Polynomials and Representations XXIV

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.

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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. ♦

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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:

sample_filling

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).

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