is commutative
=
Let
I
61n
=6 n|
f
E Regn(H,B)I
f(1@
- --
1)
= 1]
Reg n(H,B): Reg n(HB) -+Regn+l(HB)
.
59.
6
nf(1 l[n])
= 1
so
-+ Regln+1 (H,B)
6 n: Reg n(H,B)
We define
6 0 b (1)
Reg
= (1
(H, B)
6 so0
andb 6
= B
so
- b)b~1 = 1 , b 6 B r
6 0: Reg,
-+
(HB)
Reg
(H,B)
.
To show the cohomology obtained from the normal complex is
the same as the previous cohomology we define maps between
the two complexes where the compositions are chain homotopic to the identities.
n: Reg, (H,B)
-+
Regn(H,B)
the natural inclusion is a morphism of complexes, i.e.,
0o
n
n+1 o 6 n
Given
f e Reg(HB)
7% = f(l
chapter I
by
- -- @l) E Br
Note
- NfN
.f*gLet
.
denote the map
h 0
-- .
hn
Nf E (hl)
Then
Regn(HB)
f
-+
-+
Regn(HB)
\
--. e (hn )
Af
6o.
s a homomorphism; as is
rn: Regn(H,B)
f
{
6nf
-.
Reg n(H,B)
-+
f *
= f *
1
n
is
even
n
is
odd
n
even
.
so
F-
A 6 nf =
Af~nnodd ,
E
n
even
6=
f f
n
odd
6 nf
This implies
7
Io r
6n
*
'1
(Aj
6nf
6n
-+ Regln(HB)
)
n
odd,
n even
1
-(A
EF) n odd
is a morphism of complexes.
n: Reg 1 n(H,B)
rn
C
{
6 n n
and that
Hence,
rn+1
n even
Clearly
is the identity.
is chain homotopic to the identity as f ollows:
61.
v n: Regn(H,B)
vn(f)(h0 ...
--
f *(n
Thus
=
6 o v * vo
*
Regn-1(H,B)
=
h
n
6 n-lV (f)
-.
Af(ho
(
- - - Qhn-1
f *Af*r-l
)
Af .
f
vn+16n(f)
n
=I*
QD
r)~
and
rw
is chain homo-
topic to the identity, and the two complexes give the same
cohomology.
Extensions
H
is a Hopf algebra
B
an algebra and
H-comodule as an algebra,
T)
which is an
the structure.
The sequence
B
->
A
4:
an algebra
A - A () H
A
-
A
H
being
is called
left exact if
is an injective algebra morphism,
1)
n
2)
Im T = A []H k .
The sequence is called split exact if:
the sequence is left exact, and there is
H
has an antipode,
a: H -. A
fying
is a morphism of right
1)
a
2)
a s Reg 1(HA) .
H-comodules
If the sequence is split exact, by theorem 1,
satis-
62.
Bo
B0
H-m
H-)
A@A
is an isomorphism of left
We further suppose
ture of a C.H.A.
H
B
)A
B-modules and right
?P: H 8 B -+ B
H-module.
A
gives
H-comodules.
B
the struc-
is called an extension of
when
A
AOH
1)
B
2)
The following diagram is commutative,
A @B
-.
--
>
-
A
is a split exact sequence,
>A
H(B
B
(2,1)
B
A
A@ I
I/
M
m
A
A
A, A '
Given extensions
of
H
by
B , where the split
exact sequences are
B
B
TB
+ A
+A
-+ AI' -+ A'
Q H
H ,
we say they are isomorphic if there is an algebra morphism
'y: A -+ A' , such that
63.
}-A Q. H
A -B
B
1@
A'
A
is commutative.
B-module morphism and right
y is a left
In this case
H-comodule morphism.
If
a: H -+ A
'ya: H - A' has the same
H-comodules and is regular then
properties.
is a morphism of right
Then the diagram
mo@
BH
mo
a
®'Ya
A?
is commutative; the horizontal and diagonal maps are isomorphisms; hence,
y
is an algebra isomorphism.
Extensions
being isomorphic is an equivalence relation and we can form
the isomorphism classes of extensions.
Examples of Extensions, the Smash Product
We now define the smash product for an algebra which
is a left H.A.
H-module.
When the algebra is a left C.H.A.
H-module the smash product is an extension whose isomorphism
class corresponds to the identity.
In a sense all other
64.
extensions are isomorphic to a smash product "altered" by
a
2-cocycle.
4:
H @ B -+ B
H-module.
Map
gives
B
Hom(B @ H, B@ H) , we define the
B&@H
map by showing how
(B @ H) 0(B @ H)
the structure of a left H.A.
B @ H
acts--from the left--on
BH@HB
I-d
B 0 H
H
(1,2,4, 3,5)
? BOH@B@H@H
I@@m
m@I
:
If
bs
'F()@H
B@B@H
.
h, e e H
e B
p(b 0h)
(3
@
7)=-b(h '
i
p
is injective since
=
Hom(B
H ,
)h
p(b®0 h)(1® 1)
"
=
b@ h
c0 H)
.End(B
has a usual algebra structure
B)H)
Im p
by means of composition,
@
is a subalgebra of End(B @ H)
.
We defp(Z bhi' ' 13h")
i
ine an associative algebra structure on B 0 H by insisting
since
p
p(b
h)p(P
=
be an algebra morphism.
be denoted
B V H.
bE b
with this structure will
The above formula shows:
E
h,.p9
(b
BQ H
@
h)(
B
()
E
@
H
=VZ
h
-0
0 (b
h
.
65.
If
B
H
right
B
is a C.H.A.
H-module and
is an extension of
H-comodule
B
H
H
B.
E
h -+
b
has an antipode,
Since
is a right
E
b
by
H
H
is a
H-comodule where
h1 "1
h1 '
E
$
is an algebra morphism so
®H
B
is
a right
H-comodule
--as an algebra.
1(
e
k
r: B
B
H
H
E
is an injective algebra morphism and
Im rj =(B
H)O
k
so
B
-
B
H
-*
B
H0H
is a left exact sequence which is split exact because
6
a=k@I:H
is a morphism of right
S
the antipode of
H .
B ® H
H-comodules and has inverse
With
b, 0 s B , h E H
k
S
,
I
66.
C3IQ
Ir
1I
A"il
d
I
8H
e
E
i
Equality holds because
commutative.
B
is commutative and
H
is co-
This is the second condition required of
extensions.
We have really proved a theorem of Kostant's about the
smash product when at the end of chapter II we showed if
H
were a split conilpotent Hopf algebra with antipode
then
He
rT(G)
is a linear isomorphism.
algebra, if' F(G)
acts on
-+
H
r(G)
He
is a group algebra Hopf
by
g - h = ghg~A , g E
G
67.
h s H
then
H
is a left H.A.
F(G)-module.
We can now
say
M
He & f(G)
is an algebra isomorphism.
In this case the smash product
has further structure.
is a Hopf algebra as is
He
1 ,
6
Hee
1
f(G)
He
I'(G)
6
generate
is a coalgebra.
e
E
m: H
follows because
@
He
e
,
so that
That it is a Hopf algebra
7~(G) -+ H
r
is a coalgebra isomor-
phism as well as algebra isomorphism.
He
F~(G)
r(G)
Thus
m
F(G)
-+ H
is an isomorphism of Hopf algebras.
We generalize the smash product as follows; suppose
B
is
and
a C.H.A.
H
H-module
*:
has an antipode.
B
H @ B -+ B
H
,
f E Hom(H (
is the vector space with
multiplication
1)
(B
f
0
H)
(B
f
0
Ic
d2
Iod
H)
1,2,5,3,6,4,7
B(
H
H(
H
B
H@H
Be
H
He
H
I~pf
B@H
M2
)
B
H ,
or
H
H, B)
q)m
)
B®B©B©H
,
68.
f
f
(b
h)
=
f
B 0
Lemma 1):
10k
TI: B
H
morphism of right
right
and
H
'
(
is an extension of
f
B
-
f (h "
bh '
k®I
a: H -
,
H
H
is a
Proof:
B
with
0
H, H
a regular
have the usual
H-comodule structures) if and only if
f
by
.
f
B
f
(B
H-comodules
h '
f E Reg 2(HB)
2-cocycle.
Suppose
2
f E Reg
(H,B)
is a
2-cocycle
then
7P o If
2)
* f
* f
o m®I
m * f 1
o I
= F0
3)
f * f
7P o I
o m@I| HZ k@ k ,
considered as maps of
inverse maps.
2)
to
f o I(m|
H
to
B
,
E
e
or
e * f o m@I
H@ k
k
.
can be
and as such they are
Thus applying both sides of the formula
h @l0 1l
f(h @ 1) = E(h) .
Let
m = f®
o I
®
, gives
Similarly
p: B0H -+End(B
action of
B@ H
removed).
p
f~ (h @ 1)
on
B
@
E(h)
be defined by the left
given in
1)
is injective since
p(b 0h)(1@ 1)
thus
f(l 0 h) = E(h) .
@H)
H
=
=
b@ h
.
(with the
f's
69.
p(b(2
h)p(p@ ()c
g)
=
p(b(
h)( zj,k
gk
4
' -jc fV
jfl gkf)
= i,
k
b (h I - [0(t i 'I - cMf((
"
gk' )])f (h, f(
ItI
=
J
1, j, k
b(h 'I - P) (h "( If'I - c)(h
- f(
"
gk
_
_
h
5
t1
91r
(4)
))f(h('"g
__
_(
(4)
(by formula 3))
=
J
1,j,k
b (h 'I - 0) (h "( fi 'I - c)f(h
ff(h ( 4e j
h
b (h I - P)f(h
= 1, ,k
(4)
- c)f(h
j)(hi
gk
(
k
I@gk
h(5) 1(4) k
=
p (zI
b(h
This shows
' - P)f (h " It
Im p
'i) 0h
is a subalgebra of
we transport the structure of
I)(c 0
1 1I
Im p
End(B
to
B
g)
@
@
H)
H
and if
we have
I )
)
70.
f
if
B
H , thus
f(l
h)
B
H
E(h) = f(h
is associative.
f
implies
@
1) =
f
1
l
1
is the unit for
@
B
H
and
if
B
@
is an algebra.
10k
f
H
TI: B
phism.
B
-
B
T(B) ={(B
H
®
is an injective algebra mor-
H
is a right
H-comodule as an algebra.
f
H)
k .
B --- )
B
H -
H
I
k
is a left exact sequence.
a: H
H-comodule morphism with inverse
*
H
fi
-
) B
a~
-
0
H
is a right
o (SDI) o d]
I o f~
a o S,
a1
* a
o f~ 1 o (I@
=
a o S *
-
[ o f 1 o (S @I) o d] *
a o S * a(h) = Z[(l
[1
if
S(hi')][(l
S) o d] ,
because
[a o S * a]
if
h "]
(dS
=
S@ S o d
i
H
cocommutative)
=
Z
=
Z
to
a o S *a
T of
1
'
h
f(S(h
')
h ''
")
S(h'
)h( 4
i
= TI
Thus
f(S(h
=
of 0 (S
o f
has right inverse
I) o d (h)
o (S
o (S@I) o d
1
@ 1)
and
o d
a1
*as
which is inverse
.
Similarly
a
71.
a o S *
which shows
[1
o f~
o (I@
S)
o d]
Thus
is regular and the two inverses equal.
a
TI f
-a BQ
B
f
B (@ HOH
and
a
is a split exact sequence and
is .a regular right
H-
comodule morphism.
f
- v1:4
e
beh' he
f -I it
.
I
(3, I~2
~.)
01' hF,
Y
I
?lI
f
,pelb
+
07
'P 0
e.0
m
_-
b(h()5)e A~
I
I
Thus
f
B 0®H
is an extension of
H
by
/
B .
f
Conversely, suppose
B
I
k
with ra: B--)k (9I
f
a: H-)
B 0 H
B
f
B @ H
is an extension of
H
by
an algebra morphism
a regular morphism of right
H-comodules.
72.
