## Schur-Weyl Duality

Throughout the article, we denote $V = \mathbb{C}^n$ for convenience.

So far we have seen:

• the Frobenius map gives a correspondence between symmetric polynomials in $x_1, x_2, \ldots$ of degree d and representations of $S_d$;
• there is a correspondence between symmetric polynomials in $x_1, \ldots, x_n$ and polynomial representations of $GL_n\mathbb C$.

Here we will describe a more direct relationship between representations of $S_d$ and polynomial representations of $GL_n\mathbb{C}.$ Recall from earlier, that $S_d$ and $GL_n\mathbb C$ act on $V^{\otimes d}$ as follows:

\begin{aligned} w\in S_d &\implies v_1 \otimes \ldots \otimes v_d \mapsto v_{w^{-1}(1)} \otimes \ldots \otimes v_{w^{-1}(d)},\\ g\in GL_n(\mathbb C) &\implies v_1 \otimes \ldots \otimes v_n \mapsto g(v_1) \otimes \ldots \otimes g(v_n),\end{aligned}

and the two actions commute, so $w\circ g = g\circ w$ as endomorphisms of $V^{\otimes d}.$

Lemma. The subspace $\text{Sym}^d V \subset V^{\otimes d}$ of all elements fixed by every $w\in S_d$ is spanned by $\{v^{\otimes d}: v\in V\}.$

Proof

Use induction on d; the case d=1 is trivial so suppose d>1. For integers $k\ge 0$, consider the binomial expansion in $\text{Sym}^d V$:

$\displaystyle(v + kw)^d = v^d + \left(\sum_{i=1}^{d-1} k^i {d\choose i} v^{d-i} w^{i}\right) + k^d w^d.$

We claim: for large k, the $(d+1)\times k$ matrix with (i, j)-entry $j^i {d\choose i}$ (where $0\le i \le d$) has rank d+1.

• Indeed, otherwise there are $\alpha_0, \ldots, \alpha_d \in \mathbb{C}$, not all zero, such that $\alpha_0 {d\choose 0} + \alpha_1 {d\choose 1} k + \ldots + \alpha_d {d\choose d} k^d = 0$ for all large k, which is absurd since this is a polynomial in k.

Hence, we can find a linear combination summing up to:

$\alpha_0 v^d + \alpha_1 (v+w)^d + \ldots + \alpha_k (v+kw)^d = vw^{d-1}, \qquad \text{ for all }v, w \in V.$

Thus $vw^{d-1}$ lies in the subspace spanned by all $v^d$. By induction hypothesis, the set of all $w^{d-1} \in \text{Sym}^{d-1} V$ spans the whole space. Hence, the set of all $v^d$ spans $\text{Sym}^d V$. ♦

This gives:

Proposition. If $f: V^{\otimes d} \to V^{\otimes d}$ is an $S_d$-equivariant map, then it is a linear combination of the image of $GL_n\mathbb{C} \hookrightarrow \text{End}(V^{\otimes d})$.

Proof

Note that since $\text{End}(V) \cong V\otimes V^\vee$ we have $\text{End}(V^{\otimes d}) \cong \text{End}(V)^{\otimes d}.$ Hence from the given condition

$f \in \text{End}(V^{\otimes d})^{S_d} = (\text{End}(V)^{\otimes d})^{S_d}.$

By the above lemma, f is a linear combination of $u^{\otimes d}$ for all $u\in\text{End}(V).$ Since $\text{GL}_n\mathbb{C} \subset \text{End}(V)$ is dense, f is also a linear combination of $u^{\otimes d}$ for $u\in \text{GL}_n\mathbb{C}$. ♦

## Main Statement

Now let U be any complex vector space and consider the complex algebra $\text{End}(U).$ Recall: if $A\subseteq \text{End}(U)$ is any subset,

$C(A) = \{a \in \text{End}_{\mathbb C}(U) : ab = ba \text{ for all }b \in A\}$

is called the centralizer of A. Clearly $C(A) \subseteq \text{End}(U)$ is a subalgebra and we have $A\subseteq C(C(A)).$

Theorem (Schur-Weyl Duality). Let $A\subseteq \text{End}(U)$ be a subalgebra which is semisimple. Then:

• $B:=C(A)$ is semisimple;
• $C(B) = A$; (double centralizer theorem)
• U decomposes as $\oplus_{\lambda} (U_\lambda \otimes W_\lambda)$, where $U_\lambda, W_\lambda$ are respectively complete lists of irreducible A-modules and B-modules.

Proof

Since A is semisimple, we can write it as a finite product $\prod_\lambda \text{End}(\mathbb{C}^{m_\lambda})$. Each simple A-module is of the form $U_\lambda := \mathbb{C}^{m_\lambda}$ for some $m_\lambda >0.$ As an A-module, we can decompose: $\displaystyle U \cong \oplus_{\lambda} U_\lambda^{n_\lambda}.$ Here $n_\lambda > 0$ since as A-modules we have:

$U_\lambda \subseteq A \subseteq \text{End}(U) \cong U^{\dim U}.$

By Schur’s lemma $\text{End}_A(U_\lambda, U_\mu) \cong \mathbb{C}$ if $\lambda = \mu$ and 0 otherwise. This gives:

