# The Unapologetic Mathematician

## Commutant Algebras in General

And in my hurry to get a post up yesterday afternoon after forgetting to in the morning, I put up the wrong one. Here’s what should have gone up yesterday, and yesterday’s should have been now.

Now we can describe the most general commutant algebras. Maschke’s theorem tells us that any matrix representation $X$ can be decomposed as the direct sum of irreducible representations. If we collect together all the irreps that are equivalent to each other, we can write $\displaystyle X\cong m_1X^{(1)}\oplus m_2X^{(2)}\oplus\dots\oplus m_kX^{(k)}$

where the $X^{(i)}$ are pairwise-inequivalent irreducible matrix representations with degrees $d_i$, respectively. We calculate the degree: $\displaystyle\deg X=\sum\limits_{i=1}^k\deg\left(m_iX^{(i)}\right)=\sum\limits_{i=1}^km_id_i$

Now, can a matrix in the commutant algebra send a vector from the subspace isomorphic to $m_iX^{(i)}$ to the subspace isomorphic to $m_jX^{(j)}$? No, and for basically the same reason we saw in the case of $X^{(i)}\oplus X^{(j)}$. Since it’s an intertwinor, it would have to send the whole $\mathbb{C}[G]$-orbit of the vector — a submodule isomorphic to $X^{(j)}$ — into the target subspace $m_jX^{(j)}$, but we know that that submodule itself has no submodules isomorphic to $X^{(i)}$.

And so any such matrix must be the direct sum of one matrix in each commutant algebra $\mathrm{Com}_G\left(m_iX^{(i)}\right)$. But we know that these matrices are of the form $M_{m_i}\boxtimes I_{d_i}$. And so we can write $\displaystyle\mathrm{Com}_G(X)=\left\{\bigoplus\limits_{i=1}^k(M_{m_i}\boxtimes I_{d_i})\bigg\vert M_{m_i}\in\mathrm{Mat}_{m_i}(\mathbb{C})\right\}$

which has dimension $\displaystyle\dim\mathrm{Com}_G(X)=\sum\limits_{i=1}^km_i^2$

Advertisements

October 7, 2010