Observe
ma
B
phisms from
B
@
H
H
@
a
H
to
B
®H, (in fact
f
is
a*a
Since
m
is
is a coalgebra
o m =
T
o f
invertible where
a o m * m o (2,1)
Hence,
B 0 H
invertible we obtain
m o a
This shows
as mor-
H
"altered" by this formula).
morphism and
f E Reg 2(H,B)
o a
is an algebra morphism implies
T
.
= q o f~
a
f
f
1
so
a = n o f * a o m
o a
f
is the unit of
1
f E Reg 2(HB)
f
((1~ h)1J))(
.
f
B
0
H
which implies
1) = 1
f(l
Finally by associativity
f
g)
f
@h) ((1
= (1
f
f
)(1
g)
implies
if
'j)f(h
iZ
i,
j,k f(h '10
1
=
t
Taking
(f
since
(hi'
e
8)
f
*
@hi '"ttj "
I1 gk')
"("Ot
f(%'@ gk'))f(h"
gkI
fh
% "gk")
"
of both sides gives
(f o m @ I)
= (7p o If)
is regular this says
f
(f o I @ m),
*
is a
2-cocycle.
Q.E.D.
When
f = E
@
E
f
E = EH
H , then
B
@
H
is
B
@ H
gk
73.
the smash product.
Lemma 2):
Suppose
then the extensions
e, f
and only if
Proof:
B e H ,
B
e
S
'y: B (DH
-B
are cocycles,
are isomorphic if
Reg 2(H,B) .
0
H
is an isomorphism
Then
e
k
I
H
are homologous in
Suppose
of extensions.
e, f E Reg 2(H,B)
B
I@
e
B
H@H
$e
BIe
ry
Ik
B
is commutative and
d
H
f(
I
H
d
B
( HH
By
y is an algebra morphism.
theorem 1 chapter II
C
H
B
H
Im(B
Ip
f
f
il(B)
f
P: B ®DH - B
H
a
-?
f
H (B
so we consider
where
0 1 D a
T1fo P = m
B (& H
f
7
e
CF
We define
f
B
v: H --
B
--
and wish to show commutivity of:
BHH
o
0
P
---- > B
.
m
f
H -- > B 0H)
74.
e
4)
B @&
IH $,
H
If
B
H®H
IT1 f 0 V,
0®2
f
B@H @B
f
B
@
I
(3,4,1,2)
f
HOBOH
fV
m
B
5)
that is
f
e
0h)
TY(b
o v(h "))(b
i
By definition
(TV ,
B
B
0 I®@ f ~1
a -1 0
H
e
1
0 h ') 0 ~-1 (h ")) .
,If o v(h)
right side of
o y o ae
so
Inserting this in
and using cocommutivity gives
5)
f
e
z
I] o d
[o( o ae)
H
6)
h I)
comodule morphisms)
CT
f
m
o I@
V
0
Ti
&)H
a f-(h
f(1
")(b (Z
e
=
T(1
y
h
()
)aCr
1
(hi ") (b
® 1)Of (hi)
In theorem 1 we showed
b
o a
=S
(2,1) o d
applying this and the second condition for extensions,
(and cocommutivity):
the
I
75.
a
(h)(b
01)
= I
(S(h ')
b @ 1)a f(h i)
i
Inserting this in the right side of
e
- b)
b 0 1)a -f
hD ')(S(h")
i y(l
e
=
6)
yields
"' )af(hi(4
)
if
h')(S(h ") - b Q 1) E (h??t
Z Ty(1
e
7)
=Z'T(1l
-b
(h ' )(S(h")
0
1)
1
Again using the second condition for extensions and that
T
7)
is a comodule morphism,
=
I
([h i ' - (S(h "') b)]
((h i
f
b C) 1)y(1
S(h it ))
f
(E(h1 ')
b
®
f
q) 1)y(l
1)'y(l
e
Q)h
"')
e
&h "f )
e
0 hj)
i
f
=
(b
0
1)T(1
e
e
0 h) = 'y(b D h).
The last equality, because
4)
Thus we have established
Applying
Ty(1
5)
ly is
to the fact
e
0 h)y(1
a left
and
"ny
e
(e)
5)
.
is an algebra morphism":
e
e
=Y ((1
(@ h) (1
@(C))
" (V(
')
if
Lef t side
z(hi'
(V
Oh
1
B-module morphism.
f
"
76.
8)
=- Sv(h ')(h1 "
W~
Right side = ry(
9)
e(h
@
')
"'® ")@ h1 (4)
h
(
) e (h "
v(h i'
Z
=
V(%'))f(h
.
h
")
41
Taking
I
@
E
of
6)
E
*
(P
(v
10)
8)
O
o v = m
Since
V
it follows
and equating gives:
I
V) * f = (v o m) * e .
6
0
('y
=
9)
,
a)
af
I
o [(y
=
f *
o a ~ ) .
10)
61 v
so
Conversely suppose
let
61
= s * f
e
'y: B
T H
1
e
e
v E Reg
,
= 1
1
implies
and
Reg, (HB).
E
e * f
=
af (1)
v(l) = 1
thus
1
1
v
By
f
11 ; and
ae(1) = 1
=
o d
is regular where
e
af(1)
)I]
f
*
Tif 0 V 1
Moreover
a e)
and
and
1
f
f
are homologous.
are homologous
(HB) .
Define
f
-
B
H
in
terms of
v
e
'y(1
1)
=
v(1) 1
by
4)
f
@ 1 = 1
f
0
l
.
m
77.
e
e
'y(b 0h)y(p 0( U
z
b(h1
')P O0%
ci
"i
)b(h "1
=, ci v(h '
=
f
f'
h11) (V(
h
Z (v (hi1')b
- ((
')p))f (h "'@tv
'I
v (e
")
(if.)
'))f(h
=
2
10)
b(h ' - p)v(h fi)
.9ci
=
TI)
3
(3)
hf 5
6 1 v = e ef~
(3)
h
I,i
(since
it)
is satisfied)
e (h "'f (
T(b(h ' - Ij e (h " (
')
Oh
(4).
(3)
ci
)
)
"t
1
i,ci
e
y([b
=
So
it
e
0 h][P @ t ]) .
ry is an algebra morphism.
is
a left
phism; hence,
From the definition of
B-module morphism and right
T
H-comodule mor-
'y is an isomorphism of extensions.
Q.E.D.
Lemma
of the form
Proof:
3)
Each extension is isomorphic to an extension
f
B ()H.
Let the extension be
sequence be
Ti
A
,
the split exact
78.
a: H
-
A
be a regular morphism of right
H-comodules.
Consider
(m o a@ a) * (a
1
o m): H
H -. A
The diagram
(m o a
a)
*
(a
o m)
H0
A
(m o a @ a) * (a 1 o m)
A (H
A
I (
k
is commutative; hence,
Im((m o a 0 a) * (a~ 1 0 m))
C Tj(B)
f: H ® H -+B where
and there is
T
o f
(m o a
=
a) * (a~
o m)
By theorem 1
f
3 0a
a: B @ H-A
is an isomorphism of left
m
(aA-->A
B-modules and right
H-comodules
so the diagram
f
f
H
BB
a
B
T1
is commutative.
A
In particular
B
H
H
Ian
I
4-
---+ A (DH
,
w
79.
if
1B
k
I
A
11
is commutative.
a
A
Since
a
is bijective if we show it is
if
implies B ( H is an algebra and an
multiplicative, it
extension; and it implies
a
is an isomorphism of exten-
sions.
f
()
f
®
a(b
h)a(
= r(b)a(h)TI(0)a( )
(by second extension condition and
=
i
i
i(b(h
f
,o
TI(b(h ' -*
=
a comodule morphism)
- @))a(h ")(
f
(by definition of
a
*a o m =m o a
) (h "
')a(h
a
til
if
a(b(h.
-Z
13*
h
1,j
f
=a[(b
f
h)(0
t )].
Q.E.D.
Theorem
3:
When
H
has an antipode there is a
natural correspondence between the isomorphism classes of
extensions and
e s H (H,B)
H (HB) .
Under this correspondence
corresponds to the smash product
B 0 H
8o.
By lemmas 1, 2 there is a natural bijective
Proof:
correspondence between
H2(H,B)
and those isomorphism
f
classes containing an extension of the form B Q H , by
lemma 3 all isomorphism classes are included.
Since the smash product is
B
@
H
the last
statement holds.
Q.E.D.
Suppose
H
=
7(G)
B
where
is a finite Galois extension of
G
is the Galois group.
B
k
and
is an H.A.
H-module where the structure is induced by G . Then exf
tensions of the form B (D H are central simple algebras
over
k
.
This is treated in
[1] ,
Rings with Minimum
Condition, under the name "crossed products".
Their
notion of "factor set" is precisely our 2-cocycle. Since
f
B @ H is central simple, and since the extensions equif
valent to B ® H are isomorphic as algebras; they are
algebras in the same class--element--in the Brauer Group.
That distinct classes of extensions correspond to distinct
elements of the Brauer Group follows from their Corollary
8.50 which says the classes of extensions correspond to
those elements of the Brauer Group with splitting field
The smash product -B
B
E
®H
is a total matrix algebra (over
k), this is their Corollary 8.4G.
In the Brauer Group the multiplication corresponds
to the tensor product of algebras, we give a method for
81.
forming the product of two extensions so the correspondence
3
of theorem
Let
is a group isomorphism.
A, A
be extensions of
H
by
B
where the split
exact sequences are:
1
-+
B
A
A@H
-A
T1
A(
B
In
4®I) -
T = Ker(((l,3,2) o
let
A®A
H
a subalgebra of
A
®
A
is
since it is the kernel of the dif-
ference of two algebra morphisms.
T
T
The defining property of
is expressed in the commutivity of:
LQ
()AA
T
A®
A
b@I
-A.
A®H@A
AH
(1,3,2)
let
Im
~
ST
T C T
T @ H
(1,3,2)
-+A (D A
(T
H
=
0
(1,9 3,2,v4) o
b®~
I0$
so that we consider
Ker([((1,3,2)
1I - I
0 1Ie@ I o
Next we show
I A OA OH).
o
T I
I)
$T: T
I
T
T@H
I),
oST
0$
ST-I(
-+
0
1
$T
82.
(1,3,2,1) o
501
®I I 0 (1,3,2) o
$
I @®I
d 0 (1,3,2) 0
4
10I d)[(1,93,2)
(I0
which is zero applied to
bT:
and it
clear
is
T
T
-+
T
is an
@ I - I () I
0
these as maps to
H
: B
A@A
I C T
it is clear
I
in
We propose
T
Im(ri0'k) C T ~ Im(k @
T .
)
a
2-
we consider
> TOH08
q0k
I
I
is
The diagram
TOB
TOT f
d oI0
-
to show the right ideal generated by
Note
I1o
H-comodule (as an algebra).
I = Im(n @ k - k 0 ')
sided ideal.
01@
-
T , thus
gk -k
If
I
>AHOA0B
A-A A A.,
ieI
0I
|{2,
, 1,3)
'
H 80A OA
0arl
8 (R ffA'vX
(1,3,1)
'
n*T
iL
H0OT
11.
IV
~
^10
A0A@A AA
A eAe&
momm
m~T
MTmi
TT
T~T
TO&T
83.
is commutative where the inner pentagon commutes by the
L~ J
second extension condition, and the
commute from the definitions.
Similarly the diagram:
I
T®B
(TD
I
T
@ (k
T
B
1%
0
(2,3,1)
TO
H
T
H
B
B
T
T
(k@ 7)@ I
I
T®
(D
mT
is commutative.
$T Q)
B
T
Thus
To
I@(Tk-k
H
T 0T
HB
I
B @T
9L
mT
B
(2,3,1)
T
(Ti @k-kOTI)q® I
T
T
mT
is commutative which shows
TI C IT
and the right ideal
84.
is
2-sided.
morphism so
D
=
D
I
is a sub comodule and
IT
H-comodule (as an algebra).
I ,
T
-
T
D
7D = (B
is commutative, let
Since
A
: B -+ D
is an extension we have
a: H
-+
A
a regu-
P: A
-+
B
where
A
(See theorem 1.)
A
H
a
we have
H-comodules and
I0 a-l
A
$
-*
D
H k
lar morphism of right
7 o P = A
T-* D) .
-
Im 11D C D
is an algebra morphism, and
A
Let
From the defi-
be the comodule structure.
