# The Unapologetic Mathematician

## 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 $j$th 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 $e_t$.

## 6 Comments	»

1. [...] for the last couple posts I’ve talked about using Garnir elements to rewrite nonstandard polytabloids — those [...]

Pingback by | January 20, 2011 | Reply

2. [...] must be a row descent — we’ve ruled out column descents already — and so we can pick our Garnir element to write as the sum of a bunch of other polytabloids , where in the column dominance order. But [...]

Pingback by | January 21, 2011 | Reply

3. There seems to be a typo at the end of the 1st sentence—shouldn’t it be $g_{A,B}e_T=0$ there?

Comment by | April 3, 2011 | Reply

4. I’m not sure I understand your question.

Comment by | April 3, 2011 | Reply

• oops, I meant “at the end of the 2nd sentence”

Comment by | April 3, 2011 | Reply

5. oh, I dropped the “=0″, sorry.

Comment by | April 3, 2011 | Reply

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