Finite dimensional algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 6 November 2012

Finite dimensional algebras

The trace, tr(a), of a d×d matrix a=aij is the sum of its diagonal elements, tr(a)=iaii A trace t on an algebra A is a -linear map t:A such that for all a,bA,

t(ab)= t(ba). (2.1)

Every representation V of A determines a trace tV on A given by tV(a)=tr(V(a)) where aA. A trace is nondegenerate if for each aA, a0, there exists bA such that t(ba)0. A trace t on A determines a symmetric bilinear form , on A given by

a,b= t(ab). (2.)

Suppose A is finite dimensional and let B={b1,b2,,bs} be a basis of A. A basis B*={b1*,b2*,,bs*} of A is dual to B with respect to the form , if

bi*,bj =δij.

The Gram matrix of A is the matrix

G= bi,bj . (2.3)

Suppose that B* exists and that C=cij an s×s matrix such that

bi*=k cikbk. (2.4)

Then

bi*,nj =kcik bk,bj =δij.

In matrix notation this says that CG=Is. So C must be G-1. Conversely, if C=G-1 then defining bi* by (2.4) determines a dual basis B*. This shows that B* exists if and only if G is invertible and that if it exists it is unique.

(2.5) Proposition. If t is a trace on a finite dimensional algebra A with basis B={b1,b2,,bs} and , is given by (2.2) then the Gram matrix G is invertible if and only if is nondegenerate.

Proof.

The trace t is degenerate if and only if there exists a,bA such that t(ab)=0 for all aA. This is the same as saying that t(b1b)=0 for each basis element bi. If b=jcjbj, cj, we have that the cj satisfy the system of equations

t(bib)= t(bibj) cj=0.

This system has a nontrivial solution if and only if the matrix G=t(bibj) is singular.

Symmetrization

Let A be a finite dimensional algebra with a nondegenerate trace t and let B be a basis of A. Let B* be the dual basis to B with respect to the form , given by (2.2). For gB let g* denote the element of B* such that t(gg*)=1. Let V1 and V2 be representations of A of dimensions d1 and d2 respectively.

(2.6) Proposition. Let C be any d1×d2 matrix with entries in . If

[C]= gBV1(g) CV2(g*),

then, for any aA,

V1(a)[C]= [C]V2(a).

Proof.

Let aA. Then

V1(a)[C] = gV1(ag) CV2(g*) = gBV1 ( hB ag,h* h ) CV2(g*) = g,hB ag,h* V1(h)C V2(g*) = g,hB V1(h)Ct (agh*) V2(g*) = hBV1 (h)CgB t(h*ag) V2(g*) = hBV1(h) CV2 ( gB h*a,g g* ) = hB V1(h)C V2(h*a) = [C]V2(a).

If V1 and V2 are irreducible then Schur's lemma gives that [C]=0 if V1 and V2 are inequivalent and that if V1=V2 then [C]=cId1 for some c.

Let A be a finite dimensional algebra. The action of A on itself by multiplication on the left turns A into an A-module. The resulting representation is the regular representation of A and we denote it by A. The set is the same as the set A, but we distinguish elements of A by writing aA. As usual we denote the algebra of this representation by A(A). We denote the trace of this representation by tr. Notice that the trace tr of the regular representation can be given by

tr(a)= gBag g, (2.7)

(2.8) Theorem. If A is a finite dimensional algebra such that the regular representation A has nondegenerate trace then every representation V of A is completely decomposable.

Proof.

Let tr denote the trace of the regular representation. Let B be a basis of A and for each gB let g* denote the element of the dual basis to B with respect to the trace tr such that tr(gg*)=1.

Let V be a representation of A of dimension d and let V1 be an irreducible invariant subspace of V. Let P:VV be an arbitrary projection of V onto V1. Define

P1=gBV (g)PV(g*).

Then, by (2.6), we know that

V(a)P1= P1V(a).

Since V1 is an A-invariant subspace, P1VV1. Since V1 is irreducible P1V is either 0 or V1.

