Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 27 August 2013

Lecture 19

3$$Centralizer Algebras

Definition 3.1 If $A$ is an algebra, let $M_d\left(A\right)$ denote the $d\times d$ 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 $$\bar{A}=\{b\in M_d\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}ab=ba\hspace{0.17em}\forall a\in A\}\text{.}$$ Note that $A\subseteq \overline{\stackrel{‾}{A}},$ although the inclusion may be proper.

Proposition 3.3 For any subset $A\subseteq M_d\left(ℂ\right),$ $\bar{A}$ is an algebra over $ℂ\text{.}$

		Proof.
	The set $\bar{A}$ inherits a ring structure from $M_d\left(ℂ\right),$ since for all $b_1,b_2\in \bar{A},$ and all $a\in A$ $$\begin{array}{ccc}(b_1+b_2)a& =& a(b_1+b_2)\\ \left(b_1{b}_2\right)a& =& a\left(b_1{b}_2\right)\text{.}\end{array}$$ Moreover, $cI=Z\left(A\right)$ and so must lie in the center of $\bar{A}\text{.}$ Hence $ℂ\cong cI$ acts as scalars on $\bar{A}$ 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 $\overline{A_1}\supseteq \overline{A_2}\text{.}$

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

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

Proposition 3.6 $\overline{M_d\left(ℂ\right)}=Z\left(M_d\left(ℂ\right)\right)=ℂI_d,$ $\overline{ℂI_d}=M_d\left(ℂ\right)$

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 $\overline{V\left(A\right)},$ 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 $$\overline{V\left(A\right)}\cong {\text{End}}_A\left(V\right)\stackrel{\text{def}}{=}\{b\in \text{End}\left(V\right)\hspace{0.17em}|\hspace{0.17em}ba\left(v\right)=ab\left(v\right)\hspace{0.17em}\forall \hspace{0.17em}a\in A,v\in V\}\text{.}$$

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

		Proof.
	Let $B\in \overline{W\left(A\right)}\text{.}$ Then for all $a\in A$ $$BW\left(a\right)=W\left(a\right)B$$ 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 $\overrightarrow{A}$ given by $a_1\overrightarrow{a_2}=\overrightarrow{a_1{a}_2}\text{.}$ What is ${\text{End}}_A\left(\overrightarrow{A}\right)\text{?}$

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

Let $B\in {\text{End}}_A\left(\overrightarrow{A}\right),$ and let $b\in A$ be defined by $B\overrightarrow{1}=\overrightarrow{b}\text{.}$ Then for all $a\in A,$ $$B\overrightarrow{a}=Ba\overrightarrow{1}=aB\overrightarrow{1}=a\overrightarrow{b}=\overrightarrow{ab}\text{.}$$ Hence $B=R_b$ acts on $A$ as right multiplication by $b,$ and so ${\text{End}}_A\left(\overrightarrow{A}\right)=R_A\text{.}$

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 $d_1\times d_2$ matrix and $Q$ be a $r_1\times r_2$ matrix, both over the same field (or algebra). The tensor product of $P$ and $Q,$ $P\otimes Q$ is the $d_1{r}_1\times d_2{r}_2$ matrix with entries in $ℂ$ described by the block structure $$P\otimes Q=\left(\begin{array}{cccc}{p}_{1,1}Q& p_{1,2}Q& & p_{1,d_2}Q\\ p_{2,1}Q& p_{2,2}Q& & p_{2,d_2}Q\\ & & \ddots & \\ p_{d_1,1}Q& p_{d_1,2}Q& & p_{d_1,d_2}Q\end{array}\right)$$ For example, $E_{i,j}\otimes Q$ is zero other than the $r_1\times r_2$ block in the $(i,j)$ position, where it agrees with $Q\text{.}$

Problem 3.10 Suppose $P_1$ and $P_2$ are $d_1\times d_2$ and $d_2\times d_3$ matrices, respectively, and suppose also that $Q_1$ and $Q_2,$ respectively, are $r_1\times r_2$ and $r_2\times r_3$ (so that we may form the products $P_1{P}_2$ and $Q_1{Q}_2\text{).}$

Check that $(P_1\otimes Q_1)(P_2\otimes Q_2)=P_1{P}_2\otimes Q_1{Q}_2\text{.}$

Let $A$ be a subalgebra of $M_d\left(ℂ\right)\text{.}$ We defined $M_s\left(A\right)$ to be $s\times s$ 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 $\overline{M_s\left(A\right)},$ which also lies inside $M_{sd}\left(ℂ\right)\text{.}$

Suppose $B\in \overline{M_s\left(A\right)},$ and write $B={\sum }_{1\le i,j\le s}{E}_{i,j}\otimes b_{i,j}$ where each $b_{i,j}$ is a $d\times d$ complex matrix. Then $B$ commutes with all elements of $M_s\left(A\right),$ i.e. for all choices of $s^2$ matrices $a_{k,\ell }\in A,$ $$(\sum _{1\le i,j\le s}{E}_{i,j}\otimes b_{i,j})(\sum _{1\le k,\ell \le s}{E}_{k,\ell }\otimes a_{k,\ell })=(\sum _{1\le k,\ell \le s}{E}_{k,\ell }\otimes a_{k,\ell })(\sum _{1\le i,j\le s}{E}_{i,j}\otimes b_{i,j})\text{.}$$

If we multiply this out, the left hand side becomes $$\sum _{i,k,\ell }{E}_{i,\ell }\otimes b_{i,k}{a}_{k,\ell }=\sum _{i,\ell }{E}_{i,\ell }\otimes \sum _k{b}_{i,k}{a}_{k,\ell }\text{.}$$ Performing the same calculation on the right hand side and setting it equal to the previous computation gives $$\sum _{i,\ell }{E}_{i,\ell }\otimes {\sum }_k{b}_{i,k}{a}_{k,\ell }=\sum _{k,j}{E}_{k,j}\otimes \sum _i{a}_{k,i}{b}_{i,j}\text{.}$$ Comparing coefficients of $E_{m,n},$ we see that $$\begin{array}{cc}\sum _k{b}_{m,k}{a}_{k,n}=\sum _i{a}_{m,i}{b}_{i,n}& \text{(3.11)}\end{array}$$ 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 $$b_{m,n}=\sum _i{a}_{m,i}{b}_{i,n}=\sum _k{b}_{m,k}{a}_{k,n}=0\text{.}$$

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 $$b_{m,m}=b_{m,m}{a}_{m,n}=a_{m,n}{b}_{n,n}=b_{n,n}\text{.}$$

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 $$b{a}_{m,n}=b_{m,m}{a}_{m,n}=a_{m,n}{b}_{n\text{.}n}=a_{m,n}b$$ and so $b\in \bar{A}\text{.}$ Thus $B\in I_s\otimes \bar{A},$ hence $\overline{M_s\left(A\right)}\subseteq I_s\otimes \bar{A}$ (where we mean the set of matrices of the form $I_s\otimes b$ for $b\in \bar{A}\text{).}$ 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 $\overline{M_s\left(A\right)}=I_s\otimes \bar{A}\text{.}$

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.

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