Coefficient Space Properties and a Schur Algebra Generalization
Except where reference is made to the work of others, the work described in
this dissertation is my own or was done in collaboration with my
advisory committee. This thesis does not include
proprietary or classifled information.
David P. Turner
Greg A. Harris Randall R. Holmes, Chair
Associate Professor Associate Professor
Department of Mathematics Department of Mathematics
William D. Ullery Eugene J. Clothiaux
Professor Professor
Department of Mathematics Department of Physics
Stephen L. McFarland
Acting Dean
Graduate School
Coefficient Space Properties and a Schur Algebra Generalization
David P. Turner
A Dissertation
Submitted to
the Graduate Faculty of
Auburn University
in Partial Fulflllment of the
Requirements for the
Degree of
Doctor of Philosophy
Auburn, Alabama
December 16, 2005
Coefficient Space Properties and a Schur Algebra Generalization
David P. Turner
Permission is granted to Auburn University to make copies of this dissertation at its
discretion, upon request of individuals or institutions and at their expense.
The author reserves all publication rights.
Signature of Author
Date of Graduation
iii
Vita
David Presnell Turner, son of Thomas E. Turner and Mary Elizabeth Shope Turner, was
born June 24, 1960, in Lancaster, Pennsylvania. He graduated from North Gallia High
School in Vinton, Ohio in 1978. He entered Rio Grande College in September 1978. He
graduated with the degree of Bachelor of Science in mathematics in June 1986. He entered
Indiana University in Bloomington, Indiana in August 1986. He graduated with the degree
of Master of Arts in mathematics in May 1988. He entered Purdue University in West
Lafayette, Indiana in August 1989. He graduated with the degree of Master of Science
in mathematics in May 1994. He enrolled in the Ph.D. program in the Department of
Mathematics, Auburn University in September, 1993. He worked as Adjunct Instructor of
Mathematics for Lexington Community College in Lexington, Kentucky from August 1991
to June 1993. He has taught mathematics and physics at Faulkner University in Mont-
gomery, Alabama from August 1993 to the present. He married Brenda White, daughter
of Fred and Ruth (Dillon) White, on August 31, 1984.
iv
Dissertation Abstract
Coefficient Space Properties and a Schur Algebra Generalization
David P. Turner
Doctor of Philosophy, December 16, 2005
(M.S., Purdue University, 1994)
(M.A., Indiana University, 1988)
(B.S., Rio Grande College, 1986)
54 Typed Pages
Directed by Randall R. Holmes
Let K be an inflnite fleld and ? = GLn(K). If we linearly extend the natural action of ?
on the set E of n-dimensional column vectors over K to the group algebra K?, then E
becomes a K?-module. We then construct theK?-module E?r, the r-fold tensor product of
E. The image Sr(?) of the corresponding representation of K? is called the Schur algebra.
If E is replaced by a difierent K?-module L, the same construction results in an algebra
Sr;L. The subalgebra A(n) of K? generated by the coordinate functions cfifl : ? ! K with
1 ? fi; fl ? n is a bialgebra. A(n) has a subcoalgebra Ar which consists of homogeneous
polynomials of total degree r in the indeterminants cfifl. Classically, the dual A?r of Ar is an
algebra isomorphic to Sr(?) and Ar is the coe?cient space of E?r. We identify Sr;L with
the dual A?r;L of the coe?cient space Ar;L of L?r and give a description of Ar;L.
v
Acknowledgments
This work is dedicated to my mother Mary Elizabeth Shope Turner (1927-1985) and to
Brenda White Turner, my wife for the past 21 years. Words cannot express my indebtedness
to each of them and my love for both of them. Many thanks to Dr. Randall Holmes, my
major professor, for his algebraic expertise and patient guidance, and to the other members
of my committee, Dr. William Ullery, Dr. Greg Harris, and Dr. Eugene Clothiaux, for
their excellent classroom instruction in mathematics and physics. Also thanks to Faulkner
University for sabbatical leave during my residency year at Auburn and for numerous
scheduling accommodations.
vi
The style manual used is Auburn University Graduate School Guide to Preparation and
Submission of Theses and Dissertations. The bibliography uses the style in LATEX: A
Document Preparation System by Leslie Lamport.
The computer software packages used are LATEX2" and AMS-LATEX with diagrams gener-
ated by XY-pic. All packages are under a MiKTEX implementation with WinEdt used as
the text editor. The flnal output was printed from Adobe Acrobat.
vii
Table of Contents
List of Figures ix
1 Preliminaries 1
1.1 Modules, Algebras, and Group Rings . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Representations and Characters . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.4 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2 Algebras and Coalgebras 16
2.1 Algebras and Commutative Diagrams . . . . . . . . . . . . . . . . . . . . . 16
2.2 Coalgebras and Bialgebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3 Results in Schur Algebras 27
3.1 Polynomial Functions and Coe?cient Space . . . . . . . . . . . . . . . . . . 27
3.2 Schur Algebras and Group Actions . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Bibliography 43
Index 44
viii
List of Figures
Figure 1: Tensor Product Universal Property 6
Figure 2: Associative Law and Unitary Property 16
Figure 3: Tensor Product of K-algebras 18
Figure 4: Coassociative Law and Counitary Property 19
Figure 5: Dual of a K-Coalgebra 20
Figure 6: Dual of a Finite-Dimensional K-Algebra 21
Figure 7: Tensor Product of K-coalgebras 23
Figure 8: Coalgebra Homomorphism 25
Figure 9: Bialgebra Equivalent Conditions 26
Figure 10: T3;2(g) 35
ix
Chapter 1
Preliminaries
Deflnitions and statements of standard results in the theory of modules, algebras, group
rings, tensor products, representations, characters, and linear functionals have been drawn
from [1-8, 10].
1.1 Modules, Algebras, and Group Rings
1 DEFINITION. Let R be a ring. A left R-module is an additive abelian group M
together with a function R ?M ! M ((r; m) 7! rm) which satisfles the module axioms
(i) r(m+n) = rm+rn, (ii) (r +s)m = rm+sm, and (iii) r(sm) = (rs)m for all r; s 2 R
and m; n 2 M. A right R-module has a similar deflnition with r on the right. Let M be a
(left) R-module. M is called unitary if R has an identity 1R and 1R?m = m for all m 2 M.
N is called an R-submodule of M if N is a subgroup of M and rn 2 N for all r 2 R and
n 2 N. If N is an R-module, a function f : M ! N such that f(m + n) = f(m) + f(n)
and f(rm) = rf(m) for all m; n 2 M and r 2 R is called an R-module homomorphism.
The set of all R-modules homomorphisms from M to N is denoted HomR(M; N). Let K
be a fleld. A unitary K-module V, a K-submodule of V, and a K-module homomorphism
are called a K-space, a K-subspace, and a K-linear map, respectively.
\Module" means \left module" unless otherwise noted. K always represents a fleld. Since
K is commutative, a K-space V can be viewed as a right K-space by deflning kv = vk
for all k 2 K and v 2 V. An injective, surjective, or bijective homomorphism is called a
monomorphism, epimorphism, or isomorphism, respectively.
1
2 EXAMPLES. Let R be a ring and f : M ! N an R-module homomorphism. Then
kerf = f?1(f0g) is an R-submodule of M, imf is an R-submodule of N, and the quotient
group M=N = fm+N j m 2 Mg is an R-module called a quotient module.
3 THEOREM (First Isomorphism Theorem). If f : M ! N is an R-module homomor-
phism then M=kerf ?= imf.
Proof. See [5, p. 172].
4 DEFINITION. A K-algebra is a ring A with identity such that A is a K-space (with
addition via the ring structure) satisfying the algebra condition k(ab) = (ka)b = a(kb) for all
k 2 K and a; b 2 A. A K-subalgebra of a K-algebra is a subring that is also a K-subspace.
If A and B are K-algebras, then a K-algebra homomorphism is a ring homomorphism
? : A ! B mapping 1A to 1B such that ?(ka) = k?(a) for all k 2 K and a 2 A.
5 LEMMA. Let A be a ring with identity. Then A is a K-algebra if and only if there is
a ring homomorphism f : K ! A such that f(K) cent(A) and f(1K) = 1A.
Proof. (=)) Deflne f : K ! A by f(k) = k1A. We have that f is a ring homomorphism
since f(jk) = (jk)1A = j(k1A) = j(k(1A1A)) = j(1A(k1A)) = (j1A)(k1A) = f(j)f(k) and
f(j + k) = (j + k)1A = j1A + k1A = f(j) + f(k) (j; k 2 K) by the algebra condition
and module axiom (ii). Also f(k)a = (k1A)a = k(1Aa) = ka = k(a1A) = a(k1A) = af(k)
(k 2 K, a 2 A) implies f(K) cent(A), and f(1K) = 1K1A = 1A since A is unitary.
((=) Deflne k?a = f(k)a (k 2 K, a 2 A) where f(k)a is the multiplication in the ring A.
Note f(k)a = af(k) since f(K) cent(A). Let j; k 2 K and a; b 2 A. Since A satisfles
ring distributive and associative laws, and f is a ring homomorphism,
(i) k?(a+b) = f(k)(a+b) = f(k)a+f(k)b = k?a+k?b,
(ii) (j +k)?a = f(j +k)a = (f(j)+f(k))a = f(j)a+f(k)a = j ?a+k?a,
(iii) j ?(k?a) = f(j)(f(k)a) = (f(j)f(k))a = f(jk)a = (jk)?a,
2
(iv) 1K ?a = f(1K)a = 1Aa = a, (v) k?(ab) = f(k)(ab) = (f(k)a)b = (k?a)b, and
(vi) (k?a)b = (f(k)a)b = (af(k))b = a(f(k)b) = a(k?b).
Thus A is a K-space by (i) - (iv), and satisfles the algebra condition by (v) and (vi).
6 NOTATION. Let ? = ?n (n 2 Z+) denote the general linear group GLn(K) and put
K? := ff j f : ? ! Kg.
7 EXAMPLES. The following are K-algebras: (a) K, (b) the set MatnK of all n ? n
matrices over K, (c) the set EndK(V) of all K-linear maps from a K-space V to itself, and
(d) K? with pointwise addition and multiplication, and identity 1K?(g) = 1K for all g 2 ?.
8 DEFINITION. Let G be a group and R a commutative ring with identity 1R 6= 0R.
The group ring RG of G over R is the set of all (formal) sums Pg2G rgg where only flnitely
many rg 2 R satisfy rg 6= 0R. The equation Pg2G rgg +Pg2G sgg = Pg2G(rg +sg)g deflnes
addition while ?Pg2G rgg??Ph2G shh? = Pg;h2G(rgsh)(gh) = Pg2G?Ph2G rgh?1sh?g deflnes
multiplication where rgsh is the product in R and gh is the product in G.
RG is a ring. By the deflnition of multiplication, RG is commutative if and only if G is
abelian. We may consider G as a subset of RG by identifying g 2 G with 1Rg. Similarly,
R RG by identifying r 2 R with r1G. Thus, by restriction, any KG-module may be
viewed as a K-space. Further, KG is a K-space with scalar multiplication given by the
ring multiplication (viewing K KG).
9 LEMMA. Let H be a group. Then KH is a K-algebra.
Proof. KH is a ring by the preceding remark. It has identity 1K1H. Deflne f : K ! KH
by f(k) = k1H (k 2 K). So f(j + k) = (j + k)1H = j1H + k1H = f(j) + f(k) (j; k 2 K)
and, by the deflnition of multiplication in KH,
f(jk) = (jk)1H = (jk)(1H1H) = (j1H)(k1H) = f(j)f(k).
3
Consequently f is a ring homomorphism. For k 2 K and s 2 KH, we have
f(k)s = (k1H)s = ks = sk = s(k1H) = sf(k),
so f(K) cent(KH). Also f(1K) = 1K1H. Lemma (5) implies KH is a K-algebra.
10 THEOREM. Let H be a group, A a K-algebra, and A? the multiplicative group of
invertible elements of A. Then every group homomorphism ? : H ! A? has a unique
extension to a K-algebra homomorphism ? : KH ! A.
Proof. Suppose ? : H ! A? is a group homomorphism. We deflne ? : KH ! A by
??Ph2H ahh? = Ph2H ah?(h). Then
?
X
h2H
ahh+
X
h2H
bhh
?
= ?
X
h2H
(ahh+bhh)
?
=
X
h2H
(ah +bh)?(h)
=
X
h2H
ah?(h)+
X
h2H
bh?(h) = ?
X
h2H
ahh
?
+?
X
h2H
bhh
?
and
?
?X
h2H
ahh
??X
h2H
bhh
??