D -+ D 0 H
B
For
is an algebra
is a sub comodule and the algebra
T/IT is an
nition of
ST
P
,
.
P,
P
m
A
-+
.
are morphisms of left
B-modules.
a@ o d
Consider
lies in
H
)AD
@A ,
T , let
d
aD
the image of this map
a @Na
is an invertible morphism of right
H-comodules where
the inverse is
-1
aD
Then we have
a1
d
:H
)4HH
P : D - D QH k
where
"-1
T--D
85.
m
D@(D ->D
H
PD=D-)D
D
D
kE.
D
The diagram:
P
P
I
E
T -- >D
B
IPD
T
)D DH
It can be rewritten
is commutative.
) D
T
m oP
k
P
PD
H k.
JD
B
TID
H
Since
Im P
Im TD = D
are left
DH k ,
D
H~~
Clearly
EH k.
B-module morphisms
is a factoring
Q
m oP @ P
is surjective, it follows
I C Ker m o P (FP,
IT C Ker m o P
,
T
Thus
T -+ D
D
@
since
P ,
P,
P
or there
86.
(R@k
B
)
-
T
D
)
B
is commutative.
have shown
The horizontal composite is
is
TrD
injective.
11D
B
VD
hence we
Thus
$D
D
D
-
)H
is a split exact sequence.
The second condition for extensions is satisfied.
Thus
D
If
is an extension of
y: A -+ A, ,
H
by
y: A -+ A1
B
are equivalences of exten-
sions then:
,y() Y: A®() A
y
y@ ~y|T
factors to
of extensions.
~yIT: T
A
-T
yD: D-+ D
and
yD
is
an equivalence
Thus the method of producting extensions
gives the isomorphism classes of extensions a product struc-
f
ture.
Direct calculation shows the product of
f
f
B
H
is extension isomorphic to
B
B
H ,
f'
H ,
so that
under the correspondence of theorem 3, the multiplicative
structures correspond.
This shows the isomorphism classes
form a group and the correspondence is a group isomorphism.
pop--
87.
Chapter IV
Cohomology
We treat the dual situation to the previous chapter.
We outline the definition of abelian cohomology groups
H(BH)i
i=0,1,2... , when
algebra,
B
H
is a commutative Hopf
a cocommutative coalgebra which is a right
H-comodule satisfying certain conditions.
We also outline
the dual extension theory, defining when a. coalgebra is an
extension of
B
by
H
and when extensions are isomorphic.
The isomorphism classes of extension again correspond to
H(O2 , we define the dual extension to B f) H of the
H(B,H)
previous chapter; this indicates the correspondence beH(B,H) 2
tween
B
is
is
a right
and the isomorphism classes of extensions.
42.
and
d: B
-BO)B
s: B
->k ,
B
5: B
B ®B
H.A.
is a right C.H.A.
H-comodule,
is a right H.A.
H
B .
is a right
B
are comodule morphisms.
B
a Hopf algebra
H-comodule structure for
H-comodule under
if
H
a coalgebra,
B
-* B(
H
is an
H-comodule if
H-comodule
is cocommutative
is commutative.
For a coalgebra
C
and an algebra
0 < n e Z
let
88.
Hom(C,A)n = Hom(C,A En)
Then
C.H.A.
f
Reg(B,H)n
B
Let
B
B
is a
Reg(B,H)O = Reg(B,k).
be a right C.H.A.
Reg(BH)
E
Reg(C,A)n = (Hom(C,A)n)r
is an abelian group when
H-comodule.
Let
Let
.
H-comodule.
For
consider
4k@H=
>B@H
H.
Define
60
Reg(B,H)
=
6 0f
we are considering
-+ Reg(B,H) 1
f: B
-+
f * fI
k C H .
'I
When
o
),
n > 0
define
6 n: Reg(B,H)n -+ Reg(B,H)n+l
6nf = k O f *
o fn+1 i *
I
d
I
i=1
Io
Then
6n+1an = E,
6
b
.
is a group homomorphism so we have
a complex
61
6n
Reg(B,H) 0
Reg(B,H)
) -
and define the cohomology groups
H(B,H)n
=
Ker 6 n/Im
H (B, H)O
=
Ker 6 0
6
n-1
n > 1
89.
As in the previous theory we have a normal complex:
n > 0
= ff 6 Reg(B,H )n
Reg(B,H)
B
[n]
-
k
is commutative
Sn+ =
n|iReg(B,H)+n: Reg(B,H)+n -*Reg(B,H)+n+1
Reg(B,H)+ 0
Reg(B,H) 0
=
60+ = 60: Reg(B,H)+ 0
-+ Reg(B,H)+ 1
The cohomology computed by this complex is the same.
Extension Theory
H
is a Hopf algebra
A, B
H-module (as a coalgebra),
structure.
called
A
a left
*: H(@ A -+ A , being the module
The sequence
H(
is
coalgebras,
A
) A -- -B
,
right exact if
1)
r
2)
Coim r = k OH A
is a surjective coalgebra morphism
Ker Tr = H
i.e.
A
The sequence is called split exact when:
H
has an antipode,
90.
the sequence is right exact, and there is a regular left
H-module morphism
d
A
)A
A
P: A -+ H
P g r
-
H(
comodules and left
give
A
B
B
In this case
.
is an isomorphism of right
H-modules by theorem 2.
the structure of a right C.H.A.
is called an extension of
4
r
A
-+ B
B
by
H
H-comodule, then
when:
HD
@A
2)
the following diagram is commutative,
d
A->
is a split exact sequence,
I@7r
A@A
A@B
-
--4
d
A
$: B -+B
Let
1)
-+
B-
A
B
B
H
H
where the
A*
A
ir~I
Given two extensions
A, A'
of
B
by
split exact sequences are
4
H@ A
HwA'
-+
ri
Al -+
B
A'a
B ,
we say they are isomorphic if there is a coalgebra morphism
'y: A -+ A' ,
such that,
H
'1
91.
H@
10
A
A
-
B
y
Y
H@
)A'
Being isomorphic is an equivalence rela-
is commutative.
tion and we can form isomorphism classes of extensions.
An example:
Given
g e Reg(B,HH)
2
a
2-cocycle, form
H
®B
which
g
H®@ B
is
H-module.
as a left
I
E
7r: H
B
BB
'
g
I@
P: H
E
B
0 H
.
g
d@ g)
g
B
H
B
)H
$
I
B
B
g
(1,3,5,2,4,6,7)
H 0H
H
H
B
H
B
H(
B
H
H
H
B
'I
m2 (
m
H
(H
defines the coalgebra structure on
H
@
g
H
H
B
g
B
-H
g
B
B .
( B)
g
Then
(H
@ B)
g
92.
is split exact, the second condition for extensions is
satisfied so
H a B is an extension of B by H . Two
g
such extensions are isomorphic if and only if the 2-cocycles
are homologous.
There is such an extension in each isomor-
phism class of extensions.
tive correspondence between
classes of extensions.
This yields the natural bijecH(B,H) 2
and the isomorphism
93.
Chapter V
Cocommutative, Coconnected Hopf Algebras
We have shown a split conilpotent Hopf algebra with
antipode is a smash product
E
B (D H
where
Hopf algebra acting as automorphisms of
split Hopf algebra where
G(B) = 1j.
B
H
.
is a group
B
is a
We shall study such
Hopf algebras which-are cocommutative.
In characteristic
zero we prove such a Hopf algebra is a universal enveloping algebra--u.e.a.--of a Lie algebra, in characteristic
p > 0
it contains a restricted u.e.a.--r.u.e.a.--of a res-
tricted Lie algebra.
(We assume familiarity with univer-
sal enveloping algebras and restricted universal enveloping algebras as can be found in
[2]
Lie Algebras.)
We define divided powers as sequences of elements
1
=
I 0 '2l''''
N
n
n < N
din = Z
and show that in a
i=0
class of cocommutative split Hopf algebras H , where
where for
G(H) =
[1
the coalgebra structure of
H
is characterized
by divided powers.
A Hopf algebra
G(H) =1j.
phisms of
Since
k
to
H
G(H)
H
,
is coconnected if it is split and
corresponds to the coalgebra mor-
coconnectivity is equivalent to:
94.
being split, and
morphism of
k
TI:
k -+ H
is the unique coalgebra
H .
to
all coconnected Hopf algebras have antipodes.
By
All Hopf Algebras Henceforth Are Assumed To Be
Coconnected Unless Otherwise Specified.
A primitive element
x
in a Hopf algebra
H
is one
where
dx
Then
x E H1
=
xEl
+
1@ x
e(x) = 0 .
and
Let
under
[
,
] ; if characteristic
=
dxp
(binomially expand
+
1@ x
L(H)
H .
space of primitive elements in
(dx)P), so
.
L
denote the
(=L)
forms a Lie algebra
p > 0
x
L
1
forms a restricted Lie
algebra.
Suppose
x E H
dx E H
dx
I * E = I
E * E = E
I
=
,
E(x) = 0 .
@ H1
+ H
10y
+
H0 ,
z
HO =
so
l
E(y) +
imply
k
z
= X
+
E(z)
= x
E (y) +
E(Z)
=
y
E(x)
=
0
95.
dx = 1QD (y + e(z)) + (z + E(y))Y
Thus
and
x
is primitive.
U
and
L C U .
U
H
x&l,
= L(q H 0 = L(0 k
denote the u.e.a. of
and the r.u.e.a. of
E(Z)
This shows
Ker efH 1= L
Let
+
l®x
=
1 - 19
L
in
L
in
char p > 0 .
.
char p =0
We consider
is a (coconnected) Hopf algebra where the co-
algebra structure is induced by specifying the elements of
L
are primitive.
Then
L(U) = L .
By the defining property of
U
there is an algebra
morphism
: U -H
which is induced by the identification of
L C H .
y
L C U
with
is also a coalgebra morphism; hence, is a mor-
phism of Hopf algebras.
That
y is injective follows from the next lemma.
Lemma 4I:)
bra,
if
V
v: A -+ C
,
A
a Hopf algebra
a coalgebra morphism.
vjL(A)
Proof:
v
C a coalge-
is injective if and only
is injective.
Suppose
a coalgebra morphism
v|L(A)
is injective.
Since
V
is
96.
EC(V(L(A))) = 0
thus
vA 1
VIAn
is injective
is injective.
injective.
C(V(l)) = 1
,
By induction we can assume
a e A n+
Suppose
v(a) = 0
@A
VI An
is
then
da=l0a+a®l +Y,
Suppose
V
n > 1 ; hence,
YEAn
A
.
then
®
0 = dv(a) = v @ voda = v(1)
v(a) + v(a)
) v(1) + (v
@
v)(y)
= (v&@v)(Y) .
Since
v
VIA
|A
n
is injective it follows
a E L(A) ; which contradicts
VIL
is injective.
The converse is obvious.
This implies
'y: U -+ H
Q.E.D.
is injective.
with its image which is the subalgebra of
L .
(Thus
U
Y = 0,
is the subalgebra of
H
We identify
H
U
generated by
generated by
L
both as algebra and coalgebra.)
We shall show if
where
J = k
C H*
H
is cocommutative and
char p = 0
U = H ,
char p > 0
U = (H*p(J))£
,
p(J) =
the ideal generated by
p(J)
a*P|a* s J]
.
and
(H*p(J))
We do this by developing
is
97.
a technique which "picks-out" subspaces of
associate with subspaces of
the polynomial algebra on
(Note:
H , subspaces of
fxaJ a basis for
work with
f
k[ f xa
k[ I xJ ]
L(H)
.
Given a vector space
9-.module.
n > 0
SnX =
n = 0
S0 X
xa]
x
we have chosen to
,
]--which makes the basis clear--rather
L .)
than the symmetric algebra on
Let
We
.
since the method of associating subspaces with sub-
spaces depends upon the basis
left
H
X
n
is a
x =®X
,
The symmetric tensors of degree
Y E X[n]aj=
Y
Y
=
are:
J
aE
k .
be an ordered basis for
k[
all
n
x
X ,
is a graded algebra.
]I
CO
k[
x a]
SX =
is isomorphic to
e
SnX
as follows:
n=0
a
Let
(i.e.
