# The Unapologetic Mathematician

## Kostka Numbers

Now we’ve finished our proof that the intertwinors $\bar{\theta}_T$ coming from semistandard tableauxspan the space of all intertwinors from the Specht module $S^\lambda$ to the Young tabloid module $M^\mu$. We also know that they’re linearly independent, and so they form a basis of the space of intertwinors — one for each semistandard generalized tableau.

Since the Specht modules are irreducible, we know that the dimension of this space is the multiplicity of $S^\lambda$ in $M^\mu$. And the dimension, of course, is the number of basis elements, which is the number of semistandard generalized tableaux of shape $\lambda$ and content $\mu$. This number we call the “Kostka number” $K_{\lambda\mu}$. We’ve seen that there is a decomposition

$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}m_{\lambda\mu}S^\lambda$

Now we know that the Kostka numbers give these multiplicities, so we can write

$\displaystyle M^\mu=\bigoplus\limits_{\lambda\trianglerighteq\mu}K_{\lambda\mu}S^\lambda$

We saw before that when $\lambda=\mu$, the multiplicity is one. In terms of the Kostka numbers, this tells us that $K_{\mu\mu}=1$. Is this true? Well, the only way to fit $\mu_1$ entries with value $1$, $\mu_2$ with value $2$, and so on into a semistandard tableau of shape $\mu$ is to put all the $i$ entries on the $i$th row.

In fact, we can extend the direct sum by removing the restriction on $\lambda$:

$\displaystyle M^\mu=\bigoplus\limits_\lambda K_{\lambda\mu}S^\lambda$

This is because when $\lambda\triangleleft\mu$ we have $K_{\lambda\mu}=0$. Indeed, we must eventually have $\lambda_1+\dots+\lambda_i<\mu_1+\dots+\mu_i$, and so we can't fit all the entries with values $1$ through $i$ on the first $i$ rows of $\lambda$. We must at the very least have a repeated entry in some column, if not a descent. There are thus no semistandard generalized tableaux with shape $\lambda$ and content $\mu$ in this case.

February 17, 2011