= ?
X
g2H
?X
h2H
agh?1bh
?
g
?
=
X
g2H
X
h2H
agh?1bh
?
?(g)
=
X
h2H
X
g2H
agh?1bh
?
?(g) =
X
h2H
X
g2H
agbh
?
?(gh)
=
X
h2H
ah?(h)
? X
h2H
bh?(h)
?
= ?
X
h2H
ahh
?
?
X
h2H
bhh
?
show ? is a ring homomorphism. Also ?(1K1H) = 1K?(1H) = 1K1A = 1A. Now let k 2 K
and Ph2H ahh 2 KH. Then
?
k
X
h2H
ahh
?
= ?
X
h2H
(kah)h
?
=
X
h2H
(kah)?(h) = k
X
h2H
ah?(h) = k?
X
h2H
ahh
?
.
Consequently, ? is a K-algebra homomorphism. Finally, we establish uniqueness. Suppose
that ? : KH ! A is a K-algebra homomorphism such that ?jH = ?. Then ? = ? since
?
X
h2H
ahh
?
=
X
h2H
ah?(h) =
X
h2H
ah?(h) = ?
X
h2H
ahh
?
.
4
1.2 Tensor Products
In this section, K-spaces are assumed to be flnite-dimensional.
11 DEFINITION. Let fv1; v2; ??? ; vng and fw1; w2; ??? ; wmg be bases for K-spaces V
and W, respectively. Then the tensor product of V and W, denoted V ?W, is the K-space
with basis fvi ?wj j 1 ? i ? n; 1 ? j ? mg. For arbitrary v 2 V and w 2 W, we may
write v = Pi fiivi and w = Pj fljwj. We deflne v ?w := Pi;j fiifljvi ?wj 2 V ?W.
12 REMARKS. Let V and W be K-spaces. (a) dim(V ?W) = (dimV)(dimW) follows
from the deflnition. (b) Let v 2 V. Then v?0 = v?(0+0) = v?0+v?0. Since 0 is the
only element of a group that satisfles x + x = x, we have v ?0 = 0. Similarly, 0?v = 0.
(c) The tensor product of V1 ?????Vn of n K-spaces V1; :::; Vn is deflned similarly. We
have v1 ?????vn = 0 if any vi = 0.
13 LEMMA. Let V and W be K-spaces. Suppose u 2 V ?W. Then there is a positive
integer n, a linearly independent subset fv1; :::; vng of V and a subset fw1; :::; wng of W
such that u = Pni=1 vi ?wi.
Proof. Let fvfigfi2I be a basis of V. Write u = Pni=1 xi ? yi (xi 2 V; yi 2 W). Thus
xi = ki1vfi1 +???+kinvfin (kij 2 K, vfij 2 V, 1 ? i; j ? n). Then
u =
nX
i=1
(ki1vfi1 +???+kinvfin)?yi =
nX
i=1
[(ki1vfi1 ?yi)+???+(kinvfin ?yi)]
=
nX
i=1
[(vfi1 ?ki1yi)+???+(vfin ?kinyi)]
= (vfi1 ?k11y1 +???+vfin ?k1ny1)+???+(vfi1 ?kn1yn +???+vfin ?knnyn)
= [vfi1 ?(k11y1 +???+kn1yn)]+???+[vfin ?(k1ny1 +???+knnyn)]
=
nX
i=1
vfii ?(k1iy1 +???+kniyn).
The result follows since each k1iy1 +???+kniyn 2 W.
5
14DEFINITION. IfR isacommutativeringwith1R, M1; :::; Mn, andLareR-modules,
and, for all r; r0 2 R and m1; :::; mn; m0i 2 M, f : M1 ?????Mn ! L satisfles
f(m1; :::; mi?1; rmi +r0m0i; mi+1; :::; mn) = rf(m1; :::; mn)+r0f(m1; :::; m0i; :::; mn)
then f is called n-multilinear (or bilinear when n = 2).
15 EXAMPLES. (a) Let V and W be K-spaces. Deflne fl : V ? W ! V ? W by
fl(v; w) = v?w (v 2 V; w 2 W. Then for all v; v1; v2 2 V, w; w1; w2 2 W, and k1; k2 2 K,
we have
fl(k1v1 +k2v2; w) = (k1v1 +k2v2)?w = k1v1 ?w +k2v2 ?w
= k1(v1 ?w)+k2(v2 ?w) = k1fl(v1; w)+k2fl(v2; w)
and, similarly, fl(v; k1w1 +k2w2) = k1fl(v; w1)+k2fl(v; w2). Thus fl is bilinear. fl is called
the canonical bilinear map. (b) We generalize (a). Let V1; :::; Vn be K-spaces. Deflne
fl : V1 ????? Vn ! V1 ????? Vn by fl(v1; :::; vn) = v1 ????? vn (vi 2 Vi, 1 ? i ? n).
Similar to (a), fl is bilinear. fl is called the canonical n-multilinear map. (c) Similar to (a),
t : V ? W ! W ? V, p1 : V ? K ! V and p2 : K ? V ! V given by t(v; w) = w ? v,
p1(v; k) = vk and p2(k; v) = kv (v 2 V, w 2 W, k 2 K) are bilinear.
16 THEOREM. Suppose U, V, and W are K-spaces and let
f : U ?V ! W be bilinear. Then there exists a unique K-linear
map ?f : U ?V ! W such that ?f ?fl = f, where fl is the canonical
bilinear map.
U ?V
fl
??
f // W
U ?V
?f
;;ww
ww
w
Figure 1: Tensor Product
Universal Property
Proof. See [5, p. 211].
17 LEMMA. Let M, N, P, and Q be K-spaces and let f : M ! P and g : N ! Q be
K-linear maps. Then there exists a unique K-linear map f ?g : M ?N ! P ?Q such
that (f ?g)(m?n) = f(m)?g(n) for all m 2 M and n 2 N.
6
Proof. Deflne h : M ? N ! P ? Q by h(m; n) = f(m) ? g(n). Then h is bilinear. By
Theorem (16) there exists a unique K-linear map f ? g : M ? N ! P ? Q such that
(f ?g)?fl = h where fl is the canonical bilinear map. Then for all m 2 M and n 2 N,
(f ?g)(m?n) = (f ?g)(fl(m; n)) = [(f ?g)?fl](m; n) = h(m; n) = f(m)?g(n).
18 DEFINITION. Let V and W be K-spaces with bases V and W, respectively. By
Theorem (16), the map t of Example (15c) induces the K-linear map ? : V ?W ! W ?V
given by ?(v ?w) = w ?v for all v 2 V and w 2 W. ? is called the twist map. Similarly,
for all v 2 V and k 2 K, the maps p1 and p2 of Example (15c) induce the K-linear maps
?1 : V ?K ! V and ?2 : K ?V ! V given by ?1(v ?k) = vk and ?2(k ?v) = kv. ?1
and ?2 are called the canonical projections. ?1 : V ! V ?K and ?2 : V ! K ?V given by
?1(v) = v ?1K and ?2(v) = 1K ?v are called the canonical injections.
19 LEMMA. Let V and W be K-spaces, ? : V ?W ! W ?V and ?0 : W ?V ! V ?W
twist maps, ?1 and ?2 the canonical projections, and ?1 and ?2 the canonical injections. (a)
?0?? = 1V?W, ???0 = 1W?V, ?1??1 = 1V, ?1??1 = 1V?K, ?2??2 = 1V, and ?2??2 = 1K?V. (b)
?, ?1, ?2, ?1, and ?2 are K-space isomorphisms. (c) Let v1; v2; v3 2 V and w1; w2; w3 2 W.
Deflne ? : V ?W ?V ?W ?V ?W ! V ?V ?V ?W ?W ?W by
?(v1 ?w1 ?v2 ?w2 ?v3 ?w3) = v1 ?v2 ?v3 ?w1 ?w2 ?w3.
Then ? is a K-space isomorphism.
Proof. a. (?0 ??)(v ?w) = ?0(w ?v) = v ?w for all v 2 V, w 2 W. So ?0 ?? = 1V?W.
Similarly, ? ??0 = 1W?V. (?1 ??1)(v) = ?1(v ?1K) = v1K = v = 1V(v) for all v 2 V.
Thus ?1 ??1 = 1V. Similarly ?2 ??2 = 1V. For all v 2 V and k 2 K, we have
(?1 ??1)(v ?k) = ?1(vk) = vk?1K = v ?k1K = v ?k = 1V?K(v ?k).
Thus ?1 ??1 = 1V?K. Similarly ?2 ??2 = 1K?V.
7
b. The indicated maps are all K-linear by the preceding remarks. They are K-space
isomorphisms by (a).
c. Similar to the proof that ? is a K-space isomorphism.
Let U, V, and W be K-spaces. The technique proving ? is a K-space isomorphism may
be applied to show that the natural identiflcation of (U ?V)?W with U ?(V ?W) is a
K-space isomorphism. Thus the tensor product is associative.
1.3 Representations and Characters
In this section, K-spaces are assumed to be flnite-dimensional. Also, KG-modules are
assumed to be flnite-dimensional as K-spaces.
20 DEFINITION. Suppose V and W are K-spaces. Denote by GL(V) the group of
invertible K-linear maps from V to itself. If G is a flnite group and ? : G ! GL(V) is a
group homomorphism, then ? is called a representation of G. Let B = fv1; ??? ; vng and
C = fw1; ??? ; wmg be ordered bases of V and W, respectively, and f : V ! W a K-linear
map. For 1 ? j ? n, we may write f(vj) = Pmi=1 fiijwi for unique fiij 2 K. The m ? n
matrix [fiij] is called the matrix of f relative to the bases B and C. Let ? : G ! GL(V) be a
representationand[fiij(g)]thematrixof?(g)(relativetoB)foreachg 2 G. ThenT : G ! ?
given by T(g) = [fiij(g)] is a group homomorphism called the matrix representation of G
afiorded by V relative to B.
Suppose V is a K-space. We establish a correspondence between representations of G and
KG-modules. Let ? : G ! GL(V) be a representation. Then V becomes a KG-module
when we deflne gv = ?(g)(v) for g 2 G and v 2 V and extend linearly to all of KG via
(Pg2G kgg)v = Pg2G kg(gv) = Pg2G kg?(g)(v) (cf. Theorem (10)). Conversely, suppose V
8
is a KG-module. We then deflne ? : G ! GL(V) by ?(g)(v) = gv. For g 2 G, ?(g) is a
linear map by the module axioms. Further
(?(g)?(g?1))(v) = ?(g)[?(g?1)(v)] = g(g?1v) = (gg?1)v = v = 1V(v) (v 2 V).
Hence ?(g)?(g?1) = 1V and ?(g) 2 GL(V). Consequently ? is well-deflned. Finally for
g; h 2 G; v 2 V, ?(gh)(v) = (gh)v = g(hv) = ?(g)(hv) = ?(g)?(h)(v) since V is a KG-
module. Thus ? is a group homomorphism. It follows that ? is a representation of G by
deflnition. We call ? the representation afiorded by V.
21 DEFINITION. Let A = [aij] 2 MatnK, and B 2 MatpK. The trace of A is the scalar
trA = a11 + a22 +???+ ann. The Kronecker product of A and B, denoted by A?B, is a
block matrix in MatnpK whose (i; j)-block is aijB.
22 THEOREM. (a) If A; B; C 2 MatnK with C nonsingular, then tr(AB) = tr(BA)
and tr(C?1AC) = trA. (b) If A 2 MatnK and B 2 MatpK, then tr(A?B) = (trA)(trB).
Proof. a. Let A = [aij] and B = [bij]. Then
tr(AB) = tr
nX
k=1
aikbkj
?
=
nX
i=1
nX
k=1
aikbki =
nX
k=1
nX
i=1
bkiaik
= tr
nX
i=1
bkiail
?
= tr(BA).
So tr(C?1AC) = tr([C?1A]C) = tr(C[C?1A]) = tr([CC?1]A) = tr(IA) = trA.
b. Let A = [aij] and B = [bk?]. Consequently A ? B =
2
66
66
64
a11B ??? a1nB
... ... ...
an1B ??? annB
3
77
77
75 and
tr(aiiB) = aii(b11 +???+bpp) for 1 ? i ? n imply
tr(A?B) = a11(b11 +???+bpp)+???+ann(b11 +???+bpp)
= (a11 +???+ann)(b11 +???+bpp) = (trA)(trB).
9
Let V be a K-space, f : V ! V a K-linear map, and A the matrix of f relative to some
basis B of V. Deflne trf = trA. If a difierent basis B0 is chosen, the matrix of f relative
to B0 is C?1AC, where C is the change-of-basis matrix that changes B0 coordinates to B
coordinates. So trf is well-deflned by Theorem (22a).
