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


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


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  5. Hi, I have a question, if Specht modules are cyclic, why do we need to find a basis? I know you have just explain that and maybe this is a basic question, but I have troubles with that idea. If one polytabloid generate all Specht module, then what is the point to find a basis?

    Comment by Mari | August 28, 2014 | Reply

  6. The module may be generated as an S_n-module by a single element, but not as a vector space.

    In fact, that’s exactly what makes the representation theory of (the group algebra of) S_n interesting while the representation theory of a field is boring: the only interesting thing about a field representation (vector space) is its dimension — the number of generators — and each generator behaves the same as every other. S_n-modules have a lot more structure than that.

    Comment by John Armstrong | August 28, 2014 | Reply

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