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I am teaching myself representation theory on $GL(V)$ and $S_n$ using my friend's lecture notes, and have reached a proof of the Schur-Weyl Duality theorem; on reading through I'm struggling to make my way through the first part of the proof though, and was hoping you might be able to explain a few of the things I'm confused about. I'm not sure how much of this proof is standard (a fair amount, I suspect) so it's possible some questions aren't going to be answerable and will simply be the nuances of the way the lecturer has presented the material. However, if you could help me as much as you can with my understanding I would greatly appreciate it. (I will write out the first part of this section of the notes but I won't present the full proof in its entirety.)

A forewarning: the lecture notes have quite a few errors in; the last step of the proof of Schur-Weyl duality for example is totally wrong, though was easily fixed via earlier results in the notes. Also, please forgive any stupid questions I ask, almost all of my representation theory knowledge is self-taught, so I occasionally overlook the blindingly obvious. My questions are I think simple, although I have quite a bit that needs clarifying.

Let $V$ be an $m$-dimensional vector space over $\mathbb{C}$. $V^{\otimes n}$ is a $\mathbb{C}S_n$ module, via $\sigma(v_1 \otimes \ldots \otimes v_n) = v_{\sigma^{-1}(1)} \otimes \ldots \otimes v_{\sigma^{-1}(n)}$, and $g(v_1 \otimes \ldots \otimes v_n) = gv_1 \otimes \ldots \otimes gv_n$ (I presume this is for $g \in GL(V)$ though not explicitly stated.)

We can define $\phi: \mathbb{C}S_n \to \operatorname{End}_\mathbb{C}(V^{\otimes n}),\,\sigma \mapsto (v \mapsto \sigma v)$. $V$ can also be regarded as representations of $GL(V)$ in a natural way, so $V^{\otimes n}$ becomes a $\mathbb{C}GL(V)$ module, so $\exists \, \psi: \mathbb{C}GL(V) \to \operatorname{End}_{\mathbb{C}GL(V)}(V^{\otimes n}), \eta \mapsto \eta^{\otimes n}$.

Theorem (Schur-Weyl Duality): The images of $\mathbb{C}S_n$ and $\mathbb{C}GL(V)$ in $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ are each others centralizers; that is, $\psi(GL(V)) = \operatorname{End}_{\mathbb{C}S_n}(V^{\otimes n})$ and $\phi(\mathbb{C}S_n) = \operatorname{End}_{\mathbb{C}GL(V)}(V^{\otimes n})$.

In order to prove this, we define the $\mathbb{C}$-algebra $S_\mathbb{C}(m,n)$, the "Schur algebra", to be the subalgebra of $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ consisting of endomorphisms commuting with the image of $\mathbb{C}S_n$: that is , $S_\mathbb{C}(m,n) = \operatorname{End}_{\mathbb{C}S_n}(V^{\otimes n})$. Now $\mathbb{C}S_n$ is semisimple and $V^{\otimes n}$ is a finite dimensional $\mathbb{C}S_n$ module, therefore $S(m,n)$ is a semisimple $\mathbb{C}$-algebra. Put $W = \operatorname{End}_\mathbb{C}(V)$: this is a $\mathbb{C}GL(V)$ module by conjugation.

Question 1: Is there a reason why we would want a $\mathbb{C}GL(V)$ module by conjugation? I can't see any point during any of the proof at which conjugation is used, and I wouldn't have thought this would be the natural action to choose.

Lemma: There exists an isomorphism $\alpha: W^{\otimes n} \to \operatorname{End}_\mathbb{C}(V^{\otimes n})$, sending $f_1 \otimes \ldots \otimes f_n \mapsto \left(v_1 \otimes \ldots \otimes v_n \mapsto f_1(v_1) \otimes \ldots \otimes f_n(v_n)\right)$, an isomorphism of $\mathbb{C}GL(V)$-modules and of $\mathbb{C}S_n$-modules.

Proof: $W^{\otimes n} = W \otimes \ldots \otimes W \cong (V \otimes V^*) \otimes \ldots \otimes (V \otimes V^*)$ $ \cong (V \otimes V \ldots \otimes V) \otimes (V^* \otimes \ldots \otimes V^*) \cong V^{\otimes n} \otimes (V^*)^{\otimes n} \cong \operatorname{End}_\mathbb{C}(V^{\otimes n})$.

