# The Unapologetic Mathematician

## Young’s Natural Representation

Now that we have a canonical basis for our Specht modules composed of standard polytabloids it gives us a matrix representation of $S_n$ for each $\lambda\vdash n$. We really only need to come up with matrices for the swaps $(k\,k+1)$, for $1\leq k\leq n-1$, since these generate the whole symmetric group.

When we calculate the action of the swap on a polytabloid $e_t$ associated with a standard Young tableau $t$, there are three possibilities. Either $k$ and $k+1$ are in the same column of $t$, they’re in the same row of $t$, or they’re not in the same row or column of $t$.

The first case is easy. If $k$ and $k+1$ are in the same column of $t$, then $(k\,k+1)\in C_t$, and thus $(k\,k+1)e_t=-e_t$.

The third case isn’t much harder, although it’s subtler. I say that if $k$ and $k+1$ share neither a row nor a column, then $(k\,k+1)t$ is again standard. Indeed, swapping the two can’t introduce either a row or a column descent. The entries to the left of and above $k+1$ are all less than $k+1$, and none of them are $k$, so they’re all less than $k$ as well. Similarly, all the entries to the right of and below $k$ are greater than $k$, and none of them are $k+1$, so they’re all greater than $k+1$ as well.

Where things get complicated is when $k$ and $k+1$ share a row. But then they have to be next to each other, and the swap introduces a row descent between them. We can then use our Garnir elements to write this polytabloid in terms of standard ones.

Let’s work this out explicitly for the Specht module $S^{(2,1)}$, which should give us our well-known two-dimensional representation of $S_3$. The basis consists of the polytabloids associated to these two tableaux:

$\displaystyle\begin{array}{cc}1&2\\3&\end{array}\qquad\begin{array}{cc}1&3\\2&\end{array}$

We need to come up with matrices for the two swaps $(1\,2)$ and $(2\,3)$. And the second one is easy: it just swaps these two tableaux! Thus we get the matrix

$\displaystyle X^{(2,1)}\left((2\,3)\right)=\begin{pmatrix}0&1\\1&0\end{pmatrix}$

The action of $(1\,2)$ on the second standard tableau is similarly easy. Since $1$ and $2$ are in the same column, the swap acts by multiplying by $-1$. Thus we can write down a column of the matrix

$\displaystyle X^{(2,1)}\left((1\,2)\right)=\begin{pmatrix}?&0\\?&-1\end{pmatrix}$

As for the action on the first tableau, the swap induces a row descent. We use a Garnir element to straighten it out. With the same abuse of notation as last time, we write

$\displaystyle\begin{array}{cc}2&1\\3&\end{array}=\begin{array}{cc}1&2\\3&\end{array}-\begin{array}{cc}1&3\\2&\end{array}$

and so we can fill in the other column:

$\displaystyle X^{(2,1)}\left((1\,2)\right)=\begin{pmatrix}1&0\\-1&-1\end{pmatrix}$

From here we can write all the other matrices in the representation as products of these two.