(x 1 )1
e 1 +--+
a
Let
Let
---
Y =
1el
x
em
(tm)
em =n,
<
ei e Z ,
xa
k[
]n
-
a 1 <.<am).
a em]
m
9G a
=
n
O
a e2]
2
..
(M)
be a monomial of
.
=
the isotropy group of
Y
Ie
(a., ... a m).-left cosets--corresponds to the
n
m
-1
98.
orbit of
Y , let
a e
ax((x 1) 1
a
em
a s @'n
By linearity
a: k[
P: SX - k[ fxal ]
fxaj
] -+ SX .
n(aeli
a
am
is an isomorphism,
denotes the inverse isomorphism.
All Hopf Algebras Henceforth Are Assumed To Be Cocommutative (as well as coconnected) Unless Otherwise
Specified.
Let
H
be a Hopf algebra,
] .
K = k[ { x
and
K(h)
L = L(H)
,
a basis for
x
We shall now define an element
E K
h e H .
for each
This is not a linear map.
For
K(h)
For
n > 0
(d
dh
E
denotes
projection.
P(h) E KO
Z H
I
-
,
E[n] o dn-lh
HO
-
+--+
-
h F Hn
I)
Hn@ H 0 +- --
e
E
=
denotes
dn-lh
1)
If
let
h E HO (=k)
so
H
H
en = n
.
HO = Ker E
L[n]
and
EJH
since in
H
-+L
a
L
99.
1)
any summand except
Ker(E [n)
-1--®
H
H1
lies in
and
--- @ Hl) C L[n]
E[n] (H
By cocommutivity and coassociativity
E
If
a1 ,...,an
E J
each summand of
1)
o dn-lh
then
a
Thus
<a *
*
a 1 *@
a *
--
a
vanishes on
and on this summand'
takes the same value as
o E
-a
* an *,h> =
-..
L
H
except
a
ES
<a 1
- - @®
= <a
-
nd
h>
a0n*,qE [n]o dn-lh>
This proves:
0
h E Hn - Hn-l
For
h e H
if
0 := E[n] o dn-lh E SnL .
let
- H
K(h) =P 0oE
Then we have
and only if
n]o dn-l(h)
Kn
H-J/yJVK
h
-+ K(h)
which is not linear.
morphism we have,
But by
and since
0
is an iso-
100.
if
h E H
, g E H
and
K(h) - K(g) = 0
h -g
For a subspace
K(X)
let
the subspace of
K(H) I K .
In general
Xn = X
Recall for a subspace
K(Xn) = K(X) n (Kn a
Lemma 5:)
- --
0 + K(h)
fh
C Xn ci
n - 1 .
? 1K(h 1 ) +---
then it is clear
K0 )
X, Y
We show by induction
then
=
($
1
n Hn
K(X) = K(Y)
n = 0 , suppose true for
7 1(p o
K
Given subspaces
X 0 =YO ; then
Proof:
0
spanned by
K
S X.
(K(x)|x
X C Y,
and
H n-1*
E
X C H
=
g e Hn
then
of
H
where
X = Y.
implies
X D Yn .
Let
True for
h e Yn -n-l'
+ ?mK(hm) ,
where
A
E k ,
Thus
[n] o d nl(h )) + +
oo E[n] o d n-(?h
which implies
g = N h
+
+ Am(P o E[n] o dn-l(hM
+ Nmhm)
-
+
mhm
E Xn - Xn-1
and
101.
K(g)
x
K(h)
.
- g E Y n-
=h
=
h
Observe if
=
X, Y
By
so that
C X
Q.E.D.
x + g e X .
are both subalgebras, subcoalgebras,
XO =
ideals or coideals the condition
is automatically
satisfied.
0)
If
x e Hn
[
o dn-l(X)
p
Y
where
Hq
(xa)
a
1
a
E
]o d
f
. 0 ( x ' m)
1
(o
<---< am
0 < ei, f
then
e
1
o E[n+q] o d n+q-
E Z
+ f
e
m
a em + fm
~1
... (Xm)
This is a purely combinatoric result and the proof is
Q denotes the rational numbers.
combinatoric.
Hopf algebra
B
be
Q[x
.x m]
, ..
Let the
as an algebra, and the
diagonal and augmentation are induced by making
primitive.
L(B)
the isomorphism
induction an
PB: SL(B) -* Q[x
e 1 + -- * + em = n
a
2)
o dn-((X)
Bo 0 E[n]
eI
1*
, we let
{x I
has basis
e1
*0
,..
.x
].
xa)
denote
PB
We prove by
that
a
( em
em
)
e
(x
e
...
(m)
m
ON
102.
n = 1 , say true for
True for
e
+
+ em = n
*-
(we suppose without loss that
e -l
a
1
2e
e1 > 0)
emae-
2
2e
a
e
1 (x
m
2
1
] o dn
E
n - 1 , then if
---
B*
is a right
B-module where
<b* - a, b> = <b*, ab>
If
x E L(B)
then
x
b* e B* , a, b E B
acts as a derivation of
a basis of monomials in
, within
xi
basis (not necessarily a basis for
basis for
B
x aim
a i*]
Let
.
so
(x M)
<a *,
i=l
B*
B*
.
B
has
choose a dual
B*) to the monomial
be the dual elements to
x a>
=69
e .
E
a
a *
2)
<a1 *
6
x
it suffices to show
a ei
(x )
gigm
e1 .
*--*am*
1
<a
<(a
**
*
*
*
1
if
O
otherwise.
g (a
a*m (x
am*m) ,
m--
g,= e
l
a1
the unit of
a em
xM m
er.,
m
>
B*-
where
To verify
g
+ ---
,...gm= em
a em
m)
>
a1 e1 -1
ejz1
a
e 21
e2
a
e.
m.
em
+ gm
n
103.
a
<
m
.al,
* a2*
e-l
a
e).x,.)e>
x
<a--
(X
e2
a
em
m
2
by induction
Thus
1
if
0
otherwise .
2)
gi-1 =e-l,
g2
o d
1B 0 E
gm
2
1
a m
- - -1)
1
em
Also that
is established.
a
e
f
fm
*.(x M)
m
Then
B o E[n+q] o1
B
e
+
L
(xd)
a
1
m,
m
e em
1
a
m
(m)
m)B o E[n+q] o dn+q-1(
1
)
a
...
by 2)
e 1+
(
f
e1 1)
This establishes
e
em+fm
(x M) (em+fm '
m+f
1
...
em
0
.
1f
)
(m
The following theorem is
M+fm
ea
a special-
ization of a theorem of Kostant.
Theorem 4.
In characteristic
0
a coconnected
m
1o4.
cocommutative Hopf algebra
L(H)
H
is
U the u.e.a. of
U ,
.
Proof:
then by
xa]
K = k[
If
a basis for
x
,
if
a
x
el
a m
LX..
m
=
(x m)
K(x) = (x a )
lemma 5
U
.
e
em
K(U) = K .
which shows
,
=
U
E
H .
U
=
U C (H*p(J))
elements from
L ,
monomial
i
e L .
b* * a*P
form
<b*
H
act on
<a* - hg> = <a*, hg> .
H* .
*
a*P,
p > 0
(1*p(J))
: Let
derivation of
By
Q.E.D.
We now study the case when characteristic
by:
L ,
U
If
.
h
1
Let
acts as a
and monomials of
A
1--
m
m
be a
is spanned by elements of the
H*p(J)
a*
E J.
- - Am > = <(b* *
= <(b*
+ <b* *
=
from the right
h e L
is spanned by
1 E (H*p(J))
b* E H
H*
<(b*
a*P) -A
,
2 ''
m>
A1 ) * a*P, £2
l),
(a*P
A1)
*
Am>
2
a*P, £2
Am>
'
m>
105.
= 0
by induction on the length of the monomial;
U C (H*p(J))
hence,
UO = (H*p(J))0
Letting f xa)
K = k[ rx"I]
Let
V
fal
xe
k .
be an ordered basis for
L
Q
then by
b tD
K(U)
=
HO0
m = 0,1,...
am
p. --
a
<- - < a
0 < e < p
I
be the subspace of
spanned by the right side.
K
We show
K((H*p(J))
U C
Since
V C
c:
)
v .
(H*p(J))
it follows from
K(U) C: K((H*p(J))
)
that
V
Let
=
K(U)
=
K((H*p(J))
C
V
I)
h E (H*p(J))
e
a
K(h)
=
Z A\i x
1
A
---
a
x m
A
E
k .
1
Suppose for some
d,ual basis to
xa
t, j
et I> p
m
0Then
Let
a
CJ
be a
1 =1
a* = al*
m
e
H*p(J)
and
<a*,h>
=
t
106.
Xt = 0
which implies
and
K(h) E V .
Thus we are done
by lemma 5.
We can draw the corollary that in characteristic
p > 0
if
U = H
if and only if for each
H
is cocommutative and coconnected then
follows because
Since
U
H*p(J) = 0
a*
H
generated by the primitives in
We shall show
is
J
a*P = 0 .
This
p(J) = 0
if and only if
may not equal
we examine the ideal
H .
=a*p Ia*
LH = HL = p(H*)
E
e
LH = p(H*)
H*]
HL = p(H*)
the proof
the same (up to reflection).
HL c p(H*)
:
Let
<a* - h,g> = <a*, hg>
2
act on
H*
from the right by
a* e H* ,
Then for
.
<a*P,
since
H
ih> = <a*P -
,h>
=
E L , h E H ,
0
acts as a derivation.
If fx
is an ordered basis for
L
K = k[ { xl] ,
K D W = the space spanned by
e
1
)
a
(x
m
00(a
(x
m
=
1,2,...
a1 < - -< am
for some
then
K(p(H*)
) C W .
Since if
e. p &ei
h E p(H*)
107.
K(h) =
Z
A (x
i = t
Suppose for
a
a1 eii
plet
m
C J
, then letting
...pet
1
a*
eim
)
m
x i m
be a dual basis to
i=1
et
et
a* = a
1
*--x
implies
t
=
am*
0
<a*,h> = At
and
K(LH) D W n K(H) .
Suppose
K(h) s W , we shall show there is
K(g) = K(h) .
which
K(p(H*) ) C W .
and
Next we show
where
m e p(H*)
g E LH
h s H
- H
where
Let
ei
K(h)
=
a
M
I A (xa)
i=1
0
a
a*(x) = <a*,x> .)
1 < n(i)
_<
m
al <--< am '
shall act as functions
H*
(For the present the elements of
i.e.
,
E k .
A
+
m
(x'm)
---
For each
i=l,.. .M
there is
where
p
da* - h
By cocommutivity
j
.
e i
By
©
- h
i"
=
Z h '
a*
=
I
a* o d2h e H 0 H 0 k = H 0 H .
I
dn- 2 (a* - h) = a* @
1 [n-]
o dn-l (h) .
108.
If
a* e J
then
I[n-1] o (Z
Hf1
H
9
f1+-+fn = n
f +---fn-1
some
some
f
= 0
f
n-
=n-1
=
0
and hence
od ~ (a* - h) = E [n-i] o a*® I [n-1] o a
a*
=
Thus if
mC
I a *9 i=1
With
=
J
s
-h)=
where
p
e
X
I
j=1
o a*(&
dual to
M
K(a* n()
[n-1 1] o a(K(h))
Ij
a*
a* - h s Hn-1 - Hn-2
K(a* - h)
:10)
6 J
[n-1] o a(K(h))
:
a
(x
f0
if
ei
=0
if
ei
> 0;
By
so
e
and
a1
(xi)
M
-
(pJ
(x m)
e
(- Xm)
a
=i A e
(x
Jn(i)
)
)
a Mj
1
1n(1)
M
j=1 J
a
(xan( 1))e n(1)
)
K(x n(1)(a*n(i) - h))
(K(h))
-
(Xa) ejM
m
e
I
109.
Thus if
M= 1
a
ell
a
...
