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