The Unapologetic Mathematician

Mathematics for the interested outsider

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 ith 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.

About these ads

February 17, 2011 - Posted by | Algebra, Representation Theory, Representations of Symmetric Groups

2 Comments »

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

    Pingback by More Kostka Numbers « The Unapologetic Mathematician | February 18, 2011 | Reply

  2. Your math is far beyond my comprehension, but I love reading this blog daily.

    Comment by Obi Wan Canubi | February 19, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: