The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Algebra, Group theory, Representation Theory | 1 Comment

   

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