Double centralizer nonsense

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

Last update: 6 November 2012

Double centralizer nonsense

Tensor products

If $P$ and $Q$ are two matrices with entries from $ℂ,$ then the tensor product of $P$ and $Q$ is the matrix

$$\begin{array}{cc}P\otimes Q=\mid p_{ij}Q\mid ,& \text{(4.1)}\end{array}$$

where $p_{ij}$ denotes the $(i,j)\text{th}$ entry in $P\text{.}$ If $V$ and $W$ are two vector spaces with bases $B_V=\left\{v_i\right\}$ and $B_W=\left\{w_i\right\}$ respectively, the tensor product $V\otimes W$ is the vector space consisting of the linear span of the words $v_i{w}_j\text{.}$ If $V$ is dimension $n$ and $W$ is dimension $nm\text{.}$ then $V\otimes W$ is dimension $nm\text{.}$ In general, for any $v\in V$ and $w\in W,$ the word $vw$ can be expressed in terms of the words $v_i{w}_j$ by using linearity, i.e. for all $c,d\in ℂ$ $v_i,v_j\in B_V$ and $w_r,w_s\in B_W,$

$$\begin{array}{ccc}(c{v}_i+d{v}_j)w_r& =& c{v}_i{w}_r+d{v}_j{w}_r\text{and}\\ v_i(c{w}_r+d{w}_s)& =& c{v}_i{w}_r+d{v}_i{w}_s\text{.}\end{array}$$

Suppose that $A$ and $C$ are two arbitrary algebras. We can define an algebra structure on the vector space $A\otimes C$ (we distinguish the tensor product of algebras from the vector space case by writing $(a,c)$ instead of $ac$ for a word in $A\otimes C,$ $a\in A,$ $c\in C\text{)}$ by defining multiplication of elements of $A\otimes C$ by

$$\begin{array}{cc}(a_1,c_1)(a_2,c_2)=(a_1{a}_2,c_1{c}_2),& \text{(4.2)}\end{array}$$

for all $a_1,a_2\in A$ and $c_1,c_2\in C,$ amd extending linearly.

Suppose that $V$ and $W$ are representations of $A$ and $C$ respectively. Define an action of $A\otimes C$ on the vector space $V\otimes W$ by

$$\begin{array}{cc}(a,c)\left(vw\right)=\left(av\right)\left(cw\right)& \text{(4.3)}\end{array}$$

for all $(a,c)$ and $vw,$ $a\in A,$ $c\in C,$ $v\in V,$ $w\in W\text{.}$ This defines a representation of $A\otimes C$ on $V\otimes W$ under which the action of $(a,c),$ $a\in A,$ $c\in C$ on $V\otimes W$ is given by the matrix

$$V\left(a\right)\otimes W\left(c\right)\text{.}$$

Centralizer of a completely decomposable representation

Let $V$ be a completely decomposable representation of an algebra $A\text{.}$ Assume that

$$V\cong {\oplus }_{\lambda =1}^n{W}_{\lambda }^{\oplus m_{\lambda }},$$

where the $W_i$ are nonisomorphic irreducible representations of $V\text{.}$ This means that we can decompose $V$ into irreducible subspaces $V_{\lambda j}m$ $1\le \lambda \le n,$ $1\le j\le m_{\lambda },$ so that

$$V={\oplus }_{\lambda ,j}{V}_{\lambda j},$$

where for each $\lambda$ and $j,$ $V_{\lambda j}\cong W_{\lambda }\text{.}$ Let $d_{\lambda }=\text{dim}{W}_i\text{.}$ Choosing a basis on each of the $V_{\lambda j}$ gives a basis of $V$ which we denote $ℬ\text{.}$ Using the basis $ℬ$ of $V,$ the algebra of the representation $V$ is

$$\begin{array}{cc}V\left(A\right)=\left(\begin{array}{ccccc}{I}_{m_1}\left(M_{d_1}\left(ℂ\right)\right)& 0& 0& \dots & 0\\ 0& I_{m_2}\left(M_{d_2}\left(ℂ\right)\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& I_{m_n}\left(M_{d_m}\left(ℂ\right)\right)\end{array}\right)\text{.}& \text{(4.4)}\end{array}$$

(1.11) shows that the algebra of matrices that commute with all matrices in $V\left(A\right),$ is

$$\overline{V\left(A\right)}=\left(\begin{array}{ccccc}{M}_{m_1}\left(\overline{W_1\left(A\right)}\right)& 0& 0& \dots & 0\\ 0& M_{m_2}\left(\overline{W_2\left(A\right)}\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& M_{m_n}\left(\overline{W_n\left(A\right)}\right)\end{array}\right)\text{.}$$