$\displaystyle B = C(A) = \text{End}_A(U) = \text{End}_A\left(\prod_\lambda U_\lambda^{n_\lambda} \right) \cong \prod_\lambda \text{End}(\mathbb{C}^{n_\lambda})$

which is also semisimple. Now each simple B-module $W_\lambda$ has dimension $n_\lambda$. From the action of B on U, we can write $U \cong \oplus_\lambda U_\lambda ^{n_\lambda} \cong \oplus_\lambda (U_\lambda \otimes W_\lambda)$ where A acts on the $U_\lambda$ and B acts on the $W_\lambda$. Expressed as a sum of simple B-modules, we have $U \cong \oplus_\lambda W_\lambda^{m_\lambda}$; thus repeating the above with A replaced by B gives:

$C(B) \cong \prod_\lambda \text{End}(\mathbb C^{m_\lambda})\cong A.$

From $A\subseteq C(C(A))$ we thus have $A= C(B).$ This proves all three properties. ♦

Note

From the proof, we see that

• $U = \oplus_\lambda (U_\lambda \otimes W_\lambda)$ as complex vector spaces,
• $A \cong \prod_\lambda \text{End}_{\mathbb{C}}U_\lambda$ acts on the $U_\lambda$, and
• $B\cong \prod_\lambda \text{End}_{\mathbb{C}} W_\lambda$ acts on the $W_\lambda$.

Thus the correspondence between $U_\lambda$ and $W_\lambda$ works as follows:

\begin{aligned}\text{Hom}_A(U_\lambda, U) &= \text{Hom}_{\prod \text{End}(U_\mu)}(U_\lambda, \oplus_\mu (U_\mu \otimes W_\mu))\\ &\cong \text{Hom}_{\text{End}(U_\lambda)} (U_\lambda, U_\lambda^{n_\lambda})\\ &\cong W_\lambda.\end{aligned}

The nice thing about this point-of-view is that the construction is now functorial, i.e. for any A-module M, we can define the corresponding: $F: M \mapsto\text{Hom}_A(M, U).$ This functor is additive, i.e. $F(M_1 \oplus M_2) \cong F(M_1) \oplus F(M_2)$, since the Hom functor is bi-additive.

## The Case of Sd and GLnC

Now for our main application.

Consider $S_d$ and $GL_n\mathbb C$ acting on $V^{\otimes d}$; their actions span subalgebras $A, B\subseteq \text{End}_{\mathbb C}(V)$. Now A is semisimple since it is a quotient of $\mathbb{C}[S_d]$. From the lemma, we have B = C(A) so Schur-Weyl duality says A = C(B), B is semisimple and

$V^{\otimes d} \cong \oplus_\lambda (U_\lambda \otimes W_\lambda)$

where $U_\lambda, W_\lambda$ are complete lists of simple A– and B-modules respectively. Since A is a quotient of $\mathbb{C}[S_d]$, the $U_\lambda$ are also irreps of $S_d$ so they can be parametrized by $\lambda \vdash d$.

Proposition. If $U_\lambda$ is the irrep for $S_d$ isomorphic to $V_\lambda$, then $W_\lambda$ is the irrep for $GL_n\mathbb{C}$ corresponding to $V(\lambda).$

Proof

It suffices to show: if $\mathbb{C}[X_\lambda]$ corresponds to $W'$ via the functor in the above note, then

$W'\cong \text{Sym}^\lambda V = \text{Sym}^{\lambda_1} V \otimes \ldots \otimes \text{Sym}^{\lambda_l} V.$

By definition $W' = \text{Hom}_{S_d}(\mathbb{C}[X_\lambda], V^{\otimes d}).$ Recall that $X_\lambda$ is a transitive $S_d$-set; picking a point $A=(A_i) \in X_\lambda$, any map $f:\mathbb{C}[X_\lambda] \to V^{\otimes d}$ which is $S_d$-equivariant is uniquely defined by the element $f(A)\in V^{\otimes d}$, as long as this element is invariant under the stabilizer group:

$H := \{w\in S_d : w(A) = A\} \cong S_{\lambda_1} \times S_{\lambda_2} \times \ldots \times S_{\lambda_l}.$

Thus, the coefficients $c_{i_1\ldots i_d}$ of $e_{i_1}\otimes\ldots\otimes e_{i_d}$ in $f(A)$ remain invariant when acted upon by $\prod_i S_{\lambda_i}$. So we have an element of $\text{Sym}^\lambda V.$ ♦

Theorem. The set of irreps $V_\lambda$ of $S_d$ occurring in $V^{\otimes d}$ is:

$\{ V_\lambda : \lambda \vdash d, l(\lambda) \le n\}.$

Proof

The following is the complete set of $GL_n\mathbb{C}$-irreps of degree d:

$\{V(\lambda) : \lambda\vdash d, l(\lambda) \le n\}$

We claim that this is also the set of all irreps in $V^{\otimes d}.$ Clearly, each irrep in $V^{\otimes d}$ is of degree d; conversely, $V^{\otimes d}$ has

$\psi = (x_1 + \ldots + x_n)^d = h_\mu(x_1, \ldots, x_n), \ \mu = (1,1,\ldots, 1).$

Clearly $K_{\lambda\mu} > 0$ so $V^{\otimes d}$ contains all $V(\lambda)$ of degree d. Now apply the above proposition. ♦

Example

The simplest non-trivial example follows from the decomposition

$V^{\otimes 2} = (\text{Sym}^2 V) \oplus (\text{Alt}^2 V).$

The action of $S_2$ is trivial on the first component and alternating on the second.

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