# The Unapologetic Mathematician

## The Sign Lemma

As we move towards proving the useful properties of Specht modules, we will find the following collection of results helpful. Through them all, let $H\subseteq S_n$ be a subgroup, and also consider the $S_n$-invariant inner product on $M^\lambda$ for which the distinct Young tabloids form an orthonormal basis.

First, if $\pi\in H$, then

$\displaystyle\pi H^-=H^-\pi=\mathrm{sgn}(\pi)H^-$

where $H^-$ is the alternating sum of the elements of $H$. The proof basically runs the same as when we showed that $\pi e_t=\mathrm{sgn}(\pi)e_t$ where $t$ has shape $(1^n)$.

Next, for any vectors $u,v\in M^\lambda$ we have

$\displaystyle\langle H^-u,v\rangle=\langle u,H^-v\rangle$

Indeed, we can calculate

$\displaystyle\begin{aligned}\langle H^-u,v\rangle&=\sum\limits_{\pi\in H}\langle\mathrm{sgn}(\pi)\pi u,v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi)\pi^{-1}v\rangle\\&=\sum\limits_{\pi\in H}\langle u,\mathrm{sgn}(\pi^{-1})\pi^{-1}v\rangle\\&=\sum\limits_{\tau\in H}\langle u,\mathrm{sgn}(\tau)\tau v\rangle\\&=\langle u,H^-v\rangle\end{aligned}$

where we have used the facts that $\mathrm{sgn}(\pi)=\mathrm{sgn}(\pi^{-1})$, and that as $\pi$ runs over a group, so does $\tau=\pi^{-1}$.

Next, if the swap $(b\,c)\in H$, then we have the factorization

$\displaystyle H^-=k(1-(b\,c))$

for some $k\in\mathbb{C}[S_n]$. To see this, consider the subgroup $K=\{1,(b\,c)\}\subseteq H$, and pick a transversal. That is, write $H$ as a disjoint union:

$\displaystyle H=\biguplus\limits_ik_iK$

but then we can write the alternating sum

$\displaystyle\begin{aligned}H^-&=\sum\limits_{\pi\in H}\mathrm{sgn}(\pi)\pi\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i+\mathrm{sgn}(k_i(b\,c))k_i(b\,c)\right)\\&=\sum\limits_i\left(\mathrm{sgn}(k_i)k_i-\mathrm{sgn}(k_i)k_i(b\,c)\right)\\&=\sum\limits_i\mathrm{sgn}(k_i)k_i\left(1-(b\,c)\right)\\&=\left(\sum\limits_i\mathrm{sgn}(k_i)k_i\right)\left(1-(b\,c)\right)\end{aligned}$

as we stated.

Finally, if $t$ is some tableau with $b$ and $c$ in the same row, and if the swap $(b\,c)\in H$, then

$\displaystyle H^-\{t\}=0$

Our hypothesis tells us that $(b\,c)\{t\}=\{t\}$. We can thus use the above factorization to write

$\displaystyle\begin{aligned}H^-\{t\}&=k(1-(b\,c))\{t\}\\&=k\{t\}-k(b\,c)\{t\}\\&=k\{t\}-k\{t\}\\&=0\end{aligned}$

## 4 Comments	»

1. [...] of the Sign Lemma The results we showed last time have a few immediate consequences we will have use [...]

Pingback by | December 31, 2010 | Reply

2. [...] entries in tells us that we must have some pair of and in the same row of . Thus the swap . The sign lemma then tells us that . Since this is true for every summand of , it is true for [...]

Pingback by | January 18, 2011 | Reply

3. [...] we’ve used the sign lemma. So any two polytabloids coming from tableaux in the same column equivalence class are scalar [...]

Pingback by | January 21, 2011 | Reply

4. [...] the other hand, assume in the same column of . Then . But then the sign lemma tells us that is a factor of , and thus [...]

Pingback by | February 8, 2011 | Reply

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