Since, by Schur's Lemma, $\overline{W_{\lambda }\left(A\right)}=I_{d_{\lambda }}\left(ℂ\right),$ such that

$$\begin{array}{cc}\overline{V\left(A\right)}=\left(\begin{array}{ccccc}{M}_{m_1}\left(I_{d_1}\left(ℂ\right)\right)& 0& 0& \dots & 0\\ 0& M_{m_2}\left(I_{d_2}\left(ℂ\right)\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& M_{m_n}\left(I_{d_n}\left(ℂ\right)\right)\end{array}\right)\text{.}& \text{(4.5)}\end{array}$$

(4.6) Theorem. If a representation $V$ of an algebra $A$ is completely decomposable in the form

$$V\cong {\oplus }_{\lambda =1}^n{W}_{\lambda }^{\oplus m_{\lambda }},$$

where the $W_{\lambda }$ are nonisomorphic irreducibles, then the centralizer $\overline{V\left(A\right)}$ of $V\left(A\right)$ is semsimple and

$$\overline{V\left(A\right)}\cong {\oplus }_{\lambda =1}^n{M}_{m_{\lambda }}\left(ℂ\right)\text{.}$$

		Proof.
	By a change of basis on $V$ we can put the matrices of (4.5) in the form $$\begin{array}{cc}\left(\begin{array}{ccccc}{I}_{d_1}\left(M_{m_1}\left(ℂ\right)\right)& 0& 0& \dots & 0\\ 0& I_{d_2}\left(M_{m_2}\left(ℂ\right)\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& I_{d_n}\left(M_{m_n}\left(ℂ\right)\right)\end{array}\right)\text{.}& \text{(4.7)}\end{array}$$ These matrices are of exactly the same form as those in (4.4) except that the $d_{\lambda }\text{s}$ and $m_{\lambda }\text{s}$ are switched!! (4.7) shows that $\overline{V\left(A\right)}\cong {\oplus }_{\lambda =1}^n{M}_{m_{\lambda }}\left(ℂ\right)$ as algebras. $\square$

Let $B$ be an algebra with an action on $B$ such that $V\left(B\right)=\overline{V\left(A\right)}\text{.}$ Let $B^{\prime }$ be the kernel of the act $B$ on $V$ and let $C$ be the quotient $B/B^{\prime }$ so that the induced action of $C$ on $V$ is injective. $C\cong V\left(B\right)=\overline{V\left(A\right)}\text{.}$

From (4.5) we see that with respect to the basis $ℬ$ on $V$ the action of an element $q\in C$ is given by a matrix of the form

$$\begin{array}{cc}\left(\begin{array}{ccccc}{Q}_1\otimes I_{m_1}& 0& 0& \dots & 0\\ 0& Q_2\otimes I_{m_2}& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& Q_n\otimes I_{m_n}\end{array}\right)\text{.}& \text{(4.8)}\end{array}$$

where $Q_{\lambda }\in M_{m_{\lambda }}\left(ℂ\right)\text{.}$ This action determines a map

$$\begin{array}{cccc}\varphi :& C& ⟶& {\oplus }_{\lambda }{M}_{m_{\lambda }}\left(ℂ\right)\\ & q& ⟼& \left(\begin{array}{cccc}{Q}_1& 0& \dots & 0\\ 0& Q_2& \dots & 0\\ \dots & \dots & \dots & \dots \\ 0& 0& \dots & Q_n\end{array}\right),\end{array}$$

which, by Theorem (4.6), is an isomorphism. Note that, for each $\lambda ,$ the map

$$\begin{array}{cc}\begin{array}{cccc}{C}_{\lambda }:& C& ⟶& M_{m_{\lambda }}\left(ℂ\right)\\ & q& ⟼& Q_{\lambda }\end{array}& \text{(4.9)}\end{array}$$

is an irreducible representation of $C\text{.}$

Let $E_{ij}^{\lambda }$ denote the matrix in ${\oplus }_{\lambda }{M}_{m_{\lambda }}\left(ℂ\right)$ that is 1 in the $(i,j)\text{th}$ entry of the $\lambda \text{th}$ block and 0 everywhere else. Define a set of matrix units $e_{ij}^{\lambda },$ $1\le \lambda \le n,$ $1\le i,j\le m_{\lambda }$ in $C$ by

