## Lectures in Representation Theory

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 $\text{tr}$ of the regular representation of $A$ is nondegenerate. 2 Every representation $V$ of $A$ is completely decomposable. 3 $A\simeq \underset{{d}_{\lambda }\in \stackrel{ˆ}{A}}{⨁}{M}_{{d}_{\lambda }}\left(ℂ\right)\text{.}$ 4 $\sqrt{A}=0$

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 $\stackrel{\to }{t}$ be a trace on $A,$ a finite dimensional algebra. We say that $\stackrel{\to }{t}$ is nondegenerate if for every nonzero $a\in A$ there exists $b\in A$ such that $\stackrel{\to }{t}\left(ba\right)\ne 0\text{.}$

Example 1.43 Let $\stackrel{\to }{t}$ be the ordinary trace $\text{tr}$ on ${M}_{d}\left(ℂ\right),$ and let $a\in {M}_{d}\left(ℂ\right)$ be the matrix unit ${E}_{ij}\text{.}$ Then for $b={E}_{ji},$ one has that $\stackrel{\to }{t}\left(ba\right)=\text{tr}\left({E}_{jj}\right)\ne 0\text{.}$

Definition 1.44 Given an algebra $A$ with basis $\left\{{g}_{1},\dots ,{g}_{d}\right\},$ and let $\stackrel{\to }{t}$ be a trace on $A\text{.}$ The Gram matrix is the matrix $M=\left(⟨{g}_{i},{g}_{j}⟩\right)=\left(\stackrel{\to }{t}\left({g}_{i}{g}_{j}\right)\right)\text{.}$

Proposition 1.45 Given an algebra $A$ with basis $G=\left\{{g}_{1},\dots ,{g}_{d}\right\},$ and let $\stackrel{\to }{t}$ be a trace on $A\text{.}$ The following are equivalent:

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

 Proof. $\left(3\right)↔\left(2\right) \text{.}$ Suppose ${G}^{*}=\left\{{g}_{1}^{*},\dots ,{g}_{d}^{*}\right\}$ is a dual basis for $A$ with respect to $\stackrel{\to }{t}\text{.}$ Let $C=\left({c}_{jk}\right)$ be the transition matrix between the bases $G$ and ${G}^{*}$ (that is, ${{g}_{j}}^{*}=\sum {c}_{jk}{g}_{k}\text{).}$ Then by duality, $δij= ⟨gi,gj*⟩= ⟨gi,∑kcjkgk⟩ =∑kcjk⟨gi,gk⟩.$ This yields the matrix equation $C{M}^{t}={I}_{d}=CM$ (since $\stackrel{\to }{t}$ is symmetric, ${M}^{t}=M\text{).}$ 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 ${g}_{i}^{*}$ as above. The invertibility of $C$ guarantees that the ${g}_{i}^{*}$ will form a basis, while the above calculation will show that the basis is dual to the original one. $\left(3\right)↔\left(2\right)$ $\square$ $\left(1\right)↔\left(2\right) \text{.}$ Let $a$ be a nonzero element of $A,$ and suppose that $\stackrel{\to }{t}$ is degenerate at $a\text{;}$ i.e. for every $b$ in $A$ one has $\stackrel{\to }{t}\left(ba\right)=0\text{.}$ In particular, let $b$ run through the basis $G,$ so that $\stackrel{\to }{t}\left({g}_{i}a\right)=0$ for all $i\text{.}$ Consider $a=\sum {a}_{j}{g}_{j},$, for ${a}_{j}\in ℂ\text{.}$ Now we have that $0= t→(gia)= ∑jt→(giajgj)= ∑j⟨gi,gj⟩aj =0$ This yields the matrix equation $M(a⋮ad) =0,$ whence that the Gram matrix $M$ is not invertible. Conversely, if $M$ is not invertible, one can produce $\stackrel{\to }{a}$ as above in the kernel of $M\text{;}$ the trace $\stackrel{\to }{t}$ is degenerate at $a=\sum {a}_{j}{g}_{j}\text{.}$ $\square$

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

Let $\stackrel{\to }{t}$ be the trace of the regular representation of $V\text{.}$ Then for $a\in A,$ and $G=\left\{\stackrel{\to }{{g}_{1}},\dots ,\stackrel{\to }{{g}_{d}}\right\}$ a basis for $\stackrel{\to }{A},$ one has $t→(a)= ∑gi→∈Ga· gi→|gi→,$ where this last $a·\stackrel{\to }{{g}_{i}}{|}_{\stackrel{\to }{{g}_{i}}}$ is meant to denote the coefficient of $\stackrel{\to }{{g}_{i}}$ in the image of $a·\stackrel{\to }{{g}_{i}}\text{.}$ In the case that we have a dual basis, one can replace $a·\stackrel{\to }{{g}_{i}}{|}_{\stackrel{\to }{{g}_{i}}}$ with $⟨a{g}_{i},{g}_{i}^{*}⟩,$ and the above formula becomes $t→(a)= ∑gi→∈Ga· gi→|gi→ =∑gi→∈G ⟨agi,gi*⟩.$

