## Lectures in Representation Theory

Last update: 19 August 2013

## Lecture 7

 Continued proof. We would like to show that the basis ${E}_{ij}$ of matrix units of ${M}_{{d}_{\lambda }}\left(ℂ\right)$ is contained in ${W}^{\lambda }\left(A\right),$ whence that ${M}_{{d}_{\lambda }}\left(ℂ\right)={W}^{\lambda }\left(A\right)\text{.}$ We need to produce scalars ${t}_{\lambda }\in ℂ$ 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, $∑gdλWjiλ (g*) Wijλ(g) = ∑gdλWjiλ (g*)Idλ Wijλ(g) = ∑gdλ Wjiλ(g*) ∑m=1dλ EmmWλ(g) = ∑g∑m=1dλ dλEmjWλ (g*)Eim Wλ(g) = ∑m=1dλdλ Emjδimdλ Idλ = ∑m=1dλ Emjδim =Eij$ Thus we have shown that all of the matrix units of ${M}_{{d}_{\lambda }}\left(ℂ\right)$ are in ${W}^{\lambda }\left(A\right),$ that is ${M}_{{d}_{\lambda }}\left(ℂ\right)={W}^{\lambda }\left(A\right),$ and thus $A≃V(A)≃ ⨁Wλ(A)≃ ⨁Mdλ(ℂ).$ $\square$ $1\to 3$

Homework Problem 1.50 Where in this proof that the nondegeneracy of $\text{tr}$ 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 $\text{tr}$ of the regular representation is nondegenerate.

Definition 1.52 Let $\varphi :A\to {M}_{d}\left(ℂ\right)$ be a faithful representation of $A,$ and let $\stackrel{\to }{t}:A\to ℂ$ be the trace of $\varphi \text{.}$ The radical of $A,$ $\sqrt{A},$ is the set of elements of $A$ at which the trace $\stackrel{\to }{t}$ is degenerate; that is, $A= { a∈A | t→ (ba)=0 ∀b∈A }$

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

Theorem 1.53 Given a faithful representation $\varphi :A\to {M}_{d}\left(ℂ\right)$ of $A,$ and the associated trace and radical,

 1 $\sqrt{A}$ is an ideal of $A\text{.}$ 2 If $J⊲A$ is an ideal of nilpotent elements, then $J\subseteq \sqrt{A}\text{.}$ (Recall that an element $a\in A$ is nilpotent provided ${a}^{d}=0$ for some positive integer $d\text{.)}$ 3 Every element of $\sqrt{A}$ is nilpotent.

 Proof. 1: Let $a\in \sqrt{A},$ and $c\in A\text{.}$ Then for every $b\in A,$ $t→(b ac)= t→(ba c)= t→(c ba)= t→(cb a)=0.$ The second equality holds since $\stackrel{\to }{t}$ is a trace; the last since $a$ was chosen to be in the radical.) Thus $\sqrt{A}$ is invariant under right multiplication from $A\text{.}$ Similar arguments show that $\sqrt{A}$ is invariant under left multiplication, and closed under addition. 2: Let $J⊲A$ be an ideal of nilpotent elements, and let $a\in J\text{.}$ To show that $a\in \sqrt{A},$ consider $ba,$ for any $b\in A\text{.}$ Since $J$ is an ideal, we have $ba\in J,$ whence $ba$ is nilpotent. Thus $\varphi \left(ba\right)$ is a nilpotent matrix in ${M}_{d}\left(ℂ\right),$ and so its trace $\text{tr}\left(\varphi \left(ba\right)\right)=0\text{.}$ Thus $\stackrel{\to }{t}\left(ba\right)=0,$ and so $a\in \sqrt{A},$ as desired. 3: Given an element $a\in \sqrt{A},$ show that $a$ is nilpotent. Note that for every $k>0,$ we have $\stackrel{\to }{t}\left({a}^{k}\right)=\stackrel{\to }{t}\left({a}^{k-1}a\right)=0\text{.}$ Thus we have that the trace $\text{tr}\left(\varphi {\left(a\right)}^{k}\right)=\text{tr}\left(\varphi \left({a}^{k}\right)\right)=0,$ for all $k>0\text{.}$ We want to show that this implies that the matrix $\varphi \left(a\right)$ is nilpotent. I can’t think without symmetric functions. – A.Ram We know that $\varphi \left(a\right)$ is similar to an upper triangular matrix, and what the $k\text{-th}$ 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 ${\lambda }_{1}^{k}+\dots +{\lambda }_{d}^{k}=0\text{.}$ Define $Pk(λ1,…,λd) =λ1k+…+λdk,$ and $ek(λ1,⋯,λd) =∑1≤i1<⋯ One has the following Newton identity (c.f. Macdonald Symmetric functions and Hall polynomials): $ek(λ1,⋯,λd)= ∑m1⋯mk∈ℕ∑imi=k (-1)k-∑mi (P1(λ))m1⋯ (Pk(λ))mk 1m1 1m2⋯ kmkm1! m2!⋯mk! .$ For our current purposes, the point is that ${P}_{r}\left(\lambda \right)=0$ for all $r>0,$ and therefore all of the ${e}_{k}\left(\lambda \right)$ are zero. Now consider the characteristic polynomial of $\varphi \left(a\right)\text{.}$ $det(ϕ(a)-tI) = det ( λ1-t* ⋱ 0λd-t ) = (λ1-t)⋯ (λd-t) = (-t)d+e1 (λ)(-t)d-1 +e2(λ) (-t)d-2+… +ed(λ) = (-t)d$ Thus the characteristic polynomial of the matrix $\varphi \left(a\right)$ is ${\left(-t\right)}^{d},$ and so we have that $\varphi {\left(a\right)}^{d}=0\text{.}$ Once $\varphi \left(a\right)$ is nilpotent, we know that a is nilpotent as well. $\square$

Corollary 1.54 The radical $\sqrt{A}$ is the largest ideal of nilpotent elements in $A,$ whence is independent of the choice of the faithful representation $\varphi \text{.}$

Corollary 1.55 The trace $\text{tr}$ of the regular representation is nondegenerate if and only if $\sqrt{A}=0\text{.}$

## 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.