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 20

More generally, let $\hat{V}$ be a finite index set and suppose $A_{\lambda }\subseteq M_{d_{\lambda }}\left(ℂ\right),$ where $d_{\lambda }\in \mathbb{P}$ for each $\lambda \in \hat{V}\text{.}$ Let $d={\sum }_{\lambda }{d}_{\lambda }$ and let $A={⨁}_{\lambda \in \hat{V}}{A}_{\lambda }\text{.}$

Elements of $A$ are matrices of the form $$\left(\begin{array}{cc}\begin{array}{c}{a}_{\lambda }\\ & a_{\mu }\end{array}& 0\\ 0& \begin{array}{c}\ddots \\ & a_{\delta }\end{array}\end{array}\right)$$ where $a_{\rho }\in A_{\rho }$ is a $d_{\rho }\times d_{\rho }$ matrix with complex entries. So $A$ is a subalgebra of $M_d\left(ℂ\right)$ consisting of block diagonal matrices. Note that we do not insist that $A_{\lambda }=M_{d_{\lambda }}\left(ℂ\right),$ so $A$ is merely block diagonal and not necessarily semisimple.

Proposition 3.13 The centralizer of an algebra $A$ of block diagonal matrices is the direct sum of the centralizers of each of the blocks. That is, $$\overline{\underset{\lambda \in \stackrel{ˆ}{V}}{⨁}{A}_{\lambda }}=\underset{\lambda \in \hat{V}}{⨁}\overline{A_{\lambda }}\text{.}$$

		Proof.
	Since block diagonal matrices multiply componentwise, it is obvious that $\overline{\left({\oplus }_{\lambda \in \stackrel{ˆ}{V}}{A}_{\lambda }\right)}\supseteq {⨁}_{\lambda \in \hat{V}}\overline{A_{\lambda }}\text{.}$ Suppose, on the other hand, that $B={\sum }_{\mu ,\nu \in \hat{V}}{E}_{\mu ,\nu }\otimes b_{\mu ,\nu }\in \bar{A}\text{.}$ Elements of $A$ may be written as ${\sum }_{\lambda \in \hat{V}}{E}_{\lambda ,\lambda }\otimes a_{\lambda }$ where $a_{\lambda }\in A_{\lambda }\text{.}$ Then $B$ commutes with all such elements, hence $$(\sum _{\lambda \in \hat{V}}{E}_{\lambda ,\lambda }\otimes a_{\lambda })(\sum _{\mu ,\nu \in \hat{V}}{E}_{\mu ,\nu }\otimes b_{\mu ,\nu })=(\sum _{\mu ,\nu \in \hat{V}}{E}_{\mu ,\nu }\otimes b_{\mu ,\nu })(\sum _{\lambda \in \hat{V}}{E}_{\lambda ,\lambda }\otimes a_{\lambda })\text{.}$$ Multiplying out, we obtain $$\sum _{\lambda ,\nu \in \hat{V}}{E}_{\lambda ,\nu }\otimes a_{\lambda }{b}_{\lambda ,\nu }=\sum _{\mu ,\lambda \in \hat{V}}{E}_{\mu ,\lambda }\otimes b_{\mu ,\lambda }{a}_{\lambda }\text{.}$$ Comparing coefficients of $E_{\lambda ,\nu }$ on each side, we obtain $$\begin{array}{cc}{a}_{\lambda }{v}_{\lambda ,\nu }=b_{\lambda ,\nu }{a}_{\nu }& \text{(3.14)}\end{array}$$ for all $\lambda ,\nu \in \hat{V}$ and for all choices of elements $a_{\mu }\in A_{\mu }\text{.}$ Taking $\mu =\lambda$ we obtain $$a_{\lambda }{b}_{\lambda ,\lambda }=b_{\lambda ,\lambda }{a}_{\lambda }$$ for all $a_{\lambda }\in A_{\lambda }\text{;}$ hence, $b_{\lambda ,\lambda }\in \overline{A_{\lambda }}$ for all $\lambda \in \hat{V}\text{.}$ Fix $\lambda \in \hat{V}$ and choose $a_{\lambda }=I$ and $a_{\mu }=0$ for $\mu \ne \lambda \text{.}$ Then 3.14 becomes $$b_{\lambda ,\nu }=a_{\lambda }{b}_{\lambda ,\nu }=b_{\lambda ,\nu }{a}_{\nu }=0$$ so $b_{\lambda ,\nu }=0$ whenever $\lambda \ne \nu \text{.}$ Thus $B={\sum }_{\mu \in \hat{V}}{E}_{\mu ,\mu }\otimes b_{\mu }$ where $b_{\mu }\in \overline{A_{\mu }}\text{.}$ Therefore, $\bar{A}\subseteq {\oplus }_{\lambda \in \hat{V}}\overline{A_{\lambda }}\text{.}$ $\square$

