Kostka Numbers
Now we’ve finished our proof that the intertwinors 
coming from semistandard tableauxspan the space of all intertwinors from the Specht module 
to the Young tabloid module 
. 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 in . And the dimension, of course, is the number of basis elements, which is the number of semistandard generalized tableaux of shape 
and content 
. This number we call the “Kostka number” 
. We’ve seen that there is a decomposition
Now we know that the Kostka numbers give these multiplicities, so we can write
We saw before that when 
, the multiplicity is one. In terms of the Kostka numbers, this tells us that 
. Is this true? Well, the only way to fit 
entries with value 
, 
with value 
, and so on into a semistandard tableau of shape is to put all the 
entries on the th row.
In fact, we can extend the direct sum by removing the restriction on :
This is because when 
we have 
. Indeed, we must eventually have 
, and so we can't fit all the entries with values through on the first rows of . 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 and content in this case.
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February 17, 2011 -
Posted by John Armstrong |
Algebra, Representation Theory, Representations of Symmetric Groups
[...] First let’s mention a few more general results about Kostka numbers. [...]
Pingback by More Kostka Numbers « The Unapologetic Mathematician | February 18, 2011 |
Your math is far beyond my comprehension, but I love reading this blog daily.