## Lectures in Representation Theory

Last update: 27 August 2013

## 3$\phantom{\rule{1em}{0ex}}$Centralizer Algebras

My Favorite Game
—A. Ram

Definition 3.1 If $A$ is an algebra, let ${M}_{d}\left(A\right)$ denote the $d×d$ matrices with entries in $A\text{.}$

Definition 3.2 Let $A$ be a subset of ${M}_{d}\left(ℂ\right)\text{.}$ The centralizer or commutant of $A$ is the set $A‾= {b∈Md(ℂ) | ab=ba ∀a∈A} .$ Note that $A\subseteq \stackrel{‾}{\stackrel{‾}{A}},$ although the inclusion may be proper.

Proposition 3.3 For any subset $A\subseteq {M}_{d}\left(ℂ\right),$ $\stackrel{‾}{A}$ is an algebra over $ℂ\text{.}$

 Proof. The set $\stackrel{‾}{A}$ inherits a ring structure from ${M}_{d}\left(ℂ\right),$ since for all ${b}_{1},{b}_{2}\in \stackrel{‾}{A},$ and all $a\in A$ $(b1+b2)a = a(b1+b2) (b1b2)a = a(b1b2).$ Moreover, $cI=Z\left(A\right)$ and so must lie in the center of $\stackrel{‾}{A}\text{.}$ Hence $ℂ\cong cI$ acts as scalars on $\stackrel{‾}{A}$ which gives the $ℂ$ vector space and algebra structures. $\square$

Proposition 3.4 If ${A}_{1}\subseteq {A}_{2}\subseteq {M}_{d}\left(ℂ\right)$ are subsets then $\stackrel{‾}{{A}_{1}}\supseteq \stackrel{‾}{{A}_{2}}\text{.}$

Proposition 3.5 $\stackrel{‾}{A}=\stackrel{‾}{\stackrel{‾}{\stackrel{‾}{A}}}\text{.}$

 Proof. Since $A\subseteq \stackrel{‾}{\stackrel{‾}{A}},$ the previous proposition implies that $\stackrel{‾}{A}\supseteq \stackrel{‾}{\stackrel{‾}{\stackrel{‾}{A}}}\text{.}$ But $\stackrel{‾}{A}\subseteq \stackrel{‾}{\stackrel{‾}{\left(\stackrel{‾}{A}\right)}}=\stackrel{‾}{\stackrel{‾}{\stackrel{‾}{A}}}\text{.}$ $\square$

Proposition 3.6 $\stackrel{‾}{{M}_{d}\left(ℂ\right)}=Z\left({M}_{d}\left(ℂ\right)\right)=ℂ{I}_{d},$ $\stackrel{‾}{ℂ{I}_{d}}={M}_{d}\left(ℂ\right)$

Let $V$ be a representation of an algebra $A,$ so $V:A\to {M}_{d}\left(ℂ\right)\text{.}$ The image of $A$ is the algebra of the representation $V\left(A\right)$ is a subalgebra of ${M}_{d}\left(ℂ\right)\text{.}$ We are interested in $\stackrel{‾}{V\left(A\right)},$ the centralizer algebra of the representation $V\text{.}$

From the module perspective, suppose $V$ is an $A\text{-module}$ and let $\text{End}\left(V\right)$ be the set of all linear transformations of $V\text{.}$ Then $V(A)‾≅ EndA(V)=def { b∈End(V) | ba(v)=ab(v) ∀ a∈A,v∈V } .$

Example 3.7 Let $A$ be an arbitrary algebra and let $W$ be an irreducible representation of $A\text{.}$ We claim $\stackrel{‾}{W\left(A\right)}=ℂI\text{.}$

 Proof. Let $B\in \stackrel{‾}{W\left(A\right)}\text{.}$ Then for all $a\in A$ $BW(a)=W(a)B$ and so by Schur’s lemma, $B=cI$ for some $c\in C\text{.}$ $\square$

Example 3.8 Let $A$ be an (associative) algebra. Recall the left regular representation $\stackrel{\to }{A}$ given by ${a}_{1}\stackrel{\to }{{a}_{2}}=\stackrel{\to }{{a}_{1}{a}_{2}}\text{.}$ What is ${\text{End}}_{A}\left(\stackrel{\to }{A}\right)\text{?}$

The associative law $a\left(bc\right)=\left(ab\right)c$ shows that ${\text{End}}_{A}\left(\stackrel{\to }{A}\right)\supseteq {R}_{A},$ the algebra of transformations given by right multiplication by elements of $A\text{.}$

Let $B\in {\text{End}}_{A}\left(\stackrel{\to }{A}\right),$ and let $b\in A$ be defined by $B\stackrel{\to }{1}=\stackrel{\to }{b}\text{.}$ Then for all $a\in A,$ $Ba→=Ba1→=a B1→=ab→= ab→.$ Hence $B={R}_{b}$ acts on $A$ as right multiplication by $b,$ and so ${\text{End}}_{A}\left(\stackrel{\to }{A}\right)={R}_{A}\text{.}$

We are interested in the centralizer algebras of arbitrary semisimple algebras, $A\cong {\oplus }_{\lambda }{M}_{{d}_{\lambda }}\left(ℂ\right),$ so we will develop some notation for block matrix algebras.