23 DEFINITION. Let G be a flnite group, V a KG-module, and ? the representation
afiorded by V. Then ? : G ! K given by ?(g) = tr?(g) (g 2 G) is called the character
of G afiorded by V (or by ?). If V is simple (meaning V 6= 0 and 0 and V are the only
submodules of V), then ? is an called an irreducible character.
24 REMARK. We may extend the deflnition of the tensor product. Let V and W
be KG-modules with respective K-bases fv1; ??? ; vng and fw1; ??? ; wmg. Recall from
Deflnition (11) that the tensor product V ? W of V and W is the K-space with basis
fvi?wj j 1 ? i ? n; 1 ? j ? mg and for arbitrary v = Pi fiivi 2 V and w = Pj fljwj 2 W
we deflne v ?w := Pi;j fiifljvi ?wj 2 V ?W. V ?W becomes a KG-module by deflning
g(v ?w) = gv ?gw for all g 2 G, v 2 V, and w 2 W, and then extending linearly to KG
via (Pg kgg)(v ?w) = Pg kg(gv ?gw).
25 LEMMA. Let U, V, X, and Y be (flnite-dimensional) K-spaces and let f : U ! X
and g : V ! Y be K-linear maps. Then the Kronecker product of matrices representing f
and g is a matrix representing f ?g.
Proof. Let B1 = fu1; ??? ; umg and B2 = fv1; ??? ; vng be ordered bases of U and V, re-
spectively. Also, let C1 = fx1; ??? ; xpg and C2 = fy1; ??? ; yqg be ordered bases of X and
Y, respectively. Then B = fui ? vj j 1 ? i ? m; 1 ? j ? ng is a basis of U ? V and
C = fxi ? yj j 1 ? i ? p; 1 ? j ? qg is a basis of X ? Y by Remark (24). Now let
f(ui) = Ppk=1 fikixk and g(vj) = Pq?=1 fl?jy? where each fiki; fl?j 2 K. Then
(f ?g)(ui ?vj) = f(ui)?g(vj) =
pX
k=1
fikixk
?
?
qX
?=1
fl?jy?
?
=
pX
k=1
qX
?=1
fikifl?j(xk ?y?) (1)
10
Note that A = [fiki] is the matrix of f and B = [fl?j] is the matrix of g relative to the given
bases. We now order B into m ordered lists with the ith list being ui ?v1; ??? ; ui ?vn and
similarly order C into p ordered lists with the kth list being xk ?y1; ??? ; xk ?yq. So (1)
determines the column entries for the corresponding matrix C of f ?g. Since C is a block
matrix whose (k; ?)-block is fik?B, we have C = A?B.
26 THEOREM. Let V and W be KG-modules. Suppose V and W afiord the characters
? and ?, respectively. Then V ?W afiords the character ??.
Proof. Let R be the matrix representation of G afiorded by V relative to the basis A,
and let S be the matrix representation of G afiorded by W relative to the basis B. Then
C = fv?w j v 2A; w 2Bg is a basis for V ?W as in Remark (24). Then T = R?S deflned
by T(g) = R(g)?S(g) is the matrix representation of G afiorded by V ?W relative to the
basis C by Lemma (25). Let ! be the character afiorded by V ?W. Then for each g 2 G,
!(g) = tr(T(g)) = tr(R(g)?S(g)) = [tr(R(g))][tr(S(g))] = ?(g)?(g). Consequently,
V ?W afiords the character ??.
1.4 Linear Functionals
27 DEFINITION. If A is an R-module, then the set A? of all R-module homomorphisms
from A to R is called the dual module of A and the elements of A? are called linear
functionals.
28 EXAMPLES.
a. The trace is a linear functional on MatnK since
tr(cA+B) =
nX
i=1
(cAii +Bii) = c
nX
i=1
Aii +
nX
i=1
Bii = ctrA+trB.
b. The function ? : K? ! K given by ?(?) = ?(1K) (? 2 K?) is a K-linear map.
11
c. Recall ? := GLn(K). Deflne ? : K? ! (K?)? by ?(f)?Pg2? figg? = Pg2? figf(g).
Clearly, ? is K-linear. Suppose f 2 ker?. Then f(g) = ?(f)(g) = 0 for each g 2 ?.
Consequently, f = 0. Hence ker? = 0 and ? is injective. Next let f 2 (K?)?. Then
deflne f = fj?. Thus ? is surjective since ?(f) = ?(fj?) = f. Therefore ? is a
K-isomorphism.
29 LEMMA. Let V be a (possibly inflnite-dimensional) K-space. (a) If V is flnite-
dimensional then V ?= V?. (b) ? : V??V? ! (V ?V)? given by ?(f?g)(x?y) = f(x)g(y)
where f; g 2 V? and x; y 2 V is a K-monomorphism. (c) If V is flnite-dimensional then ? is
bijective. (d) If f1; ??? ; fn 2 V? and x1; ??? ; xn 2 V then : V??????V? ! (V ?????V)?
given by (f1 ????? fn)(x1 ????? xn) = f1(x1)???fn(xn) is a K-linear map, which is a
K-space isomorphism if V is flnite-dimensional.
Proof. a. Let fv1; v2; ??? ; vng be a basis of V. For each i, deflne v?i : V ! K by
v?i (vj) = ?ij (Kronecker delta). Then v?i is a linear functional for 1 ? i ? n. Suppose
Pn
i=1 fiiv
?
i = 0. In particular, fij =
Pn
i=1 fii?ij =
Pn
i=1 fiiv
?
i (vj) = 0 for 1 ? j ? n.
Linear independence of fv?1; v?2; ??? ; v?ng now follows. Next let v? 2 V? be arbitrary.
Then for arbitrary v = Pni=1 fiivi we have
v?(v) = v?
nX
i=1
fiivi
?
=
nX
i=1
fiiv?(vi) =
nX
i=1
v?i (v)v?(vi) =
nX
i=1
v?(vi)v?i
?
(v).
Thus fv?1; v?2; ??? ; v?ng spans V? and is a basis for V?. Hence dimV = dimV?. Recall
that, for K-spaces V and W, V ?= W if and only if dimV = dimW. So V ?= V?.
b. Suppose f; f1; f2; g; g1; g2 2 V?, x; y 2 V, and k; k1; k2 2 K are arbitrary. Deflne
r(f; g) : V ? V ! K by [r(f; g)](x; y) = f(x)g(y). Clearly, r(f; g) is bilinear. By
Theorem (16) we obtain an induced map V ? V ! K and hence an element of
(V ?V)?, which we also denote by r(f; g). We have r(f; g)(x?y) = f(x)g(y). Then
12
r(k1f1 +k2f2; g)(x?y) = (k1f1 +k2f2)(x)g(y) = (k1f1(x)+k2f2(x))g(y)
= k1f1(x)g(y)+k2f2(x)g(y) = (k1r(f1; g)+k2r(f2; g))(x?y)
and similarly r(f; k1g1 + k2g2) = k1r(f; g1) + k2r(f; g2). So r is bilinear. So by
Theorem (16), r induces a K-linear map ? : V??V? ! (V ?V)? such that ??fl = r
where fl is the canonical bilinear map. Thus ? is given by
?(f ?g)(x?y) = [?(fl)(f; g)](x?y) = [(??fl)(f; g)](x?y) = [r(f; g)](x?y)
= f(x)g(y).
Leth 2 Ker?. Then byLemma (13), wemaywriteh = Pni=1 fi?gi whereff1; :::; fng
is a linearly independent subset of V? and fg1; :::; gng V?. Then for all u; v 2 V,
0 = ?(h)(u; v) = ?
nX
i=1
fi ?gi
?
(u; v) =
nX
i=1
fi(u)gi(v) =
nX
i=1
gi(v)fi
?
(u).
Thus Pni=1 gi(v)fi = 0 for all v 2 V. Consequently, gi(v) = 0 (v 2 V, 1 ? i ? n) since
ff1; :::; fng is a linearly independent subset of V?. So h = Pni=1 fi ?gi = 0 and ? is
injective.
c. Let fv1; :::; vng be a basis of V. Then fvij j 1 ? i; j ? ng is a basis for V ?V, where
vij := vi ?vj. We have ?(v?i ?v?j)(vk?) = v?i (vk)v?jv? = ?ik?j? = ?(i;j);(k;?) = v?ij(vk?). So
?(v?i ?v?j) = v?ij and ? is a K-isomorphism.
d. Apply induction to Lemma (17), (b), and (c).
30 DEFINITION. Let V and W be K-spaces and ? : V ! W a K-linear map. If
?(v) = 0 implies v = 0, then ? is called non-singular. The annihilator of S V is the set
S0 of all linear functionals f on V such that f(fi) = 0 for all fi 2 S. The dual of ? is the
map ?? : W? ! V? deflned by [??(f)](v) = f(?(v)) 2 K.
13
31 LEMMA. Let V be a K-space. (a) If W V, then W0 is a subspace of V?. (b) If
W ? V, then W? ?= V?=W0 and W0 ?= (V=W)?. (c) If V and W are subspaces of a K-space
and W ? V, then W0 ? V 0.
Proof. a. Let w 2 W. Then fwg0 = ff 2 V? j w 2 kerfg by deflnition. So fwg0 is a
subspace of V?. Since W0 =
\
w2W
fwg0, it follows that W0 is a subspace of V?.
b. First, deflne ? : V? ! W? by ?(f) = f jW . Then ? is a K-space epimorphism
with ker? = W0. So W? ?= V?=W0 by the First Isomorphism Theorem. Now deflne
? : W0 ! (V=W)? by ?(f)(v+W) = f(v). Then ? is both well-deflned and injective
since, for f 2 W0,
u+W = v +W , u?v 2 W , f(u)?f(v) = f(u?v) = 0 , f(u) = f(v)
, ?(f)(u+W) = ?(f)(v +W).
Let f 2 (V=W)? and v +W 2 V=W. Recall ? : V ! V=W given by ?(v) = v +W is
a K-space epimorphism. Put f = f ??. Then f 2 W0 and
?(f)(v +W) = f(v) = f(?(v)) = f(v +W).
Thus f = ?(f) and ? is surjective. Finally, ? is a K-space isomorphism since for all
u; v 2 V and k 2 K:
?(f)((u+W)+(v +W)) = ?(f)((u+v)+W) = f(u+v) = f(u)+f(v)
= ?(f)(u+W)+?(f)(v +W),
?(f)(k(v +W)) = ?(f)(kv +W) = f(kv) = kf(v) = k?(f)(v +W).
c. Let f 2 V 0. Then f(w) = 0 for all w 2 W. Hence f 2 W0.
32 LEMMA. If V and W are K-spaces andh; i : V ?W ! K is non-singular and bilinear,
then V? and W are isomorphic.
Proof. Deflne ? : W ! V? by [?(w)](v) = hv; wi. Note that ? is well-deflned since
hfiv; wi = fihv; wi and hv1 +v2; wi = hv1; wi+hv2; wi imply that ?(w) 2 V?. Also, since
14
[?(w1 +w2)](v) = hv; w1 +w2i = hv; w1i+hv; w2i = [?(w1)](v)+ [?(w2)](v) and similarly
for scalar multiplication, ? is a K-linear map. Let x 2 ker?. Then hv; xi = 0 for all
v 2 V. Hence x = 0 since h; i is non-singular. So ker? = 0. Thus ? is injective. Finally
suppose fv1; ??? ; vng V is linearly independent and fw1; ??? ; wmg is a basis of W. By
the injectivity of ?, n ? m. Assume n > m. Put cij = hvj; wii. Recall (linear algebra)
there exist a1; a2; ??? ; an 2 K not all of which are zero such that Pj ajcij = 0 for all i
since n > m. So v := Pj ajvj 6= 0. We show hv; wi = 0 for all w 2 W. Thus we must show
hv; wii = 0 for each i. Then hv; wii = ?Pj ajvj; wifi = Pj ajhvj; wii = Pj ajcij = 0 for all
i since h; i is bilinear, contrary to h; i being non-singular. Therefore n = m.
15
Chapter 2
Algebras and Coalgebras
Deflnitions and statements of standard results in the theory of algebras and coalgebras have
been drawn from [9-11].
2.1 Algebras and Commutative Diagrams
33 THEOREM. A is a K-algebra if and only if A is a K-space and there exist K-linear
maps ? : A?A ! A and ? : K ! A such that the diagrams (Figure 2) commute.
Associative Law Unitary Property
A?A?A
1A??
