The Unapologetic Mathematician

Polytabloids

Given any collection $H\subseteq S_n$ of permutations, we define two group algebra elements.

\displaystyle\begin{aligned}H^+&=\sum\limits_{\pi\in H}\pi\\H^-&=\sum\limits_{\pi\in H}\mathrm{sgn}(\pi)\pi\end{aligned}

Notice that $H$ doesn’t have to be a subgroup, though it often will be. One particular case that we’ll be interested in is

$\displaystyle\kappa_t=C_t^-=\sum\limits_{\pi\in C_t}\mathrm{sgn}(\pi)\pi$

where $C_t$ is the column-stabilizer of a Young tableau $t$. If $t$ has columns $C_1,\dots,C_k$, then $C_t=S_{C_1}\times\dots\times S_{C_k}$. Letting $\pi$ run over $C_t$ is the same as letting $\pi_i$ run over $S_{C_i}$ for each $i$ from $1$ to $k$. That is,

\displaystyle\begin{aligned}\kappa_t&=\sum\limits_{\pi_1\in S_{C_1}}\dots\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_1\dots\pi_k)\pi_1\dots\pi_k\\&=\sum\limits_{\pi_1\in S_{C_1}}\dots\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_1)\dots\mathrm{sgn}(\pi_k)\pi_1\dots\pi_k\\&=\left(\sum\limits_{\pi_1\in S_{C_1}}\mathrm{sgn}(\pi_1)\pi_1\right)\dots\left(\sum\limits_{\pi_k\in S_{C_k}}\mathrm{sgn}(\pi_k)\pi_k\right)\end{aligned}

so we have a nice factorization of this element.

Now if $t$ is a tableau, we define the associated “polytabloid”

$\displaystyle e_t=\kappa_t\{t\}$

Now, as written this doesn’t really make sense. But it does if we move from just considering Young tabloids to considering the vector space of formal linear combinations of Young tabloids. This means we use Young tabloids like basis vectors and just “add” and “scalar multiply” them as if those operations made sense.

As an example, consider the tableau

$\displaystyle t=\begin{array}{ccc}4&1&2\\3&5&\end{array}$

Our factorization lets us write

$\displaystyle\kappa_t=\left(e-(3\,4)\right)\left(e-(1\,5)\right)$

And so we calculate

$\displaystyle e_t=\begin{array}{ccc}\cline{1-3}4&1&2\\\cline{1-3}3&5&\\\cline{1-2}\end{array}-\begin{array}{ccc}\cline{1-3}3&1&2\\\cline{1-3}4&5&\\\cline{1-2}\end{array}-\begin{array}{ccc}\cline{1-3}4&5&2\\\cline{1-3}3&1&\\\cline{1-2}\end{array}+\begin{array}{ccc}\cline{1-3}3&5&2\\\cline{1-3}4&1&\\\cline{1-2}\end{array}$

Now, the nice thing about $e_t$ is that if we hit it with any permutation $\pi\in C_t$, we get $\pi e_t=\mathrm{sgn}(\pi)e_t$.

December 23, 2010