# The Unapologetic Mathematician

## More Commutant Algebras

We continue yesterday’s discussion of commutant algebras. But today, let’s consider the direct sum of a bunch of copies of the same irrep.

Before we get into it, let’s discuss a bit of notation. Given a representation $X$ we write $mX$ for the direct sum of $m$ copies of $X$. We say that $m$ is the “multiplicity” of $X$.

Now, let’s let $X^{(1)}$ be an irrep of degree $d$, and let $X=2X^{(1)}$. Our analysis proceeds exactly as yesterday — with $X^{(2)}=X^{(1)}$ — up until we write down our four equations. Now they read: \displaystyle\begin{aligned}T_{1,1}X^{(1)}&=X^{(1)}T_{1,1}\\T_{1,2}X^{(1)}&=X^{(1)}T_{1,2}\\T_{2,1}X^{(1)}&=X^{(1)}T_{2,1}\\T_{2,2}X^{(1)}&=X^{(1)}T_{2,2}\end{aligned}

This time, Schur’s lemma tells us that each $T_{i,j}$ is an intertwinor between $X^{(1)}$ and itself. And so we conclude that each of the blocks is a constant times the identity: $T_{i,j}=c_{i,j}I_d$. That is: $\displaystyle T=\begin{pmatrix}c_{1,1}I_d&c_{1,2}I_d\\c_{2,1}I_d&c_{2,2}I_d\end{pmatrix}$

We can recognize this as a Kronecker product of two matrices: $\displaystyle T=\begin{pmatrix}c_{1,1}&c_{1,2}\\c_{2,1}&c_{2,2}\end{pmatrix}\boxtimes I_d$

which is the matrix version of the tensor product of two linear maps. If you don’t know much about the tensor product, don’t worry; we’ll refresh more as we go. You can also review tensor products in the context of vector spaces and linear transformations here. What we want to think of here is that the matrix $\left(c_{i,j}\right)$ shuffles around the two copies of the irrep $X^{(1)}$, and the identity matrix $I_d$ stands for the trivial transformation on an irreducible representation.

Since any values are possible for the $c_{i,j}$, the first matrix can take any value in the algebra $\mathrm{Mat}_2(\mathbb{C})$ of $2\times2$ complex matrices. We say that $\displaystyle\mathrm{Com}_G(X)=\{M_2\boxtimes I_d\vert M_2\in\mathrm{Mat}_2(\mathbb{C})\}$

In more generality, if $X=mX^{(1)}$, where $X^{(1)}$ is an irrep of degree $d$, then we find $\displaystyle\mathrm{Com}_G(X)=\{M_m\boxtimes I_d\vert M_m\in\mathrm{Mat}_m(\mathbb{C})\}$

The degree of the representation $X$ is $md$ — we get $d$ for each of the $m$ copies of $X^{(1)}$ — and the dimension of the commutant algebra is the dimension of the matrix algebra $\mathrm{Mat}_m(\mathbb{C})$, which is $m^2$.

October 5, 2010 -

## 1 Comment »

1. […] these summands, we can pick a basis for and and use the same sorts of methods we did to calculate commutant algebras. We find that if — — then there are no -morphisms at all, even if we include […]

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