??
??1A // A?A
?
??
A?A ? // A
K ?A
?2 %%JJJJJ
JJJJ
J
??1A // A?A
?
??
A?K1A??oo
?1yytttttt
tttt
A
Figure 2: Associative Law and Unitary Property
Proof. (=)) Deflne m : A?A ! A by m(a; b) = ab for all a; b 2 A. Then m is bilinear.
So by Theorem (16), m induces a K-linear map ? : A ? A ! A such that ? ? fl = m
where fl is the canonical bilinear map. Then ?(a ? b) = (? ? fl)(a; b) = m(a; b) = ab
for all a; b 2 A. Deflne ? : K ! A by ?(k) = k1A. Then for all fi; fl; k 2 K, we have
?(fi + fl) = (fi + fl)1A = fi1A + fl1A = ?(fi) + ?(fl) and ?(kfi) = (kfi)1A = k(fi1A) = k?(fi).
Consequently ? is also a K-linear map. Let a; b; c 2 A and k 2 K. The algebra condition
k(ab) = (ka)b = a(kb) implies a(k1A) = k(a1A) = ka = k(1Aa) = (k1A)a. Then
(??(??1A))(a?b?c) = ?(?(a?b)?1A(c)) = ?(ab?c) = (ab)c = a(bc)
= ?(1A(a)??(b?c)) = (??(1A ??))(a?b?c),
16
(??(??1A))(k?a) = ?(?(k)?1A(a)) = ?(k)1A(a) = (k1A)a = ka = ?2(k?a),
and similarly (??(1A ??))(a?k) = ?1(a?k). Thus the diagrams commute.
((=) Let a; b; c; 2 A and k 2 K. Deflne a product in A by ab := ?(a?b). The product
is associative. Indeed, by the Associative Law diagram commutativity we have
a(bc) = ?(a?bc) = ?(1A(a)??(b?c)) = (??(1A ??))(a?b?c)
= (??(??1A))(a?b?c) = ?(?(a?b)?1A(c)) = ?(ab?c) = (ab)c.
Next, (a+b)c = ?((a+b)?c) = ?(a?c+b?c) = ?(a?c)+?(b?c) = ac+bc. Similarly,
c(a+b) = ca+cb, so the product distributes over addition. Deflne 1A := ?(1K). The (left)
Unitary Property diagram yields
ka = ?2(k?a) = (??(??1A))(k?a) = ?(k1A ?a) = (k1A)a (1)
Similarly, the (right) Unitary Property diagram yields ak = a(k1A). Thus
k(ab) = (k1A)(ab) = ((k1A)a)b = (ka)b = (ak)b = (a(k1A))b = a((k1A)b) = a(kb).
This establishes the algebra condition. Finally, by (1), 1Aa = (1K1A)a = 1Ka = a and
similarly a1A = a. So 1A is an identity. Therefore A is a K-algebra by deflnition.
Theorem (33) permits (A; ?; ?) to denote a K-algebra A and its structure maps ? and ?,
which are respectively called the multiplication map and unit map.
34 THEOREM. The tensor product of K-algebras is a K-algebra.
Proof. Suppose (A; ?A; ?A) and (B; ?B; ?B) are K-algebras, ? : A?B ! B ?A the twist
map, and ?1 : K ! K ?K the canonical injection. Put ?A?B = ?A ??B ?(1A ?? ?1B)
and ?A?B = (?A ??B)??1. We verify the Associative Law and Unitary Property (Figure 3).
17
A?B ?A?B ?A?B
1A?B??A?B
??
?A?B?1A?B//
A?B ?A?B
?A?B
??
A?B ?A?B ?A?B // A?B
Associative Law
K ?A?B
?2 ((QQQQQQ
QQQQQ
QQ
?A?B?1A?B//
A?B ?A?B
?A?B
??
A?B ?K
1A?B??A?Boo
?1vvmmmmmmm
mmmmm
m
A?B
Unitary Property
Figure 3: Tensor Product of K-algebras
Let A and B be bases for A and B, respectively. Then we have for all a1; a2; a3; a 2 A,
b1; b2; b3; b 2B and k 2 K that
(?A?B ?(?A?B ?1A?B))((a1 ?b1)?(a2 ?b2)?(a3 ?b3))
= ?A?B((?A ??B)?(1A ?? ?1B)(a1 ?(b1 ?a2)?b2)?(a3 ?b3))
= ?A?B((?A ??B)(((a1 ?a2)?(b1 ?b2))?(a3 ?b3)))
= ?A?B((a1a2 ?b1b2)?(a3 ?b3))
= ((?A ??B)?(1A ?? ?1B))(a1a2 ?(b1b2 ?a3)?b3)
= (?A ??B)(a1a2 ?(a3 ?b1b2)?b3) = (a1a2)a3 ?(b1b2)b3 = a1(a2a3)?b1(b2b3),
similarly (?A?B ?(1A?B ??A?B))((a1 ?b1)?(a2 ?b2)?(a3 ?b3)) = a1(a2a3)?b1(b2b3),
(?A?B ?(?A?B ?1A?B))(k?a?b) = ?A?B(?A?B(k)?1A?B(a?b))
= ?A?B((?A ??B)(k?1K)?a?b) = ?A?B(?A(k)??B(1K)?a?b)
= (?A ??B ?(1A ?? ?1B))(?A(k)?(1B ?a)?b)
= (?A ??B)(?A(k)?a?1B ?b) = ?A(?A(k)?a)??B(1B ?b)
= ?A(k)a?1Bb = ka?b = ?2(k?a?b),
and similarly (?A?B ?(1A?B ??A?B))(a?b?k) = ?1(a?b?k). Extend linearly. Apply
Theorem (33).
18
2.2 Coalgebras and Bialgebras
35 DEFINITION. If C is a K-space, ?C : C ! C ?C and "C : C ! K are K-linear
maps, and ?1 and ?2 the canonical injections, then (C; ?C; "C) is called a K-coalgebra
whenever the diagrams (Figure 4) commute. ?C and "C are respectively called the comul-
tiplication and counit maps and together are called the structure maps of C.
Coassociative Law Counitary Property
C ?C ?C C ?C
?C?1Coo
C ?C
1C??C
OO
C
?C
OO
?C
oo
K ?C C ?C
"C?1Coo 1C?"C//
C ?K
C
?2
eeKKKK
KKKK
KK?C
OO
?1
99sss
ssss
sss
Figure 4: Coassociative Law and Counitary Property
A K-subspace D of a K-coalgebra (C; ?C; "C) that satisfles ?C(D) D ?D is called a
K-subcoalgebra of C whose structure maps are the restrictions of ?C and "C to D.
36 EXAMPLE. Let H be a group. A := KH ?KH is a K-algebra by Theorem (34).
Deflne ? : H ! A? by ?(g) = g ?g. Then ?(gh) = gh?gh = (g ?g)(h?h) = ?(g)?(h)
for all g; h 2 H. Thus the group homomorphisms ? and ? : H ! K? given by ?(g) = 1K
respectively extend uniquely to K-algebra homomorphisms ? : KH ! A and " : KH ! K
by Theorem (10). Then (KH; ?; ") is a K-coalgebra since
((1KH ??)??)
X
g2H
agg
?
= (1KH ??)
X
g2H
agg ?g
?
=
X
g2H
agg ?(g ?g)
=
X
g2H
ag(g ?g)?g =
X
g2H
ag?(g)?1KH(g),
= ?
X
g2H
(??1KH)(agg)
?
= (??(??1KH))
X
g2H
agg
?
,
(("?1KH)??)
X
g2H
agg
?
= ("?1KH)
X
g2H
agg ?g
?
=
X
g2H
ag1K ?g
= 1K ?
X
g2H
agg
?
= ?2
X
g2H
agg
?
,
and similarly ((1KH ?")??)(Pg2H agg) = ?1(Pg2H agg).
19
37 THEOREM. The dual of a K-coalgebra is a K-algebra.
Proof. Let (C; ?; ") be a K-coalgebra. By Deflnition (30), ?? : (C ?C)? ! C? is given
by [??(f)](c) = f(?(c)) for c 2 C. Deflne ? : C? ? C? ! C? and ? : K ! C? by
?(f ? g)(c) = [?? ? ?](f ? g)(c) and ?(k)(c) = k"(c) for f; g 2 C?, c 2 C, and k 2 K
where ? : C??C? ! (C?C)? is the K-space isomorphism of Lemma (29c). We verify the
Associative Law and Unitary Property (Figure 5).
Associative Law Unitary Property
C? ?C? ?C?
1C???
??
??1C?//
C? ?C?
?
??
C? ?C? ? // C?
K ?C?
??1C? //
?2 &&MMMMM
MMMM
MM C
? ?C?
?
??
C? ?K
1C???oo
?1xxqqqqqq
qqqq
q
C?
Figure 5: Dual of a K-Coalgebra
For c 2 C, write ?(c) = Pi ci ? di, ?(ci) = Pj aij ? bij, ?(di) = Pj eij ? fij, and let
: C? ?C? ?C? ! (C ?C ?C)? be the 3-fold analog of ? (see Lemma (29d)). Then
?(f ?g)(c) = [?? ??](f ?g)(c) = ?(f ?g)(?(c)) =
X
i
f(ci)g(di)
for f; g 2 C? and c 2 C. This implies that for f; g; h 2 C? and c 2 C we have
(??(??1C?))(f ?g ?h)(c) = (?(?(f ?g)?h))(c) =
X
i
?(f ?g)(ci)h(di)
=
X
i;j
f(aij)g(bij)h(di) = (f ?g ?h)((??1C)??)(c)
= (f ?g ?h)((1C ??)??)(c) =
X
i;j
f(ci)g(eij)h(fij)
=
X
i
f(ci)?(g ?h)(di) = (1C? ??)
X
i
f(ci)(g ?h)(di)
?
= (1C? ??)(?(f ?(g ?h))(c)) = ((1C? ??)??)(f ?g ?h)(c)
This establishes the Associative law. Next, for any c 2 C, the commutativity of the
Counitary Property diagrams and Lemma (19c) yields Pi "(ci)di = c = Pi ci"(di) from
20
c = 1C(c) = (?2 ??2)(c) = (?2 ?("?1C)??)(c) = ?2 ?("?1C)
X
i
ci ?di
?
= ?2
X
i
"(ci)?di
?
=
X
i
"(ci)di
and similarly c = Pi ci"(di). Then for all k 2 K, f 2 C?, and c 2 C,
(??(??1C?))(k?f)(c) = ?(?(k)?f)(c) =
X
i
?(k)(ci)f(di) =
X
i
k"(ci)f(di)
= kf
X
i
"(ci)di
?
= kf(c) = ?2(k?f)(c)
and similarly (??(1C???))(f?k)(c) = ?1(f?k)(c). This establishes the Unitary Property
and (C?; ?; ?) is a K-algebra by Theorem (33).
38 THEOREM. The dual of a flnite-dimensional K-algebra is a K-coalgebra.
Proof. Suppose (A; ?; ?) is a flnite-dimensional K-algebra. Then ?? : A? ! (A ? A)? is
given by [??(f)](a?b) = f(?(a?b)) and ?? : A? ! K? is given by ??(f)(k) = f(?(k)) for
f 2 A?, a 2 A, and k 2 K by Deflnition (30). Recall ? : K? ! K given by ?(?) = ?(1K)
for ? 2 K? is a K-linear map . We may now deflne ?A? : A? ! A??A? and "A? : A? ! K
by ?A?(f)(a) = [??1 ???](f)(a) and "A?(f)(k) = [? ???(f)](k) for f 2 A?, a 2 A? ?A?,
where ? : A??A? ! (A?A)? is the K-space isomorphism of Lemma (29c) (dimA < 1 is
required). We verify the Coassociative Law and Counitary Property (Figure 6).
Coassociative Law Counitary Property
A? ?A? ?A? A? ?A?
?A??1A?oo
A? ?A?
1A???A?
OO
A?
?A?
OO
?A?
oo
K ?A? A? ?A?
"A??1A?oo 1A??"A?//
A? ?K
A?
?2
ffMMMM
MMMM
MMM?A?
OO
?1
88qqq
qqqq
qqqq
Figure 6: Dual of a Finite-Dimensional K-Algebra
Write ?A?(f) = Pi gi ?hi, ?A?(gi) = Pj mi;j ?ni;j, and ?A?(hi) = Pj pi;j ?qi;j where
gi; hi; mi;j; ni;j; pi;j; qi;j 2 A?. Then:
21
(?A? ?1A?)?A?(f) = (?A? ?1A?)
X
i
gi ?hi
?
