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.

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

3 Comments »

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

Comment by John Armstrong | February 15, 2011 | Reply
[…] 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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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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