Lemma 1.47 Given an algebra $A$ with basis $G=\left\{{g}_{1},\dots ,{g}_{d}\right\},$ and $\stackrel{\to }{t}$ a non- degenerate trace on $A\text{.}$ (Note that by the previous proposition, $G$ has a dual basis ${G}^{*}\text{.)}$ Let ${W}_{1}:A\to {M}_{{d}_{1}}\left(ℂ\right)$ and ${W}_{2}:A\to {M}_{{d}_{2}}\left(ℂ\right)$ be representations of $A\text{.}$ Let $C$ be any ${d}_{1}×{d}_{2}$ matrix, and let $\left[C\right]=\sum {W}_{1}\left({g}_{i}^{*}\right)C{W}_{2}\left({g}_{i}\right)\text{.}$ (“symmetrize $C\text{”)}$ Then for all $a$ in $A,$ $W1(a)[C]= [C]W2(a),$ so $\left[C\right]$ is an algebra homomorphism.

 Proof. $W1(a)[C] = W1(a) ∑gi∈G W1(gi*)C W2(gi) = ∑giW1(agi*) CW2(gi) since W1 is an alg hom = ∑giW1 ( ∑hk ⟨agi*,hk⟩ hk* ) CW2(gi) expanding gi* in terms of a basis hk* = ∑hkW1(hk*) CW2 ( ∑gi ⟨agi*,hk⟩ gi ) change order of summation, and move scalars around = ∑hkW1(hk*) CW2 ( ∑gi ⟨hk,gi*⟩ gi ) since t→(agi*hk) =t→(hkagi*) = ∑hkW1(hk*) CW2(hka) =[C]W2(a)$ $\square$

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

Proof.

Let $V$ be a representation of $A\text{.}$

case 1: $V$ is irreducible.

Done.

case 1: $V$ is not irreducible.

Since $V$ is not irreducible, there exists a submodule ${V}_{1}$ in $V,$ and without loss of generality we may presume that ${V}_{1}$ is irreducible. We want to produce an $A\text{-submodule}$ ${V}_{2}$ such that $V\simeq {V}_{1}\oplus {V}_{2}\text{.}$

Let $P:V\to V$ be a projection onto ${V}_{1}$ (i.e. $P$ is a $C\text{-space}$ morphism such that $P{v}_{1}={v}_{1}$ for all ${v}_{1}\in {V}_{1}$ and $PV\subseteq {V}_{1}\text{).}$ One may produce examples of projections as follows: pick a basis for ${V}_{1},$ and extend to a basis for $V\text{.}$ Define $P$ by mapping the basis elements of ${V}_{1}$ to themselves, and the other basis elements of $V$ to any arbitrary vectors in ${V}_{1}\text{.}$

Idea of proof: Let ${V}_{2}=\left(I-P\right)V\text{.}$ Then $V=PV+\left(I-P\right)V={V}_{1}+{V}_{2}\text{.}$ This doesn’t quite work, since there is no reason for this ${V}_{2}$ to be an $A\text{-module.}$ The $\left[P\right]$ will play a useful role here.

Let ${P}_{1}:V\to V$ be the map given by $P1=[P]= ∑giV(gi*) PV(gi).$

Claim 1: ${P}_{1}$ is a projection onto ${V}_{1}\text{.}$

Proof.

${P}_{1}V\subseteq {V}_{1}\text{:}$ Note that for each $i,$ $PV\left({g}_{i}\right)V\subseteq PV\subset {V}_{1},$ and that $V\left({g}_{i}^{*}\right){V}_{1}\subseteq {V}_{1}\text{.}$ This latter containment follows since ${V}_{1}$ is $A\text{-stable.}$

${P}_{1}{v}_{1}={v}_{1}\phantom{\rule{1em}{0ex}}\forall {v}_{1}\in {V}_{1}\text{:}$ This part only works if we are using the regular trace; we’ll see how that goes in the claim below.

$P1v1 = ∑giV (gi*)PV (gi)v1 = ∑giV(gi*) V(gi)v1 since P is a projection = ∑giV (gi*gi)v1$

Claim: $\sum _{{g}_{i}}{g}_{i}^{*}{g}_{i}=1$ if ${g}_{i}^{*}$ is constructed from tr, the trace of the regular representation.

 Proof. 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.