The Unapologetic Mathematician

Mathematics for the interested outsider

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)}

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February 18, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups

1 Comment »

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

    Comment by Wouter | January 15, 2012 | Reply


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