4.18. Lecture 18

4.18.1. Irreducibility of Specht modules

The goal is to prove the first part of Theorem 4.17.12. Recall that for any Young tableau $t$ , $C(t)\subseteq S_{n}$ is the set of permutations that preserve the columns of $t$ .

Lemma 4.18.1.

For any $t\in\mathbf{YT}^{\lambda}$ , $C(t)$ is a subgroup of $S_{n}$ , and for any $\sigma\in S_{n}$ ,

\sigma C(t)\sigma^{-1}=C(\sigma\cdot t).

Proof.

This is Problem 4.18.2. ∎

Recall that for $t\in\mathbf{YT}^{\lambda}$ , we define ${\bf e}_{t}=\sum_{\sigma\in C(t)}\operatorname{sgn}(\sigma)[\sigma\cdot t]$ , and ${\mathcal{S}}^{\lambda}=\mathrm{span}\{{\bf e}_{t}\,|\,t\in\mathbf{YT}^{% \lambda}\}\subseteq{\mathcal{M}}^{\lambda}$ .

Proposition 4.18.2.

For any $\sigma\in S_{n}$ and $t\in\mathbf{YT}^{\lambda}$ , $\pi(\sigma){\bf e}_{t}={\bf e}_{\sigma\cdot t}$ . As a consequence, $(\pi,{\mathcal{S}}^{\lambda})$ is a subrepresentation of $(\pi,{\mathcal{M}}^{\lambda})$ .

Proof.

We have

\pi(\sigma)\mathbf{e}_{t}=\sum_{\tau\in C(t)}\mathrm{sgn}(\tau)[\sigma\cdot(% \tau\cdot t)]=\sum_{\tau\in C(t)}\mathrm{sgn}(\tau)[\sigma\tau\sigma^{-1}\cdot% (\sigma\cdot t)].

Using the previous lemma, we may rewrite this as follows:

\pi(\sigma)\mathbf{e}_{t}=\sum_{\tau\in C(\sigma\cdot t)}\mathrm{sgn}(\sigma^{% -1}\tau\sigma)[\tau\cdot(\sigma\cdot t)]=\sum_{\tau\in C(\sigma\cdot t)}% \mathrm{sgn}(\tau)[\tau\cdot(\sigma\cdot t)]=\mathbf{e}_{\sigma\cdot t}.

∎

Given $t\in\mathbf{YT}^{\lambda}$ , define $P_{t}:{\mathcal{M}}^{\lambda}\rightarrow{\mathcal{M}}^{\lambda}$ by

P_{t}=\sum_{\sigma\in C(t)}\operatorname{sgn}(\sigma)\cdot\pi(\sigma).

Observe that $P_{t}[t]={\bf e}_{t}$ and if $(\pi,W)$ is a subrepresentation of $(\pi,{\mathcal{M}}^{\lambda})$ , then $P_{t}{\bf w}\in W$ for all $t\in\mathbf{YT}^{\lambda}$ and ${\bf w}\in W$ .

Lemma 4.18.3.

Let $\langle\cdot,\cdot\rangle$ be any $S_{n}$ -invariant inner product on ${\mathcal{M}}^{\lambda}$ . Then $P_{t}$ is self-adjoint with respect to $\langle\cdot,\cdot\rangle$ , i.e.

\langle P_{t}{\bf u},{\bf v}\rangle=\langle{\bf u},P_{t}{\bf v}\rangle\qquad% \forall{\bf u},{\bf v}\in{\mathcal{M}}^{\lambda}.

Proof.

Since $\{[s]\}_{s\in\mathbf{YTD}^{\lambda}}$ spans ${\mathcal{M}}^{\lambda}$ , by linearity it suffices to show that $\langle P_{t}[s],[r]\rangle=\langle[s],P_{t}[r]\rangle$ for all $r,s,t\in\mathbf{YT}^{\lambda}$ . By the definition of $P_{t}$ , we then have

	$\displaystyle\langle P_{t}[s],[r]\rangle=$	$\displaystyle\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma)\langle\pi(\sigma)[s],[r% ]\rangle=\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma)\langle[s],\pi(\sigma^{-1})[% r]\rangle$
		$\displaystyle=\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma^{-1})\langle[s],\pi(% \sigma^{-1})[r]\rangle=\sum_{\sigma^{-1}\in C(t)}\langle[s],\mathrm{sgn}(% \sigma)\pi(\sigma)[r]\rangle$
		$\displaystyle=\sum_{\sigma\in C(t)}\langle[s],\mathrm{sgn}(\sigma)\pi(\sigma)[% r]\rangle=\langle[s],P_{t}[r]\rangle;$

the second to last equality holding since $C(t)$ is a subgroup of $S_{n}$ . ∎

Proposition 4.18.4.

For any $t\in\mathbf{YT}^{\lambda}$ and ${\bf v}\in{\mathcal{M}}^{\lambda}$ , $P_{t}{\bf v}\in{\mathbb{C}}{\bf e}_{t}$ .

Proof.

