Lectures in Representation Theory

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

Last update: 19 August 2013

Lecture 7

Continued proof.

We would like to show that the basis Eij of matrix units of Mdλ() is contained in Wλ(A), whence that Mdλ()=Wλ(A). We need to produce scalars tλ such that Eij=g tλWλ(g*) Wλ(g).

By analogy with the way the matrix units arise in the case of the regular representation (do either of you want me to type that up as well?), wherein the dual g* are constructed using the ordinary trace tr, one may hope that Eij=gdλ Wjiλ(g*) Wijλ(g) will hold in this more general situation. In fact, gdλWjiλ (g*) Wijλ(g) = gdλWjiλ (g*)Idλ Wijλ(g) = gdλ Wjiλ(g*) m=1dλ EmmWλ(g) = gm=1dλ dλEmjWλ (g*)Eim Wλ(g) = m=1dλdλ Emjδimdλ Idλ = m=1dλ Emjδim =Eij

Thus we have shown that all of the matrix units of Mdλ() are in Wλ(A), that is Mdλ()=Wλ(A), and thus AV(A) Wλ(A) Mdλ().


Homework Problem 1.50 Where in this proof that the nondegeneracy of tr implies that A is isomorphic to a direct sum of matrix rings does the Jacobson density theorem occur?

Homework Problem 1.51 Prove the converse: that if A is isomorphic to a direct sum of matrix rings, then the trace tr of the regular representation is nondegenerate.

Definition 1.52 Let ϕ:AMd() be a faithful representation of A, and let t:A be the trace of ϕ. The radical of A, A, is the set of elements of A at which the trace t is degenerate; that is, A= { aA|t (ba)=0bA }

A priori, the definition of the radical would seem to be dependent on the choice of nondegenerate representation ϕ; this turns out to not be the case, as we shall see below.

Theorem 1.53 Given a faithful representation ϕ:AMd() of A, and the associated trace and radical,

1. A is an ideal of A.
2. If JA is an ideal of nilpotent elements, then JA. (Recall that an element aA is nilpotent provided ad=0 for some positive integer d.)
3. Every element of A is nilpotent.


1: Let aA, and cA. Then for every bA, t(bac)= t(bac)= t(cba)= t(cba)=0.

The second equality holds since t is a trace; the last since a was chosen to be in the radical.) Thus A is invariant under right multiplication from A. Similar arguments show that A is invariant under left multiplication, and closed under addition.

2: Let JA be an ideal of nilpotent elements, and let aJ. To show that aA, consider ba, for any bA. Since J is an ideal, we have baJ, whence ba is nilpotent. Thus ϕ(ba) is a nilpotent matrix in Md(), and so its trace tr(ϕ(ba))=0. Thus t(ba)=0, and so aA, as desired.

3: Given an element aA, show that a is nilpotent. Note that for every k>0, we have t(ak)=t(ak-1a)=0. Thus we have that the trace tr(ϕ(a)k)=tr(ϕ(ak))=0, for all k>0. We want to show that this implies that the matrix ϕ(a) is nilpotent.

I can’t think without symmetric functions. – A.Ram

We know that ϕ(a) is similar to an upper triangular matrix, and what the k-th powers of such a matrix are: ϕ(a) ( λ1* 0λd ) And thus ϕ(a)k ( λ1k* 0λdk )

Thus for every k>0 we know that λ1k++λdk=0.
Define Pk(λ1,,λd) =λ1k++λdk, and ek(λ1,,λd) =1i1<<ikd λi1λik

One has the following Newton identity (c.f. Macdonald Symmetric functions and Hall polynomials):

ek(λ1,,λd)= m1mkimi=k (-1)k-mi (P1(λ))m1 (Pk(λ))mk 1m1 1m2 kmkm1! m2!mk! .

For our current purposes, the point is that Pr(λ)=0 for all r>0, and therefore all of the ek(λ) are zero.

Now consider the characteristic polynomial of ϕ(a).

det(ϕ(a)-tI) = det ( λ1-t* 0λd-t ) = (λ1-t) (λd-t) = (-t)d+e1 (λ)(-t)d-1 +e2(λ) (-t)d-2+ +ed(λ) = (-t)d

Thus the characteristic polynomial of the matrix ϕ(a) is (-t)d, and so we have that ϕ(a)d=0.

Once ϕ(a) is nilpotent, we know that a is nilpotent as well.

Corollary 1.54 The radical A is the largest ideal of nilpotent elements in A, whence is independent of the choice of the faithful representation ϕ.

Corollary 1.55 The trace tr of the regular representation is nondegenerate if and only if A=0.

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.

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