Polynomials and Representations XXXVI

V(λ) as Schur Functor

Again, we will denote V := \mathbb{C}^n throughout this article. In the previous article, we saw that the Schur-Weyl duality can be described as a functor:

  • given a \mathbb{C}[S_d]-module M, the corresponding GL_n\mathbb{C}-module is set as \text{Hom}_{S_d}(M, V^{\otimes d}).

Definition. The construction

F_M(V) := \text{Hom}_{S_d}(M, V^{\otimes d})

is functorial in V and is called the Schur functor when M is fixed.

Here, functoriality means that any linear map V\to W induces a linear F_M(V) \to F_M(W).

For example, when M = \mathbb{C}[S_d], the functor F_M is the identity functor. By Schur-Weyl duality, when M is irreducible as an S_d-module, the resulting F_M(V) is either 0 or irreducible. We will see the Schur functor cropping up in two other instances.


Following the reasoning as in S_d-modules, we have for partitions \lambda\vdash d and \mu := \overline\lambda,

\begin{aligned}\text{Sym}^\lambda V &:= \bigotimes_i \text{Sym}^{\lambda_i} V = V(\lambda) \oplus\left( \bigoplus_{\nu\trianglerighteq \lambda, \nu\ne\lambda} V(\nu)^{\oplus K_{\nu\lambda}}\right),\\ \text{Alt}^{\mu} V &:= \bigotimes_j \text{Alt}^{\mu_j}V = V(\lambda) \oplus \left(\bigoplus_{\nu\trianglelefteq \lambda, \nu\ne\lambda} V(\nu)^{\oplus K_{\overline\nu\mu}} \right).\end{aligned}

Since the only common irrep between the two representations is V(\lambda), any non-zero G-equivariant f: \text{Sym}^\lambda V\to \text{Alt}^\mu V must induce an isomorphism between those two components. We proceed to construct such a map.

For illustration, take \lambda = (3, 1) and pick the following filling:


To construct the map, we will take \text{Sym}^d V and \text{Alt}^d as subspaces of V^{\otimes d}. Thus:

\begin{aligned}\text{Sym}^d V \subseteq V^{\otimes d},\quad& v_1 \ldots v_d \mapsto \sum_{w\in S_d} v_{w(1)} \otimes \ldots \otimes v_{w(d)},\\ \text{Alt}^d V\subseteq V^{\otimes d},\quad &v_1 \wedge \ldots \wedge v_d \mapsto \sum_{w\in S_d} \chi(w) v_{w(1)} \otimes \ldots \otimes v_{w(d)},\\ V^{\otimes d}\twoheadrightarrow \text{Sym}^d V, \quad &v_1 \otimes \ldots\otimes v_d \mapsto v_1 \ldots v_d, \\ V^{\otimes d} \twoheadrightarrow \text{Alt}^d V, \quad &v_1 \otimes \ldots \otimes v_d \mapsto v_1 \wedge \ldots \wedge v_d.\end{aligned}

Let us map \text{Sym}^\lambda V \to V^{\otimes 4} according to the above filling, i.e. \text{Sym}^3 V goes into components 1, 4, 2 of V^{\otimes 4} while V goes into component 3. Similarly, we map V^{\otimes 4} \to \text{Alt}^\mu V by mapping components 1, 3 to \text{Alt}^2 V, components 4 and 2 to the other two copies of V. In diagram, we have:


This construction is clearly functorial in V. Hence, if f:V\to W is a linear map of vector spaces, then this induces a linear map f(\lambda) : V(\lambda) \to W(\lambda).


Young Symmetrizer Revisited

Another means of defining the Schur functor is by the Young symmetrizer. Here we shall let GL_n\mathbb{C} act on V^{\otimes d} on the left and S_d act on it on the right via:

w\in S_d \implies(v_1 \otimes \ldots \otimes v_d)w := v_{w(1)} \otimes \ldots \otimes v_{w(d)}.

Now given any (left) \mathbb{C}[S_d]-module M, consider:

V(M) := V^{\otimes d} \otimes_{\mathbb{C}[G]} M,

a left GL_n\mathbb{C}-module. We shall prove that V(M) corresponds to the Schur-Weyl duality, i.e. M = V_\lambda \implies V(M) \cong V(\lambda). Once again, by additivity, we only need to consider the case M = \mathbb{C}[X_\lambda]. This gives M \cong \mathbb{C}[G]a_T where T is any filling of shape λ and thus:

V(M) = V^{\otimes d} \otimes_{\mathbb{C}[G]} \mathbb{C}[G]a_T \cong V^{\otimes d}a_T.

From here, it is clear that V(M) \cong \text{Sym}^\lambda V and so V\mapsto V(M) is yet another expression of the Schur functor.

Recall that the irreducible S_d-module V_\lambda can be written as \mathbb{C}[S_d]c_T where c_T is the Young symmetrizer for a fixed filling of shape λ. Hence, the irrep V(\lambda) can be written as:

V^{\otimes d} \otimes_{\mathbb{C}[G]} V_\lambda \cong V^{\otimes d}\otimes_{\mathbb{C}[G]} \mathbb{C}[G]c_T \cong V^{\otimes d}c_T.


Example: d=3

For d=3, and \lambda = (2,1), let us take the Young symmetrizer:

c_T = a_T b_T = (e + (1,2))(e - (1,3)) = e + (1,2) - (1,3) - (1,3,2).

If e_1, \ldots, e_n is the standard basis for V= \mathbb{C}^n, then V^{\otimes d}c_T is spanned by elements of the form:

\alpha_{i,j,k} := e_i \otimes e_j \otimes e_k + e_j \otimes e_i \otimes e_k - e_k \otimes e_j \otimes e_i - e_k \otimes e_i \otimes e_j, \ 1 \le i, j, k\le n.

These satisfy the following:

\alpha_{j,i,k} = \alpha_{i,j,k},\quad \alpha_{i,j,k} + \alpha_{j,k,i} + \alpha_{k,i,j} = 0.

By the first relation, we only include those \alpha_{i,j,k} with i \le j. By the second relation, we may further restrict to the case i<k since if i=k we have \alpha_{i,j,k} = 0 and if k <i\le j we replace \alpha_{i,j,k} = \alpha_{k,j,i}+ \alpha_{k,i,j}. We claim that the resulting spanning set \{\alpha_{i,j,k} : i\le j, i<k\} forms a basis. Indeed the number of such triplets (ijk) is:

d = \sum_{i=1}^n (n-i+1)(n-i) = \frac{n(n+1)(n-1)}3.

On the other hand, we know that V^{\otimes 3} has one copy of \text{Sym}^3, one copy of \text{Alt}^3 and two copies of V(\lambda) so

2\dim V = n^3 - \frac{(n+2)(n+1)n}6 - \frac{n(n-1)(n-2)}6 =\frac{2n(n+1)(n-1)}3.

Thus \dim V is the cardinality of the set and we are done. ♦


Observe that the set \{(i,j,k) \in [n]^d : i\le j, i<k\} corresponds to the set of all SSYT with shape (2, 1) and entries in [n] (by writing ij in the first row and k below i). This is an example of our earlier claim that a basis of V(\lambda) can be indexed by SSYT’s with shape \lambda and entries in [n]. For that, we will explore V(\lambda) as a quotient module of \otimes_j \text{Alt}^{\mu_j} V in the next article. This corresponds to an earlier article, which expressed S_d-irrep V_\lambda as a quotient of \mathbb{C}[S_d]b_T.


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