=
X
i;j
mi;j ?ni;j ?hi,
(1A? ??A?)?A?(f) = (1A? ??A?)
X
i
gi ?hi
?
=
X
i;j
gi ?pi;j ?qi;j.
Note that for all f 2 A? and a; b 2 A, we have
f(ab) = [??(f)](a?b) = [??(??1 ???)(f))](a?b) = [?(?A?(f))](a?b)
=
?
?
X
i
gi ?hi
??
(a?b) =
X
i
gi(a)hi(b) (1)
Recall : A? ?A? ?A? ! (A?A?A)? given by (u?v ?w)(a?b?c) = u(a)v(b)w(c)
where u; v; w 2 A? and a; b; c 2 A is a K-space isomorphism by Lemma (29d). It follows
from the deflnition of and (1) that
?
X
i;j
mi;j ?ni;j ?hi
??
(a?b?c) =
X
i;j
mi;j(a)ni;j(b)hi(c) =
X
i
gi(ab)hi(c) = f(abc)
=
X
i
gi(a)hi(bc) =
X
i;j
gi(a)pi;j(b)qi;j(c) =
?
X
i;j
gi ?pi;j ?qi;j
??
(a?b?c)
Since is injective,Pi;j mi;j?ni;j?hi = Pi;j gi?pi;j?qi;j. Consequently, the Coassociative
Law holds. Next, for all f 2 A?, we have
(("A? ?1A?)??A?)(f) = ("A? ?1A?)
X
i
gi ?hi
?
=
X
i
("A? ?1A?)(gi ?hi)
=
X
i
("A?(gi)?hi) =
X
i
(? ???(gi)?hi) =
X
i
(??(gi)(1K)?hi) =
X
i
(gi(?(1K))?hi)
=
X
i
(1Kgi(1A)?hi) =
X
i
(1K ?gi(1A)hi) = 1K ?
X
i
gi(1A)hi = 1K ?f = ?2(f).
For the penultimate inequality, we have used that f(a) = f(1Aa) = Pi gi(1A)hi(a) (a 2 A).
Similarly, ((1A? ? "A?) ? ?A?)(f) = ?1(f). Thus the check of the Counitary Property is
complete and (A?; ?A?; "A?) is a K-coalgebra by Deflnition (35).
39 NOTATION. Let (C; ?; ") be a K-coalgebra. We write ?(c) = Pi ci(1) ? ci(2) for
each c 2 C or succinctly as ?(c) = c(1) ?c(2) with summation implicit.
22
40 LEMMA. Let (C; ?; ") be a K-coalgebra with ?(c) = c(1) ?c(2) for all c 2 C.
a. c(1)(1) ?c(1)(2) ?c(2) = c(1) ?c(2)(1) ?c(2)(2). b. c = "(c(1))c(2) = c(1)"(c(2)).
Proof. a. By the Coassociative Law
c(1)(1) ?c(1)(2) ?c(2) = ?(c(1))?c(2) = (??1C)(c(1) ?c(2)) = ((??1C)??)(c)
= ((1C ??)??)(c) = (1C ??)(c(1) ?c(2)) = c(1) ??(c(2)) = c(1) ?c(2)(1) ?c(2)(2).
b. Since ?1(c) = c?1C and ?2(c) = 1C ?c, by the Counitary Property we have:
1C ?c = (("?1C)??)(c) = "(c(1))?c(2) = 1C ?"(c(1))c(2),
c?1C = ((1C ?")??)(c) = c(1) ?"(c(2)) = c(1)"(c(2))?1C.
Therefore c = "(c(1))c(2) = c(1)"(c(2)).
41 THEOREM. The tensor product of K-coalgebras is a K-coalgebra.
Proof. Suppose (C; ?C; "C) and (D; ?D; "D) are K-coalgebras and ? is the twist map. Put
?C?D = (1C???1D)??C??D and "C?D = ?2?("C?"D). We will verify the Coassociative
Law and Counitary Property (Figure 7).
C ?D?C ?D?C ?D C ?D?C ?D
?C?D?1C?D
oo
C ?D?C ?D
1C?D??C?D
OO
C ?D
?C?D
OO
?C?D
oo
Coassociative Law
K ?C ?D C ?D?C ?D
"C?D?1C?Doo 1C?D?"C?D//
C ?D?K
C ?D
?2
hhRRRRR
RRRRR
RRR ?C?D
OO
?1
66llll
lllll
llllCounitary Property
Figure 7: Tensor Product of K-coalgebras
For all c 2 C and d 2 D, we have
((?C?D ?1C?D)??C?D)(c?d) = [(?C?D ?1C?D)?(1C ?? ?1D)?(?C ??D)](c?d)
= [(?C?D ?1C?D)?(1C ?? ?1D)](c(1) ?c(2) ?d(1) ?d(2))
23
= (?C?D ?1C?D)(c(1) ?d(1) ?c(2) ?d(2))
= [(1C ?? ?1D)?(?C ??D)(c(1) ?d(1))]?(c(2) ?d(2))
= [(1C ?? ?1D)(c(1)(1) ?c(1)(2) ?d(1)(1) ?d(1)(2))]?(c(2) ?d(2))
= c(1)(1) ?d(1)(1) ?c(1)(2) ?d(1)(2) ?c(2) ?d(2)
and similarly ((1C?D ??C?D)??C?D)(c?d) = c(1) ?d(1) ?c(2)(1) ?d(2)(1) ?c(2)(2) ?d(2)(2).
Recall the K-space isomorphism ? of Lemma (19c). Then by Lemma (40a),
?(((?C?D ?1C?D)??C?D)(c?d)) = ?(c(1)(1) ?d(1)(1) ?c(1)(2) ?d(1)(2) ?c(2) ?d(2))
= c(1)(1) ?c(1)(2) ?c(2) ?d(1)(1) ?d(1)(2) ?d(2)
= c(1) ?c(2)(1) ?c(2)(2) ?d(1) ?d(2)(1) ?d(2)(2)
= ?(c(1) ?d(1) ?c(2)(1) ?d(2)(1) ?c(2)(2) ?d(2)(2))
= ?(((1C?D ??C?D)??C?D)(c?d)).
Consequently, ((?C?D ?1C?D)??C?D)(c?d) = ((1C?D ??C?D)??C?D)(c?d) since ? is
a K-space isomorphism. Then extending linearly establishes the Coassociative Law. Next,
for all c 2 C and d 2 D, applying Lemma (40b) yields
(("C?D ?1C?D)??C?D)(c?d) = [("C?D ?1C?D)?(1C ?? ?1D)?(?C ??D)](c?d)
= [("C?D ?1C?D)?(1C ?? ?1D)](c(1) ?c(2) ?d(1) ?d(2))
= ("C?D ?1C?D)(c(1) ?d(1) ?c(2) ?d(2))
= [(?2 ?("C ?"D))(c(1) ?d(1))]?(c(2) ?d(2)) = "C(c(1))"D(d(1))?c(2) ?d(2)
= 1K ?"C(c(1))c(2) ?"D(d(1))d(2) = 1K ?c?d = ?2(c?d)
and similarly ((1C?D?"C?D)??C?D)(c?d) = ?1(c?d). Extend linearly. Thus the Counitary
Property holds and (C ?D; ?C?D; "C?D) is a K-coalgebra.
24
42 DEFINITION. Suppose that (C; ?C; "C) and (D; ?D; "D) are K-coalgebras and
there exists a K-linear map f : C ! D such that ?D ?f = (f ?f)??C and "D ?f = "C
(Figure 8). Then f is called a K-coalgebra homomorphism.
C f //
?C
??
D
?D
??
C ?C f?f // D?D
C f //
"C ??@@@
@@@
@ D
"D
??
K
Figure 8: Coalgebra Homomorphism
43 EXAMPLE. Put L := K ? K. We have that (K; ?K; "K) is a K-coalgebra with
?K : K ! L and "K : K ! K given by ?K(k) = k ? 1K and "K(k) = 1K for all
k 2 K. Let ? : L ! L be the twist map. We may now deflne ?L : L ! L ? L and
"L : L ! K by ?L(k ? ?) = (1K ? ? ? 1K) ? (?K ? ?K)(k ? ?) = k ? ? ? 1K ? 1K and
"L(k??) = (?1 ?("K ?"K))(k??) = ?1(1K ?1K) = 1K for all k; ? 2 K. Then (L; ?L; "L)
is a K-coalgebra by Theorem (41). Deflne ?K : L ! K and ?K : K ! K by ?K(k??) = k?
and ?K(k) = k for all k; ? 2 K. Then ?K is a K-coalgebra homomorphism since
(?K ??K)(k??) = ?K(k?) = k??1K = (?K ??K)(k???1K ?1K)
= ((?K ??K)?(?L)(k??))
and
("K ??K)(k??) = "K(k?) = 1K = "L(k??)
for all k; ? 2 K. Similarly since ?K ??K = (?K ??K)??K and "K ??K = "K, it follows that
?K is a K-coalgebra homomorphism.
44 THEOREM. Let B be a K-space, (B; ?; ?) a K-algebra, and (B; ?B; "B) a K-
coalgebra. The following are equivalent: (a) ? and ? are K-coalgebra homomorphisms, (b)
?B and "B are K-algebra homomorphisms, (c) ?B(bc) = ?B(b)?B(c), ?B(1B) = 1B ?1B,
"B(bc) = "B(b)"B(c), and "B(1B) = 1K for all b; c 2 B.
25
Proof. Consider the following four diagrams:
i. ??? = (???)?(1B ?? ?1B)???? ii. ??? = (???)??1
B ?B ? //
???
??
B
?
??
B ?B ?B ?B
1B???1B
??
B ?B ?B ?B ??? // B ?B
K ? //
?1
??
B
?
??
K ?K ??? // B ?B
iii. "?? = ?2 ?("?") iv. "?? = 1K
B ?B ? //
"?"
??
B
"
??
K ?K ?2 // K
K ? //
1K ??AAA
AAAA
A B
"
??
K
Figure 9: Bialgebra Equivalent Conditions
We have that ? = ?B is a K-algebra homomorphism when (i) and (ii) are satisfled, "B is
a K-algebra homomorphism when (iii) and (iv) hold, ? is a K-coalgebra homomorphism
when (i) and (iii) are satisfled, and ? is a K-coalgebra homomorphism when (ii) and (iv)
hold. So (a) is equivalent to (b). (b) is equivalent to (c) by Deflnition (4).
45 DEFINITION. Let (B; ?; ?) be a K-algebra and (B; ?; ") a K-coalgebra. If any
condition of Theorem (44) is satisfled then (B; ?; ?; ?; ") is called a K-bialgebra.
46 EXAMPLES.
a. (K; ?K; ?K; ?K; "K) is a K-bialgebra. See Examples (7) and (43).
b. Let H be a group. Recall (KH; ?; ?) is a K-algebra by Lemma (9) and Theorem
(33) and (KH; ?; ?) is a K-coalgebra and ? and ? are K-algebra homomorphisms
by Example (36). Thus (KH; ?; ?; ?; ?) is a K-bialgebra.
26
Chapter 3
Results in Schur Algebras
Deflnitions and statements of standard results in the theory of Schur algebras have been
drawn from [12, 13].
3.1 Polynomial Functions and Coe?cient Space
47 DEFINITION. Let E be the set of n-dimensional column K-vectors. For g 2 ? and
x 2 E, deflne gx by usual matrix multiplication. We may extend linearly to all of K? via
(Pg2? kgg)x = Pg2? kg(gx)(kg 6= 0 for flnitely many g 2 ? assumed throughout). Then E
is called the standard or natural K?-module.
We write I(n; r) := fi = (i1; i2; ??? ; ir) j 1 ? ik ? n for 1 ? k ? rg. Suppose
fe1; e2; ??? ; eng is the standard basis for E. Deflne g(v1 ? ??? ? vr) = gv1 ? ??? ? gvr
for g 2 ?. Consequently E?r = E ? ??? ? E (r factors) becomes a K?-module with
K-basis fei = ei1 ? ??? ?eir j i 2 I(n; r)g.
48 PROPOSITION. Let v; w 2 E. (a) ? : E?2 ! E?2 given by ?(v ? w) = w ? v
is a K?-module homomorphism. (b) The sets S2(E) = fx 2 E?2 j ?(x) = xg and
^2(E) = fx 2 E?2 j ?(x) = ?xg are K?-submodules of E?2. (c) S2(E) = (1 + ?)(E?2),
^2(E) = (1??)(E?2), and E?2 = S2(E)u^2(E) if charK 6= 2.