Let e=gBgg*. If aA then

tr(ae) = tr ( gBagg* ) = gB ag,g* = gBag g = tr(a).

This shows that tr(a(e-1))=0 for all aA. Since tr is nondegenerate we have that

e=gBgg* =1. (2.9)

Now let vV1. Then since V(g*)vV1 we have

P1v= = gBV(g) P(V(g*)v) = gBV(g) V(g*)v = V ( gBgg* ) v = V(1)v = v.

So P1V=V1 and P1P1V=P1V.

Let P1=Id-P1 and let V2=P1V. Notice that V(a)P1=P1V(a) for all aA. So V2 is an A-invariant subspace of V. Since, for every vV, v=P1v+ (Id-P1)v= P1v+P1v, we have V=P1V+P1V. If vP1VP1V then v=P1v=P1 P1v=P (Id-P1)v=0. So P1VP1V=0. Thus we see that V=P1VP1V.

If P1V is irreducible then we are done. If not apply the same process again with P1V in place of V. Since V is finite dimensional continuing this process will eventually produce a decomposition of V into irreducible representations.

Now let A be a finite dimensional algebra such that the trace tr of the regular representation A of A is nondegenerate. Let B be a basis of A and for each gB let g* denote the element of the dual basis to B with respect to the trace tr such that tr(gg*)=1. Let V be a faithful representation of A. By (2.8) we know that V can be completely decomposed into irreducible representations. Choose a maximal set {Wλ} of nonisomorphic irreducible representations appearing in the decomposition of V. Let dλ=dimWλ and define Md()λ Mdλ(). We view Md() as an algebra of block diagonal matrices with one dλ×dλ block for each λ. V(A)λWλ(A) is a subalgebra of Md() in a natural way. Let Eijλ denote the d×d matrix with 1 in the (i,j) entry of the λth block and 0 everywhere else and let Iλ be the matrix which is the identity on the λth block and 0 everywhere else.

For each gB let Wijλ(g*) denote the (i,j) entry of the matrix Wλ(g*). Then

kth row of ( Wjiλ (g*)Wλ(g) ) =jth row of ( Wλ(g*) EikλWλ (g) ) .

So

kth row of ( gB Wjiλ (g*)Wλ(g) ) = jth row of ( g Wλ(g*) EikλWλ (g) ) = jth row of (cIλδik) . (2.10)

So the ith row of gWjiλ(g*)Wλ(g) is all zeros except for c in the jth spot and all other rows of gBWjiλ(g*)Wλ(g) are zero. So

gWjiλ (g*)Wλ(g) =cEijλ (2.11)

for some c. We can determine c by setting i=k to get

cdλ = tr (cIλδii) = tr ( gWλ (g*) EiiλWλ (g) ) = tr ( g Wλ(g) Wλ(g*) Eii ) = tr ( Wλ (ggg*) Eiiλ ) .

Since the trace of the regular representation was used to construct the g* we have, (2.9), that ggg*=1, giving

tr ( Wλ (ggg*) Eiiλ ) = tr ( Wλ(1) Eiiλ ) = tr (IλEiiλ) = 1.

So cdλ=1 and we can write (2.11) as

dλg Wjiλ (g*)Wλ (g)=Eijλ.

Since we have expressed each Eijλ as a linear combination of basis elements of V(A) we have that EijλV(A) for every i and j. But the Eijλ form a basis of Md(). So Md()V(A). Then AV(A)=Md(). We have proved the following theorem.

(2.12) Theorem. (Artin-Wedderburn) If A is a finite dimensional algebra such that the trace of the regular representation of A is nondegenerate, then, for some set of positive integers dλ,

Aλ Mdλ().

