Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 19 August 2013

Lecture 5

The previous results yield a great amount of information concerning semisimple algebras. How can one tell if an algebra is semisimple?

Theorem 1.41 Let A be a finite dimensional algebra. The following are equivalent:

1. The trace tr of the regular representation of A is nondegenerate.
2. Every representation V of A is completely decomposable.
3. A dλAˆ Mdλ().
4. A=0

We will build the proof (and the new definitions) of this theorem over the next few lectures/pages.

Our previous definition of nondegenerate trace used the hypothesis that the algebra A was semisimple. The following is an alternate definition of nondegeneracy which does not assume semisimplicity.

Definition 1.42 Let t be a trace on A, a finite dimensional algebra. We say that t is nondegenerate if for every nonzero aA there exists bA such that t(ba)0.

Example 1.43 Let t be the ordinary trace tr on Md(), and let aMd() be the matrix unit Eij. Then for b=Eji, one has that t(ba)=tr(Ejj)0.

Definition 1.44 Given an algebra A with basis {g1,,gd}, and let t be a trace on A. The Gram matrix is the matrix M=(gi,gj) =(t(gigj)).

Proposition 1.45 Given an algebra A with basis G={g1,,gd}, and let t be a trace on A. The following are equivalent:

1. t is a nondegenerate trace.
2. The Gram matrix is invertible.
3. A has a dual basis with respect to (the inner product generated by) t.


(3)(2). Suppose G*={g1*,,gd*} is a dual basis for A with respect to t. Let C=(cjk) be the transition matrix between the bases G and G* (that is, gj*=cjkgk). Then by duality, δij= gi,gj*= gi,kcjkgk =kcjkgi,gk. This yields the matrix equation CMt=Id=CM (since t is symmetric, Mt=M). Thus the existence of a dual basis implies the invertibility of the Gram matrix.

Conversely, if the Gram matrix is invertible, let C be its inverse, and define the gi* as above. The invertibility of C guarantees that the gi* will form a basis, while the above calculation will show that the basis is dual to the original one.


(1)(2). Let a be a nonzero element of A, and suppose that t is degenerate at a; i.e. for every b in A one has t(ba)=0. In particular, let b run through the basis G, so that t(gia)=0 for all i.

Consider a=ajgj,, for aj. Now we have that 0= t(gia)= jt(giajgj)= jgi,gjaj =0

This yields the matrix equation M(aad) =0, whence that the Gram matrix M is not invertible.

Conversely, if M is not invertible, one can produce a as above in the kernel of M; the trace t is degenerate at a=ajgj.

Definition 1.46 We can think of the C-algebra A as a vector space, denoted A, and consider the action of A on A given by a·b=ab. This is the (left) regular action, and we will say that A is the (left) regular representation of A.

Let t be the trace of the regular representation of V. Then for aA, and G={g1,,gd} a basis for A, one has t(a)= giGa· gi|gi, where this last a·gi|gi is meant to denote the coefficient of gi in the image of a·gi. In the case that we have a dual basis, one can replace a·gi|gi with agi,gi*, and the above formula becomes t(a)= giGa· gi|gi =giG agi,gi*.

Lemma 1.47 Given an algebra A with basis G={g1,,gd}, and t a non- degenerate trace on A. (Note that by the previous proposition, G has a dual basis G*.) Let W1:AMd1() and W2:AMd2() be representations of A. Let C be any d1×d2 matrix, and let [C]=W1(gi*)CW2(gi). (“symmetrize C”) Then for all a in A, W1(a)[C]= [C]W2(a), so [C] is an algebra homomorphism.

W1(a)[C] = W1(a) giG W1(gi*)C W2(gi) = giW1(agi*) CW2(gi) sinceW1is an alg hom = giW1 ( hk agi*,hk hk* ) CW2(gi) expandinggi*in terms of a basis hk* = hkW1(hk*) CW2 ( gi agi*,hk gi ) change order of summation, and move scalars around = hkW1(hk*) CW2 ( gi hk,gi* gi ) since t(agi*hk) =t(hkagi*) = hkW1(hk*) CW2(hka) =[C]W2(a)

Theorem 1.48 Let A be a finite dimensional algebra. Suppose that tr, the trace of the regular representation, is nondegenerate. Then every representation V of A is completely decomposable.


Let V be a representation of A.

case 1: V is irreducible.


case 1: V is not irreducible.

Since V is not irreducible, there exists a submodule V1 in V, and without loss of generality we may presume that V1 is irreducible. We want to produce an A-submodule V2 such that VV1V2.

Let P:VV be a projection onto V1 (i.e. P is a C-space morphism such that Pv1=v1 for all v1V1 and PVV1). One may produce examples of projections as follows: pick a basis for V1, and extend to a basis for V. Define P by mapping the basis elements of V1 to themselves, and the other basis elements of V to any arbitrary vectors in V1.

Idea of proof: Let V2=(I-P)V. Then V=PV+(I-P)V=V1+V2. This doesn’t quite work, since there is no reason for this V2 to be an A-module. The [P] will play a useful role here.

Let P1:VV be the map given by P1=[P]= giV(gi*) PV(gi).

Claim 1: P1 is a projection onto V1.


P1VV1: Note that for each i, PV(gi)VPVV1, and that V(gi*)V1V1. This latter containment follows since V1 is A-stable.

P1v1=v1v1V1: This part only works if we are using the regular trace; we’ll see how that goes in the claim below.

P1v1 = giV (gi*)PV (gi)v1 = giV(gi*) V(gi)v1 sincePis a projection = giV (gi*gi)v1

Claim: gigi*gi=1 if gi* is constructed from tr, the trace of the regular representation.


Continued next lecture.

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

page history