Proof. a. ?(gx) = ?(g(x1 ?x2)) = ?(gx1 ?gx2) = gx2 ?gx1 = g(x2 ?x1) = g?(x) for all
g 2 ? and x = x1 ?x2 2 E?2. Extend linearly.
27
b. Let g 2 ?, x 2 S2(E), and y 2 ^2(E). Note that ?(gx) = g?(x) = gx and that
?(gy) = g?(y) = ?gy by (a). Extend linearly.
c. First, let x 2 S2(E). Then x = x2 + x2 = x2 +? ?x2? = (1+?)?x2?2 (1+?)(E?2). Thus
S2(E) (1+?)(E?2). Conversely, let x 2 (1+?)(E?2). Then x = y+?(y) for some
y 2 E?2. We have ?(?(y)) = y by linear extension. It then follows that
?(x) = ?(y +?(y)) = ?(y)+?(?(y)) = ?(y)+y = x.
Thus x 2 S2(E). Consequently, (1 + ?)(E?2) S2(E) and the flrst equality is
shown. The second equality is established similarly. Suppose that x 2 ^2(E). Then
x = x2 + x2 = x2 ?? ?x2? = (1??)?x2? 2 (1??)(E?2). Thus ^2(E) (1??)(E?2).
Conversely, any x 2 (1??)(E?2) may be written as x = y ??(y) for some y 2 E?2.
We also have ?(?(y)) = y by linear extension. Consequently,
?(x) = ?(y ??(y)) = ?(y)??(?(y)) = ?(y)?y = ?x.
Thus x 2^2(E). So (1??)(E?2) ^2(E) and the second equality also holds. Finally,
it is clear that S2(E)\^2(E) = f0g. Let x 2 E?2. Then
x = 12[x+?(x)+x??(x)] = 12(1+?)(x)+ 12(1??)(x).
Since 12(x+?(x)) 2 S2(E) and 12(x??(x)) 2^2(E), then E?2 = S2(E)+^2(E).
49 DEFINITION. Let gfifl denote the (fi; fl)-entry of the matrix g. cfifl : ? ! K where
cfifl(g) = gfifl for all g 2 ? is called a coordinate function. Suppose n := f1; 2; ??? ; ng and
An := fcfifl j fi; fl 2 ng. We will denote by A(n) the K-subalgebra of K? generated by
An. A(n) is called the algebra of polynomial functions and the elements of A(n) are called
polynomial functions on ?. fy1; ??? ; yqg in a K-algebra is called algebraically independent
over K if no nonzero polynomial p 2 K[x1; ??? ; xq] exists such that p(y1; ??? ; yq) = 0.
28
50 LEMMA. If K is inflnite then every subset of An is algebraically independent over K.
Proof. This result is well-known (see [13, page 9]). We present a proof of the case S An
with jSj = 1. Then S = fcijg for some flxed i and j. Let p(x) 2 K[x] with p(cij) = 0K.
Assume p(x) is not the zero polynomial. Suppose i 6= j. We may choose a nonzero
fi 2 K with p(fi) 6= 0K since K is inflnite. Construct matrix g where ghh = 1K for
1 ? h ? n, gij = fi, and g?m = 0K for all other pairs (?; m) with ?; m 2 n. Then g 2 ?
but 0K = p(cij)(g) = p(fi). Contradiction. So i = j. Again choose a nonzero fi 2 K with
p(fi) 6= 0K. Construct matrix g where ghh = 1K for 1 ? h ? n with h 6= i, gii = fi, and
g?m = 0K for all other pairs (?; m) with ?; m 2 n. Then g 2 ? but 0K = p(cii)(g) = p(fi).
Contradiction. Thus p is the zero polynomial. So S is algebraically independent over K.
51 DEFINITION. Let V be a K?-module and T : ? ! ? the matrix representation
afiorded by V relative to the basisfv1; ??? ; vngof V. So T(g) = [fiij(g)] for unique fiij 2 K?
with gvj = Pi fiij(g)vi (g 2 ?). We extend linearly by T(Pg2? kgg) = Pg2? kgT(g). The
K-space cf(V) spanned by the fiij is called the coe?cient space of V.
52 EXAMPLES. Put e1 =
2
641
0
3
75, e
2 =
2
640
1
3
75, and g =
2
64a b
c d
3
75.
a. Let ? : ? ! GL2(K) be the matrix representation corresponding to the natural K?-
module E relative to the basis fe1; e2g of E. Consequently ? satisfles ?(g) = g since
ge1 = ae1 +ce2 and ge2 = be1 +de2. Thus cf?E?1? = hc11; c12; c21; c22i.
b. We use the convention that ci1i3;i2i4 = ci1i2ci3i4. E?2 has basis fe11; e12; e21; e22g
where eij = ei ?ej. Then by a calculation similar to (c) below,
cf?E?2? = hc211; c212; c11c12; c11c21; c11c22; c12c21; c12c22; c221; c222; c21c22i
= hc11;11; c11;22; c11;12; c12;11; c12;12; c12;21; c12;22; c22;11; c22;22; c22;12i
c. S2(E) has basis fe11; e12 +e21; e22g. Then:
29
ge11 = ge1 ?ge1 = (ae1 +ce2)?(ae1 +ce2) = a2e11 +ac(e12 +e21)+c2e22,
g(e12 +e21) = ge1 ?ge2 +ge2 ?ge1
= [(ae1 +ce2)?(be1 +de2)]+[(be1 +de2)?(ae1 +ce2)]
= 2abe11 +(ad+bc)(e12 +e21)+2cde22,
ge22 = ge2 ?ge2 = (be1 +de2)?(be1 +de2) = b2e11 +bd(e12 +e21)+d2e22.
Hence
cf(S2(E)) = hc211; 2c11c12; c212; c11c21; c11c22 +c12c21; c12c22; c221; 2c21c22; c222i
= hc11;11; 2c11;12; c11;22; c12;11; c12;12 +c12;21; c12;22; c22;11; 2c22;12; c22;22i.
d. Similarly, ^2(E) has basis fe12 ?e21g and cf(^2(E)) = hc12;12 ?c12;21i.
53 NOTATION. Let ? 2Pr. Denote ?ci;j := ci;j? where j? = (j?(1); ??? ; j?(r)).
54 PROPOSITION. If ? = (12) 2Pr, then (1??)cf?E?2? = cf(1??)?E?2?.
Proof. cf?E?2? = hc11;11; c11;22; c11;12; c12;11; c12;12; c12;21; c12;22; c22;11; c22;22; c22;12i
by Example (52b). Note that:
(1+?)(c11;11) = c11;11 +c11;11 = 2c11;11, (1+?)(c11;22) = c11;22 +c11;22 = 2c11;22,
(1+?)(c11;12) = c11;12 +c11;21 = c11c12 +c12c11 = 2c11c12 = 2c11;12,
(1+?)(c12;11) = c12;11 +c12;11 = 2c12;11, (1+?)(c12;12) = c12;12 +c12;21,
(1+?)(c12;21) = c12;21 +c12;12, (1+?)(c12;22) = c12;22 +c12;22 = 2c12;22,
(1+?)(c22;11) = c22;11 +c22;11 = 2c22;11, (1+?)(c22;22) = c22;22 +c22;22 = 2c22;22,
(1+?)(c22;12) = c22;12 +c22;21 = c21c22 +c22c21 = 2c21c22 = 2c22;12.
Thus by Example (52c) and Proposition (48c),
(1+?)?cf?E?2?? = hc11;11; c11;22; c11;12; c12;11; c12;12 +c12;21; c12;22; c22;11; c22;22; c22;12i
= cf(S2 (E)) = cf?(1+?)?E?2??.
Similarly, note that:
30
(1??)(c11;11) = c11;11 ?c11;11 = 0, (1??)(c11;12) = c11;12 ?c11;21 = c11c12 ?c12c11 = 0,
(1??)(c12;12) = c12;12 ?c12;21, (1??)(c11;22) = c11;22 ?c11;22 = 0,
(1??)(c12;11) = c12;11 ?c12;11 = 0, (1??)(c12;21) = c12;21 ?c12;12,
(1??)(c12;22) = c12;22 ?c12;22 = 0, (1??)(c22;11) = c22;11 ?c22;11 = 0,
(1??)(c22;12) = c22;12 ?c22;21 = c21c22 ?c22c21 = 0, (1??)(c22;22) = c22;22 ?c22;22 = 0.
Applying Example (52d) and Proposition (48c) yields
(1??)?cf?E?2?? = hc12;12 ?c12;21i = cf(^2 (E)) = cf?(1??)?E?2??.
55 NOTATION.
a. A polynomial is called homogeneous when each of its terms has the same degree. We
let K be inflnite hereafter. By Lemma (50), A(n) may be viewed as the polynomial
algebra over K in the indeterminants cfifl. Let Ar (r ? 0) denote the K-subspace of
A(n) generated by the homogeneous polynomial functions of total degree r.
b. Let I = ff j f : r ! ng. G = Pr acts on I via i? = (i?(1); :::i?(r)) and G acts on
I ?I by (i; j)? = (i?; j?) for i; j 2 I and ? 2 G. For i; j 2 I, deflne (i; j) ? (p; q)
for (i; j); (p; q) 2 I ?I when p = i? and q = j? for some ? 2 G. Let R(n; r) denote
a set of representatives for the equivalence classes of I ?I under ?.
56 REMARK. For flxed g 2 ? and with E viewed as a K-space, deflne t0g : E?r ! E?r
by t0g(x1; ??? ; xr) = g(x1 ?????xr) for all x1; ??? ; xr 2 E. Then t0g is r-multilinear and
induces a K-linear map tg : E?r ! E?r (Theorem (16) and induction) such that t0g = tg?fl
where fl is the canonical r-multilinear map. Then tg gives rise to a matrix representation
T0n;r : ? ! GLn(E?r) given by T0n;r(g) = tg. Extending linearly to K? and using the
standard basis fei j i 2 Ig of E?r yields the matrix representation Tn;r : K? ! MatIK
given by Tn;r(?) = [gi;j] for i; j 2 I where ?ej = Pi2I gi;jei. Similarly, ci;j may be extended
linearly to K?.
31
57 LEMMA. Let r be a nonnegative integer. Then
rX
i=0
n?2+i
i
?
=
n+r?1
r
?
.
Proof. We proceed by induction on r. The result is obvious for r = 0. Recall [14, p. 8] that
Pascal?s Rule says
n
k?1
?
+
n
k
?
=
n+1
k
?
for 1 ? k ? n. Then, using the induction
hypothesis, we have
rX
i=0
n?2+i
i
?
=
?r?1X
i=0
n?2+i
i
??
+
n?2+r
r
?
=
n?2+r
r?1
?
+
n?2+r
r
?
=
n+r?1
r
?
.
58 THEOREM. (a) C = fci;j = ci1j1 ???cirjr j (i; j) 2 R(n; r)g is a K-basis for Ar. (b)
dimAr =
n2 +r?1
r
?
. (c) Ar = cf(E?r).
Proof. a. Ar is spanned as a K-space by the monomials fci;j j i; j 2 Ig. Now since
ci;j = ck;? if and only if (i; j) ? (k; ?), we have that this set equals C. So C spans Ar,
and the elements of C are distinct. Thus C is linearly independent by Lemma (50).
Consequently C is a K-basis for Ar.
b. We show that the number of distinct monomials xr11 ???xrmm in the m commuting
indeterminants xi with Pi ri = r is
m+r?1
r
?
. We proceed by induction on m.
The result is obvious for m = 1. Let w? be the number of distinct monomials with
P
i ri = r such that rm = ?. The number in question is w =
P
? w?. By Lemma (57),
w = w0 +???+wr
=
m?1+r?1
r
?
+
m?1+r?1?1
r?1
?
+???+
m?1+0?1
0
?
=
rX
i=0
m?2+r?i
r?i
?
=
rX
i=0
m?2+i
i
?
=
m+r?1
r
?
.
The claim now follows from (a).
c. By Remark(56), gej = Pi2I ci;j(g)ei. Thus cf(E?r) = Pi;j2I Kci;j = Ar.
32
Deflne F : K? ?K? ! K??? by [F(f; g)](u; v) = f(u)g(v) (f; g 2 K?, u; v 2 ?). There
exists a unique K-linear map ' : K? ?K? ! K??? given by ['(f ?g)](u; v) = f(u)g(v)
by Theorem (16) since F is bilinear. ' is injective by an argument similar to that given in
the proof of Lemma (29b). So, we may consider K? ?K? as a K-subspace of K???.