K(x n(l)(a* n( )
- h)) = A e n(1) (xa1l1
1
) - h)
xan(1) (a*
p
e
K(g)
g =
so
1
= K(h)
smaller than
E LH
e n(1)
By induction on
and
we can assume for values
M
g E LH
there is
M
M)
K(g) = K(h)
where
Let
xan(M)(a*n(M)
en(M)
M
K(9)
j=1l
en(M)
E
LH
where
M-1
K(h-g)
J=1
= K(h)
g + g s LH
a
1
em
(x m)
.
e ( (M))
Mn(M)
and
otherwise
a )j
e
e
and by induction there is
Then
E LH
in which case we have exhibited
K(h)
K(g)
e
1x)
eMn(M
This may equal
g
a
h)
.
1)
g E LH
(.Xm)
1
where
K(g + g) = K(h)
So we have shown:
K(LH) D w n K(H) D K(p(H*)
) ,
.
m
K(g) = K(h -'9)
110.
and
LH
W
Thus
n
C
K(p(H*)
)
= K(LH)
K(H)
(LH)O =o)
)
K(p(H*)
(p(H*) )0
=
so by lemma 5
we are done.
Observe in
we showed
K(LH) = W ; but
Thus if
---
=
V
K)= W
and in
n (K®
-. e-
Kp)
U+ = U n Ker e , then
U
K(U
v n (K 1$
K(U)
)=
C (LH)p
V n (K Q ..-(D
so by lemma 5
U
= ((LH) )O =
(U +)O
=
K)= W n (K
coideal; that is
and
- - OKP) = K((LH)p
LH)p
Since the elements of
algebra morphism the
0)
L
are primitive and
2-sided ideal
dLH C H 0
LH + LH
LH
@
is a
H .
d
is an
2-sided
Hence,
H/LH
has a natural algebra and coalgebra structure by which it
is a Hopf algebra.
We shall now show
H/LH
with its vec-
tor space structure altered is naturally isomorphic to a
sub Hopf algebra of
H .
This is the first major step to-
ward obtaining the coalgebra structure of
We assume now that
space over
k
let
X
k
is perfect.
H.
If
X
is a vector
denote the vector space which
group theoretically is
X
?-x
and for
=-
p
?A e k
x
E
X
111.
For vector spaces
f = f
is linear where
If
A
X, Y
is
where
f: X -+ Y
then
f: X - Y
set theoretically.
an algebra (over k) then
A
has multipli-
cative structure
A
m:
A -+ A
and we define the unit
A
k
kQA-:
With this structure
A
a coalgebra over
then
diagonal map is
k
d
k
C
H
If
H
is a Hopf
We shall freely
X, Y
are vector
then
XY=
There is a natural map
x E_ X
is
is a Hopf algebra where the algebra and
make obvious identifications such as if
X D Y
C
1
[c (c)]JP
coalgebra structure are as indicated above.
spaces
If
.
is a coalgebra where the
is perfect this makes sense.
algebra then
k
and
e_ c)
since
is an algebra over
X/Y
i
(X)*
-
(X*)
where if
a* e (X)*
,
112.
.
<ix (a*), x> = <a*,x>
is injective and surjective since
i
x
(i
is linear.)
If
(C)*
,
is perfect.
is a coalgebra so that
C
(c
coalgebra and
k
ic: (C)*
C
is
a
are algebras then
-+
(C*)
is an algebra isomorphism.
If
A
is an algebra
m
A®A
-+
A
AOA
tm
=
A&
m
induce
A*
-+
(A
A-
A
A)*
tm
(A)*
~> (A® A)*
then
(A*)
t
-CLMI
-
.)
[(@
A
(A_)*
(A0A)*
=
is
If
(A@ A)*
which shows
1A
A@ A
I
t
is commutative.
A)*]
A
=
m
is finite dimensional so that
(A)*@ (A)*
then
is a coalgebra morphism
iA
A
=
iAiA
(the augmentation
preserved).
We identify
(X)*
with
(
through
ix ; by the
113.
foregoing if
X
is an algebra, coalgebra or Hopf algebra
the structure morphisms of
For a Hopf algebra
Let
p
(X)*
H
and
a*
-
(
(H)*
=
-+ a*P
t p: (H)** -+ H** .
H*
We shall show
Choose an ordered basis
f ha}
Im(tp|H) C H .
H , then by co-
for
for all
and we have
P o dn~l (h) e k[ Ih
M
11
dn-1 (h) = Z X
i=l
or
a(ha
x e H
.
let *(x)
]n
1-1
e 1 + - -- + e
where
x E H
is commutative.
dn-l(h) C S nH
commutivity and coassociativity
For
H, H, H*, (H*) = (H)*
we have
it is an algebra morphism because
h e H
are preserved.
be the algebra morphism,
p: H*
Then
(X*)
ha
e m)
= n , a 1<- -
be the minimal integer
By choosing the basis for
choose a basis for
H0
extend to a basis for
H1
extend to a basis for
H2
H
,V X
E k ,
m
am
t
where
as follows:
114.
and then ordering it
we have
M
dn-(h) = z N
1=1
where
+
ei
# (h)
For
-+
a(ha
- h-ah)
ii
am
= n
e
a
#(h
) + 1 %
> e
<- -
am
<
and
+ ei #(ha
m
m
a* e H*
<p(a*),h> = <a*[P] , dp-
(h) >
e
M
= <a*[P]
e1
a(h,
As a symmetric tensor in
h
am
the number of terms
involved in
ii
a(ha
divisible by
p
)
m
am
e
---
1 < t < m
h
is of the form
a
m
i=l,...Mh e
1
then for
i e I
ht
Let
e
I
is
unless
e
ha
for
--- h
x1
let
-- * ham
m
n(i)
e
---
ei
h m
am
p
= ha
t
be such that
hi
an (1)
1
< t < m
I
,
115.
If
i f I
<a*
S-eha m )> = 0
am
e
(consider the number of terms in
a(h
a1
,,a(ha
a
e
m
Thus
=
<a*[P],
z
N a(h
)>
an(i)
isI
=
Z
ieI
A <a*, ha
n(i)
1
3)
This shows
t1(h)
=
E
ha
Let
Observe
C
Im( tp(H)
or
V = tpIH: H
#(h)
-+
= f(h)
(or
>
-
H
H .
H
= (H)n ).
H , for
the "extending" basis for
-#(h)
E H ,
nii
isl
p #(h
an (i)
Since we choose
iEI
)
and so
p #(V (h))
We wish to show
V
< # (h_) .
is
a morphism of Hopf algebras,
first we show it is a coalgebra morphism.
Being an algebra morphism
116.
pH P
H*®( H*
Ip
and
k
(H)*
-*(H)*
m
m
(H)*
(H)*
p
are commutative.
Dualizing gives commutative diagrams,
t
t0
( H)**
(H)**
t
H**
1IV
and
k
t P
H**
t
m
S(_ )]
tTi
t(P
In the left diagram
H c
(H)**
V
t
tp|H:H
-+
H C H** ,
t 1 1|H: H
-+
k
t|H
H
Q
I
H:
k
which implies commutivity of
H
.
V
H
In the right diagram
H®@_H
<(a)*
m
()*
[(H)*@
c
,
(H)*]*
=<(a)*h><()*,>
where
(H* @ H*)*
p)
p.-
117.
H @H
similarly
Then
tp|H = V ,
C: (H*® H*)*
tH
.
= d
t(p D p) H® H = V
@V
,
tmH = dH
and
V
dH
H
H
H
H
-
H
so
V@V
V
is a morphism of coalgebras.
Now to show
V
we have
t(h)
is a morphism of algebras.
as a basis for
3)
1
=
Ai ha
n(1)
1iI
In choosing our "extending" basis for
{1
At
HO
thus by
H
we can choose
3) ,
H
k
Ti
H( H
is a coalgebra and one main property of a Hopf
algebra is that
Since
mH: H 0
A
is commutative.
HD
@H
H -+ H
mH: H0R H -+ H
is a coalgebra
is a coalgebra morphism.
(HQ H)*
is an algebra,
being a coalgebra morphism implies
118.
tm
H*sH
( Ha@H)*
is an algebra morphism.
If
r:
(H
@
H)*
r
=
(
P
-+(Y*
Y*
then
((H C H)*)
-+
is an algebra morphism and
tm
H*H
(H
4/
r
P
(H)*
t
)(H
mH
t
mH).
0H)*
mH
is commutative, (since tmH
t
H)*
an algebra morphism and
Then passing to the duals and transpose maps
and restricting to
H, H, H
@ H,
H® H
gives commutivity
of
m
H
H-H
V 0
Nle
H (D H
so that
V
V
V
m
7 H
is an algebra morphism.
Hopf algebra morphism.
Since
p = tV
it follows
Ker V = (Im p)A,
Thus
V: H -+ H
is
a
r
119.
Thus by
Ker V = LH , and there is a factoring
@
>H
H
I
T \
V
H/LH
where
H/LH
V
=
is an injective Hopf algebra morphism.
H/LH
so
V:H/LH
H
-+
is an injective Hopf algebra morphism.
Within
H
we have the sub Hopf algebra
Im V
We
define inductively a nested sequence of sub Hopf algebras:
VO = H ,
V c H ,
V
is formed
vi)
V1+1
We have
if
V
DV
D
let
CO
V
=
nv
i=o
.
Being the intersection of subalgebras
V Mo
is a subalgebra.
We show it is a subcoalgebra as follows:
each
hence
Vi
is a subcoalgebra
(Vi)
C H*
is a
2-sided ideal,
120.
I-L
(VO)
I.
c (v1
C (v2)
aU
1=0
(c H*)
V
This implies
Vc
is
is a
=X
2-sided ideal.
.
a subcoalgebra,
we present a simple
proof.
(V )*
is naturally isomorphic to
H*/X
as a vector
space where
L
H*
H*/X
H (V
are transpose maps.
td
(H @ H)*
7
)H*
d
H @ H (-H
<
)H*/X
i
V
are transpose maps so
Im(d o i)
=
Ker(r o td)
Ker(r o td) D H* @ X + X @ H*
Im(d o 1)
Thus
V
which implies
c (H* @ X + X D H*)
=V
is also a sub Hopf algebra of
n
Vn = Im(H-*
V-
n-1
H
3
0---
V
2
1
H--+ H
-
V
H
)H)
V
V.
00
A
121.
we have the transpose map
n
so
let
V
1
n = [Ker(p 0 - -pn denote
]
op)
We shall for simplicity
.
n
0***
0
P)
n
Yn
denote
V o -.. o V
and
x
=
-+
pn: H*
denote
x
.
As a map
()*
n
a*-+
Then
(Vn)
= Ker(pn)
(a*)p
.
We define
L
=
L
=L n v,
L n Vi
1=0,1,...
so we have a decreasing sequence of restricted Lie algebras
L
=L
0
D
L 1D
CO
n Li = LO.
1=0
----
,
122.
Divided Powers
In any characteristic suppose we have a (possibly
infinite) sequence of elements
xo' x
where for any
x2
n
n
dxn = Z
in-1
i=0
Then
tivity
0
0
dx0
0
so
1 '
x0
dx 1 =
x0
l + x e
= 10D x
so that
is grouplike and by coconnec-
x
is primitive.
+ x, (
1
Such a sequence is called a
sequence of divided powers of
n th divided power of _
x0.
xi.
xn
is called an
.
1
Example
In characteristic
and
xi
=
xi
then
Given a sequence
because
x1
n > 0 .
if
x
x 0 ' x1 ' x2 ,...
of divided powers of
E(xn) = 0
0
is a primitive element
is an infinite sequence
x .
x 0 ' x1 ,...
of divided powers of
This follows by induction:
is primitive, for
n > 1
E(x1 )
xi
=
0
,
123.
n
E
sx*
E
-
(x n
E(x
=
)0
E(xni)
i=o
(by induction)
2
-
(x)
Suppose characteristic
1
d
(xn
Z
Xe1 0
Xe
e 1+*'+ep
0
V(xn)
For
e
E
=
Z
p > 0 , since
p
p
n
n
lei
let
Z
E
satisfy
let + 1
lei
< e < p
p
Then if we have a finite sequence of divided powers of
X 0 .9 X 1 .9 00
x 1 E Vie
Xe
e = pn+
and if
x
1
<e
i
xi E
If we have an infinite sequence of divided powers of
x
then
x
E V
Theorem 5
If
divided powers of
0
X
E
Ln
there is a sequence of
x
l''''p
-1n+1
124.