$$e_{ij}^{\lambda }={\varphi }^{-1}\left(E_{ij}^{\lambda }\right)\text{.}$$

The action of the element $e_{ii}^{\lambda }$ on $V,$ is given by the matrix $E_{ii}^{\lambda }\otimes I_{\overrightarrow{m}}\in \overline{V\left(A\right)}\text{.}$ The action of this matrix on $V$ is the projection $p:V\to V_{\lambda i};$

$$V_{\lambda i}=e_{ii}^{\lambda }V\text{.}$$

Conversely, if $\left\{e_{ij}^{\lambda }\right\}$ is a set of matrix units of $C,$ then, since $1=\sum _{\lambda ,i}{e}_{ii}^{\lambda }$ as an element of $C,$ we have a decomposition

$$V=1·V=\left(\sum _{\lambda ,i}{e}_{ii}^{\lambda }\right)V=\sum _{\lambda ,i}{e}_{ii}^{\lambda }V\text{.}$$

Since the action of $A$ on $V$ commutes with the action of $C$ we have that $a{e}_{ii}^{\lambda }V=e_{ii}^{\lambda }aV\subset e_{ii}^{\lambda }V$ for all $a\in A,$ showing that each of the spaces $e_{ii}^{\lambda }V$ is $A\text{-invariant.}$ Since, §1 Ex. 5, $e_{ii}^{\lambda }V\cap e_{jj}^{\mu }=\varnothing$ unless $\lambda =\mu$ and $i=j,$ the decomposition given above is a direct sum decomposition of $V\text{.}$ This decomposition is a decomposition of $V$ into irreducible subspaces under the action of $A,$

$$\begin{array}{cc}V={\oplus }_{\lambda ,i}{e}_{ii}^{\lambda }V\text{.}& \text{(4.10)}\end{array}$$

Define an action of $C\otimes A$ on $V$ by

$$(q,a)v=qav$$

where $(q,a)\in C\otimes A$ and $v\in V\text{.}$ Since the actions of $C$ and $A$ on $V$ commute this action is well defined and makes $V$ into an $C\otimes A$ representation. Theorem (4.6) shows that the irreducible representations of $C$ are in one to one correspondence with the irreducible representations of $A$ appearing in the decomposition of $V\text{.}$ Let $C_{\lambda }$ denote the irreducible representation of $C$ corresponding to $A\text{.}$

(4.11) Theorem. As $C\otimes A$ representations,

$$V\cong {\oplus }_{\lambda =1}^n{C}_{\lambda }\otimes W_{\lambda }\text{.}$$

		Proof.
	With respect to the basis $ℬ$ of $V$ the action of $(q,a)\in C\otimes A$ on $V$ is given by the matrix product $$\left(\begin{array}{ccccc}{Q}_1\otimes I_{m_1}& 0& 0& \dots & 0\\ 0& Q_2\otimes I_{m_2}& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& Q_n\otimes I_{m_n}\end{array}\right)\left(\begin{array}{ccccc}{I}_{m_1}\otimes W_1\left(a\right)& 0& 0& \dots & 0\\ 0& I_{m_2}\otimes W_2\left(a\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& I_{m_n}\otimes W_n\left(a\right)\end{array}\right),$$ which is equal to $$\begin{array}{cc}\left(\begin{array}{ccccc}{Q}_1\otimes W_1\left(a\right)& 0& 0& \dots & 0\\ 0& Q_2\otimes W_2\left(a\right)& 0& \dots & 0\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & 0& Q_n\otimes W_n\left(a\right)\end{array}\right)\text{.}& \text{(4.12)}\end{array}$$ Recalling (4.9) we see that the action of each block of (4.12) is by the representation $C_{\lambda }\otimes W_{\lambda }\text{.}$ $\square$

