The Dominance Lemma for Tabloids

If $k<l$ , and $k$ appears in a lower row than $l$ in the Young tabloid $\{t\}$ , then $(k\,l)\{t\}$ dominates $\{t\}$ . That is, swapping two entries of $\{t\}$ so as to move the lower number to a higher row moves the tabloid up in the dominance relations.

Let the composition sequences of $\{t\}$ and $(k\,l)\{t\}$ be $\lambda^i$ and $\mu^i$ , respectively. For $i<k$ and $i\geq l$ we automatically have $\lambda^i=\mu^i$ . For $k\leq i<l$ there is a difference between the two: the entry $k$ has been added in a different place. Let $k$ and $l$ be in rows $q$ and $r$ of $\{t\}$ , respectively. In $\lambda^i$ , the entry $k$ is added to row $q$ , while in $\mu^i$ it’s been added to row $r$ . That is, $\lambda^i$ is the same as $\mu^i$ with part $q$ increased by one and part $r$ decreased by one. Our assumption that $k$ is in a lower row than $l$ in $\{t\}$ is that $q>r$ . Therefore, since the lower row in $\lambda^i$ is less than in $\mu^i$ , we find that $\lambda^i\triangleleft\mu^i$ . And we conclude that $\{t\}\trianglelefteq(k\,l)\{t\}$ , as asserted.

January 11, 2011 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

3 Comments »

[…] any such pair, the dominance lemma tells us that . That is, by “untwisting” the column inversion, we can move up the […]

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[…] one thing at least will make our life simpler: it should be clear that we still have a dominance lemma for column dominance. To be explicit: if , and appears in a column to the right of in the column […]

Pingback by The Column Dominance Order « The Unapologetic Mathematician | January 20, 2011 | Reply
[…] of course have a dominance lemma: if , occurs in a column to the left of in , and is obtained from by swapping these two […]

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