# The Unapologetic Mathematician

## Centers of Commutant Algebras

We want to calculate the centers of commutant algebras. We will have use of the two easily-established equations: \displaystyle\begin{aligned}(B_1\oplus B_2)\circ(A_1\oplus A_2)&=(B_1\circ A_1)\oplus(B_2\circ A_2)\\(B_1\otimes B_2)\circ(A_1\otimes A_2)&=(B_1\circ A_1)\otimes(B_2\circ A_2)\end{aligned}

Where $A_1:U_1\to V_1$, $A_2:U_2\to V_2$, $B_1:V_1\to W_1$, and $B_2:V_2\to W_2$ are linear functions. In particular, this holds where $A_1$ and $A_2$ are $m\times m$ matrices representing linear endomorphisms of $\mathbb{C}^m$, and $B_1$ and $B_2$ are $n\times n$ matrices representing linear endomorphisms of $\mathbb{R}^n$.

Now let $X$ be a matrix representation and consider a central matrix $C\in Z_{\mathrm{Com}_G(X)}$. That is, for all $T\in\mathrm{Com}_G(X)$, we have $\displaystyle CT = TC$

Let’s further assume that we can write $\displaystyle X=m_1X^{(1)}\oplus\dots\oplus m_kX^{(k)}$

where each $X^{(i)}$ is an irreducible representation of degree $d_i$. Then we know that we can write \displaystyle\begin{aligned}T&=\bigoplus\limits_{i=1}^k\left(T_{m_i}\boxtimes I_{d_i}\right)\\C&=\bigoplus\limits_{i=1}^k\left(C_{m_i}\boxtimes I_{d_i}\right)\end{aligned}

Thus we calculate: \displaystyle\begin{aligned}CT&=\left(\bigoplus\limits_{i=1}^k(C_{m_i}\boxtimes I_{d_i})\right)\left(\bigoplus\limits_{i=1}^k(T_{m_i}\boxtimes I_{d_i})\right)\\&=\bigoplus\limits_{i=1}^k\left((C_{m_i}\boxtimes I_{d_i})(T_{m_i}\boxtimes I_{d_i})\right)\\&=\bigoplus\limits_{i=1}^k(C_{m_i}T_{m_i}\boxtimes I_{d_i})\\TC&=\bigoplus\limits_{i=1}^k(T_{m_i}C_{m_i}\boxtimes I_{d_i})\end{aligned}

This is only possible if for each $i$ we have $C_{m_i}T_{m_i}=T_{m_i}C_{m_i}$ for all $T_{m_i}\in\mathrm{Mat}_{m_i}$. But this means that $C_{m_i}$ is in the center of $\mathrm{Mat}_{m_i}$, which implies that $C_{m_i}=c_iI_{m_i}$. Therefore a central element can be written $\displaystyle C=\bigoplus\limits_{i=1}^k\left(c_iI_{m_i}\boxtimes I_{d_i}\right)=\bigoplus\limits_{i=1}^kc_iI_{m_id_i}$

As a concrete example, let’s say that $X=2X^{(1)}\oplus X^{(2)}$, where $\deg\left(X^{(1)}\right)=3$ and $\deg\left(X^{(2)}\right)=4$. Then the matrices in the commutant algebra look like: $T=\left(\begin{tabular}{ccc|ccc|cccc}a&0&0&b&0&0&0&0&0&0\\{0}&a&0&0&b&0&0&0&0&0\\{0}&0&a&0&0&b&0&0&0&0\\\hline c&0&0&d&0&0&0&0&0&0\\{0}&c&0&0&d&0&0&0&0&0\\{0}&0&c&0&0&d&0&0&0&0\\\hline{0}&0&0&0&0&0&x&0&0&0\\{0}&0&0&0&0&0&0&x&0&0\\{0}&0&0&0&0&0&0&0&x&0\\{0}&0&0&0&0&0&0&0&0&x\end{tabular}\right)$

and the dimension of the commutant algebra is evidently $m_1^2+m_2^2=2^2+1^2=5$.

The central matrices in the commutant algebra, on the other hand, look like: $C=\left(\begin{tabular}{ccc|ccc|cccc}a&0&0&0&0&0&0&0&0&0\\{0}&a&0&0&0&0&0&0&0&0\\{0}&0&a&0&0&0&0&0&0&0\\\hline{0}&0&0&a&0&0&0&0&0&0\\{0}&0&0&0&a&0&0&0&0&0\\{0}&0&0&0&0&a&0&0&0&0\\\hline{0}&0&0&0&0&0&x&0&0&0\\{0}&0&0&0&0&0&0&x&0&0\\{0}&0&0&0&0&0&0&0&x&0\\{0}&0&0&0&0&0&0&0&0&x\end{tabular}\right)$

And the dimension is $k=2$.

October 8, 2010