$W^{\otimes n}$ has natural structure of a $\mathbb{C}S_n$-module and $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ inherits its structure from $V^{\otimes n}$. It is easy to see $\alpha$ is a $\mathbb{C}S_n$-homom; this completes the lemma.

Question 2: I follow the symbolic part of the proof, and I guess $W^{\otimes n} = (\operatorname{End}_\mathbb{C}(V))^{\otimes n}$ has the "natural" structure of a $\mathbb{C}S_n$ module via $\sigma(f_1 \otimes \ldots \otimes f_n) = f_{\sigma^{-1}(1)} \otimes \ldots \otimes f_{\sigma^{-1}(n)}$? I don't know what is meant by $\operatorname{End}_\mathbb{C}(V^{\otimes n})$ inheriting its structure from $V^{\otimes n}$: does this mean $\sigma$ acts by mapping $\left(v_1 \otimes \ldots \otimes v_n \mapsto f_1(v_1) \otimes \ldots \otimes f_n(v_n)\right)$ to $\left(v_1 \otimes \ldots \otimes v_n \mapsto f_{\sigma^{-1}(1)}\left(v_{\sigma^{-1}(1)}\right) \otimes \ldots \otimes f_{\sigma^{-1}(n)}\left(v_{\sigma^{-1}(n)}\right)\right)$ in the same way as was stated in the first line?

If this is correct however, then I can't see how $\alpha$ commutes with $\sigma$: the latter has indices consistent between $f_i$ and $v_i$ whereas the former has $f_{\sigma^{-1}(i)}(v_i)$. But then how is $\alpha$ a $\mathbb{C}S_n$-homomorphism? What have I misunderstood? Back to the notes:

Lemma: $S_\mathbb{C}(m,n) = \alpha(T^nW_{sym})$, where $T^nW_{sym}$ denotes the n-th symmetric power.

Proof: Observe $S_\mathbb{C}(m,n) = \{x \in \operatorname{End}_\mathbb{C}(V^{\otimes n}): \sigma x = x \, \forall \, \sigma \in S_n\}$, and $T^nW_{sym} = \{y \in W^{\otimes n}: \sigma y = y \, \forall \, \sigma \in S_n\}$, completing the proof.

Question 3: I'm probably being stupid, but why does $S_\mathbb{C}(m,n) = \{x \in \operatorname{End}_\mathbb{C}(V^{\otimes n}): \sigma x = x \, \forall \, \sigma \in S_n\}$? It's meant to be endomorphisms commuting with the image of $\mathbb{C}S_n$, but wouldn't that give something like $\sigma x = x \sigma$, rather than $\sigma x = x$?

That's all for now; apologies for the lengthy/multiple questions, but I didn't see any benefit to posting the same lecture notes 3 times in separate questions. Thank you in advance for your help.

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And for the interested reader: to complete the proof, we go on to observe that $T^n W_{sym} =$ span$(\varphi \otimes \ldots \otimes \varphi: \varphi \in GL(V))$ by considering Zariski closed sets in affine space $\mathbb{A}^n$. Equivalently we have said $S_\mathbb{C}(m,n) = $span$(\varphi^{\otimes n}: \varphi \in GL(V))$, and the result follows from all the above results (and a simple technical lemma about surjective maps). –  Spyam May 18 '12 at 5:16
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2 Answers

up vote 4 down vote accepted
+50

I hope the follows help clear up things a bit. First of all, some corrections/suggestions to your typos/writings.

(1) (just a personal preference, but I think this is a better practice) the $V^{\otimes n}$ is a right $\mathbb{C}S_n$-module. And $\phi: \mathbb{C}S_n\to \mathrm{End}_{\mathbb{C}}(V^{\otimes n})$ is given by $\sigma \mapsto (v \mapsto v\sigma)$.

(2) $\psi:\mathbb{C}GL(V) \to \mathrm{End}_{\mathbb{C}}(V^{\otimes n})$ is given by $g \mapsto (v\mapsto gv)$, where $gv$ is given in your first paragraph.

