Properties of Garnir Elements from Tableaux 1
Pick a Young tableau 
, and sets 
and 
as we did last time. If there are more entries in 
than there are in the 
th column of — the one containing — then 
. In particular, if we pick and by selecting a row descent, letting be the entries below the left entry, and letting be the entries above the right entry, then this situation will hold.
As a first step, I say that 
. That is, if we allow all the permutations of entries in these two sets (along with signs) then everything cancels out. Indeed, let 
be any column-stabilizing permutation. Our hypothesis on the number of entries in tells us that we must have some pair of 
and 
in the same row of 
. Thus the swap 
. The sign lemma then tells us that 
. Since this is true for every summand 
of 
, it is true for 
itself.
Now, our assertion is not that this is true for all of 
, but rather that it holds for our transversal 
. We use the decomposition
This gives us a factorization
And so we conclude that 
.
But now we note that 
. So if 
we use the sign lemma to conclude
Thus 
, and so
which can only happen if , as asserted.
This result will allow us to pick out a row descent in and write down a linear combination of polytabloids that lets us rewrite in terms of other polytabloids. And it will turn out that all the other polytabloids will be “more standard” than .
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January 18, 2011 -
Posted by John Armstrong |
Algebra, Representation Theory, Representations of Symmetric Groups
The Unapologetic Mathematician 
[…] for the last couple posts I’ve talked about using Garnir elements to rewrite nonstandard polytabloids — those […]
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[…] 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 […]
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There seems to be a typo at the end of the 1st sentence—shouldn’t it be
there?
I’m not sure I understand your question.
oops, I meant “at the end of the 2nd sentence”
oh, I dropped the “=0”, sorry.