The Unapologetic Mathematician

Independence of Intertwinors from Semistandard Tableaux

Let’s start with the semistandard generalized tableaux $T\in T_{\lambda\mu}^0$ and use them to construct intertwinors $\bar{\theta}_T:\hom(S^\lambda,M^\mu)$. I say that this collection is linearly independent.

Indeed, let’s index the semistandard generalized tableaux as $T_1,\dots,T_m$. We will take our reference tableau $t$ and show that the vectors $\bar{\theta}_{T_i}(e_t)\in M^\mu$ are independent. This will show that the $\bar{\theta}_{T_i}$ are independent, since any linear dependence between the operators would immediately give a linear dependence between the $\bar{\theta}_{T_i}(v)$ for all $v\in S^\lambda$.

Anyway, we have

$\displaystyle\bar{\theta}_{T_i}(e_t)=\theta_{T_i}\left(\kappa_t\{t\}\right)=\kappa_t\theta_{T_i}(\{t\})$

Since we assumed $T_i$ to be semistandard, we know that $[T_i]\triangleright[S]$ for all summands $S\in\theta_{T_i}(\{t\})$. Now the permutations in $\kappa_t$ do not change column equivalence classes, so this still holds: $[T_i]\triangleright[S]$ for all summands $S\in\kappa_t\theta_{T_i}(\{t\})$. And further all the $[T_i]$ are distinct since no column equivalence class can contain more than one semistandard tableau.

But now we can go back to the lemma we used when showing that the standard polytabloids were independent! The $\kappa_t\theta_{T_i}(\{t\})=\bar{\theta}(e_t)$ are a collection of vectors in $M^\mu$. For each one, we can pick a basis vector $[T_i]$ which is maximum among all those having a nonzero coefficient in the vector, and these selected maximum basis elements are all distinct. We conclude that our collection of vectors in independent, and then it follows that the intertwinors $\bar{\theta}_{T_i}$ are independent.

February 9, 2011

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, $k$ 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.

February 9, 2011