The Unapologetic Mathematician

Mathematics for the interested outsider

Intertwinors from Semistandard Tableaux Span, part 1

Now that we’ve shown the intertwinors that come from semistandard tableaux are independent, we want to show that they span the space \hom(S^\lambda,M^\mu). This is a bit fidgety, but should somewhat resemble the way we showed that standard polytabloids span Specht modules.

So, let \theta\in\hom(S^\lambda,M^\mu) be any intertwinor, and write out the image

\displaystyle\theta(e_t)=\sum\limits_Tc_TT

Here we’re implicitly using the fact that \mathbb{C}[T_{\lambda\mu}]\cong M^\mu.

First of all, I say that if \pi\in C_t and T_1=\pi T_2, then the coefficients of T_1 and T_2 differ by a factor of \mathrm{sgn}(\pi). Indeed, we calculate

\displaystyle\pi\left(\theta(e_t)\right)=\theta\left(\pi(\kappa_t\{t\})\right)=\theta(\mathrm{sgn}\kappa_t\{t\})=\mathrm{sgn}(\pi)\theta(e_t)

This tells us that

\displaystyle\pi\sum\limits_Tc_TT=\mathrm{sgn}(\pi)\sum\limits_Tc_TT

Comparing coefficients on the left and right gives us our assertion.

As an immediate corollary to this lemma, we conclude that if T has a repetition in some column, then c_T=0. Indeed, we can let \pi be the permutation that swaps the places of these two identical entries. Then T=\pi T, while the previous result tells us that c_T=\mathrm{sgn}c_T=-c_T, and so c_T=0.

February 11, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 2 Comments

   

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