## Specht Modules

Till now, our description of the irreps $\{V_\lambda\}_{\lambda \vdash d}$ of $S_d$ are rather abstract. It would be helpful to have a more concrete construction of these representations – one way is via Specht modules. First write $\displaystyle \mathbb{C}[X_\lambda] \cong V_\lambda \oplus \left( \bigoplus_{\nu\trianglerighteq \lambda,\, \nu\ne\lambda} V_\nu^{K_{\nu\lambda}}\right)$ \displaystyle \begin{aligned} \implies \mathbb{C}[X_\mu] \otimes \chi \cong \left( V_\mu \otimes\chi\right) \oplus \left( \bigoplus_{\nu\trianglerighteq \mu,\, \nu\ne\mu} (V_\nu \otimes\chi)^{K_{\nu\mu}}\right) \cong V_{\overline\mu} \oplus \left( \bigoplus_{\nu\trianglelefteq \overline\mu,\, \nu\ne\overline \mu} V_\nu^{K_{\overline\nu\mu}}\right) \end{aligned}

Thus if $\mu = \overline\lambda$, the only common irrep between $\mathbb{C}[X_\lambda]$ and $\mathbb{C}[X_{\overline\lambda}] \otimes \chi$ is $V_\lambda.$ This suggests that we should look at those two representations in more concrete terms. ## General Result

Lemma. Let $G$ be a finite group acting transitively on a non-empty finite set X. Pick $x_0\in X$ and let $H := \{g\in G : gx_0 = x_0\}$ be its stabilizer group. Then for any one-dimensional character $\psi : G\to \mathbb{C}^*$, $\mathbb{C}[X]\otimes \psi \cong \mathbb{C}[G]a,$

as left $\mathbb{C}[G]$-modules, where $a:= \sum_{g\in H} \psi(g)g \in \mathbb{C}[G].$

Proof

First note that if $h\in H$, then $ha = \psi(h)^{-1}a$ since: $\displaystyle a = \sum_{g \in H}\psi(hg)hg = \psi(h)\sum_{g\in H} \psi(g)hg = \psi(h)ha.$

On the other hand, if $h\not\in H$, $ha$ is not a multiple of $a$ since the coefficient of e is 0.

Now define the map $\mathbb{C}[X] \otimes\psi \to \mathbb{C}[G]a$ as follows: for $x=gx_0 \in X,$ take it to $\psi(g)^{-1}ga \in \mathbb{C}[G]a.$ The map is well-defined and injective since: \begin{aligned} gx_0 = g'x_0 &\iff g^{-1}g' \in H \iff g^{-1}g' a = \psi(g^{-1}g')^{-1} a\\ &\iff \psi(g')g'a = \psi(g)ga.\end{aligned}

Surjectivity is clear; it is also G-equivariant since: $\begin{matrix} & x = gx_0 & \mapsto & & \psi(g)^{-1}ga, \\ \substack{\text{twisted action}\\ \text{of }g'\in G} &\downarrow & & &\downarrow &\substack{\text{left-mult}\\ \text{by } g'\in G} \\ & \psi(g') g'gx_0 & \mapsto & \psi(g')\psi(g'g)^{-1} g'ga = &\psi(g)^{-1}g'ga\end{matrix}$

Hence we get an isomorphism of left $\mathbb{C}[G]$-modules. ♦ ## Young Symmetrizer

In particular, consider the case where $G = S_d$ and let $\psi$ be either the trivial or the alternating character $\chi : S_d \to \{-1, +1\}.$ Given a partition $\lambda\vdash d$, fix a filling $T$ of  the Young diagram of $\lambda$ with $\{1,\ldots, d\}$ such that each element occurs exactly once. E.g. for $\lambda = (3,2,1)$ let us pick: Now let $R(T)$ (resp. $C(T)$) be the set of all $g\in G$ which takes every element in the i-th row (resp. column) to an element of the i-th row (resp. column). Thus, in our above example: $R(T) = S_{\{3,6,5\}} \times S_{\{1,4\}} \times S_{\{2\}}, \qquad C(T) = S_{\{3,1,2\}} \times S_{\{6,4\}} \times S_{\{5\}},$

where $S_A$ is the group of all permutations of set A. By the above general lemma, we have: \begin{aligned}\displaystyle \mathbb{C}[X_\lambda] \cong \mathbb{C}[G]a_T, \qquad &a_T := \sum_{g\in R(T)} g\in \mathbb{C}[G], \\ \mathbb{C}[X_{\overline\lambda}] \otimes\chi \cong \mathbb{C}[G]b_T, \qquad &b_T := \sum_{g\in C(T)} \chi(g)g \in \mathbb{C}[G].\end{aligned}

Now let us define a common component of $\mathbb{C}[G]a_T$ and $\mathbb{C}[G]b_T.$

Definition. The Young symmetrizer of the filling $T$ is defined by: $c_T := a_T b_T \in \mathbb{C}[G].$

Note that $\mathbb{C}[G]c_T = \mathbb{C}[G]a_T b_T$ is contained in both $\mathbb{C}[G]a_T$ and $\mathbb{C}[G]b_T$ since $\mathbb{C}[G]a_T \stackrel{v\mapsto vb_T}\twoheadrightarrow \mathbb{C}[G]a_T b_T \subseteq \mathbb{C}[G]b_T.$

Hence $\mathbb{C}[G]c_T$ is either 0 or $V_\lambda.$ It has to be the latter since the coefficient of e in $c_T$ is 1 (because $R(T) \cap C(T) = \{e\}$). Similarly, we see that $\mathbb{C}[G]b_T a_T$ is also isomorphic to $V_\lambda.$

## Example

Suppose d=3 and $\lambda = (2,1)$. Pick the filling T by placing 1, 2 at the top row, then 3 below. We have R(T) = {e, (1,2)} and C(T) = {e, (1,3)}. This gives: $c_T = e + (1,2) - (1,3) - (1,3,2).$

The module $\mathbb{C}[G]c_T$ has a basis $\{c_T, (1,3)c_T\}$ where $(1,3)c_T = (1,3) + (1,2,3) - e-(2,3).$ This entry was posted in Uncategorized and tagged , , , , . Bookmark the permalink.