The Unapologetic Mathematician

Mathematics for the interested outsider

Some Commutant Algebras

We want to calculate commutant algebras of matrix representations. We already know that if X is an irrep, then \mathrm{Com}_G(X)=\mathbb{C}, and we’ll move on from there.

Next, let X^{(1)} and X^{(2)} be two inequivalent matrix irreps, with degrees d_1 and d_2, respectively, and consider the representation X=X^{(1)}\oplus X^{(2)}. As a matrix, this looks like:

\displaystyle X(g)=\begin{pmatrix}X^{(1)}(g)&0\\{0}&X^{(2)}(g)\end{pmatrix}

Where we’ve broken the d_1+d_2 rows and columns into blocks of size d_1 and d_2. Now let’s determine the algebra of matrices T commuting with each such matrix X(g). Let’s break down T into blocks like X.

\displaystyle T=\begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix}

The nice thing about this is that when the block sizes are the same, and when we break rows and columns into the same blocks, the rules for multiplication are the same as for regular matrices:

\displaystyle\begin{aligned}TX(g)&=\begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix}\begin{pmatrix}X^{(1)}(g)&0\\{0}&X^{(2)}(g)\end{pmatrix}\\&=\begin{pmatrix}T_{1,1}X^{(1)}(g)&T_{1,2}X^{(2)}(g)\\T_{2,1}X^{(1)}(g)&T_{2,2}X^{(2)}(g)\end{pmatrix}\\X(g)T&=\begin{pmatrix}X^{(1)}(g)&0\\{0}&X^{(2)}(g)\end{pmatrix}\begin{pmatrix}T_{1,1}&T_{1,2}\\T_{2,1}&T_{2,2}\end{pmatrix}\\&=\begin{pmatrix}X^{(1)}(g)T_{1,1}&X^{(1)}(g)T_{1,2}\\X^{(2)}(g)T_{2,1}&X^{(2)}(g)T_{2,2}\end{pmatrix}\end{aligned}

If these are to be equal, we have four equations to satisfy:

\displaystyle\begin{aligned}T_{1,1}X^{(1)}(g)&=X^{(1)}(g)T_{1,1}\\T_{1,2}X^{(2)}(g)&=X^{(1)}(g)T_{1,2}\\T_{2,1}X^{(1)}(g)&=X^{(2)}(g)T_{2,1}\\T_{2,2}X^{(2)}(g)&=X^{(2)}(g)T_{2,2}\end{aligned}

And we can apply Schur’s lemma to all of them. In the middle two equations, we see that both T_{1,2} and T_{2,1} must be either be invertible or zero. But if either one is invertible, then it gives an equivalence between the matrix irreps X^{(1)} and X^{(2)}. But since we assumed that these are inequivalent, we conclude that T_{1,2} and T_{2,1} are both the appropriate zero matrices. And then the first and last equations are handled just like single irreps were last time. Thus we must have

\displaystyle T=\begin{pmatrix}c_1I_{d_1}&0\\{0}&c_2I_{d_2}\end{pmatrix}

And so \mathrm{Com}_G(X)=\mathbb{C}\oplus\mathbb{C}, where the multiplication is handled component by component. Similarly, the direct sum of n pairwise-inequivalent irreps X=X^{(1)}\oplus\dots\oplus X^{(n)} has commutant algebra \mathrm{Com}_G(X)=\mathbb{C}^n, with multiplication handled componentwise. The degree of the representation X is the sum of the degrees of the irreps, and the dimension of the commutant is n.

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October 4, 2010 - Posted by | Algebra, Group theory, Representation Theory

2 Comments »

  1. [...] 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 [...]

    Pingback by More Commutant Algebras « The Unapologetic Mathematician | October 5, 2010 | Reply

  2. [...] the subspace isomorphic to to the subspace isomorphic to ? No, and for basically the same reason we saw in the case of . Since it’s an intertwinor, it would have to send the whole -orbit of the [...]

    Pingback by Commutant Algebras in General « The Unapologetic Mathematician | October 7, 2010 | Reply


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