Direct proof that the centralizer of $GL(V)$ acting on $V^{\otimes n}$ is spanned by $S_n$ - MathOverflow

Direct proof that the centralizer of $GL(V)$ acting on $V^{\otimes n}$ is spanned by $S_n$

2012-03-03T00:43:41Z

Let $V$ be a finite dimensional vector space over a field of characteristic zero. Let $A$ be the space of maps in $\mathrm{End}(V^{\otimes n})$ which commute with the natural $GL(V)$ action. Clearly, any permutation of the tensor factors is in $A$. I am looking for an elementary proof that these permutations span $A$.

If $\dim V \geq n$, there is a very simple proof. Take $e_1$, $e_2$, ..., $e_n$ in $V$ linearly independent and let $\alpha \in A$. Then $\alpha(e_1 \otimes e_2 \otimes \cdots \otimes e_n)$ must be a $t_1 t_2 \cdots t_n$ eigenvector for the action of the matrix $\mathrm{diag}(t_1, t_2, \ldots )$ in $GL(V)$. So $\alpha(e_1 \otimes \cdots \otimes e_n) = \sum_{\sigma \in S_n} c_{\sigma} e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)}$ for some constants $c_{\sigma}$. It is then straightforward to show that $\alpha$ is given by the corresponding linear combination of permutations.

I feel like there should be an elementary, if not very well motivated, extension of the above argument for the case where $\dim V < n$, but I'm not finding it.

Motivation: I'm planning a course on the combinatorial side of $GL_N$ representation theory -- symmetric polynomials, jdt, RSK and, if I can pull it off, some more modern things like honeycombs and crystals. Since it will be advertised as a combinatorics course, I want to prove a few key results that give the dictionary between combinatorics and representation theory, and then do all the rest on the combinatorial side. Based on the lectures I have outlined so far, I think this will be one of the few key results.

The standard proof is to show that the centralizer of $k[S_n]$ is spanned by $GL(V)$, and then apply the double centralizer theorem. Although the double centralizer theorem (at least, over $\mathbb{C}$) doesn't formally involve anything I won't be covering, I think it is pretty hard to present it to people who aren't extremely happy with the representation theory of semi-simple algebras. So I am looking for an alternate route.

Answer by Mark Wildon for Direct proof that the centralizer of $GL(V)$ acting on $V^{\otimes n}$ is spanned by $S_n$

2012-03-05T22:23:08Z

Let $W$ be a vector space of dimension $n$ containing $V$. Let $\alpha$ be an endomorphism of $V^{\otimes n}$ commuting with the action of ${\rm GL}(V)$. Suppose that $\alpha$ can be extended to an endomorphism $\beta$ of $W^{\otimes n}$ that commutes with the action of ${\rm GL}(W)$. Then, by the argument given by David Speyer in the question, there exist scalars $c_\sigma \in \mathbf{C}$ such that

$$ \beta = \sum_{\sigma \in S_n} c_\sigma \sigma $$

and this also expresses $\alpha$ as a linear combination of place permutations of the tensor factors. (As I noted in my comment, this expression is, in general, far from unique.)

Any proof that such an extension exists must use the semisimplicity of $\mathbf{C}S_n$, since otherwise we get an easy proof of general Schur-Weyl duality. If we assume that ${\rm GL}(W)$ acts as the full ring of $S_n$-invariant endomorphisms of $W^{\otimes n}$ then a fairly short proof is possible. I think it is inevitable that it uses many of the same ideas as the double-centralizer theorem. A more direct proof would be very welcome.

Let $U$ be a simple $\mathbf{C}S_n$-module appearing in $V^{\otimes n}$. Let

$$ X = U_1 \oplus \cdots \oplus U_a \oplus U_{a+1} \oplus \cdots \oplus U_b $$

be the largest submodule of $W^{\otimes n}$ that is a direct sum of simple $\mathbf{C}S_n$-modules isomorphic to $U$. We may choose the decomposition so that $X \cap V^{\otimes n} = U_1 \oplus \cdots \oplus U_a$. Each projection map $W^{\otimes n} \rightarrow U_i$ is $S_n$-invariant, and so is induced by a suitable linear combination of elements of ${\rm GL}(W)$. Hence each $U_i$ for $1 \le i \le a$ is $\alpha$-invariant. Similarly, for each pair $i$, $j$ there is a isomorphism $U_i \cong U_j$ induced by ${\rm GL}(W)$; these isomorphisms are unique up to scalars (by Schur's Lemma). Using these isomorphisms we get a unique ${\rm GL}(W)$-invariant extension of $\alpha$ to $X$.