Pro )f:
We shall use an induction within an induction
in this I roof.
The theoirem is
true for
n = 0
1
Xi
xi=
since for
x
E LO
i=0,1,...p-l
is the desired sequence of divided powers of
by induction the result is true for
Include
a basis for
........ ...
,
Let
x
in a basis for
n(a)
x
xal
.
n
if
=
maximal
Ln
Extend the basis to
fxa]
for
E
Z
where
Order
L
n
xa E Lt
By the induction assumption we know that if
there is a sequence of divided powers of
a
a l'
xO'
Ln- 2 '
LO = L .
x"* part of the basis for
otherwise
t
Suppose
.
n - 1 .
Ln-l , then extend to a basis for
obtaining finally a basis
the basis
n
The "outer" or first induction is on
n(a) < n
xa
x apn(a)+l1_
We shall construct sequences of divided powers for
the
xa
where
n(a) = n .
The construction has
2
dis-
tinct parts which we separate.
I)
Suppose for each
i)
there is
a
where
a sequence of divided powers for
x a 0'... x a t-l
i
n(a) = n ,
0 < t < pn+1
xa
-1
125.
ii)
iii)
O < e < t - 1
xae E V 1
there is an element
K(yat)
and
n I
y t
where
yat
at
=
then each such sequence can be extended to a sequence of
divided powers of
xa
xa O'
satisfying
xa
0<
E Vn-|eI
Proof of I).
xaO''''xat-1
a
x t
< t .
Say we are given the sequence
Suppose we have
to extend.
2 < s < t
y s
satisfying:
E
o ds-1 (yas)
=2
a
a
fs
0 < f
fl + ' - - s = t
y
let
s
for example,
VnIt
yat= y t ;
ya s-l1
we shall construct
E[s-1] o d s-2 (y
)
=
z
0
< f
f
D
...
8
n-It|
x
s-1
f1 +-.-+--1 = t
y s-1
satisfying
aisyn
'
126.
Consider
E [S-1]
Z
f i>o
xa f
1
o ds-2
a
@ - - - @ xa
s-
Y
H+[S-1]
E
f +.-+f s-1
We show
= Ker e)
(H
Ker(E @ E o d)
E
= H
Y e L
H+[s-2]
Since
.
it suffices to show
E 0 I[s-2] o d (
I[s-2](Y) = 0
or it suffices to show
a
E @ E @ I[s-2] o d 0I[s-2]( Z Xa
f > 0
i
f +- -±
--
t
E ® E ® I[s-2] o d 0 I[s-2]( xaf
f i>0
1
f +- --
+ Y)
s-1
=
s
)
t.
The left hand side (top)
=
E
E0I
[s2
o d
I[s-2] o E[Sl1
(since
=
Els]
E0E od oE =EOEod)
a
o ds-1(Yas)
Xfi
f i> 0
f +- .
- - fs= t
a
which is the right hand side, (bottom).
ing
Y E H
* .*
o ds-2 (as
L
- --.@ H+
so
By similar reason-
Y e L[s-1]
127.
Thus
~1
E
o dss-2]
as=
e.
a1
a
m
+ 2 Aa(x
z
f > 0
f1 +- -.+f
where
1
_
s-1
= t
e 11+- o*+
= S - 1
e
a1 <.-< am .
Now we wish to
n(ar) < n --
show that we can assume--when
nf(arr)n+| t|+1
for all
q, r -Suppose
not, that for some
qr
n(ar)-n+|t|+1
e
.
> p
r
qr =1
by the choice of an extending basis,
n(1 )+1
xa
Without loss we assume
so we can choose a dual basis
ai*)m C J
where we assume
a
E
so
Since
ya
E
1
m
(Vna )+L -
(Vn(a)+1)
(a*) na
]
x
to
=
Ker p
)+1
= 0
let
Vn-ItI(()
S(a*n
and
z E H
=
y .
)-n+|t|+
where
Let
128.
eF -pnda)-n+ t|+1
a*= a1 *
1
I.
e1
*
am
m
*
Note for
b* a H* ,
h e H
- h)
Vn(pn(b*)
m
b* - V"(h)
since
<c*,
Vn(pn(b*)
<pn(c*
=
- h)>
* b*),
<pn(c*),
h>
=
->
pn(b*)
<c* * b*,
Vn(h)>
= <c*, b* - Vn(h)>
Consider
0 =
<pn-It|(al*Pn(a
)-n+|t|+1
a )
[n-|t|
)P
z >
n(a 1 )-n+It|+1
=
<a
Vn- t| (Pn-|t| a*
*P
n(a)-n+|t|+1
, a* - ya >
= <a *P
e
=
=
<a
e1
*
*
+
>00
f
.*
aIm .
<a*,
m .
af>1
a S>
X
a
e
> --1-
<am*
f 1+-0- -s-1
For a term in the right sum not to be zero we must have
af
129.
since
n -
have
I
a1 * E (Vn(a )+1)
|fj|
Since
.
_< n(< )
(
f
1 a fn(a )+1'''
x
a
)+1
E
V
i=l,...el 1 , or
, we must
n - n(a1 ) < If i
which implies
n-n(a )
i=1,...e
1
.
it
follows
Suppose this happens then since
f 1+
e
s
-
1
e
+ fs-1
t
n-n(a )
p
+ [s -1-
e
.
]'_<t
and we are assuming that
1
ei1
p
)-n+|tI+1
1
so the above inequality yields
n(a 1)-n+|t|+1
n-n(x1)
p
p
pt|
or
Thus
eq
+1 < t
a contradiction.
<pn-'t| (a *P n(a)-n+|t|+1
which implies
W1 = 0
<pn (ar)-n+| t|+1
t < p n+
implies
- el ] _< t
+ [
,
),n-|t| (a*)] * >
=
and hence we can assume
when
tI < n
n(ar) < n .
so that
e
In particular
< pi
when
130.
n (ar)
Now examine a typical term in
e
A(
1
a
--- (x )
a((x )
say
),
e
i Ai
e
am m
a(xCl
Look at
x
.
n(ai) = n
If
since
< s - 1 < s < t
e
i
there is by hypothesis
Sai
S
e
Y
of
i)
I
I
)
I C Vn-I
n (a +1
If
n(a i) < n
e
by the o uter inducai
. Since
tion hypothesis of the theorem there is x
since
< p
there is actually a sequence
Sai
x0
it
Sa
x i 1 0 00&
xai
n (ai )+1
- 1
p
9
*
follows
x i
e iE Vn(a )-Ie
< n(a
We have
)-n+|tl+1
| '
which implies
i
lel
1I
n(ai) - n + It|
n -
It!
< n(ai ) -
or
le 1
131.
so
x
Vn(a)-|e
E
e
n-|t|
1
Then by
1 ll
a.
since
K(x el
If
1
am ) = NJ(xa
elm
x a1
K(A
z
x 1
z
a
1e
e Vn-|t
then
'
z
z
as
for all
E V
E V
2
E[s-1] o ds- as-1
+..f
and
fs-1
s-l=t
So we have completed the induction step of
stop at
E
ya2
wE o d y
Then let
where
V-it
a
t1a
a
2
xat
i
t-i
2
-
2
then
j
n-|t|
fl
f
m
x)
)
a
x m
el
Similarly we form
-a
i
)=(x
am
---
2)
*(ya
I)
and finally
132.
xat
E (X
n-|It|1
t-l
E@ E o dxat =
x
1=1
and
dxat = E
x
E o dxat + I®@
xat
a
Since
xat + xt
@I
o dxat - E®E o dx
@ 1 + d E (xat)
a
q xt-i
is the desired
x t E V
t-
o dxat + e
t-1
aaa
=Z x i®x t-i + 10
t
Zxt
i=0
0
t
.
tL
divided power of
Thus we have proved
Now so that
I)
xa
,
and
I .
can operate we must produce the
yat's
II:
For
xa e Ln
be simple to obtain the other
is
z E H
where
ya n .
we wish to find
Vn()
=
Yt 's
is
ya n
P
E Ln
x
will then
so there
x"
First we prove that if there is
Vn(Z)
It
=
a
Say we are given such
e
M
a 1L
'-K(z) = x
+iA
x
z E H n
p
where
then there
z , then
e
a
m
x m
133.
a
<
-< am
+ ---
e
+ e
Then for all
ei
< p
e1
,
r
r
1
x
e
where
ei
r
=
pn
< pn(ar) +
ei
,
,because
a
n(z) = x+
For all
= n
m
1
i.e.
lei I < n(ar)
r
n(ar) < n
In showing this we assume
since otherwise the
n (ar )+l
result has just been shown.
say
e
and
Let
1
m=1
e1
.r
> p
fxij m
be a dual basis to
1=1
a 1*
a*n
ai*
Suppose some
E
(V (a )+l)
(a1 )+1
= 0 ,
(.
.Then
(note
xa
n
but
n (a 1)+1 )
e
<a
1 *..-*a
m, z>=
which shows
?l = 0
and
we can assume
lei I
_< n(ar)
r
Thus when
n (ar)
<
n
by the "outer " induction hypo-
thesis of the theorem, we have
x r
eM
the
r
power.
When
n(ar) = n
eM
r
<pn ; since
L
M
divided
a
x r
V c Vn-1
r
134.
by the outer induction hypothesis we have
ai
1
'' em
y = AMX a
am
e
am
y
=
Let
Then
.
a
a
K(y)
a
x r .
r
M(x
--
) by
.
Since p
e
Mm
ee
e
r
ysMrn
%
M-1
a1 1
am m
Ker Vn .So
K(z -y) = Z X x
x
and
i=1
Vn(z-x)
S
= x
=Vn(z)
there is
z
x
.
M
This shows by induction on
that
where
V"(z)=
xa
and
pn
K(z)
(xa
=
)
.
Clearly
z e Vn-jpn
= VO
ya n
so let
p
Second we prove that for all
where
z
a
Vn(z
where
#(z)
>p
e
al 1
Z
=
Vn)
is minimal and
x
z E H
p
there is
Suppose not; given such
M
K(z)
xa
=
xa
x
choose
.
Then
e
a
m
xm
i=l
a1 <---< am
and we can assume
0 < M is minimal.
be a dual basis to
x
p
because
.
eM'
t
Let
a
We claim for some
nt
C J
eMt
N
135.
e
a1
K (n ( 1 /pfn
i
-- xm
i
whe re
pn
pnJe.
n1
Thus we can assume
eM
Note if
eM
pn
=
n(a1 ) < n
eM
Then
0< b< p n
apn + b,
then as usual
=0 pn + b , and since
M ,
Thus if
e /pn
am m
/pfl
pn
$
$
< n(a1 ) so that
eM
0< b <p n
eM
1
n(a1 ) < n
b
=
eM
and by
< n(a,)
x e
the overall induction assumption there is
aeM
divided power of x
n(a1 ) = n , since
If
tion assumption there is
x
x b
Consider
M
b
x
z) =
a
bxa
K(a 1 *b*
e iK(a1 *b
divided power of
a
b-
divided
z) , by
e.
b
e12
1a 2
x 2
ei
_i
-
1
1
a
m
1=1
where
eM
b < pn , by the overall induc-
So in either case there is
power of
an
< b
-
otherwise .
x
136.
If we let the usual binomial coefficient
when
c < d
0
then by
a e1i
M
K(x
1
b(a 1*b
a1
)
0Z)
(c)d = 0
i
1=1
a
..
bly
ei
m
m
eM
Now
4
( b )
0
(mod p)
e
b
(x a(a
so if
y
x l
-^ b E Ker Vn
=
*b
so
M-1
Z
1=1
then
1)
Vn(z-Y) = xa
VUnand
y e Ker
Cx1ei1
a
xl
b
e(1)
A
(1
a af1 m
--- x
where
z E H
p
n(a) = n
j=0,1, . . . n .
and hence there is
,
we have
Then
K(y
p
)
ya
y t
M , hence we can
y n
p
Let
yp
n-1 pJ1
For
S,m
)
which contradicts the minimality of
assume
....