59 LEMMA. A(n) is a K-bialgebra, and Ar is a K-subcoalgebra of A(n).
Proof. A(n) is a K-algebra as it is a K-subalgebra of K?. Then ? : A(n)?A(n) ! A(n)
and ? : K ! A(n) given by ?(ci;j ?ck;?) = ci;jck;? and ?(k) = k1 are the structure maps for
A(n) by the proof of Theorem (33). Deflne ? : K? ! K??? by [?(f)](u; v) = f(uv) and
" : K? ! K by "(f) = f(1?) for all f 2 K?, u; v 2 ?. Since for all f; g 2 K?, u; v 2 ?,
and k 2 K, we have
(i) [?(f +g)](u; v) = (f +g)(uv) = f(uv)+g(uv) = [?f](u; v)+[?g](u; v),
(ii) [?(fg)](u; v) = (fg)(uv) = f(uv)g(uv) = [?f](u; v)[?g](u; v),
(iii) [?(kf)](u; v) = (kf)(uv) = kf(uv) = k[?f](u; v),
(iv) [?(1K?)](u; v) = 1K?(uv) = 1K = 1K???(u; v),
(v) "(f +g) = (f +g)(1?) = f(1?)+g(1?) = "(f)+"(g),
(vi) "(fg) = (fg)(1?) = f(1?)g(1?) = "(f)"(g),
(vii) "(kf) = (kf)(1?) = kf(1?) = k"(f), and
(viii) "(1K?) = 1K?(1?) = 1K,
? and " are K-algebra homomorphisms by (i) - (iv) and (v) - (viii), respectively. Now
restrict ? and " to A(n). Then ?(cfifl) = Pn =1 cfi ?c fl and "(cfifl) = ?fifl for 1 ? fi; fl ? n.
We next verify the Coassociative Law and Counitary Property. Then
((??1)??)(cfifl) = (??1)
X
cfi ?c fl
?
=
X
;?
(cfi? ?c? )?c fl
=
X
;?
cfi? ?(c? ?c fl) = (1 ??)
X
?
cfi? ?c?fl
?
= ((1 ??)??)(cfifl)
33
(("?1)??)(cfifl) = ("?1)
X
cfi ?c fl
?
=
X
"(cfi )?c fl = 1K ?
X
"(cfi )c fl
= 1K ?
X
?fi c fl = 1K ?cfifl = ?2(cfifl),
and similarly ((1 ?")??)(cfifl) = ?1(cfifl). So (A(n); ?; ") is a K-coalgebra. By Theorem
(44), (A(n); ?; ?; ?; ") is a K-bialgebra. Finally, note Ar is a K-subspace of A(n) by the
deflnition of Ar. Let ci;k = ci1k1 ???cirkr 2 Ar. Then, using the fact that ? is a K-algebra
homomorphism, we flnd that ?(ci;k) = Pj2I ci;j ?cj;k 2 Ar ?Ar. Thus ?(Ar) Ar ?Ar.
Ar is a K-subcoalgebra of A(n) by Deflnition (35).
3.2 Schur Algebras and Group Actions
Let f 2 K? and ? = P?gg 2 K?. Deflne f(?) = P?gf(g). Then f is a unique linear
extension of f. Let V be a flnite-dimensional K?-module with basis fvb j b 2 Bg. If
? acts as gvb = PB fiab(g)va (as in the deflnition of coe?cient space), then K? acts as
?vb = Pa fiab(?)va for all ? 2 K? and all b 2 B. Let ? : K? ! EndK(V) be the
representation afiorded by V, and let Y = ker?.
60 LEMMA. Let f 2 K? and ? 2 K?. Then (a) ? 2 Y if and only if f(?) = 0 for all
f 2 cf(V) and (b) f 2 cf(V) if and only if f(?) = 0 for all ? 2 Y.
Proof. a. Let ? 2 Y and f 2 cf(V). Then f = Pa;b dabfiab for some dab 2 K. Since
fiab(?) = 0 for all a; b 2 B, we have f(?) = Pa;b dabfiab(?) = 0. Conversely, let
f(?) = 0 for all f 2 cf(V). Since fiab 2 cf(V) for all a; b 2 B, we have fiab(?) = 0 for
all a; b 2 B. So ?(?)(vb) = ?vb = Pa fiab(?)va = 0 for all a; b 2 B. So ? 2 Y.
b. Let N := ?(K?). Deflne h; i : Y 0 ? N ! K by hf; ?i = f(?) for all f 2 Y 0 and
? = ?(?) 2 N. Suppose ?(?) = ?(?) for some ?; ? 2 K? and let f 2 Y 0. Since ?
is a homomorphism, then ?(? ? ?) = 0. Hence ? ? ? 2 ker?. Thus f(? ? ?) = 0
34
since f 2 Y 0. So f(?)?f(?) = 0 since f is linear. Hence f(?) = f(?). Thus h; i is
well-deflned. Now suppose hf; ?i = 0 for every ? 2 N. So f(?) = 0 for every ? 2 K?.
In particular, 0 = f(1Kg) = f(g) for every g 2 ?. So f = 0. Now let ? = ?(?) 2 N.
Supposehf; ?i = 0 for every f 2 Y 0. Then f(?) = 0 for every f 2 Y 0 by the deflnition
of h; i. Note Y = (Y 0)0 = fx j f(x) = 0 for every f 2 Y 0g. Hence ? 2 Y. So ? = 0.
So h; i is non-singular. By (a), cf(V) Y 0. Observe if ? = ?(?) 2 N such that
f(?) = hf; ?i = 0 for all f 2 cf(V), then ? 2 Y by (a). Hence ? = 0. This implies
h; i restricted to cf(V)?N is non-singular. So cf(V) ?= N? ?= Y 0 by two applications
of Lemma (32). Thus dimcf(V) = dimY 0. Therefore cf(V) = Y 0.
61 EXAMPLE. Let g be the 3?3 matrix with g11 = g12 = g22 = g33 = 1 and 0 elsewhere.
We compute T3;2(g). Note ei = ei1 ? ei2 since i = (i1; i2). We write ejk := ej ? ek and
gjk;?m := fi(g)(j;k);(?;m). A few calculations are included:
ge11 = g(e1 ?e1) = ge1 ?ge1 = e1 ?e1 = e11
) g11;11 = 1 and gi;11 = 0 for i 6= (1; 1);
ge12 = ge1 ?ge2 = e1 ?(e1 +e2) = e11 +e12
) g11;12 = g12;12 = 1 and gi;12 = 0 for i 6= (1; 1); (1; 2);
ge13 = ge1 ?ge3 = e1 ?e3 = e13
) g13;13 = 1 and gi;13 = 0 for i 6= (1; 3);
ge21 = ge2 ?ge1 = (e1 +e2)?e1 = e11 +e21
) g11;21 = g21;21 = 1 and gi;21 = 0 for i 6= (1; 1); (2; 1).
We eventually obtain T3;2(g) = g ?g (Figure 10).
2
66
66
66
66
66
66
66
66
66
66
66
66
66
64
1 1 0 1 1 0 0 0 0
0 1 0 0 1 0 0 0 0
0 0 1 0 0 1 0 0 0
0 0 0 1 1 0 0 0 0
0 0 0 0 1 0 0 0 0
0 0 0 0 0 1 0 0 0
0 0 0 0 0 0 1 1 0
0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 1
3
77
77
77
77
77
77
77
77
77
77
77
77
77
75
Figure 10: T3;2(g)
62 DEFINITION. The Schur algebra, denoted by Sr or Sr(?), is the image of K? under
Tn;r with identity 1Sr = [?i;j] where ?i;j = ?i1j1 ????irjr.
Note that [?i;j] is just the identity matrix.
35
63 THEOREM. (a) h; i : Ar ?Sr ! K given by hci;k; Tn;r(?)i = ci;k(?) is well-deflned,
non-singular, and bilinear where ci;k 2 Ar, and ? 2 K?. (b) A?r and Sr are isomorphic as
K-spaces. (c) Sr is a K-algebra with dimSr =
n2 +r?1
r
?
.
Proof. (a) First, if Tn;r(?) = Tn;r(?0), then ???0 2 KerTn;r. So ci;k(???0) = 0 (Theorem
(58c) and Lemma (60b)) and ci;k(?) = ci;k(?0). Consequently, the form is well-deflned.
Suppose 0 = hci;k; Tn;r(?)i = ci;k(?) for all ? 2 K?. Thus ci;k = 0. Now suppose
ci;k(?) = hci;k; Tn;r(?)i = 0 for all ci;k 2 Ar. Then ? 2 kerTn;r by Lemma (60a). Hence
Tn;r(?) = 0. Thus h; i is non-singular. Next for all ch;i; cj;k 2 Ar, ?; ? 2 K?, and x; y 2 K,
we have
hxch;i +ycj;k; Tn;r(?)i = (xch;i +ycj;k)(?) = xch;i(?)+ycj;k(?)
= xhch;i; Tn;r(?)i+yhcj;k; Tn;r(?)i
and
hci;k; xTn;r(?)+yTn;r(?)i = hci;k; Tn;r(x?+y?)i = ci;k(x?+y?) = xci;k(?)+yci;k(?)
= xhci;k; Tn;r(?)i+yhci;k; Tn;r(?)i.
Thus h; i is bilinear. (b) dimA?r is flnite by Theorem (58b). Then A?r and Sr are isomorphic
as K-spaces by Lemma (32). (c) Sr is a homomorphic image of the K-algebra K? so it is a
K-algebra. Moreover, dimSr = dimA?r = dimAr =
n2 +r?1
r
?
by Theorem (58b).
64 LEMMA. Let ?; ? 2 Sr and i; j 2 I.
a. hci;j; ?i = the (i; j)th entry of ?. b. hci;j; ??i = Ph2Ihci;h; ?ihch;j; ?i.
Proof. a. Note that ? : E?r ! E?r is a linear map. We write the matrix of ? relative
to the basis fei j i 2 Ig as [?ij]. We must show that hcij; ?i = ?ij. That is, we
must show that ?(ej) = Pihcij; ?iei. Suppose that ? = Tn;r(g) for some g 2 ?. By
Theorem (63), we must show that ?(ej) = Pi cij(g)ei. But this is clear since
36
?(ej) = Tn;r(g)(ej) = gej1 ?????gejr =
nX
i1=1
ci1j1(g)ei1 ?????
nX
ir=1
cirjr(g)eir
=
X
i
cij(g)ei1 ?ei2 ?????eir =
X
i
cij(g)ei.
Since Tn;r and cij are linear, we obtain Tn;r(?)(ej) = Pi cij(?)(ej) for each ? 2 K?,
and the claim follows.
b. By (a), hcij; ??i = (i; j)th entry of ?? = Ph2I ?i;h?h;j = Ph2Ihcih; ?ihchj; ?i.
65 THEOREM. ? : Sr(?) ! A?r given by ?(?)(f) = hf; ?i is a K-algebra isomorphism.
Proof. First, ? is a K-space isomorphism by the proofs of Theorem (63b) and Lemma (32).
Multiplication in the algebra A?r is deflned by (fifl)(c) = Pfi(ci)fl(di) (fi; fl 2 A?r, c 2 Ar),
where ?(c) = Pi ci ?di. Indeed,
(fifl)(ci;j) = [??(fi?fl)](ci;j) = (fi?fl)(?(ci;j)) = (fi?fl)
X
h2I
ci;h ?ch;j
?
=
X
h2I
fi(ci;h)fl(ch;j).
Now let ?; ? 2 Sr(?) and i; j 2 I. Then by the deflnition of ? and Lemma (64b)
?(??)(ci;j) = hci;j; ??i =
X
h2I
hci;h; ?ihch;j; ?i =
X
h2I
?(?)(ci;h)?(?)(ch;j).
Since the ci;j span Ar, we have ?(??)(c) = Pi2I ?(?)(ci)?(?)(di) for all c 2 Ar. Next, let
fi = ?(?) and fl = ?(?). Consequently,
?(?)?(?)(c) = (fifl)(c) =
X
i2I
fi(ci)fl(di) =
X
i2I
?(?)(ci)?(?)(di) = ?(??)(c)
for all c 2 Ar. Thus ? is a homomorphism. Therefore ? is an algebra map since
?(1Sr(?))(ci;j) = hci;j; 1Sr(?)i = ?i;j = "(ci;j).
66 DEFINITION. Let S be a set, G a group, and e the identity of G. An action of G
on S is a function G?S ! S given by (g; x) 7! gx such that ex = x and (gh)x = g(hx)
37
for all x 2 S and g; h 2 G. A right action of G has a similar deflnition with g appearing
on the right. S is called a (right) G-set when an (right) action exists.
