## 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 since if $i=k$ we have $\alpha_{i,j,k} = 0$ and if $k 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 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. ♦

Note

Observe that the set $\{(i,j,k) \in [n]^d : i\le j, i 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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