The Unapologetic Mathematician

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 $c_T=0$.

2 Comments	»

1. [...] continue our proof that the intertwinors that come from semistandard tableaux span the space of all such [...]

Pingback by | February 12, 2011 | Reply

2. [...] we are ready to finish our proof that the intertwinors coming from semistandard generalized tableaux span the space of all [...]

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