Examples

  1. Let 𝒜={ai} and = b i be two bases of A and let 𝒜*= a i * and *= b i * be the associated dual bases with respect to a nondegenerate trace t on A. Then b i = j s ij a j ,and b i *= j t ij a j *,and for some constants s ij and t ij . Then δ ij = b i b j * = k s ik a k l t jl a l * = k,l s ik t jl a k a l * = k,l s ik t jl δ kl = k s ik t jk . In matrix notation this says that the matrices S= s ij and T= t ij are such that S T t =I. Then, in the setting of Proposition 2.6, i V 1 b i C V 2 b i * = i j s ij V 1 a j C k t ik V 2 a k * = j,k i s ij t ik V 1 a j C V 2 a k * = j,k δ jk V 1 a j C V 2 a k * = j V 1 a j C V 2 a j *. This shows that the matrix C of proposition 2.6 is independent of the choice of basis.
  2. Let A be the algebra of elements of the form c 1 + c 2 e, c 1 , c 2 , where e 2 =0. A is commutative and t defined by t c 1 + c 2 e = c 1 + c 2 is a nondegenerate trace on A. The regular representation A of A is not completely decomposable. The subspace e A is invariant and its complemenetary subspace is not. The trace of the regular representation is given explicitly by tr 1 =2 and tr e =0. tr is degenerate. There is no matrix representation of A that has trace given by t .
  3. Suppose G is a finite group and that A=G is its group algebra. The the group elements gG form a basis of A. So, using 2.7, the trace of the regular representation can be expressed in the form tr a = gG ag | g = gG a | 1 = G a | 1 , where 1 denotes the identity in G and a | g denotes the coefficient of g in a. Since tr g -1 g = G 0 for each gG, tr is nondegenerate. If we set t a =a | 1 then t is a trace on A and g -1 gG is the dual basis to the basis g gG with respect to the trace.
  4. Let t be the trace of a faithful realisation φ of an algebra A (ie for each aA, t a is given by the styandard trace of φ a where φ is an injective homomorphism φ:A M d ). Let A = aA| t ab =0for allbA . A is an ideal of A. Let a A . Then tr a k-1 a =tr a k =0 for all k. If λ 1 ,, λ d are the eigenvalues of φ a then t a k = λ 1 k + λ 2 k ++ λ d k = p k λ =0 for all k>0 , where p k represents the k -th power symmetric functions [Mac1979]. Since the power symmetric functions generate the ring of symmetric functions this means that the elementary symmetric functions e k λ =0 for k>0 , [Mac1979] p17, 2.14. Since the characteristic polynomial of φ a can be written in the form char φ a t = t d - e 1 λ t d-1 + e 2 λ t d-2 +± e d λ , we get that char φ a t = t d . But then the Cayley-Hamilton theorem implies that φ a d =0. Since φ is injective we have that a d =0. So a is nilpotent. Let J be an ideal of nilpotent elements and suppose that aJ. For every element bA,baJ and ba is nilpotent. This implies that φ ba is nilpotent. By noting that a matrix is nilpotent only if in Jordan block form the diagonal contains all zeros we see that t ba =0. Thus a A . So A can be defined as the largest ideal of nilpotent elements. Furthermore, since the regular representation of A is always faitful, A is equal to the set aA| tr ab =0for allbA where tr is the trace of the regular representation of A.
  5. Let 𝒜 be the basis and t the trace of a faithful realisation of an algebra A as in Ex3 and let G 𝒜 be the Gram matrix with respect to the basis 𝒜 and the trace t as given by 2.2 and 2.3. If is another basis of A then G = P t G 𝒜 P, where P is the change of basis matrix from 𝒜 to . So the rank of the Gram matrix is independent of the choice of the basis 𝒜.

    Choose a basis a 1 , a 2 ,, a k of A ( A defined in Ex 3) and extend this basis to a basis a 1 a 2 a k b 1 b s of A. The Gram matrix with respect to this basis is of the form 0 0 0 G B where G B denotes the Gram matrix on b 1 , b 2 b s . So the rank of the Gram matrix is certainly less than or equal to s .