67 NOTATION. We set N := EndK(E?r). Then N has basis fei;j j i; j 2 Ig where
ei;j : E?r ! E?r is given by ei;j(ek) = ?j;kei.
68 EXAMPLES. Let G = Pr, 2 G, i = (i1; i2; ??? ; ir), and recall I := I(n; r).
a. Consider i = (i ?1(1); ??? ; i ?1(r)). This makes I a G-set since
(1)i = (i(1)?1(1); ??? ; i(1)?1(r)) = (i1; ??? ; ir) = i, and
(?i) = (i??1(1); ??? ; i??1(r)) = (j1; ??? ; jr) = (j ?1(1); ??? ; j ?1(r))
= (i??1( ?1(1)); ??? ; i??1( ?1(r))) = (i( ?)?1(1); ??? ; i( ?)?1(r)) = ( ?)i,
where jk := i??1(k).
b. Next, consider E?r and ei = e i = e i1 ? ??? ? e ir. Then (1)ei = e1i = ei and
(?ei) = (e?i) = e (?i) = e( ?)(i) = ( ?)ei by (a). Hence E?r is a G-set with the
above action extended linearly.
c. Deflne ei;j by (ei;j )(e) = ei;j( e) (e 2 E?r). Consequently, extending this ac-
tion linearly yields that N a right G-set because (ei;j1)(e) = ei;j(1e) = ei;j(e)
and ((ei;j )?)(e) = (ei;j )(?e) = ei;j(( ?)e) = (ei;j( ?))(e). Moreover, we have
(ei;j )ek = ei;j( ek) = ei;j(e k) = ?j; kei = ? ?1j;kei = (ei; ?1j)ek. It therefore follows
that ei;j = ei; ?1j.
d. Arguing as in (c), we flnd that N? is a G-set with action ( e?i;j)ek;? = e?i;j(ek;? ) where
e?i;j(ek;?) = ?i;k?j;?. Moreover,
( e?i;j)ek;? = e?i;j(ek;? ) = e?i;j(ek; ?1?) = ?i;k?j; ?1? = ?i;k? j;? = e?i; j(ek;?)
by (c). So we have e?i;j = e?i; j.
38
3.3 Main Results
69 NOTATION. Suppose ? is a character of G = Pr. We set t? := P 2G ?( ) 2 KG,
L := t?E?r, NL := EndK(L), and AL := cf(L).
70 DEFINITION. Let T : K? ! N be the representation corresponding to the K?-
module E?r. We deflne TL : K? ! NL by TL(?) = T(?)flflL.
The action of ? on E?r clearly commutes with the action of G. So T(?)(L) L and TL is
well-deflned.
71 LEMMA. ? : (imT)t? ! imTL given by ?(T(?)t?) = T(?)flflL is a K-isomorphism.
Proof. We have that ? is well-deflned and injective since,
T(?)t? = T(?)t? , T(?)t?(e) = T(?)t?(e) for all e 2 E?r
, T(?)(t?e) = T(?)(t?e) for all e 2 E?r
, T(?)flflL = T(?)flflL , ?(T(?)t?) = ?(T(?)t?).
? is also surjective since ?(T(?)t?) = T(?)flflL for any T(?)flflL 2 imTL. Finally ? is a K-space
isomorphism since for all k 2 K and for all T(?)flflL; T(?)flflL 2 imTL:
?(T(?)t? +T(?)t?) = ?(T(?+?)t?) = T(?+?)flflL = T(?)flflL +T(?)flflL
= ?(T(?)t?)+?(T(?)t?),
?(kT(?)t?) = ?(T(k?)t?) = T(k?)flflL = kT(?)flflL = k?(T(?)t?.
72 REMARK. The dual of ? in Lemma (71) is the map ?? : (imTL)? ! ((imT)t?)?
deflned by ??(f)(T(?)t?) = f(?(T(?)t?)) = f(T(?)flflL) by Deflnition (30). Also since
N is a right G-set by Example (68c), it follows that HomK(K?; N) is a right G-set by
(f )(?) = f(?) . In particular, (Tt?)(?) = T(?)t?.
39
73 LEMMA. (a) : AL ! (imTL)? given by ( (a))(TL(?)) = a(?) is a K-isomorphism.
(b) If ? = ?? ? then ?(a)?(Tt?) = a as functions from K? to K for every a 2 AL. (c)
AL = he?i;j ?Tt?i.
Proof. a. By Lemma (60b) we have AL = (kerTL)?. Note that there exists an isomor-
phism F : (kerTL)? ! (K?=kerTL)? by Lemma (31b). Similarly, there exists an
isomorphism G : (K?=kerTL)? ! (imTL)? by the First Isomorphism Theorem. Now
deflne = G?F. Consequently is an isomorphism with
(a)(TL(?)) = G(F(a))(TL(?)) = F(a)(?+kerTL(?)) = a(?).
b. Let ? 2 K?. Then by (a) (?(a))(T(?)t?) = ??( (a))(T(?)t?) = (a)(TL(?)) = a(?)
(T(?)t? 2 (imT)t?). Thus a(?) = (?(a))(T(?)t?) = ?(a)?(Tt?)(?)? = (?(a)?Tt?)(?)
by the last sentence of Remark (72). Consequently, ?(a)?(Tt?) = a.
c. First, AL he?i;j ?Tt?i since, using (b),we have for each a 2 AL
a = ?(a)?Tt? = (?? ? )(a)?Tt? =
X
i;j
ai;je?i;j
flfl
fl(
imT)t?
?
?Tt?
=
X
i;j
ai;j(e?i;j ?Tt?) 2he?i;j ?Tt?i.
where we have used that ((imT)t?)? is spanned by the restrictions of the e?i;j to
(imT)t? to express (?? ? )(a) as indicated. For the converse, let ? 2 kerTL. Conse-
quently, T(?)t? = ??1(T(?)flflL) = ??1(0) = 0 by Lemma (71). Then since
(e?i;j ?Tt?)(?) = e?i;j(Tt?(?)) = e?i;j(T(?)t?) = e?i;j(0) = 0;
we may conclude that e?i;j ?Tt? 2 ((imT)t?)? = AL. Thus he?i;j ?Tt?i AL.
40
74 LEMMA. There exists a well-deflned K-endomorphism t? of Ar with the property
t?ci;j = P 2G ?( )ci; j (i; j 2 I).
Proof. Since Ar is spanned by the ci;j, it is enough to check that the assignment is well-
deflned. Suppose ci;j = ck;?. Then k = i? and ? = j? for some ? 2 G (see Notation (55b)).
Then ck; ? = ci?; (j?) = c??1i; ??1j = ci;? ??1j. So
t?ck;? =
X
2G
?( )ck; ? =
X
2G
?( )ci;? ??1j =
X
?2G
?(??1??)ci;?j =
X
?2G
?(?)ci;?j = t?ci;j,
where we have used the fact that characters are constant on conjugacy classes.
75 NOTATION. If E is replaced by L, the same construction (see Remark (56) and
Deflnition (62)) which yielded Sr results in a K-algebra, which we denote by Ss;L. Put
As;L := cf(L?s).
Theorem (76), Theorem (78), and Theorem (80) below are the main results. Theorem (76)
generalizes Proposition (54), Theorem (78) generalizes Theorem (58c), and Theorem (80)
generalizes Theorem (63b).
76 THEOREM. cf(t?E?r) = t?cf(E?r).
Proof. Note
(e?i;j ?Tt?)(?) = e?i;j((Tt?)(?)) = e?i;j((T(?)t?) = (t?e?i;j)(T(?)) = ((t?e?i;j)?T)(?).
Thus e?i;j ? Tt? = t?e?i;j ? T. Now t?e?i;j = P 2G ?( ) e?i;j = P 2G ?( )e?i; j. Then by
Lemma (58c), Lemma (73c), Lemma (74), and since ci;j = e?i;j ?T, we have
cf(t?E?r) = AL = he?i;j ?Tt?i = h(t?e?i;j)?Ti =
? X
2G
?( )e?i; j
?
?T
=
?X
2G
?( )(e?i; j ?T)
=
?X
2G
?( )ci; j
= ht?ci;ji
= t?hci;ji = t?Ar = t?cf(E?r).
41
77 NOTATION. Let r1; ??? ; ru 2Z+. For each i, let ?i be a character of Pri and put
L?i = t?iE?ri. We write Qi t?iAri to mean the set of all products Qi ci with ci 2 t?iAri.
78 THEOREM. cf?Ni L?i? = Qi t?iAri.
Proof. The matrix representation of a tensor product of modules is the Kronecker product
of the matrix representations of the factors (see the proof of Theorem (26)). By Theorem
(76), cf?Ni L?i? = Qi cf?L?i? = Qi cf?t?iE?ri? = Qi t?icf(E?ri) = Qi t?iAri.
79 COROLLARY. cf(L?s) = (t?Ar)s for any s 2Z+.
Proof. Immediate from Theorem (78).
80 THEOREM. Ss;L ?= A?s;L.
Proof. Let TL : K? ! End(L?s) be the representation afiorded by L?s (extended to
K?). Then K?=kerTL ?= imTL = Ss;L by the First Isomorphism Theorem. Therefore
As;L = cf(L?s) = (kerTL)0 ?= (K?=kerTL)? = S?s;L by Lemma (60b) and Lemma (31b).
42
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[11] D.G. Northcott. Multilinear Algebra. Cambridge University Press, Cambridge, UK,
1984.
[12] James A. Green. Polynomial Representations of GLn. Springer-Verlag, New York, NY,
1980.
[13] Stuart Martin. Schur Algebras and Representation Theory. Cambridge University
Press, Cambridge, UK, 1993.
[14] David M. Burton. Elementary Number Theory. 5th ed., McGraw-Hill, New York, NY,
2002.
43
Index
algebra, 2
condition, 2
deflnition
diagram version, 16
homomorphism version, 2
dual, 21
examples, 3
group ring, 3{4
homomorphism, 2
extension property, 4
polynomial functions, 28, 33
Schur, 35
dimension, 36
example, 35
tensor product, 17
algebraic independence, 28
annihilator, 13
properties, 14
bialgebra, 26
equivalent deflnitions, 25
examples, 26
group ring, 26
polynomial functions, 33
bilinear map, 6
canonical, 6
examples, 6
canonical injection, 7
isomorphism, 7
canonical projection, 7
isomorphism, 7
character, 10
irreducible, 10
product, 11
coalgebra, 19
dual, 20
group ring, 19
homomorphism, 25
polynomial functions, 34
tensor product, 23
coe?cient space, 29
examples, 29{30
comultiplication map, 19
coordinate functions, 28
basis, 28
flnite subsets, 29
counit map, 19
dual map, 13
non-singular pairing, 14
dual module, 11
coalgebra, 20
flnite-dimensional algebra, 21
tensor product, 12
G-set, 38
right, 38
group
action, 37{38
examples, 38
right, 38
general linear, 3
invertible K-linear maps, 8
ring, 3
K-space, 3
algebra, 3
bialgebra, 26
coalgebra, 19
dual, 12
homomorphism
algebra, 2
extension property, 4
44
coalgebra, 25
example, 25
module, 1
image, 2
kernel, 2
K-linear map, 1
dual, 13
matrix, 8
non-singular, 13
trace, 10
Kronecker product, 9
example, 35
matrix representation, 10
trace, 9
K-space, 1
subspace, 1
linear functional, 11
examples, 11{12
module
examples, 2
First Isomorphism Theorem, 2
left, 1
natural or standard, 27
quotient, 2
right, 1
simple, 10
unitary, 1
multilinear map, 6
canonical, 6
multiplication map, 17
non-singular K-linear map, 13
bilinear pairing, 14
Pascal?s Rule, 32
polynomial functions, 28
algebra, 28, 33
bialgebra, 33
coalgebra, 34
homogeneous, 31
basis, 32
coe?cient space, 32
dimension, 32
dual, 36
K-subspace, 31
subcoalgebra, 33
representation, 8
matrix, 8
module correspondence, 8{9
Schur algebra, 35
dimension, 36
example, 35
structure maps
algebra, 17
coalgebra, 19
subalgebra, 2
polynomial functions, 33
subcoalgebra, 19
polynomial functions
homogeneous, 33
submodule, 1
Sweedler notation, 22
tensor product
algebras, 17
antisymmetric, 27
associative, 8
character, 11
coalgebras, 23
dimension, 5
dual, 12
elements, 5
KG-module, 10
K-linear maps, 6
K-space, 5
symmetric, 27
universal property, 6
zero element, 5
trace, 9
linear functional, 11
properties, 9
twist map, 7
isomorphism, 7
unit map, 17
universal property
tensor product, 6
45