0Z))
e
(
K(z -y)
(ap"+1)
((ap+b) (apn+(b-1))
b
(b-1)
=
So for all
= Vn-j(Ya
Yn)
ptl
= (xa)pj
n-j -
t < pn+l
and
a
137.
t = anpn + an-1Pn-1 +---+ a p
+
a
0 < a. < p
It
maximal
=
i
where
.
if we let
By
a
a
Yat
then
}
a,
=
CLn
p
(
Elt] o d(t-1at)
(
a0)
( pn +--an
(pn
0
Y
(Oa aO a(Xa
)
O
n
The coefficient is not zero (mod p).
an
~a_
y t
then
a0
(pn)
01)A )
a
t
K(yat)= (X)
7t
Vit
.
with
Thus we have
desired properties and by
powers can be started at
I)
(the sequences of divided
xa0= 1
for all
a)
we are done.
Q.E.D.
I1
138.
Corollary to the proof of theorem 5).
If
x E V
there is an infinite sequence of divided powers of
x
xo'xl'x 2 '....
Proof:
Since
V(V)
H = V
that is assume
V0
If
{xa]
there is
=
we may work within
V
Then
.
1
2
is a basis for
00'
L
=
Ln
by the previous theorem
a sequence of divided powers for each
a
x
V00 ,
x
a
0
x n+1_
''''
'
This particular sequence satisfies the hypothesis of
with respect to
n+1
,
I)
i.e.
ii) x e E Vn+l-le|
iii)
there is
a where
y
Owa
yat
n-t|
and
t
K(Yt)
=
.
The existence of
yat
follows from
II)
of the theorem and of course all conditions such as
x
E
Vn+l-e|I
are automatically satisfied.
the
"outer" induction hypothesis of the theorem is satis-
fied since for all
a
x a0 ''''
a
we have sequences
a
x n+1
p
-l
'
Moreover
139.
Thus this particular sequence
a
a
O '---
Ipn+l_
can be extended to
a
a
X0 ,..
X n+2
p
-l
Repeating this we can extend the sequence indefinitely.
Q.E.D.
Let
Z < oo .
xa E V
L
{xa] be a basis for
For
a
let
L
n(a) E Z U foo
Order
Z U Ioof
by
be maximal where
an extending basis for
We shall call jxa
.
.
if
> n
x"|In(a)
forms a basis for
Ln .
In
particular if the sequence
)
-
-
2
L1
S
stabilizes in a finite number of steps then there is an
extending basis for
L .
Between the theorem and corolxa
lary we have shown for
there is
a sequence of divided
powers
a
x
a
a
1
---
pn(a)+1_
where this denotes an infinite sequence if
When
n(a) = cc
let
pn(a)+1
=
00
.
n(a) =
cc
14o.
p = 0
In characteristic
L
x I
is a basis for
there is an infinite sequence of
xa
then for each
if
x"
x 'n =
divided powers, e.g. let
L
for
an extending basis.
a
For
We call any basis
.
we let
xa
a
a
Os X 1 s '..
''
p n(a)+1_1
denote any infinite sequence of divided powers for
and let
pn(a)+l =
The following theorem generalizes
.
co
xa
the Poincare-Birkhoff-Witt theorem.
Theorem
k
6):
The characteristic of
Hopf algebra where
each
xa
is arbitrary,
be a cocommutative coconnected
H
Let
is perfect.
k
has an extending basis
L
x a].
fix a sequence of divided powers
xa
Order the basis
x
1
xpn(a)+1 -l
{x'
.
-- x m
em
Span:
Let
a < - -<a m
1
h E H
k[ jxaj] = K
ei
e
m
a
a. 1 1
*-x m
Ni x
K(h) =
*
Then the monomials
H .
form a basis for
Proof:
a
'''
i
a1<*<a
e
< p n(a ) l
For
I
141.
r
< p
e
Then
Choose
ai*
C J
(Vn(a )+)
a
a* = a*
= 0
.
Suppose not say
e
a d ual basis to
x
n (a ) +1
a 1 *P
Then
.
l *-0* am* elm
and
= 0
<a*,h> = A
and we can assume
but if
which implies
n(a )+1
<
eq
p
.
Then let
a
x= A
Zx
K
x
a
---
x m
. By
eim
K(xi) = K(h) t
lies in the space spanned by the monomials so
(the space spanned by the monomials) D K(H)
lemma 5
and by
H = space spanned by the monomials.
Independence;
Let
Aa
a
be the subcoalgebra of
0' Xa 1
''''
a
x pn(a)+l
Put an algebra structure on
k
1
-+
Aa
A
a
x0
-1
by
H
spanned by
142.
(0
xa
if
) x
i+j < pn(a)+1
a
1ij
O
Then
Aa
otherwise.
is a Hopf algebra.
on(a)+
or in Q[x]/<xn
1)
This can be checked directly,
pn(a)+l =
(where if
CO
we mean
QIx])
x
1,
XPn(a)+1
x2
2 , 2
(pn(a)+1
generate a sub Hopf algebra
is a Hopf algebra over
k
A
over
Z .
Then
®Z
A
which is isomorphic to
k
Aa'
Now we show any finite set of monomials is linearly
independent.
a,, ... am , where
Given
Aa
@
1
--
a1 < ... < am , form
A%
m
which is a Hopf algebra and we have a natural morphism
'y: A(,
1
0- - -
a1
Aa
m
am
1m
'y is a coalgebra morphism.
basis
a
a
x
m
1
L(A
®---
A
)
has a
143.
Aa
'y|L(Aa
1,
x
11
x
[m-2]
a2
[m-1i
a
)
y
is injective, thus by lemma 4
Q.E.D.
is injective, which proves the linear independerce.
We call a Hopf algebra connected if
dimensional then
and
H*
is a Hopf algebra.
H*
H
H*
is split if and only if
H*
is commutative if and only if
pose
k .
to
H
unique algebra morphism from
s: H -> k
If
H
H
is the
is finite
is connected
H
is coconnected.
is cocommutative.
SupH*
is finite dimensional, connected, commutative;
k
is split and
is perfect, then
is coconnected and
H*
cocommutative and has an extending basis by finile dimensionThus by theorem 6:
ality.
H* ~
as a coalgebra.
a
(in the notation of the theorem)
a
m
By finite dimensionality
char p > 0
and
n(ai) < 00 .
H*
has a basis of monomials as described in theorem 6.
Let {[a T**] C (H*)*
for
H*
.
(a
to (x a
m=1
Let ta
,
be a dual basis to the monomial basis
**
m
C
a
*
be the dual basis
(the extending basis for
L(H*).)
144.
k[x 1 ]
Then
(H*)*
n-l
k[Y ]
n~a
p(
-@-m
)+1n(am)+1
3
p
as an algebra
where an isomorphism is induced by
-a1 aY **
By finite dimensionality
1=1,...0
m
H** = H
.
.
Thus we have proved
a finite dimensional commutative, connected, Hopf algebra
over a perfect field where the dual is split is of the
form:
(as an algebra)
k[Y ]1
k[Y ]
em
X1 >-2mp
If
k
is algebraically closed then by the results of the
first chapter we can drop the condition that the dual is
split since it is automatically satisfied, also
k
is
automatically perfect.
We now conclude by applying results of previous
chapters.
We show a coconnected cocommutative Hopf algebra
over a perfect field is an extension--as an algebra and a
coalgebra--of
H/LH
by
U
characteristic zero
H = U
is trivial.
k
Assume
when
and
U
is commutative.
LH = H+
In
so, the result
is perfect of characteristic
p > 0.
--4
145.
Lemma 6:
and
K = k[
Let
be an ordered basis for
{ xa?
If
xa ]I
where
h e V1
e
e
=
K(h)
1 1
(a
A
a
m
i
g
then there is
H
£
where
pei
1
pe
a
K(g)
=
Z
1
''
m
xm
and
V(g)
V(f) = h .
Say
show we can choose
choose
f
where
f
#(f)
K(f)
=
I
> pn
is minimal then
..
Pt
Suppose for some
be a dual basis t
i
,
f
)
C
x
Pt
We first
.
Suppose not
f
x1
S<...<
where
whe re
so
f
4)
f s H
#(f) > pn
#(f) = pn .
= n ,
#(h)
h
there is
h E V1
Since
Proof:
=
/
ek
the ordered basis,
> pn
+00+fit
pIf
90*
t
x kf
k=1
a* = a*
t
Let
it
ak]
f
fj /p
*.
*
1
/j
=
<p(a*), f) = <a*, v(f)> = <a*, h> .
c
However,
J
k=1
Itat
/p
-A
146.
(f)
a* 6 J
so
P
and
4)
Thus in
.
p '/f
Then
K(H)
hence, there is
K(f)C W
where
W
n
f
x e LH
, since
where
pe 1
By
.
V(f-x) = V(f)
,
= h .
and shows we can
Then
pe
f
a
Ap (x
= Z
is defined in
= K(x)
K(f)
LH = Ker V
4A(f) = pn .
where
r
W = K(LH);
This contradicts the minimality of #(f)
choose
there is
by
,
# (f-x) <#(f)
j
we can assume for each
the proof of
K(f)
<a*, h> = 0
implies
= 0 .
/t/
where
> n =4(h)
p
M
)+
f
Pt it
ii
Pi
--- x t
/
i
contained in the ordered basis,
Ix
Pi < - < Pt
f
is minimal.
If
plf M1
M
M= 0
+
and we can assume
aa*j C J
a* = a
-
p
a
*
a
M
/)M
k ,
we are done, suppose not.
As usual
.
E
M
Suppose
is a dual basis to
tI
and
x
ft = pn ,
±+*'
,(/
i
< a.
1-
p
0
--
a
t
I*
p
M
p
t-
ei
,a(x
1
a
m
ei.
m
147.
f> = <a*, v(f)> = <a*,
=<p(a*),
fM
fM
-z
Hence,
,
p
which implies
M.
r -1t
A<a P1
*I
/M
p *fM
a(xal
and contradicts the minimality of
0
=
ei
m
a
1
r .
for some
It is no loss to assume
r
r=
Then by
,0
(0
f
M
- f))= 2f
K(xP(aP *
1
x
i
f
'It
1
and
-
x P(a
x
if
=
E
fM
pe i
pei
a
K(f-x) = Z
i
then
LH
m
m
/(x a
f
f
M-1
+ Z
j=1
x e LH
of
so
M > 0
-
i)x 1
- x
it
t
V(f-x) = h , this contradicts the minimality
hence
M = 0 .
Q.E. D.
148.
Theorem 7.
U-modules.
is a regular morphism of left
lows:
for
l
choose a basis--say
(xJ]
Fix an ordered basis
(v ).
--for
for
1
Ax1a ei
K(VT) =
obtained as fol-
V
, extend to a basis for
(v1 )
a basis for
let
(V
Fix a basis
Proof:
which
P: H -+ U
There is a projection
L
am
Extend to
(v )2 ''' '
For each
V
E
V
ei
m
By lemma 6
we can
i
find and fix
a(VT)
e
H
pe i
pe
m
a
m
a1
K(a(V7)) = 7 A
where
x
.
Then by linearity
i
to
a
we can extend
a:
V1
-* H
VIY
-+ a(VY)
.
We wish to show
m o ia
U
)v
-
----
H
is a linear isomorphism.
Surjective.
H
By induction on
C Im m o 16 a .
as a basis for
Hn-1 C Im(m o 1
Since
(v1 )0 ,
a),
n
a(l) = 1
we shall show
because we chose
H 0 C Im(m o i
let
Q
h e Hn - Hn-1
a)
.
Suppose
1
149.
f ii
a
K(h)
I
=
a
J
suppose for some
x m
---
xl
+ 0
where
p0f1
,...
PIf
Then
f
Sim
1
j
where
plf 1
Of
,.-.
k
V
Z
k
V(h)=12
x am p
---
11p x
K(V(h))
Consider
vkP a(v k
k
a
K(x)
=
j
a
-- x m
1
lx
where
p fil
x e Im(m o i (D a)
s-0 *pfb
m
so that by considering
assume
a
K(h)
a
-o-
= Z y x
xm
ii
where for each
j
there is
where
ft
p
fit
h - x
we may
150.
