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

\displaystyle\theta(e_t)=\sum\limits_Tc_TT

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.

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

3 Comments »

  1. You know what? This makes no sense to me. I mean, it shouldn’t, as I don’t have the prerequisite knowledge to formulate an understanding…

    … but isn’t that a HUGE issue with mathematics? Why can’t this information be explained in laymen’s terms? Isn’t that the point of this blog?

    Why do mathematicians constantly shroud their art form?

    Comment by Haakon | February 15, 2011 | Reply

  2. Haakon, have you tried tracing back the links to the more foundational material?

    Comment by John Armstrong | February 15, 2011 | Reply

  3. [...] Now we’ve finished our proof that the intertwinors coming from semistandard tableauxspan the space of all [...]

    Pingback by Kostka Numbers « The Unapologetic Mathematician | February 17, 2011 | Reply


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