Corollary 3.15 If $\hat{A}$ is a finite index set, then $$\overline{\underset{\lambda \in \stackrel{ˆ}{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)}=\underset{\lambda \in \hat{A}}{⨁}\overline{M_{d_{\lambda }}\left(ℂ\right)}=\underset{\lambda \in \hat{A}}{⨁}ℂI_{d_{\lambda }}$$

Let $V$ be a representation of an arbitrary (associative) algebra $A$ and suppose $V$ is completely decomposable. Denote the decomposition of $V$ by $$V\cong \underset{\lambda \in \hat{V}}{⨁}{\left(V^{\lambda }\right)}^{\otimes m_{\lambda }}=\sum _{\lambda \in \hat{V}}{m}_{\lambda }{V}^{\lambda }$$ where the $V^{\lambda }$ denote, as usual, representatives of distinct isomorphism classes of irreducible $A\text{-modules.}$ Let $d_{\lambda }=\text{dim}\hspace{0.17em}{V}^{\lambda }\text{.}$

If we choose a basis ${ℬ}_{\lambda }=\{w_1^{\lambda },w_2^{\lambda },\dots w_{d_{\lambda }}^{\lambda }\}$ for each irreducible $V^{\lambda },$ then we may choose a basis for $V$ by taking unions of $m_{\lambda }$ copies of each ${ℬ}_{\lambda }\text{.}$ With respect to this basis, the matrices of $V\left(A\right)$ are of the form $$V\left(a\right)=\underset{\lambda \in \hat{V}}{⨁}\left(\begin{array}{cc}\begin{array}{c}{V}^{\lambda }\left(a\right)\\ & V^{\lambda }\left(a\right)\end{array}& 0\\ 0& \begin{array}{c}\ddots \\ & V^{\lambda }\left(a\right)\end{array}\end{array}\right)$$ where there are $m_{\lambda }$ many blocks of $V^{\lambda }$ for each $\lambda \text{.}$ Thus, in each matrix $V\left(a\right),$ each $\lambda$ block consists of the same matrix repeated $m_{\lambda }$ times, and so $$\begin{array}{cc}V\left(A\right)\cong {\oplus }_{\lambda }{I}_{m_{\lambda }}\left(V^{\lambda }\left(A\right)\right)& \text{(3.16)}\end{array}$$ as algebras.