Then as in the proof of
K(h) = K(I
A
where
hi
E k,
ai
Hn
X
E
U
OV
U
V
a left
S
Z
Sh i))
a *
Hn-1
where
Since
E H
h
so by induction there is
m o 1 0 a(xi) = ai*
is naturally a left
h i.
m o i @a
U-module.
is
U-module morphism so that
iA
0
- X
h - m o i@
Xi)
a(Ai
= X E U (V
K(h) - K(m o i
implies
L
* h
m o 1
Let
A(*
Ai(ai* - hi)
Then
so
CIm m o 10
h e Im m oi @ a
m o i @ a
.
a(X)) = 0
a(X) e Hn-
=
and
H
a
C
by
0
which
,
Im m o 10 a
Thus
is surjective.
Injective'
Since
U, V
are filtered,
U
V
is
filtered by
n
(U
Suppose
@Vi)n
m o 1
a
=
Ui
m o i@
X
)n
X
E
a r.u.e.a. of
(U
L
V
,
.
is not injective, then there is
X s (U0 V1 ) where
where
(V )n_
U
and
a(X) = 0 .
n
We choose such
is minimal.
has a basis
Since
U
is
~1
151.
el
x
a
and
em
a1
000
al <---< am
m = 0,1,...
0 < ej < P
x m
I
e
e1
aM
x
m
Ue
E
1
a1
i,j
m -
ei
m
a
x m
e
Thus we can write
+---+ e
SjV
'
a
Moreover since the sum is
0 < e r < p .
<---< a
r
finite we assume for
Vi
(appearing in the sum)
a
) e k[
K(V
x~M
xa
This is a notational convenience later.
X
n > e
+---+
e1
(U
,
implies
V1 )n
+ A(V J)
whenever
W 1 0 .
(This
uses the fact we have chosen an extending basis for
Moreover by minimality of
n = e
+ - 0+ e
Choose
q,r
+ #(V
n
there is
7q
j 0
r
where
"A
+ 0
'V
n = e
+
-*+e
+ #(V r)
and
a
V1
where
)
A
152.
# (V r) is maximal.
q,r = 1,1 #(v 1) = N ,
Say
p#(V )
a il
(m o i 9 a(x A
el +- - -+ el
i)
#(V
= N
X
Q
O
pN .
x
If equality holds then
Notice
if
1,1
so by our choice of
e
a im
'V
7 ~
-0*
= M.
1
since
n > e
i1
+- -+
ei
m
+ $(V'
T = M + pN .
Let
a,*J C H*
Since
U
has a basis of monomials let
be a dual basis and let
x kj
be the dual basis to
.*
where
<b* - h, g> = <b*, gh>
of
act as derivations.
H
<ak *
V
|j
xa
k,,
e Il
.
H*
fak*
is a right
H-module
and the primitive elements
By our choice of
I = fjl#(V J) = N
Let
a,*
m
ak*
,
1 E I .
is a linearly independent set and since we
have chosen an extending basis for
f K(V
)|j
V
,
E I
is a linearly independent set, as is
{a(K(V
Then
a(K(a(V J)))
))| j
)
E
I.
E
I
is a linearly independent set.
(Because by definition of
153.
a(V j ) if
Z
a(K(a(V 4 )))
e.
tJ
j sI
then
= 0
1 a(K(V VJ)) = 0
zF
Thus there is a set of elements
E H*
[pN]
4 sI,
where
a * [1
B
B
and
If
*)
Q
mpN-1 . H*
b *=mpN-1 (B4 *)
K(a(V
))
B *
on
b*
a(V
4 El
)
JEl
b
E E
pN
b
Moreover by
[each exponent is divisible by
that when
p1 1, 4,2
b
to
)))
is usual multiplication
j EI
j IE ,/
4,-
it follows for
so that
m*
a(V
H* -+ H*
...
is a dual basis
the form of
-...
is a dual basis to
jel
then letting
rb
',2]
n 3j
p ]
1 0 we can assume
Olf m.OJ,
'---0
sf a*Pa* s J
Thus any derivation vanishes
.
b *
a *e11
*0 -a*
m5
T
so that
e
<b *l
a1 *
1
*-m *
m
,m
mo i
Gb)
e
aA
<b * * a 1 *
*--*
a
*
m, m 0 i 0 a (
=
ei
am m
xm
jE:I
--I
154.
a 1
1
el
e1
A
a*..*
<(b *
,
m
jIE
=
... aeim
m
vl >
1
e
-
el 1
0 <
e1 k
< p
<b1 *, VT>
'
z
-**
e
.m
k = 1 , ... m
m o i 0 a(X) = 0
this contradicts
:
X1.,
0
and hence
so
m o i@ a
is injective.
U
Thus
left
m o10a
-
V
is an isomorphism (of
r @ E: U@
Then
U-modules).
H
V
-+U a projection
U-modules induces
of left
P: H a projection of left
U
P
U-modules.
is defined by commu-
tivity of
mo i@
a
U
I
H
1® e(1 ( 1) = 1 .
and since
H
F
m o i 0 a(1
is coconnected
P
1) = 1 , hence,
is regular.
P(l) = 1
Q.E.D.
This theorem establishes the difficult part of showing
H
is an extension of
H/LH
by
U
as an algebra and
155.
coalgebra.
Coalgebra Extension
Assume
* = m o i
under
7P
U
is
U
I .
H
commutative.
is
Let
4: U @ H -+ H
a Hopf algebra and
i s a left
H
is
a coalgebra,
U-module as a coalgebra.
Then
sequence
U0
is
4'
-+
H
Ti
H
H/LH
-
split exact by theorem 7.
H/LH
is a right C.H.A.
s: H/LH - H/LH @ U
d
H
4 HOH
$
I
-
=
I (
U-comodule under
k .
By cocommutivity
7r
H
H/LH
(2,1)
-
H/LH@ H
d
H
H/LH@U@H
0H
-~H/LH
OH
7rI
is
of
commutative; hence, by definition
H/LH
by
U
H
is an extension
as a coalgebra.
Algebra Extension
H
is a left
H-module under the adjoint representation
156.
~,
@I
d
H 0H
(1,3,2)
H
-
H (DH
.v
32)
H
H
H
H
H
H
m
2
H
S
denotes the antipode which exists because
1, x, y,
4
We show
coconnected.
H
is
defines a module structure.
z e H
1 - z = 1 z S(1) = z
x - (y - z) =
=
z)S(x i
xi(y
t )
x'(yj '(z)S(yj"))S(xi I)
2
i, j
=2Zx'yJ'
z S (-X
"y"
1,j
=
=
We show
H
I
(xy) 1 ' z S(xy)i"
(xy) - z .
is a left H.A.
x - 1 = Z xi'
H-module
1 S (Xg ")
=
E(x)
i
x
(yz)
;
i
x
'(yz)S(xi")
Z x'
i
y E (X1 11) z S(xi'")
157.
x 1Z
1
i
= z
z S(x'")
)x iyS(x
1
"
(xi' -
i
also is
1
z)
)(xi
.
a coalgebra morphism i.e.
H
H
H
d 0d
H
H
H
H
d
(1 3,2,4)
NJr
H
is
H
H
H
H
--p
H
commutative.
d(x - y) = d(Z x ' y S(x')
i
S(x
")
i9J
=
Z
xi
'y I S(xi ")
x
J
xi
Y
I S(x
Ill
)
"t S(xi " t )
1,j
=
o (1,3,2,4) o d
This shows--by an easy induction-Hence
H
is
H.
LCH 1 .
a left H.A.
Since
H-module
L
d(x
#-(x
generates
H - U C U .
@
y)
- y) < T (y)
U ,
H
s
Assume
C U
U
and
is
158.
commutative, if
x,y E U
x
=
'
y S(xj)
i
=Z x'
i
Thus
U+
|H @
and
U = 0
U
factors to
a left
(HU+)
S(x i")y = E
- U = 0
since
?': H/LH Q U -+ U
4p
under
H
is a left H.A.
HU+ =HL = LH
4
Under
.
H/LH-module and a left C.H.A.
y
,
U
is
H/LH-module since
H-module.
We show
4
sat-
isfies
d
I
H@ H
(1,3,2)
)
H@H®@ H ------
A,
H
H
m
* @)I
m
-----
H
x
xy =
E (x "j) = Z x
y
(x
=
If
I
H
4:
then
4
H
H &H
-
)x '"
' y
i
1
(xi' -y )xi".
H/LH -comodule by
is an
d
H-
I
) H (DH
7r
y
H
H/LH
is clearly an algebra morphism so
H/LH- comodule as an algebra.
H
is a right
From the above diagram and
159.
cocommutivity we have the commutative diagram:
d
H
I
@-U
>H @H®U
H
U
(2,1)
U
I@
H
1
i @I
H
H
m
m
-
H 0H
Using
$ = I
@7r
o d
and the definition of
H
*
we have
the commutative diagram:
H
H 0H/LH
U
U
H
U
j
U
I
V
I
m
Thus the second condition for
H
a: H/LH -+ H
m
to be an extension of
U --as an algebra--is satisfied.
By theorem 2)
H
'I
H 0H
by
H
i
H
H/LH
(2,1)
defined by
16o.
d
P1
H(
H
I
m o 1I
H
U
H
H
w
CFa
H/LH
is a morphism of right
theorem
7.)
(H/LH
V)
=
H/LH-
comodules.
It is invertible because
.
is
(P
H/LH
as in
is coconnected.
Moreover by theorem 2)
i@
a
U 0H/LH
)H
is a linear isomorphism.
@
m
H
)> H
Observe
H/LH
Dl
k .
U C H
By
theorem 1
i
H/LH
(H
a
k H/LH
O
m
)H
(D H
--
)H
H/LH
is a linear isomorphism.
U-
is a split exact sequence.
dition for
H
Hence,
U = H
H
H
®
0
k , and
H/LH
This verifies the first con-
to be an extension of
H/LH
by
U .
161.
Bibliography
[1]
E. Artin, C. Nesbitt, and R. Thrall, Rings with
Minimum Condition, University of Michigan Press,
1944.
[2]
N. Jacobson,
[3]
J. Milnor and J. Moore, On the Structure of Hopf
Algebras, Princeton University Notes.
Lie Algebras, Interscience, 1962.
162.
Index
Adjoint representation,
Algebra, 8
111
A ,
commutative, 8
Antipode, 17
Augmentation, 10, 11
Brauer Group, 80
Coalgebra, 10
C.,
111
cocommutative, 11
Coconnected, 93
Cohomology, 47, 87
6, 48
6n
,
88
H(B,H)1 , 88
H' (H,B), 56
Hom(C,A)n, 88
Homn(C,A), 48
normal complex, 58
Reg(C,A)n
88
Regn (C,A), 48
155
r
163.
Coideal,
43
Comodule, 12
(C.)H.A. comodule, 87
comodule as an algebra, 36
$2,
structure on
X © X, 36
Connected, 143
Divided power, 122
Duality, 14
Extension, 61, 62, 89,
90, 155
f
B 0) H, 67
Brauer Group, 80
H 9)
g
B,
91
isomorphism of extensions, 62, 90
left exact sequence, 61
product of extensions, 81
right exact sequence, 89
smash product, 63
split exact sequence, 61, 89
Filtration, 26
r(G),
17
Grouplike element, 24
Hopf algebra, 16
coconnected, 93
conilpotent,
22
164.
filtration, 26
24
G(H),
Hg,
25
L(H),
94
split, 21
U
95
,
J , 26
K(x),
98
L , 121
Module,
9
(C.)H.A.
H
as
module,
H*,
48
21
module as a coalgebra, 43
e3' -module,
n
2
7
structure on
Poincare-Birkhoff-Witt
X 0 X, 42
theorem, 140
Primitive element, 94
p
,
113
Restricted Universal Enveloping Algebra, 93, 95
Smash product, 63
m
He Tr(G) -+ H , 66
Symmetric tensor, 97
Universal Enveloping Algebra,
93, 95
165.
Unit, 8,
V
V,
,
13
115
119
Vector space, 7
X , 110
166.
Biography
Moss Eisenberg Sweedler was born in Brooklyn, New York,
April 29, 1942.
He was graduated from the Massachusetts
Institute of Technology with a B.S. in June 1963.
tinued study there as a graduate student.
He con-