The Unapologetic Mathematician

Mathematics for the interested outsider

Properties of Garnir Elements from Tableaux 1

Pick a Young tableau t, and sets A and B as we did last time. If there are more entries in A\uplus B than there are in the jth column of t — the one containing A — then g_{A,B}e_t=0. In particular, if we pick A and B by selecting a row descent, letting A be the entries below the left entry, and letting B be the entries above the right entry, then this situation will hold.

As a first step, I say that {S_{A\uplus B}}^-e_t=0. That is, if we allow all the permutations of entries in these two sets (along with signs) then everything cancels out. Indeed, let \sigma\in C_t be any column-stabilizing permutation. Our hypothesis on the number of entries in A\uplus B tells us that we must have some pair of a\in A and b\in B in the same row of \sigma t. Thus the swap (a\,b)\in S_{A\uplus B}. The sign lemma then tells us that {S_{A\uplus B}}^-\{\sigma t\}=0. Since this is true for every summand \{\sigma t\} of e_t=C_t^-\{t\}, it is true for e_t itself.

Now, our assertion is not that this is true for all of S_{A\uplus B}, but rather that it holds for our transversal \Pi. We use the decomposition

\displaystyle S_{A\uplus B}=\biguplus\limits_{\pi\in\Pi}\pi(S_A\times S_B)

This gives us a factorization

\displaystyle{S_{A\uplus B}}^-=\Pi^-(S_A\times S_B)^-=g_{A,B}(S_A\times S_B)^-

And so we conclude that g_{A,B}(S_A\times S_B)^-e_t=0.

But now we note that S_A\times S_B\subseteq C_t. So if \sigma\in S_A\times S_B we use the sign lemma to conclude

\displaystyle\mathrm{sgn}(\sigma)\sigma e_t=\mathrm{sgn}(\sigma)\sigma C_t^-\{t\}=C_t^-\{t\}=e_t

Thus (S_A\times S_B)^-e_t=\lvert S_A\times S_B\rvert e_t, and so

\displaystyle 0=g_{A,B}(S_A\times S_B)^-e_t=g_{A,B}\lvert S_A\times S_B\rvert e_t=\lvert S_A\times S_B\rvert g_{A,B}e_t

which can only happen if g_{A,B}e_t=0, as asserted.

This result will allow us to pick out a row descent in t and write down a linear combination of polytabloids that lets us rewrite e_t in terms of other polytabloids. And it will turn out that all the other polytabloids will be “more standard” than e_t.

January 18, 2011 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 6 Comments

   

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