# The Unapologetic Mathematician

## More Kostka Numbers

First let’s mention a few more general results about Kostka numbers.

Among all the tableaux that partition $n$, it should be clear that $(n)\triangleright\mu$. Thus the Kostka number $K_{(n)\mu}$ is not automatically zero. In fact, I say that it’s always $1$. Indeed, the shape is a single row with $n$ entries, and the content $\mu$ gives us a list of numbers, possibly with some repeats. There’s exactly one way to arrange this list into weakly increasing order along the single row, giving $K_{(n)\mu}=1$.

On the other extreme, $\lambda\triangleright(1^n)$, so $K_{\lambda(1^n)}$ might be nonzero. The shape is given by $\lambda$, and the content $(1^n)$ gives one entry of each value from $1$ to $n$. There are no possible entries to repeat, and so any semistandard tableau with content $(1^n)$ is actually standard. Thus $K_{\lambda(1^n)}=f^\lambda$ — the number of standard tableaux of shape $\lambda$.

This means that we can decompose the module $M^{(1^n)}$:

$\displaystyle M^{(1^n)}=\bigoplus\limits_{\lambda}f^\lambda S^\lambda$

But $f^\lambda=\dim(S^\lambda)$, which means each irreducible $S_n$-module shows up here with a multiplicity equal to its dimension. That is, $M^{(1^n)}$ is always the left regular representation.

Okay, now let’s look at a full example for a single choice of $\mu$. Specifically, let $\mu=(2,2,1)$. That is, we’re looking for semistandard tableaux of various shapes, all with two entries of value $1$, two of value $2$, and one of value $3$. There are five shapes $\lambda$ with $\lambda\trianglerighteq\mu$. For each one, we will look for all the ways of filling it with the required content.

$\displaystyle\begin{array}{cccc}\lambda=(2,2,1)&\begin{array}{cc}\bullet&\bullet\\\bullet&\bullet\\\bullet&\end{array}&\begin{array}{cc}1&1\\2&2\\3&\end{array}&\\\hline\lambda=(3,1,1)&\begin{array}{ccc}\bullet&\bullet&\bullet\\\bullet&&\\\bullet&&\end{array}&\begin{array}{ccc}1&1&2\\2&&\\3&&\end{array}&\\\hline\lambda=(3,2)&\begin{array}{ccc}\bullet&\bullet&\bullet\\\bullet&\bullet&\end{array}&\begin{array}{ccc}1&1&2\\2&3&\end{array}&\begin{array}{ccc}1&1&3\\2&2&\end{array}\\\hline\lambda=(4,1)&\begin{array}{cccc}\bullet&\bullet&\bullet&\bullet\\\bullet&&&\end{array}&\begin{array}{cccc}1&1&2&2\\3&&&\end{array}&\begin{array}{cccc}1&1&2&3\\2&&&\end{array}\\\hline\lambda=(5)&\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\end{array}&\begin{array}{ccccc}1&1&2&2&3\end{array}&\end{array}$

Counting the semistandard tableaux on each row, we find the Kostka numbers. Thus we get the decomposition

$\displaystyle M^{(2,2,1)}=S^{(2,2,1)}\oplus S^{(3,1,1)}\oplus2S^{(3,2)}\oplus2S^{(4,1)}\oplus S^{(5)}$

February 18, 2011 -

## 1 Comment »

1. Realy good, Complete: allows translation into programming (Mathematica).
Thanks, John.

Comment by Wouter | January 15, 2012 | Reply