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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.

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Again a question I always wanted to ask but didn't dare! (I still consider Artin-Wedderburn and all the related theory of semisimple modules untransparent and strange.) However, I fear this is related to proving Schur-Weyl in positive characteristic, and this is known to be very hard (see arxiv.org/abs/math/0610591 for a nonelementary and very indirect proof). –  darij grinberg Mar 3 '12 at 1:41
    
Maybe you can avoid the difficulties of proving the theorem by avoiding the theorem altogether, instead replacing it by the corresponding fact for Schur functors rather than Schur modules, i. e., considering $V$ as a variable rather than a fixed vector space. In that case you can more-or-less WLOG assume $\dim V\geq n$ and use your argument. Of course, the result you thus prove is weaker, but the question is whether your goal is the classical Schur-Weyl duality of $\mathrm{GL}V$ for fixed $V$, or something else where Schur-Weyl duality is just a lemma (in the latter case, chances ... –  darij grinberg Mar 3 '12 at 1:44
    
... are high that the easier functorial version of Schur-Weyl duality is already enough). –  darij grinberg Mar 3 '12 at 1:45
    
So, I have a couple of places I want to use this, but here is the first one: I want to prove that the standard inner product on symmetric functions is $\dim \mathrm{Hom}(V,W)$. If you follow Stanley and define the inner product by $\langle h_{\lambda}, m_{\mu} \rangle = \delta_{\lambda \mu}$, this comes down to computing $\mathrm{Hom}(\mathrm{Sym}^{\lambda_1} \otimes \cdots \otimes \mathrm{Sym}^{\lambda_n}, \mathrm{Sym}^{\mu_1} \otimes \cdots \otimes \mathrm{Sym}^{\mu_n})$. If you know the above claim, this turns into some very nice combinatorics, and leads into RSK in a clean way. –  David Speyer Mar 3 '12 at 13:57
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If you only prove the above result for $\dim V > n$, then you only compute the above Hom for $|\lambda| = |\mu| < n$, and there is NO finite value of $n$ for which you know that the combinatorics and the representation theory match up. That seems like a shame. Of course, I can think of other ways to prove this, but I think this one is very elegant. –  David Speyer Mar 3 '12 at 13:59

2 Answers 2

up vote 6 down vote accepted

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 $\alpha$.

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Could you be more explicit about "Using this isomorphisms we get a unique $GL(W)$-invariant extension of $\alpha$ to $X$"? I am not getting it. Thanks! –  David Speyer Mar 6 '12 at 1:43
    
@David Speyer: Suppose that $\alpha$ is defined on $U_i$ and we want to extend it to $U_j$. Let $\phi : U_i \rightarrow U_j$ be a ${\rm GL}(W)$-isomorphism. Then we define $\alpha$ on $U_j$ by $\alpha(x) = \phi\alpha\phi^{-1}(x)$. This definition is the only one possible if we want $\alpha$ to be ${\rm GL}(W)$-invariant, hence the uniqueness of $\alpha$. By the comment about Schur's Lemma it is enough to consider one non-zero map $U_i \rightarrow U_j$ for each pair $i$ and $j$, and then the extension of $\alpha$ will commute with all such maps. –  Mark Wildon Mar 6 '12 at 2:01
    
Ahh, the key point is that, for $1 \leq i < j \leq a$, the isomorphism from $U_i to U_j$ is induced by $GL(V)$, and therefore $\alpha$ acts the same way on $U_i$ and $U_j$. If the isomorphism were only induced by $GL(W)$, we don't know this. –  David Speyer Mar 6 '12 at 15:18
    
@David Speyer I think ${\rm GL}(V)$ and ${\rm GL}(W)$ should be swapped in your comment: clearly any ${\rm GL}(W)$-invariant map is ${\rm GL}(V)$-invariant. But then I agree, i.e. this 'for free' extension of maps is critical. (And my first reply addressed only the other, less important, points.) I was led to my argument by considering the corresponding inclusion of Schur algebras: this is subsumed by your Key Lemma, since one could define the Schur algebra $S(N,n)$ by $S(N,n) = \mathrm{End}_{S_n}(V^{\otimes n})$ where $V$ is an $N$-dimensional vector space. –  Mark Wildon Mar 6 '12 at 23:29

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 $S_n$.

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