Again using the fact that the tabloids $[s]$ span ${\mathcal{M}}^{\lambda}$ , we need only show that $P_{t}[s]\in{\mathbb{C}}{\bf e}_{t}$ for each $s,t\in\mathbf{YT}^{\lambda}$ . Let $i$ and $j$ be in the same column of $t$ , so $(ij)$ is in $C(t)$ . We consider two cases.

Firstly, suppose that $s$ is such that $[s]$ has the numbers $i,j$ in a single row. Let $H=\langle(ij)\rangle=\{e,(ij)\}$ . We now let $r_{1},\ldots,r_{m}\in C(t)$ be a set of representatives for $C(t)/H$ , i.e.

C(t)=r_{1}H\sqcup r_{2}H\sqcup\ldots\sqcup r_{m}H.

Since $[s]$ has $i$ and $j$ in the same row, $\pi(ij)[s]=[(ij)\cdot s]=[s]$ . We can now compute:

	$\displaystyle P_{t}[s]=\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma)\pi(\sigma)[s]=$	$\displaystyle\sum_{k=1}^{m}\mathrm{sgn}(r_{k})\pi(r_{k})\Big{(}\mathrm{sgn}(e)% \pi(e)+\mathrm{sgn}(ij)\pi(ij)\Big{)}[s]$
		$\displaystyle=\sum_{k=1}^{m}\mathrm{sgn}(r_{k})\pi(r_{k})\big{(}1+(-1)\big{)}[% s]=0.$

We now consider the remaining case, i.e. $i$ and $j$ are in different rows of $[s]$ . Letting the first column of $t$ be $a_{1,1},a_{1,2},\ldots,a_{1,M}$ , we let $\sigma_{1}\in C(t)$ be a permutation that moves each $a_{1,i}$ into the row that it is in in $[s]$ , and fixes all other elements. Similarly, let $\sigma_{2}$ be a permutation that permutes the elements of the second column into the rows they occupy in $[s]$ , $\sigma_{3}$ a permutation of the third column of $t$ , and so on. Then $[\sigma_{\lambda_{1}}\ldots\sigma_{2}\sigma_{1}\cdot t]=[s]$ , and since each $\sigma_{i}$ is just a permutation of column entries, $\widetilde{\sigma}=\sigma_{\lambda_{1}}\ldots\sigma_{2}\sigma_{1}\in C(t)$ . We now compute

	$\displaystyle P_{t}[s]=\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma)[\sigma\cdot s]=$	$\displaystyle\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma)[\sigma\widetilde{\sigma% }\cdot t]$
		$\displaystyle=\sum_{\sigma\in C(t)}\mathrm{sgn}(\sigma\widetilde{\sigma}^{-1})% [\sigma\cdot t]=\mathrm{sgn}(\widetilde{\sigma})\mathbf{e}_{t}.$

∎

Proof of the first part of Theorem 4.17.12.

Let $(\pi,W)$ be a subrepresentation of $(\pi,{\mathcal{S}}^{\lambda})$ . Suppose that there exists $t\in\mathbf{YT}^{\lambda}$ and ${\bf w}\in W$ such that $P_{t}{\bf w}\neq 0$ . Then $P_{t}{\bf w}=\lambda{\bf e}_{t}$ , with $\lambda\neq 0$ . Then ${\bf e}_{t}\in W$ , hence

{\mathcal{S}}^{\lambda}=\mathrm{span}\{{\bf e}_{s}\,|\,s\in\mathbf{YT}^{% \lambda}\}=\mathrm{span}\{{\bf e}_{\sigma\cdot t}\,|\,\sigma\in S_{n}\}=% \mathrm{span}\{\pi(\sigma){\bf e}_{t}\,|\,\sigma\in S_{n}\}\subseteq W,

i.e. $W={\mathcal{S}}^{\lambda}$ .

On the other hand, if $P_{t}{\bf w}=0$ for all $t\in\mathbf{YT}^{\lambda}$ and ${\bf w}\in W$ , then

\displaystyle 0=\langle P_{t}{\bf w},[t]\rangle=\langle{\bf w},P_{t}[t]\rangle% =\langle{\bf w},{\bf e}_{t}\rangle,

thus $W\subseteq({\mathcal{S}}^{\lambda})^{\perp}$ . Since $({\mathcal{S}}^{\lambda})^{\perp}\cap{\mathcal{S}}^{\lambda}=0$ , we then have that $W=0$ . ∎

The proof above actually gives the slightly stronger statement:

Theorem 4.18.5.

Let $(\pi,W)$ be a subrepresentation of $(\pi,{\mathcal{M}}^{\lambda})$ . Then either ${\mathcal{S}}^{\lambda}\subseteq W$ or $W\subseteq({\mathcal{S}}^{\lambda})^{\perp}$ .

4.18.2. Exercises

Problem 93. Let $\lambda\vdash n$ . Given a Young tableau $t\in\mathbf{YT}^{\lambda}$ and $\sigma\in S_{n}$ , show that $C(t)$ is a subgroup of $S_{n}$ and that

\sigma C(t)\sigma^{-1}=C(\sigma\cdot t).

Problem 94. Write out the details of the proof of Theorem 4.18.5.