show/hide this revision's text 3 some typos fixed, some clarification added

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 $S_n$. 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 $\beta$ \beta:W^{\otimes n}\to W^{\otimes n}$ to acts 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 $\alpha_i$ 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 $S_n$.
show/hide this revision's text 2 added 2 characters in body

I'm going to write up Mark Widon's 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 $S_n$. 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 Widon's 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 $\beta$ to acts 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 $\alpha_i$ 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 $S_n$.
show/hide this revision's text 1 [made Community Wiki]

I'm going to write up Mark Widon'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 $S_n$. 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 Widon'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 $\beta$ to acts 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 $\alpha_i$ 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 $S_n$.