# The Unapologetic Mathematician

## Semistandard Generalized Tableaux

We want to take our intertwinors and restrict them to the Specht modules. If the generalized tableau $T$ has shape $\lambda$ and content $\mu$, we get an intertwinor $\bar{\theta}_T\in\hom(S^\lambda,M^\mu)$. This will eventually be useful, since the dimension of this hom-space is the multiplicity of $S^\lambda$ in $M^\mu$.

Anyway, if $t$ is our standard “reference” tableau, then we can calculate

\displaystyle\begin{aligned}\bar{\theta}_T(e_t)&=\bar{\theta}_T(\kappa_t\{t\})\\&=\kappa_t\theta_T(\{t\})\\&=\kappa_t\left(\sum\limits_{S\in\{T\}}S\right)\\&=\sum\limits_{S\in\{T\}}\kappa_t(S)\end{aligned}

We can see that it will be useful to know when $\kappa_t(S)=0$. It turns out this happens if and only if $S$ has two equal elements in some column.

Indeed, if $\kappa_t(S)=0$, then

$\displaystyle S+\sum\limits_{\substack{\pi\in C_t\\\pi\neq e}}\mathrm{sgn}(\pi)\pi S=0$

Thus for some $\sigma\in C_t$ with $\mathrm{sgn}(\sigma)=-1$ we must have $S=\sigma S$. But then we must have all the elements in each cycle of $\sigma$ the same, and these cycles are restricted to the columns. Since $\sigma$ is not the identity, we have at least one nontrivial cycle and at least two elements the same.

On the other hand, assume $S(i)=S(j)$ in the same column of $S$. Then $[e-(i\,j)](S)=0$. But then the sign lemma tells us that $(e-(i\,j))$ is a factor of $\kappa_t$, and thus $\kappa_t(S)=0$.

This means that we can eliminate some intertwinors $\bar{\theta}_T$ from consideration by only working with things like standard tableaux. We say that a generalized tableau is semistandard if its columns strictly increase (as for standard tableaux) and its rows weakly increase. That is, we allow repetitions along the rows, but only so long as we never have any row descents. The tableau

$\displaystyle\begin{array}{ccc}1&1&2\\2&3&\end{array}$

is semistandard, but

$\displaystyle\begin{array}{ccc}2&1&1\\3&2&\end{array}$

is not.

February 8, 2011