Examples

1. Let $G$ be a group and let $V$ and $W$ be two representations of $G$. Define an action of $G$ on the vector space $V\otimes W$ by $$g\left(vw\right)=\left(gv\right)\left(gw\right),$$ for all $g\in G,v\in V$ and $w\in W$ (see also Section 5 Ex 4). In matrix form, the representation $V\otimes W$ is given by setting $$\left(V{\otimes }_{d}W\right)\left(g\right)=V\left(g\right)\otimes W\left(g\right),$$ for each $g\in G.$ Note, however, that if we extend this action to an action of $A=ℂG$ on $V\otimes W,$ then for a general $a\in A,$ $a\left(vw\right)$ is not equal to $\left(av\right)\left(aw\right)$ and $\left(V{\otimes }_{d}W\right)\left(a\right)$ is not equal to $V\left(a\right)\otimes W\left(a\right).$
2. Theorem 4.6 gives that there is a one-to-one correspondence between minimal central idempotents $z_{\lambda }^C$ of $C$ and characters ${\chi }_A^{\lambda }$ of irreducible representations of $A$ of $A$ appearing in the decomposition of $V$. Let ${\chi }_C^{\lambda }$ be the irreducible characters of $C$ and for each $\lambda$ set $d_{\lambda }^C={\chi }_C^{\lambda }\left(1\right),$ so that the $d_{\lambda }$ are the dimensions of the irreducible representations of $C.$ The Frobenius map is the map $$\begin{array}{rcll}F:& Z\left(C\right)& \to & R\left(A\right)\\ & \left(\frac{1}{d_{\lambda }^C}\right)z_{\lambda }^X& \mapsto & {\chi }_A^{\lambda }.\end{array}$$ Let $t:C\otimes A\to ℂ$ be the trace of the action of $C\otimes A$ on the representation $V.$ By taking traces on each side of the isomorphism in Theorem 4.11 we have that $$t\left(q,a\right)=\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right){\chi }_A^{\lambda }\left(a\right).$$ Let ${\overrightarrow{t}}_C=\left(t_{\lambda }^C\right)$ be a nondegenerate trace on $C$, let $B$ be a basis of $C$ and for each $g\in B$ let $g*$ be the element of the dual bsis to $B$ with respect to the trace ${\overrightarrow{t}}_C$ such that ${\overrightarrow{t}}_C\left(gg*\right)=1.$ Then, for any $z\in Z\left(C\right),$ the center of $C,$ $$F\left(z\right)=\sum _{g\in B}{\overrightarrow{t}}_C\left(zg*\right)t\left(g,.\right),$$ since, using 3.8 and 3.9, $$\begin{array}{rcl}F\left(\frac{z_{\mu }^C}{d_{\mu }^C}\right)& =& \sum _g\left(\frac{1}{d_{\mu }^C}\right){\overrightarrow{t}}_C\left(z_{\mu }^{C}g*\right)t\left(g,.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\chi }_C^{\mu }\left(g*\right)t\left(g,.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\chi }_C^{\mu }\left(g*\right)\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right){\chi }_A^{\lambda }\left(.\right)\\ & =& \sum _g\left(\frac{t_{\mu }^C}{d_{\mu }^C}\right){\delta }_{\mu \lambda }\left(\frac{d_{\lambda }^C}{t_{\lambda }^C}\right){\chi }_A^{\lambda }\left(.\right)\\ & =& {\chi }_{\mu }^A\left(.\right).\end{array}$$

   If we apply the inverse $F^{-1}$ of the Frobenius map to (4.13) we get $$F^{-1}\left(t\left(q,.\right)\right)=\sum _{\lambda }{\chi }_C^{\lambda }\left(q\right)\left(\frac{z_{\lambda }^C}{d_{\lambda }^C}\right).$$ Formula 3.13 shows that $$F^{-1}\left(t\left(q,.\right)\right)=\left(\sum _{\lambda }\left(\frac{t_{\lambda }^C}{d_{\lambda }^C}\right)z_{\lambda }^C\right)\left[q\right].$$ In the case that ${\overrightarrow{t}}_C$ is the trace of the regular representation $\sum _{\lambda }\left(\frac{t_{\lambda }^C}{d_{\lambda }^C}\right)z_{\lambda }^C=1$ and $F^{-1}\left(t\left(q,.\right)\right)=\left[q\right].$

Notes and References

"Double centralizer nonsense" is a term that has been used by R. Stanley in reference to Theorems (4.6) and (4.12). I have chosen to adopt this term as well. These results are originally due to I. Schur [Scl],[Sch1927], and are often referred to as the Double Commutant Theorem, or, in the special case of the representation $V^{\otimes f},$ $\text{dim}V=n$ of $\text{Gl}\left(n\right),$ Schur-Weyl duality. This was the key concept in Schur's original work on the rational representations of $\text{Gl}\left(n\right)\text{.}$

The Frobenius map given in Ex. 3 is a generalization of the classical Frobenius map [Mac1979] §1.7. In a paper [Fro1900] that demonstrates absolute genius, Frobenius used it as a tool for determining the characters of the symmetric groups.

Notes and References

This is an excerpt from the unpublished first chapter of Arun Ram's dissertation entitled Representation Theory, written July 4, 1990.

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