(3) Therefore, in the paragraph preceeding the statement of S-W duality, it should be: by commutation of action of $S_n$ and $GL(V)$, $\phi$ induces (by abusing notation) $\phi: \mathbb{C}S_n\to \mathrm{End}_{\mathbb{C}GL(V)}(V^{\otimes n})$ and similarly by swapping $S_n$ and $GL(V)$, we have $\psi:\mathbb{C}GL(V)\to \mathrm{End}_{\mathbb{C}S_n}(V^{\otimes n})$

(3.5) your $\eta$ in that paragraph should be either $\phi$ in my (1) or $\psi$ in my (2) above, depending on which correction to choose.

(4) The fact that the invariant subspace of $\mathrm{End}_{\mathbb{C}}(V^{\otimes n})$ is the same as $\mathrm{End}_{\mathbb{C}S_n}(V^{\otimes n})$ is due to a Weyl. I don't think it is obvious (?).

Question 1 and 2: Why conjugation? And what is the action of $S_n$ on $\mathrm{End}_{\mathbb{C}}(V^{\otimes n})$

In short (as always in mathematics...) this is the action that makes everything correct, and it wasn't given clearly in the proof of the Lemma of $W^{\otimes n}\cong \mathrm{End}_{\mathbb{C}}(V^{\otimes n})$. The motivation is that we need commutation: $x \theta = \theta x$ where $x$ is elements of $GL(V)$ or $S_n$ to get something in $\mathrm{End}_{\mathbb{C}G}(V^{\otimes n})$ for $G = S_n$ or $GL(V)$. This transform into $\theta = x\theta x^{-1}$. Note this is also where your guess for the action of $S_n$ on $\mathrm{End}_{\mathbb{C}}(V^{\otimes n})$ is wrong. Explicity, for $S_n$ side of the story:

Action of $S_n$ on $W^{\otimes n}$: same as you said $(f_1\otimes \cdots\otimes f_n)\cdot\sigma = f_{\sigma^{-1}(1)}\otimes\cdots\otimes f_{\sigma^{-1}(n)}$

Action of $S_n$ on $\mathrm{End}_{\mathbb{C}}(V^{\otimes n})$: $\theta\cdot\sigma = \phi(\sigma)\theta\phi(\sigma^{-1})$

Proof of commutation of $S_n$-action:

$\alpha((f_1\otimes\cdots f_n))\cdot\sigma\\ =(v_1\otimes\cdots v_n\mapsto f_1(w_1)\otimes \cdots \otimes f_n(w_n))\cdot\sigma \\ = v_1\otimes\cdots \otimes v_n\\ \quad \mapsto v_{\sigma(1)}\otimes \cdots v_{\sigma(n)}\\ \quad \mapsto f_1(v_{\sigma(1)})\otimes \cdots f_n(v_{\sigma(n)})\\ \quad \mapsto f_{\sigma^{-1}(1)} (v_{\sigma(\sigma^{-1}1)})\otimes \cdots f_{\sigma^{-1}(n)} (v_{\sigma(\sigma^{-1}n)}) = f_{\sigma^{-1}(1)} (v_1)\otimes \cdots f_{\sigma^{-1}(n)} (v_n))\\ =\alpha((f_1\otimes\cdots f_n)\cdot\sigma)$

Now you see where conjugation plays a part (in $S_n$ side of the story). So it is a good exercise for you to check the same thing in $GL(V)$.

Question 3: What's wrong?

You are correct $S(m,n)=\{ x\in \mathrm{End}_{\mathbb{C}}(V^{\otimes n})| \phi(\sigma) x = x \phi(\sigma) \; \forall \sigma \in S_n\} = \mathrm{End}_{\mathbb{C}}(V^{\otimes n})^{S_n}$ (this is the notation for saying "the subspace that is invariant under the action of $S_n$)

On the other hand, the symmetric power part is correct. Now recall the symmetric power is the invariant space under permutation action of $S_n$, therefore $S^n(W)$ (in your notation $T^n W_{sym}$) is $(W^{\otimes n})^{S_n}$. So now everything follows from the isomorphism $\alpha$. Note: $y\cdot \sigma$ in $W^{\otimes n}$ corresponds to $\phi(\sigma)\alpha(y)\phi(\sigma^{-1})$ in $\mathrm{End}_{\mathbb{C}}(V^{\otimes n})$.

