The Unapologetic Mathematician

Mathematics for the interested outsider

Intertwinors from Semistandard Tableaux Span, part 3

Now we are ready to finish our proof that the intertwinors \bar{\theta}_T:S^\lambda\to M^\mu coming from semistandard generalized tableaux T span the space of all intertwinors between these modules.

As usual, pick any intertwinor \theta:S^\lambda\to M^\mu and write


Now define the set L_\theta to consist of those semistandard generalized tableaux S so that [S]\trianglelefteq[T] for some T appearing in this sum with a nonzero coefficient. This is called the “lower order ideal” generated by the T in the sum. We will prove our assertion by induction on the size of this order ideal.

If L_\theta is empty, then \theta must be the zero map. Indeed, our lemmas showed that if \theta is not the zero map, then at least one semistandard T shows up in the above sum, and this T would itself belong to L_\theta. And of course the zero map is contained in any span.

Now, if L_\theta is not empty, then there is at least some semistandard T with c_T\neq0 in the sum. Our lemmas even show that we can pick one so that [T] is maximal among all the tableaux in the sum. Let’s do that and define a new intertwinor:

\displaystyle \theta' = \theta - c_T\bar{\theta}_T

I say that L_{\theta'} is L_\theta with T removed.

Every S appearing in \bar{\theta}_T(e_t) has [S]\trianglelefteq[T], since if T is semistandard then [T] is the largest column equivalence class in \theta_T(\{t\}). Thus L_{\theta'} must be a subset of L_\theta since we can’t be introducing any new nonzero coefficients.

Our lemmas show that if [S]=[T], then c_S must appear with the same coefficient in both \theta(e_t) and c_T\bar{\theta}_T(e_t). That is, they must be cancelled off by the subtraction. Since T is maximal there’s nothing above it that might keep it inside the ideal, and so T\notin L_{\theta'}.

So by induction we conclude that \theta' is contained within the span of the \bar{\theta}_T generated by semistandard tableaux, and thus \theta must be as well.

February 14, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 3 Comments