The Unapologetic Mathematician

Mathematics for the interested outsider

Intertwinors from Semistandard Tableaux Span, part 2

We continue our proof that the intertwinors \bar{\theta}_T:S^\lambda\to M^\mu that come from semistandard tableaux span the space of all such intertwinors. This time I assert that if \theta\in\hom(S^\lambda,M^\mu) is not the zero map, then there is some semistandard T with c_T\neq0.

Obviously there are some nonzero coefficients; if \theta(e_t)=0, then

\displaystyle\theta(e_{\pi t})=\theta(\pi e_t)=\pi\theta(e_t)=0

which would make \theta the zero map. So among the nonzero c_T, there are some with [T] maximal in the column dominance order. I say that we can find a semistandard T among them.

By the results yesterday we know that the entries in the columns of these T are all distinct, so in the column tabloids we can arrange them to be strictly increasing down the columns. What we must show is that we can find one with the rows weakly increasing.

Well, let’s pick a maximal T and suppose that it does have a row descent, which would keep it from being semistandard. Just like the last time we saw row descents, we get a chain of distinct elements running up the two columns:


We choose the sets A and B and the Garnir element g_{A,B} just like before. We find

\displaystyle g_{A,B}\left(\sum\limits_Tc_TT\right)=g_{A,B}\left(\theta(e_t)\right)=\theta\left(g_{A,B}(e_t)\right)=\theta(0)=0

The generalized tableau T must appear in g_{A,B}(T) with unit coefficient, so to cancel it off there must be some other generalized tableau T'\neq T with T'=\pi T for some \pi that shows up in g_{A,B}. But since this \pi just interchanges some a and b entries, we can see that [T']\triangleright[T], which contradicts the maximality of our choice of T.

Thus there can be no row descents in T, and T is in fact semistandard.

February 12, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 1 Comment



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