P.S. If you have a typeset version of this lecture notes, would you mind sending me a copy (with the correction if it is possible)? I am also a graduate student learning things in related area, and I like this proof more than the usual reference from Goodman-Wallach (apart from the point (4) mentioned above).

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Many thanks for your comprehensive answer. I got in touch with the lecturer and it turned out many of my issues sprung from the fact he simply wrote the wrong thing down on the board repeatedly. 2 things: Firstly, what does $(W^{\otimes n})^{S_n}$ mean? I'm not familiar with that notation. Secondly, could you please elaborate on your point 4? My understanding was that $End_H(V)$ was defined to be all of the endomorphisms commuting with $H$: what does it mean to you? Have I misunderstood the meaning? If so, if you know a reference for "due to Weyl", I'd find that very helpful... –  Spyam May 25 '12 at 4:10
    
Also, I'm afraid I don't have a typeset version of these notes: everything I have is hand-written, the old fashioned way. The lecturer sometimes types up course notes as far as I know, but this is the first time the course has ever been lectured and I'm afraid I don't think he's going to this time. –  Spyam May 25 '12 at 4:11
    
(1) In general, let $X$ be space acted upon by group $G$, the $X^G$ is the invariant subspace of $X$ under the action of $G$, hence $\{ x\in X| gx=x\forall g\in G\}$. –  Aaron May 25 '12 at 19:28
    
(2) You are right, I made a mistake by mixing up different (equivalent) definitions of Schur algebra with the one you gave here. What Weyl proved is $\psi(\mathbb{C}GL(V))=\mathrm{End}_{\mathbb{C}S_n}(V)$, which is what we have been doing the whole time. My answer to your original post is through the help of Stuart Martin's book "Schur algebra" Section 2.1. In there, the Schur algebra is defined to be the image of $\mathbb{C}GL(V)$ under $\psi$....and don't worry about the typeset notes, thanks :) –  Aaron May 25 '12 at 19:34
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This is an alternative presentation to Aaron's answer. In my opinion, all those complicated objects (the $\alpha$ isomorphism, the action by conjugation) are unnecessary for the proof of S-W duality and end up transforming a simple enough fact into an esoteric black box. I like to put it that way : S-W duality, when written out, reduces to the fact that different monomials are linearly independent (this is the "Zariski closed sets" stuff in the OP comment), see below.

Some prefer the "abstract nonsense" version as in Aaron's answer, others like me prefer writing out everything in bases. It is a matter of taste : the "abstract nonsense" version is shorter ; on the other hand, there is no mysterious step (such as the action by conjugation) in the hands-on version.

As in Aaron's answer, let us make ${\mathfrak S}_n$ act on the right, abuse notation and write : $\phi: \mathbb{C}{\mathfrak S}_n\to \mathrm{\sf End}_{\mathbb{C}}(V^{\otimes n})$ for $\sigma \mapsto (v \mapsto v\sigma)$ and $\psi:\mathbb{C}GL(V) \to \mathrm{\sf End}_{\mathbb{C}}(V^{\otimes n})$ for $g \mapsto (v\mapsto gv)$.

We want to show that $A={\sf End}_{\mathbb{C}{\mathfrak S}_n}(V^{\otimes n})$ is the same thing as $I={\sf Im}(\psi)$. In fact, all we need to show is $A \subseteq I$, because the reverse inclusion follows from the fact that the two actions commute.

What does an element $a$ of $A$ look like ? If we fix a basis ${\cal B}_1$ of $V$, we obtain a related basis ${\cal B}_2$ for $V^{\otimes n}$ : $$ {\cal B}_2=\lbrace v_1 \otimes v_2 \otimes \ldots \otimes v_n \big| v_1,v_2, \ldots ,v_n \in {\cal B}_1\rbrace $$ Note that ${\cal B}_2$ is invariant by the action of ${\mathfrak S}_n$ on $V^{\otimes n}$.