From the previous propositions, $$\begin{array}{ccc}\overline{V\left(A\right)}& =& \overline{\underset{\lambda \in \stackrel{ˆ}{V}}{⨁}{I}_{m_{\lambda }}\left(V^{\lambda }\left(A\right)\right)}\\ & =& \underset{\lambda \in \hat{V}}{⨁}\overline{I_{m_{\lambda }}\left(V^{\lambda }\left(A\right)\right)}\\ & =& \underset{\lambda \in \hat{V}}{⨁}{M}_{m_{\lambda }}\left(\overline{V^{\lambda }\left(A\right)}\right)\\ & =& \underset{\lambda \in \hat{V}}{⨁}{M}_{m_{\lambda }}\left(I_{d_{\lambda }}\left(ℂ\right)\right)\end{array}$$ So the $\lambda$ block of a matrix $b$ in the centralizer $\overline{V\left(A\right)}$ consists of $m_{\lambda }^2$ scalar matrices $c_{i,j}^{\lambda }{I}_{d_{\lambda }}\text{.}$ That is, for $b\in \overline{V\left(a\right)},$ $$b=\underset{\lambda \in \hat{V}}{⨁}\left(\begin{array}{ccc}\begin{array}{c}{c}_{1,1}^{\lambda }\\ & c_{1,1}^{\lambda }\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& c_{1,1}^{\lambda }\end{array}& \cdots & \begin{array}{c}{c}_{1,m_{\lambda }}^{\lambda }\\ & c_{1,m_{\lambda }}^{\lambda }\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& c_{1,m_{\lambda }}^{\lambda }\end{array}\\ ⋮& \ddots & ⋮\\ \begin{array}{c}{c}_{m_{\lambda },1}^{\lambda }\\ & c_{m_{\lambda },1}^{\lambda }\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& c_{m_{\lambda },1}^{\lambda }\end{array}& \cdots & \begin{array}{c}{c}_{m_{\lambda },m_{\lambda }}^{\lambda }\\ & c_{m_{\lambda },m_{\lambda }}^{\lambda }\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& c_{m_{\lambda },m_{\lambda }}^{\lambda }\end{array}\end{array}\right)$$

We next perform a (somewhat complicated) reordering of the basis of the module $V\text{.}$ In terms of representations, this exchanges $V$ for the equivalent representation $a\mapsto P^{-1}V\left(a\right)P$ where $P$ is a product of elementary matrices describing the permutation of the basis.

Recall that interchanging two columns $i$ and $j$ of a matrix is equivalent to multiplying on the right by the elementary matrix $E={\sum }_{k\ne i,j}{E}_{k,k}+E_{i,j}+E_{j,i}\text{.}$ Multiplying by this elementary matrix on the right interchanges rows $i$ and $j\text{.}$ Note that this matrix is self-invertible, $E=E^{-1},$ and so conjugating a matrix $X$ by $E$ (which is equivalent to interchanging the $i\text{th}$ and $j\text{th}$ basis vectors) corresponds to interchanging the $i\text{th}$ and $j\text{th}$ rows and columns of $X\text{.}$ We may thus describe the permutation of the basis and the resulting change on $V\left(A\right)$ and $\overline{V\left(A\right)}$ is to similarly permute their rows and columns.

We number the columns of blocks, starting from the left, from 1 to $m_{\lambda }\text{.}$ There are $d_{\lambda }$ columns within a particular column of blocks, which we label 1 up to $d_{\lambda }$ from left to right. Hence, the $n\text{th}$ column of $m\text{th}$ block of the $\lambda$ component may be labeled by the ordered pair $(m,n)\text{.}$ Since the individual blocks are square, we may label the rows similarly using the number of their block and row within that block. Thus $m$ ranges from 1 to $m_{\lambda }$ and $n$ ranges from 1 to $d_{\lambda }$ and the $\{(m_1,n_1),(m_2,n_2)\}$ entry of the $\lambda$ component is ${\delta }_{n_1,n_2}{c}_{m_1,m_2}^{\lambda }\text{.}$