Finally let $W^{\otimes n} = C \oplus D$ where $C$ is the sum of all simple $\mathbf{C}S_n$-submodules of $W^{\otimes n}$ isomorphic to a submodule of $V^{\otimes n}$ and $D$ is a complementary $\mathbf{C}S_n$-submodule. The previous paragraph extends $\alpha$ to a map $\beta$ defined on $C$. The projection map $W^{\otimes n} \rightarrow D$ is $S_n$-invariant and so is induced by ${\rm GL}(W)$. Hence we can set $\beta(D) = 0$ and obtain a ${\rm GL}(W)$-invariant extension $\beta : W^{\otimes n} \rightarrow W^{\otimes n}$ of $\alpha$.

Answer by David Speyer for Direct proof that the centralizer of $GL(V)$ acting on $V^{\otimes n}$ is spanned by $S_n$

2012-03-06T15:47:19Z

I'm going to write up Mark Wildon's proof as I understand it. As in the standard proof, we start by showing the Key Lemma that the centralizer of $k[S_n]$ is linearly spanned by $GL(V)$. Decompose $V^{\otimes n}$ into $S_n$-irreps, and let $\alpha$ be an endomorphism of $V^{\otimes n}$ commuting with $GL(V)$. For each irrep $U$ of $S_n$, let $U_1$, ..., $U_a$ be the occurrences of $U$ in $V^{\otimes n}$.

For any $U_i$, consider the endomorphism of $V^{\otimes n}$ which acts by $1$ on $U_i$ and on $0$ on all of the other summands of $V^{\otimes n}$. This commutes with $k[S_n]$ so, by the Key Lemma it is a linear combination of maps in $GL(V)$. Hence $\alpha$ commutes with it, which means that $\alpha$ takes $U_i$ to $U_i$ by some map $\alpha_i$.

Consider the endomorphism of $V^{\otimes n}$ which takes $U_i$ to $U_j$ by an $S_n$-equivariant endomorphism and acts by $0$ on every other summand of $V^{\otimes n}$. This commutes with $k[S_n]$ so, by the Key Lemma it is a linear combination of maps in $GL(V)$. Hence $\alpha$ commutes with it, which means that $\alpha_i = \alpha_j$. (Abusing equals to mean "is taken to the other along the isomorphism $U_i \to U_j$, which is unique up to scalar".) Write $\alpha(U)$ for the common value of $\alpha_1$, $\alpha_2$, ..., $\alpha_a$.

There are now two ways to finish the proof.

Standard Argument: By Maschke and Artin-Wedderburn, there is an element in $k[S_n]$ which acts on each irrep $U$ by $\alpha(U)$. This element of $k[S_n]$ induces $\alpha$.

Mark Wildon's Argument: Let $V \subset W$. We will show that we can extended $\alpha$ to an endomorphism $\beta$ of $W^{\otimes n}$ which commutes with $GL(W)$. Decompose $W^{\otimes n}$ into $S_n$ irreps, so that the previous decomposition of $V^{\otimes n}$ occurs as a subset of the summands. Let the occurrences of $U$ be $U_1 \oplus U_2 \cdots \oplus U_a \oplus \cdots \oplus U_b$. Define a linear map $\beta:W^{\otimes n}\to W^{\otimes n}$ to act on all of the $U_i$ by $\alpha(U)$ or, if $a=0$ so that $\alpha(U)$ is undefined, define $\beta$ to act on the $U_i$ by $0$.

We claim that $\beta$ commutes with $GL(W)$. Proof: Any element of $GL(W)$ commutes with $k[S_n]$. So (by Schur's lemma), it can only map $U_i$ to a linear combination of other $U_j$'s, and the component of $\alpha$ mapping $U_i$ to $U_j$ is a scalar multiple of the standard isomorphism. Clearly, $\beta$ commutes with any map of this form.

Now, by my argument in the original post, take $\dim W \geq n$ to see that $\beta$ is induced by an element of $S_n$. Then $\alpha$ is also induced by this element of $S_n$.