The Unapologetic Mathematician

Mathematics for the interested outsider

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

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October 7, 2010 - Posted by | Algebra, Group theory, Representation Theory

1 Comment »

  1. [...] each is an irreducible representation of degree . Then we know that we can [...]

    Pingback by Centers of Commutant Algebras « The Unapologetic Mathematician | October 8, 2010 | Reply


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