Permute the columns so that they appear in the order $(1,1),$ $(2,1),$ $(3,1),$ $\dots ,$ $(m_{\lambda },1),$ $(1,2),$ $(2,2),$ $\dots ,$ $(m_{\lambda },d_{\lambda })$ (i.e. interchange the block number and column number). After permuting the columns, but before performing the similar permutation of the rows, the $\lambda$ component of $b$ is of the form $$\left(\begin{array}{cccc}\begin{array}{cccc}{c}_{1,1}& c_{1,2}& \dots & c_{1,m_{\lambda }}\\ \\ \end{array}& \begin{array}{c}\\ c_{1,1}& c_{1,2}& \dots & c_{1,m_{\lambda }}\\ \end{array}& \begin{array}{c}\\ \dots \\ \end{array}& \begin{array}{c}\\ \\ c_{1,1}& c_{1,2}& \dots & c_{1,m_{\lambda }}\end{array}\\ \begin{array}{cccc}{c}_{2,1}& c_{2,2}& \dots & c_{2,m_{\lambda }}\\ \\ \end{array}& \begin{array}{c}\\ c_{2,1}& c_{2,2}& \dots & c_{2,m_{\lambda }}\\ \end{array}& \begin{array}{c}\\ \dots \\ \end{array}& \begin{array}{c}\\ \\ c_{2,1}& c_{2,2}& \dots & c_{2,m_{\lambda }}\end{array}\\ & & \ddots & \\ \begin{array}{cccc}{c}_{m_{\lambda },1}& c_{m_{\lambda },2}& \dots & c_{m_{\lambda },m_{\lambda }}\\ \\ \end{array}& \begin{array}{c}\\ c_{m_{\lambda },1}& c_{m_{\lambda },2}& \dots & c_{m_{\lambda },m_{\lambda }}\\ \end{array}& \begin{array}{c}\\ \dots \\ \end{array}& \begin{array}{c}\\ \\ c_{m_{\lambda },1}& c_{m_{\lambda },2}& \dots & c_{m_{\lambda },m_{\lambda }}\end{array}\end{array}\right)$$

It should be noted that after rearranging columns, the obvious block structure no longer consists of square blocks. Each block above is a $d_{\lambda }$ by $m_{\lambda }$ matrix. Therefore, the rows of the $\lambda$ component are still labeled by $(m,n)$ with $1\le m\le m_{\lambda }$ and $1\le n\le d_{\lambda }\text{.}$

If we now perform the permutation of the rows so that they are ordered $(1,1),$ $(2,1),$ $(3,1),$ $\dots ,$ $(d_{\lambda },m_{\lambda }),$ observe that we will collect all of the rows with nonzero entries in the first column of blocks into the first $m_{\lambda }$ rows. In fact, the $\lambda$ component becomes $$\left(\begin{array}{c}{C}^{\lambda }\\ & C^{\lambda }\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& C^{\lambda }\end{array}\right)=I_{d_{\lambda }}\otimes C^{\lambda }$$ where $C^{\lambda }=\left(c_{i,j}^{\lambda }\right)$ is an $m_{\lambda }\times m_{\lambda }$ matrix.

Since $b$ is block diagonal, we may perform this reordering simultaneously on each component. Hence, after this reordering of the basis, we may regard $b={\oplus }_{\lambda }{I}_{d_{\lambda }}\otimes C^{\lambda }\text{.}$ We have shown that $\overline{V\left(a\right)}\cong {\oplus }_{\lambda }{I}_{d_{\lambda }}\left(M_{m_{\lambda }}\left(ℂ\right)\right),$ but for any algebra $A,$ $I_{d_{\lambda }}\left(A\right)\cong A\text{.}$

Proposition 3.17 Let $A$ be an arbitrary associative algebra and let $V$ be a completely decomposable finite dimensional representation of $A\text{.}$ If $V={\oplus }_{\lambda }{m}_{\lambda }{V}^{\lambda }$ is a decomposition of $V$ into pairwise nonisomorphic irreducible modules, then $$\overline{V\left(A\right)}\cong \underset{\lambda \in \hat{V}}{⨁}{M}_{m_{\lambda }}\left(ℂ\right)$$ where $d_{\lambda }=\text{dim}\hspace{0.17em}{V}^{\lambda }\text{.}$

In particular, $\overline{V\left(A\right)}$ is semisimple and $\text{dim}\hspace{0.17em}\overline{V\left(A\right)}={\sum }_{\lambda }{m}_{\lambda }^2,$ where $m_{\lambda }$ is the multiplicity of $V^{\lambda }$ in $V\text{.}$

We have fully described $\overline{V\left(A\right)}$ in terms of complex matrices, which already yields the useful information that centralizer algebras of arbitrary finite dimensional representations are semisimple. In 3.16, we showed that $V\left(A\right)={\oplus }_{\lambda \in \hat{V}}{I}_{m_{\lambda }}\left(V^{\lambda }\left(A\right)\right)$ and would like to simplify the $V^{\lambda }\left(A\right)$ to matrices.

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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