## Corollaries of the Sign Lemma

The results we showed last time have a few immediate consequences we will have use of.

First, let and are two Young tableaux of shapes and , respectively, where and . If — where is the group algebra element we’ve defined — then dominates .

To see this, let and be two entries in the same row of . They cannot be in the same column of , since if they were then the swap would be in the column-stabilizer . Then we could conclude that , which we assumed not to be the case. But if no two entries from the same row of are in the same column of , the dominance lemma tells us that .

Now if it turns out that it’s not surprising that . Luckily in that situation we can say something interesting:

Indeed, we must have for some , basically by the same reasoning that led to the dominance lemma in the first place. Indeed, the thing that would obstruct finding such a is having two entries in some column of needing to go on the same row of , which we know doesn’t happen. And so we calculate

Now if is any vector in the Specht module, and if is a tableau of shape , then is some multiple of . Indeed, we can write

were the are -tableaux. For each one of these, we have . Thus we find

which is a multiple of , as asserted.

[…] see this, let be any vector, and let be a -tableau. Our last result showed us that for some . First, we’ll assume that there is some pair of and so that . In […]

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[…] Now, the can’t all be zero, so we must have at least one -tableau so that . But then our corollary of the sign lemma tells us that , as we […]

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