The Unapologetic Mathematician

Endomorphism and Commutant Algebras

We will find it useful in our study of $G$-modules to study not only the morphisms between them, but the structures that they form.

A couple days ago we mentioned the vector space $\hom_G(V,W)$. Today, we specialize to the case $V=W$, where we use the usual alternate name. We write $\mathrm{End}_G(V)=\hom_G(V,V)$ and call it the “endomorphism algebra” of $V$. Not only is it a vector space of $G$-morphisms, but it has a multiplication from the fact that the source and target of each one are the same and so we can compose them.

We also have an analogous concept for matrix representations. Given a matrix representation $X$, a $G$-morphism from $X$ to $X$ is given by a linear map $T$ so that $TX(g)=X(g)T$ for all $g\in G$. That is, $T$ must commute with each of the matrices $X(g)$. And so we call the algebra of such matrices the “commutant algebra” of $X$, and write it $\mathrm{Com}_G(X)$. This is the matrix analogue of the endomorphism algebra because if we get $X$ by starting with a $G$-module $(V,\rho)$, picking a basis for $V$, and writing down $X(g)$ as the matrix of $\rho(g)$ corresponding to this basis, then we find that $\mathrm{Com}_G(X)\cong\mathrm{End}_G(V)$.

Let’s start our considerations by letting $X=Y$ by any matrix irrep, and let’s calculate its commutant algebra. By definition for any $T\in\mathrm{Com}_G(X)$ we have $TX(g)=X(g)T$ for all $g\in G$. We can subtract $cIX$ from both sides of this equation to find

$\displaystyle(T-cI)X(g)=TX(g)-cX(g)=X(g)T-X(g)c=X(g)(T-cI)$

where $I$ is the identity matrix. The matrix $T-cI$ commutes with $X(g)$ for every complex scalar $c$, and so Schur’s lemma will apply to all of them.

Since $\mathbb{C}$ is algebraically closed, we must be able to find an eigenvalue $\lambda$. Letting $c$ be this eigenvalue, we see that $T-\lambda I$ commutes with $X(g)$ for all $g\in G$, and so Schur’s lemma tells us that either it’s invertible or the zero matrix. But since $\lambda$ is an eigenvalue the matrix $T-\lambda I$ can’t possibly be invertible, and so we must have $T-\lambda I=0$, and $T=\lambda I$.

Therefore the only matrices that commute with all the $X(g)$ in a matrix irrep of $G$ are scalar multiples of the identity matrix. And since the product of two such matrices is just the product of their scalars, we find that $\mathrm{Com}_G(X)=\mathbb{C}$ as a complex algebra.

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Posted by John Armstrong | Algebra, Group theory

6 Comments	»

1. [...] Commutant Algebras We want to calculate commutant algebras of matrix representations. We already know that if is an irrep, then , and we’ll move on [...]

Pingback by | October 4, 2010 | Reply

2. [...] 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 same [...]

Pingback by | October 5, 2010 | Reply

3. [...] we can describe the most general commutant algebras. Maschke’s theorem tells us that any matrix representation can be decomposed as the direct [...]

Pingback by | October 7, 2010 | Reply

4. [...] want to calculate the centers of commutant algebras. We will have use of the two easily-established [...]

Pingback by | October 8, 2010 | Reply

5. [...] approach to generalizing last week’s efforts. We’re not just going to deal with endomorphism algebras, but with all the [...]

Pingback by | October 12, 2010 | Reply

6. [...] morphism between two irreducible modules is either or an isomorphism. And, as we’ve seen in other examples involving linear transformations, all automorphisms of an irreducible module are scalars times the [...]

Pingback by | September 15, 2012 | Reply

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