Polynomials and Representations XXXV

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}. ♦

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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.

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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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