The Unapologetic Mathematician

Mathematics for the interested outsider

The Dominance Lemma

We will have use of the following technical result about the dominance order:

Let t^\lambda and s^\mu be Young tableaux of shape \lambda and \mu, respectively. If for each row, all the entries on that row of s^\mu are in different columns of t^\lambda, then \lambda\trianglerighteq\mu. Essentially, the idea is that since all the entries on a row in s fit into different columns of t, the shape of t must be wide enough to handle that row. Not only that, but it’s wide enough to handle all of the rows of that width at once.

More explicitly, we can rearrange the columns of t so that all the entries in the first i rows of s fit into the first i rows of t. This is actually an application of the pigeonhole principle: if we have a column in t that contains i+1 elements from the first i rows of s, then look at which row each one came from. Since i+1>i, we must have two entries in the column coming from the same row, which we assumed doesn’t happen.

Yes, this does change the tableau t, but our conclusion is about the shape of t, which remains the same.

So now we can figure \lambda_1+\dots+\lambda_i as the number of entries in the first i rows of t^\lambda. Since these contain all the entries from the first i rows of s^\mu, it must be greater than or equal to that number. But that number is just as clearly \mu_1+\dots+\mu_i. Since this holds for all i, we conclude that \lambda dominates \mu.


December 20, 2010 - Posted by | Algebra, Group theory, Representation Theory, Representations of Symmetric Groups

1 Comment »

  1. […] 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 […]

    Pingback by Corollaries of the Sign Lemma « The Unapologetic Mathematician | December 31, 2010 | Reply

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