The Unapologetic Mathematician

Mathematics for the interested outsider

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}

December 29, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 6 Comments

   

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