Definition 3.9 Let $P$ be a ${d}_{1}×{d}_{2}$ matrix and $Q$ be a ${r}_{1}×{r}_{2}$ matrix, both over the same field (or algebra). The tensor product of $P$ and $Q,$ $P\otimes Q$ is the ${d}_{1}{r}_{1}×{d}_{2}{r}_{2}$ matrix with entries in $ℂ$ described by the block structure $P⊗Q= ( p1,1Q p1,2Q p1,d2Q p2,1Q p2,2Q p2,d2Q ⋱ pd1,1Q pd1,2Q pd1,d2Q )$ For example, ${E}_{i,j}\otimes Q$ is zero other than the ${r}_{1}×{r}_{2}$ block in the $\left(i,j\right)$ position, where it agrees with $Q\text{.}$

Problem 3.10 Suppose ${P}_{1}$ and ${P}_{2}$ are ${d}_{1}×{d}_{2}$ and ${d}_{2}×{d}_{3}$ matrices, respectively, and suppose also that ${Q}_{1}$ and ${Q}_{2},$ respectively, are ${r}_{1}×{r}_{2}$ and ${r}_{2}×{r}_{3}$ (so that we may form the products ${P}_{1}{P}_{2}$ and ${Q}_{1}{Q}_{2}\text{).}$

Check that $\left({P}_{1}\otimes {Q}_{1}\right)\left({P}_{2}\otimes {Q}_{2}\right)={P}_{1}{P}_{2}\otimes {Q}_{1}{Q}_{2}\text{.}$

Let $A$ be a subalgebra of ${M}_{d}\left(ℂ\right)\text{.}$ We defined ${M}_{s}\left(A\right)$ to be $s×s$ matrices with entries in $A,$ so elements of ${M}_{s}\left(A\right)$ are of the form ${\sum }_{1\le i,j\le s}{E}_{ij}\otimes {a}_{i,j},$ where the ${a}_{i,j}$ are matrices in $A\text{.}$ Thus, ${M}_{s}\left(A\right)$ is a subalgebra of ${M}_{sd}\left(ℂ\right)\text{.}$ Let us compute the centralizer $\stackrel{‾}{{M}_{s}\left(A\right)},$ which also lies inside ${M}_{sd}\left(ℂ\right)\text{.}$

Suppose $B\in \stackrel{‾}{{M}_{s}\left(A\right)},$ and write $B={\sum }_{1\le i,j\le s}{E}_{i,j}\otimes {b}_{i,j}$ where each ${b}_{i,j}$ is a $d×d$ complex matrix. Then $B$ commutes with all elements of ${M}_{s}\left(A\right),$ i.e. for all choices of ${s}^{2}$ matrices ${a}_{k,\ell }\in A,$ $(∑1≤i,j≤sEi,j⊗bi,j) (∑1≤k,ℓ≤sEk,ℓ⊗ak,ℓ)= (∑1≤k,ℓ≤sEk,ℓ⊗ak,ℓ) (∑1≤i,j≤sEi,j⊗bi,j).$

If we multiply this out, the left hand side becomes $∑i,k,ℓ Ei,ℓ⊗ bi,k ak,ℓ= ∑i,ℓ Ei,ℓ⊗ ∑kbi,k ak,ℓ.$ Performing the same calculation on the right hand side and setting it equal to the previous computation gives $∑i,ℓ Ei,ℓ⊗ ∑kbi,k ak,ℓ= ∑k,j Ek,j⊗ ∑iak,i bi,j.$ Comparing coefficients of ${E}_{m,n},$ we see that $∑kbm,k ak,n=∑i am,ibi,n (3.11)$ for all choices of ${s}^{2}$ elements ${a}_{i,j}\in A,$ for each $1\le m,n\le s\text{.}$

Since $A$ is a subalgebra, we know 0 and $I$ lie in $A\text{.}$ Fix indices $m\ne n$ and choose ${s}^{2}$ elements given by ${a}_{m,m}=I$ and all others zero. Then $bm,n=∑i am,i bi,n= ∑kbm,k ak,n=0.$

Make a second choice of ${s}^{2}$ elements, with $m\ne n,$ where ${a}_{m,n}=I$ and all other ${a}_{i,j}=0\text{.}$ Equation 3.11 becomes $bm,m= bm,m am,n= am,n bn,n= bn,n.$

Hence $B={\sum }_{1\le i,j\le s}{E}_{i,j}\otimes {b}_{i,j}={\sum }_{k=1}^{s}{E}_{k,k}\otimes b$ is block diagonal. Moreover, 3.11 becomes $bam,n= bm,m am,n= am,n bn.n= am,nb$ and so $b\in \stackrel{‾}{A}\text{.}$ Thus $B\in {I}_{s}\otimes \stackrel{‾}{A},$ hence $\stackrel{‾}{{M}_{s}\left(A\right)}\subseteq {I}_{s}\otimes \stackrel{‾}{A}$ (where we mean the set of matrices of the form ${I}_{s}\otimes b$ for $b\in \stackrel{‾}{A}\text{).}$ The reverse inclusion is left as an exercise, and we claim

Proposition 3.12 If $A$ is a subalgebra of ${M}_{d}\left(ℂ\right)$ then $\stackrel{‾}{{M}_{s}\left(A\right)}={I}_{s}\otimes \stackrel{‾}{A}\text{.}$

## 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.