Dominance for Generalized Tabloids

Sorry I forgot to post this yesterday afternoon.

You could probably have predicted this: we’re going to have orders on generalized tabloids analogous to the dominance and column dominance orders for tabloids without repetitions. Each tabloid (or column tabloid) gives a sequence of compositions, and at the $i$ th step we throw in all the entries with value $i$ .

For example, the generalized column tabloid

$\displaystyle[S]=\begin{array}{|c|c|c|}2&1&1\\3&2&\multicolumn{1}{c}{}\end{array}$

gives the sequence of compositions

$\displaystyle\begin{aligned}\lambda^1&=(0,1,1)\\\lambda^2&=(1,2,1)\\\lambda^3&=(2,2,1)\end{aligned}$

while the semistandard generalized column tabloid

$\displaystyle[T]=\begin{array}{|c|c|c|}1&1&1\\2&3&\multicolumn{1}{c}{}\end{array}$

gives the sequence of compositions

$\displaystyle\begin{aligned}\mu^1&=(1,1,1)\\\mu^2&=(2,1,1)\\\mu^3&=(2,2,1)\end{aligned}$

and we find that $[S]\trianglelefteq[T]$ since $\lambda^i\trianglelefteq\mu^i$ for all $i$ .

We of course have a dominance lemma: if $k<l$ , $k$ occurs in a column to the left of $l$ in $T$ , and $S$ is obtained from $T$ by swapping these two entries, then $[T]\triangleright[S]$ . As an immediate corollary, we find that if $T$ is semistandard and $S\in\{T\}$ is different from $T$ , then $[T]\triangleright[S]$ . That is, $[T]$ is the "largest" (in the dominance order) equivalence class in $\theta_T{t}$ . The proofs of these facts are almost exactly as they were before.

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February 9, 2011 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

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Pingback by Independence of Intertwinors from Semistandard Tableaux « The Unapologetic Mathematician | February 9, 2011 | Reply
[...] would make the zero map. So among the nonzero , there are some with maximal in the column dominance order. I say that we can find a semistandard among [...]

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