We continue our proof that the intertwinors that come from semistandard tableaux span the space of all such intertwinors. This time I assert that if is not the zero map, then there is some semistandard with .
Obviously there are some nonzero coefficients; if , then
which would make the zero map. So among the nonzero , there are some with maximal in the column dominance order. I say that we can find a semistandard among them.
By the results yesterday we know that the entries in the columns of these 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 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 and and the Garnir element just like before. We find
The generalized tableau must appear in with unit coefficient, so to cancel it off there must be some other generalized tableau with for some that shows up in . But since this just interchanges some and entries, we can see that , which contradicts the maximality of our choice of .
Thus there can be no row descents in , and is in fact semistandard.