# The Unapologetic Mathematician

## Standard Polytabloids Span Specht Modules

We defined the Specht module $S^\lambda$ as the subspace of the Young tabloid module $M^\lambda$ spanned by polytabloids of shape $\lambda$. But these polytabloids are not independent. We’ve seen that standard polytabloids are independent, and it turns out that they also span. That is, they provide an explicit basis for the Specht module $S^\lambda$.

Anyway, first off we can take care of all the $e_t$ where the columns of $t$ don’t increase down the column. Indeed, if $\sigma\in C_t$ stabilizes the columns of $t$, then

$\displaystyle e_{\sigma t}=\sigma e_t=\mathrm{sgn}(\sigma)e_t$

where we’ve used the sign lemma. So any two polytabloids coming from tableaux in the same column equivalence class are scalar multiples of each other. We’ve just cut our spanning set down from one element for each tableau to one for each column equivalence class of tableaux.

To deal with column equivalence classes, start with the tableau $t_0$ that we get by filling in the shape with the numbers $1$ to $n$ in order down the first column, then the second, and so on. This is the maximum element in the column dominance order, and it’s standard. Given any other tableau $t$, we assume that every tableau $s$ with $[s]\triangleright[t]$ is already in the span of the standard polytabloids. This is an inductive hypothesis, and the base case is taken care of by the maximum tabloid $t_0$.

If $t$ is itself standard, we’re done, since it’s obviously in the span of the standard polytabloids. If not, there must be a row descent — we’ve ruled out column descents already — and so we can pick our Garnir element to write $e_t$ as the sum of a bunch of other polytabloids $e_{\pi t}$, where $[\pi t]\triangleright[t]$ in the column dominance order. But by our inductive hypothesis, all these $e_{\pi t}$ are in the span of the standard polytabloids, and thus $e_t$ is as well.

As an immediate consequence, we conclude that $\dim(S^\lambda)=f^\lambda$, where $f^\lambda$ is the number of standard tableaux of shape $\lambda$. Further, since we know from our decomposition of the left regular representation that each irreducible representation of $S_n$ shows up in $\mathbb{C}[S_n]$ with a multiplicity equal to its dimension, we can write

$\displaystyle\mathbb{C}[S_n]=\bigoplus\limits_{\lambda\vdash n}f^\lambda S^\lambda$

Taking dimensions on both sides we find

$\displaystyle \sum\limits_{\lambda\vdash n}\left(f^\lambda\right)^2=\sum\limits_{\lambda\vdash n}f^\lambda\dim(S^\lambda)=\lvert S_n\rvert=n!$

January 21, 2011 -