# 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 -

1. […] defined a bunch of objects related to polytabloids. Let’s see how they relate to […]

Pingback by Permutations and Polytabloids « The Unapologetic Mathematician | December 24, 2010 | Reply

2. […] For any partition , the Specht module is the submodule of the Young tabloid module spanned by the polytabloids where runs over the Young tableaux of shape […]

Pingback by Specht Modules « The Unapologetic Mathematician | December 27, 2010 | Reply

3. […] is the alternating sum of the elements of . The proof basically runs the same as when we showed that where has shape […]

Pingback by The Sign Lemma « The Unapologetic Mathematician | December 29, 2010 | Reply

4. […] of shapes and , respectively, where and . If — where is the group algebra element we’ve defined — then dominates […]

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

5. […] and columns are all increasing sequences. In this case, we also say that the Young tabloid and the polytabloid are […]

Pingback by Standard Tableaux « The Unapologetic Mathematician | January 5, 2011 | Reply

6. […] to the dominance order on tabloids. Specifically, if is standard and appears as a summand in the polytabloid , then […]

Pingback by The Maximality of Standard Tableaux « The Unapologetic Mathematician | January 13, 2011 | Reply

7. […] Polytabloids are Independent Now we’re all set to show that the polytabloids that come from standard tableaux are linearly independent. This is half of showing that they form a […]

Pingback by Standard Polytabloids are Independent « The Unapologetic Mathematician | January 13, 2011 | Reply

8. […] We defined the Specht module as the subspace of the Young tabloid module spanned by polytabloids of shape . But these polytabloids are not independent. We’ve seen that standard polytabloids […]

Pingback by Standard Polytabloids Span Specht Modules « The Unapologetic Mathematician | January 21, 2011 | Reply

9. This looks plagiarized from Bruce E. Sagan’s book “The Symmetric Group,” pp. 61-62. Just how unapologetic are you?

Comment by John Doe | August 27, 2013 | Reply