For any $a\in {\sf End}_{\mathbb{C}}(V^{\otimes n})$, here are $m^{2n}$ coefficients $c(v,w)$ (for $v,w\in {\cal B}_2$ ) such that $$ a(v)=\sum_{w\in {\cal B}_2}c(v,w)w $$ for any $v\in {\cal B}_2$. Then, for any $\sigma\in {\mathfrak S}_n$ we have

$$ a(v)\sigma=\sum_{w\in {\cal B}_2}c(v,w)w\sigma= \sum_{w'\in {\cal B}_2}c(v,w'\sigma^{-1})w'= \sum_{w\in {\cal B}_2}c(v,w\sigma^{-1})w $$

and

$$ a(v\sigma)=\sum_{w\in {\cal B}_2}c(v\sigma,w)w $$

So $a\in {\sf End}_{\mathbb{C}{\mathfrak S}_n}(V^{\otimes n})$ iff $c(v\sigma,w)=c(v,w\sigma^{-1})$ for any $\sigma,v,w$. This is equivalent to $c(v\sigma,w\sigma)=c(v,w)$ for any $\sigma,v,w$, i.e. $a$ is invariant with respect to the natural action of ${\mathfrak S}_n$ on ${\cal B}_2^2$ : in other words $a$ is constant on the orbits. Denote by ${\Omega}$ the set of all orbits in ${\cal B}_2^2$ for this action. For $\omega \in \Omega$, denote by ${\chi}_{\omega}$ the characteristic map of $\omega$ in ${\cal B}_2^2$ (so that ${\chi}_{\omega}(x)=1$ if $x\in\omega$, and $0$ otherwise), and define $a_{\omega}\in {\sf End}_{\mathbb{C}}(V^{\otimes n})$ by $$ a_{\omega}(v)=\sum_{w \in {\cal B}_2}\chi_{\omega}(v,w)w $$ Then ${\cal B}_3=(a_{\omega})_{\omega \in \Omega}$ constitutes a basis for ${\sf End}_{\mathbb{C}{\mathfrak S}_n}(V^{\otimes n})$.

It is now time to write out the action of $GL(V)$ on $(V^{\otimes n})$ similarly. Let $g\in GL(V)$ act on the basis ${\cal B}_1$ by

$$ g(v)=\sum_{w\in {\cal B}_1} \gamma(v,w) w \ ({\rm for \ any \ } v\in {\cal B}_1) $$

where $\gamma$ is a map ${\cal B}_1^2 \to {\mathbb C}$. Then the action of $g$ on $(V^{\otimes n})$ can be written as follows : for any $(v_1,v_2, \ldots ,v_n)\in {\cal B}_2$,

$$ g(v_1 \otimes v_2 \otimes \ldots \otimes v_n)=\sum_{(w_1,w_2, \ldots ,w_n)\in {\cal B}_2} \gamma(v_1,w_1)\gamma(v_2,w_2)\ldots \gamma(v_n,w_n) w_1 \otimes w_2 \otimes \ldots\otimes w_n $$

Let us denote by $H$ the set of $m^{2n}$ variables $\gamma(v,w)$ for $v,w\in {\cal B}_2$. Since the action of $GL(V)$ and ${\mathfrak S}_n$ commute, the extended action of $g$ on $V^{\otimes n}$ (let us denote it by $g'$) can be written the above as

$$ g'=\sum_{\omega \in \Omega}P_{\omega}(h)_{h\in H} a_{\omega} $$

where each $P_{\omega}$ is a polynomial in the variables of $H$ (and, in fact, a product of $n$ variables in $H$,multiplied by a constant). So all those $P_{\omega}$ are different monomials ; so they are linearly independent, and the span of all those $g'$ is necessarily the whole of ${\sf span}(a_{\omega})_{\omega \in \Omega}=A$, qed.

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Thank you very much for your answer. I chose Aaron's because I found it a little easier to understand notationally; I became a little confused with all your bases! Nonetheless I'm very appreciative of the effort, I will try reading it again soon and see if it makes more sense to me. Thanks for your time. –  Spyam May 25 '12 at 4:13
    
Very nice argument, Ewan, and it works for any infinite base field, not necessarily of characteristic $0$. But let me just point out for the casual reader that this proves only one half of the S-W duality (Proposition 5 in arXiv:0610591v3), which seems to be the easier of the two halves. –  darij grinberg Dec 30 '12 at 15:04
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