Understanding the proof of Schur-Weyl Duality

Solution 1:

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).

Solution 2:

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.