Schur–Weyl duality

Schur–Weyl duality forms an archetypical situation in representation theory involving two kinds of symmetry that determine each other. Consider the tensor space

${\displaystyle \mathbb {C} ^{n}\otimes \mathbb {C} ^{n}\otimes \cdots \otimes \mathbb {C} ^{n}}$ with k factors.

The symmetric group S_k on k letters acts on this space (on the left) by permuting the factors,

${\displaystyle \sigma (v_{1}\otimes v_{2}\otimes \cdots \otimes v_{k})=v_{\sigma ^{-1}(1)}\otimes v_{\sigma ^{-1}(2)}\otimes \cdots \otimes v_{\sigma ^{-1}(k)}.}$

The general linear group GL_n of invertible n×n matrices acts on it by the simultaneous matrix multiplication,

${\displaystyle g(v_{1}\otimes v_{2}\otimes \cdots \otimes v_{k})=gv_{1}\otimes gv_{2}\otimes \cdots \otimes gv_{k},\quad g\in {\text{GL}}_{n}.}$

These two actions commute, and in its concrete form, the Schur–Weyl duality asserts that under the joint action of the groups S_k and GL_n, the tensor space decomposes into a direct sum of tensor products of irreducible modules (for these two groups) that actually determine each other,

${\displaystyle \mathbb {C} ^{n}\otimes \mathbb {C} ^{n}\otimes \cdots \otimes \mathbb {C} ^{n}=\bigoplus _{D}\pi _{k}^{D}\otimes \rho _{n}^{D}.}$

The summands are indexed by the Young diagrams D with k boxes and at most n rows, and representations ${\displaystyle \pi _{k}^{D}}$ of S_k with different D are mutually non-isomorphic, and the same is true for representations ${\displaystyle \rho _{n}^{D}}$ of GL_n.

The abstract form of the Schur–Weyl duality asserts that two algebras of operators on the tensor space generated by the actions of GL_n and S_k are the full mutual centralizers in the algebra of the endomorphisms ${\displaystyle \mathrm {End} _{\mathbb {C} }(\mathbb {C} ^{n}\otimes \mathbb {C} ^{n}\otimes \cdots \otimes \mathbb {C} ^{n}).}$

Suppose that k = 2 and n is greater than one. Then the Schur–Weyl duality is the statement that the space of two-tensors decomposes into symmetric and antisymmetric parts, each of which is an irreducible module for GL_n:

${\displaystyle \mathbb {C} ^{n}\otimes \mathbb {C} ^{n}=S^{2}\mathbb {C} ^{n}\oplus \Lambda ^{2}\mathbb {C} ^{n}.}$

The symmetric group S₂ consists of two elements and has two irreducible representations, the trivial representation and the sign representation. The trivial representation of S₂ gives rise to the symmetric tensors, which are invariant (i.e. do not change) under the permutation of the factors, and the sign representation corresponds to the skew-symmetric tensors, which flip the sign.

First consider the following setup:

• G a finite group,
• ${\displaystyle A=\mathbb {C} [G]}$ the group algebra of G,
• ${\displaystyle U}$ a finite-dimensional right A-module, and
• ${\displaystyle B=\operatorname {End} _{G}(U)}$, which acts on U from the left and commutes with the right action of G (or of A). In other words, ${\displaystyle B}$ is the centralizer of ${\displaystyle A}$ in the endomorphism ring ${\displaystyle \operatorname {End} (U)}$.

The proof uses two algebraic lemmas.

Proof: Since U is semisimple by Maschke's theorem, there is a decomposition ${\displaystyle U=\bigoplus _{i}U_{i}^{\oplus m_{i}}}$ into simple A-modules. Then ${\displaystyle U\otimes _{A}W=\bigoplus _{i}(U_{i}\otimes _{A}W)^{\oplus m_{i}}}$. Since A is the left regular representation of G, each simple G-module appears in A and we have that ${\displaystyle U_{i}\otimes _{A}W=\mathbb {C} }$ (respectively zero) if and only if ${\displaystyle U_{i},W}$ correspond to the same simple factor of A (respectively otherwise). Hence, we have: ${\displaystyle U\otimes _{A}W=(U_{i_{0}}\otimes _{A}W)^{\oplus m_{i_{0}}}=\mathbb {C} ^{\oplus m_{i_{0}}}.}$ Now, it is easy to see that each nonzero vector in ${\displaystyle \mathbb {C} ^{\oplus m_{i_{0}}}}$ generates the whole space as a B-module and so ${\displaystyle U\otimes _{A}W}$ is simple. (In general, a nonzero module is simple if and only if each of its nonzero cyclic submodule coincides with the module.) ${\displaystyle \square }$

Proof: Let ${\displaystyle W=\operatorname {End} (V)}$. The ${\displaystyle W\hookrightarrow \operatorname {End} (U),w\mapsto w^{d}=d!w\otimes \cdots \otimes w}$. Also, the image of W spans the subspace of symmetric tensors ${\displaystyle \operatorname {Sym} ^{d}(W)}$. Since ${\displaystyle B=\operatorname {Sym} ^{d}(W)}$, the image of ${\displaystyle W}$ spans ${\displaystyle B}$. Since ${\displaystyle \operatorname {GL} (V)}$ is dense in W either in the Euclidean topology or in the Zariski topology, the assertion follows. ${\displaystyle \square }$

The Schur–Weyl duality now follows. We take ${\displaystyle G={\mathfrak {S}}_{d}}$ to be the symmetric group and ${\displaystyle U=V^{\otimes d}}$ the d-th tensor power of a finite-dimensional complex vector space V.

Let ${\displaystyle V^{\lambda }}$ denote the irreducible ${\displaystyle {\mathfrak {S}}_{d}}$-representation corresponding to a partition ${\displaystyle \lambda }$ and ${\displaystyle m_{\lambda }=\dim V^{\lambda }}$. Then by Lemma 1

${\displaystyle S^{\lambda }(V):=V^{\otimes d}\otimes _{{\mathfrak {S}}_{d}}V^{\lambda }}$

is irreducible as a ${\displaystyle \operatorname {GL} (V)}$-module. Moreover, when ${\displaystyle A=\bigoplus _{\lambda }(V^{\lambda })^{\oplus m_{\lambda }}}$ is the left semisimple decomposition, we have:[4]

${\displaystyle V^{\otimes d}=V^{\otimes d}\otimes _{A}A=\bigoplus _{\lambda }(V^{\otimes d}\otimes _{{\mathfrak {S}}_{d}}V^{\lambda })^{\oplus m_{\lambda }}}$,

which is the semisimple decomposition as a ${\displaystyle \operatorname {GL} (V)}$-module.

The Brauer algebra plays the role of the symmetric group in the generalization of the Schur-Weyl duality to the orthogonal and symplectic groups.

More generally, the partition algebra and its subalgebras give rise to a number of generalizations of the Schur-Weyl duality.

• Partition algebra

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• How to constructively/combinatorially prove Schur-Weyl duality?