On the trace of the regular representation of a centralizer algebra

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

Last update: 10 September 2013

Trace of the regular representation

It was shown in [Ram1994] that the trace of the regular representation of the Brauer algebra $B_m$ was related to the character of the irreducible representation of the Brauer algebra $B_{2m}$ labeled by the partition $\varnothing \text{.}$ Here we will show that this result holds in a much more general context, for any centralizer algebra.

Let $W$ be a representation of a group $G$ and let $𝒵={\text{End}}_G\left(W\right)\text{.}$ Let $W^{*}$ be the representation of $G$ which is dual to $W$ and let ${𝒵}^{\text{opp}}={\text{End}}_G\left(W^{*}\right)\text{.}$ There is a natural identification of ${𝒵}^{\text{opp}}$ with the algebra $𝒵$ with the opposite multiplication. Let $\widehat{𝒵}={\text{End}}_G(W\otimes W^{*})\text{.}$ Clearly $𝒵\otimes {𝒵}^{\text{opp}}\subseteq \widehat{𝒵}\text{.}$ Let ${𝒵}^{\varnothing }={(W\otimes W^{*})}^G$ be the $G$ invariants. ${\widehat{𝒵}}^{\varnothing }$ is a $\widehat{𝒵}\text{-submodule}$ of $W\otimes W^{*}\text{.}$ Let ${\chi }^{\varnothing }$ be the trace of $\widehat{𝒵}$ in this representation.

Let $z,y\in 𝒵\text{.}$ Let $y^{*}\in {𝒵}^{\text{opp}}$ be the element which corresponds to $y\in 𝒵\text{.}$ The bitrace of the regular representation of $𝒵$ is given by $$\text{btr}(x,y)=\sum _{z\in Z}xzy{|}_z,$$ where the sum is over a basis $Z$ of the algebra $𝒵\text{.}$

$$\text{btr}(x,y)={\chi }^{\varnothing }(x\otimes y^{*})\text{.}$$

		Proof.
	This is a trivial consequence of the chain of isomorphisms $$𝒵={\text{End}}_G\left(W\right)\simeq {(W\otimes W^{*})}^G={\widehat{𝒵}}^{\varnothing }\text{.}$$ Let $w_i$ be a basis of $W$ and let $w^i$ be a basis of $W^{*}\text{.}$ Then $z\in 𝒵$ and the corresponding element $z^{*}\in Z^{\text{opp}}$ act on $W$ and $W^{*}$ respectively by $$z{w}_i=\sum _j{z}_i^j{w}_j,\text{and}z{w}^i=\sum _j{z}_j^i{w}^j\text{.}$$ Similarly, let $g\in G$ act on $W$ and $W^{*}$ by $$g{w}_i=\sum _j{g}_i^j{w}_j,\text{and}g{w}^i=\sum _j{\hat{g}}_j^i{w}^j\text{.}$$ Then the fact that $z\in 𝒵$ and that $z\in {𝒵}^{\text{opp}}$ implies that $$\sum g_i^j{z}_j^k=\sum z_i^l{g}_l^k,\text{and}\sum {\hat{g}}_i^j{z}_j^k=\sum z_i^l{\hat{g}}_l^k\text{.}$$ Then the element ${\sum }_{i,j}{z}_i^j{w}_j{w}^i\in W\otimes W^{*}$ is invariant since $$g\sum _{i,j}{z}_i^j{w}_j{w}^i=\sum _{i,j,k,l}{\hat{g}}_l^i{z}_i^j{g}_j^k{w}_k{w}^l=\sum _{i,j,k,l}{\hat{g}}_l^i{g}_i^j{z}_j^k{w}_k{w}^l=\sum _{j,k,l}{\delta }_l^j{z}_j^k{w}_k{w}^l=\sum _{k,l}{z}_l^k{w}_k{w}^l\text{.}$$ Now let us consider the element $x\otimes y^{*}\in 𝒵\otimes {𝒵}^{\text{opp}}$ acting on the invariant in ${𝒵}^{\varnothing }={(W\otimes W^{*})}^G$ corresponding to $z\in 𝒵\text{.}$ This is the invariant in $W\otimes W^{*}$ given by $$(x\otimes y^{*})\sum _{i,j}{z}_j^i{w}_i{w}^j=\sum _{i,j,k,l}{y}_l^j{z}_j^i{x}_i^k{w}_k{w}^l,$$ which clearly corresponds to the element of $𝒵$ given by $xzy\text{.}$ $\square$

Remark. It follows that the bitrace of the regular representation of the Iwahori-Hecke algebra $H_m\left(q\right)$ of type $A$ is the same as the character of the representation of the Kosuda algebra $H_{m,m}\left(q\right)$ corresponding to the pair of partitions $[\varnothing ,\varnothing ]\text{.}$

Notes and References

This is a copy of the paper On the trace of the regular representation of a centralizer algebra by Arun Ram, Department of Mathematics, University of Wisconsin, Madison, WI 53706, February 14, 1994. This paper was supported in part by a National Science Foundation postdoctoral fellowship.

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