    Suppose that the rows of G B are linearly dependent. Then for some contants c 1 , c 2 ,, c s , not all zero, c 1 t b 1 b i + c 2 t b 2 b i ++ c s t b s b i =0 for all 1is. So t j c j b j b i =0,for alli. This implies that j c j b j A . This is a contradiction to the construction of the b j . So the rows of B B are linearly independent.

    Thus the rank of the Gram matrix is s or equivalently the corank of the Gram matrix of A is equal to the dimension of the radical A . Thus the trace tr of the regular representation of A is nondegenerate iff A = 0 .

  6. Let W be an irreducible representation of an arbitrary algebra A and let d=dimW. Denote W A by A W . Note that representation W is also an irreducible representation of A W ( W a =a for all a A W ).

    We show that tr is nondegenerate on A W , ie that if a A W ,a0 , then there exists b A W such that tr ba 0. Since a is a nonzero matrix there exists some wW such that a w0. Thus Aaw=W. So there exists some wW such that aw0. Now AawW is an A -invariant subspace of W and not 0 since aw0. Thus Aaw=W . So there exists some b A W such that baw=w. This shows that ba is not nilpotent. So tr ba 0. So tr is nondegenerate on A W . This means that A W = λ M d λ for some d λ . But since by Schur's lemma A W = I d , where d=dimW, we see that W A = A W = M d .

  7. Let A be a finite dimensional algebra and let A denote the regular representation of A. The set A is the same as the set A , but we distinguish elements of A by writing a A.

    A linear transformation B of A is in the centraliser of A if for every element aA and x A , Ba x =aB x . Let B 1 = b . Then B a = Ba 1 = aB 1 = a b = ab . So B acts on a A by right multiplication on b. Conversely it is easy to see that the action of right multiplication commutes with the action of left mutliplication since a x b=a x b , for all a,bA and x A . So the centraliser algebra of the regular represnetation is the algebra of matrices determined by the action of right multiplication of elements of A.

Notes and References

The approach to the theory of semisimple algebras that is presented in this section and the following section follows closely a classical approach to the representation theory of finite groups, see for example [Ser1977] or [Ham1962]. Once one has the analogue of the symmetrization process for finite groups, the only nontrivial step in the theory that is not exactly analogous to the theory for finite groups is formula (2.9).

I discovered this method after reading the sections of [CRl] concerning Frobenius and symmetric algebras. Frobenius and symmetric algebras were introduced by R. Brauer and C. Nesbitt, [BNe1937] and [Nes1938]. T. Nakayama [Nak1941] has a version of Theorem (2.6) and R. Brauer [Bra1945] proves analogues of the Schur orthogonality relations that are analogous to formula (2.10). Ikeda [Ike1953], and Higman [Hig1955], following work of Gaschütz [Gas1952], construct "Casimir" type elements similar to those in (2.9) and §3 Ex. 7. In [CRe1981] §9 Curtis and Reiner use a similar approach but with different proofs, communicated to them by R. Kilmoyer, to obtain theorems (3.8) and (3.9) for split semisimple algebras (over fields of characteristic 0). N. Wallach has told me that essentially the same approach works for finite dimensional Lie algebras.

This approach is useful for studying semisimple algebras that have distiguished bases. The recent interest in quantum deformations is producing a host of examples of semisimple algebras that are not group algebras but that do have distinguished bases. Some examples are Hecke algebras associated to root systems, the Brauer algebra, and the Birman-Wenzl algebra [BWe1989]. For an approach to the Hecke algebras that is essentially an application of the general theory given here see [GUn1989] and [Cr3] §68C.

I would like to thank Prof. A. Garsia for suggesting that I try to find an analogue of the symmetrization process for finite groups for the Brauer algebra. It was this problem that resulted in my discovery of this approach. I would like to thank Prof. C.W. Curtis for his helpful suggestions in locating literature with a similar approach. I would also like to thank Prof. Garsia for showing me the proofs of Exs. 4 and 5.

Notes and References

This is an excerpt from the unpublished first chapter of Arun Ram's dissertation entitled Representation Theory, written July 4, 1990.

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