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 .