# The Unapologetic Mathematician

## Inner and Outer Corners

The next thing we need to take note of it the idea of an “inner corner” and an “outer corner” of a Ferrers diagram, and thus of a partition.

An inner corner of a Ferrers diagram is a cell that, if it’s removed, the rest of the diagram is still the Ferrers diagram of a partition. It must be the rightmost cell in its row, and the bottommost cell in its column. Similarly, an outer corner is one that, if it’s added to the diagram, the result is still the Ferrers diagram of a partition. This is a little more subtle: it must be just to the right of the end of a row, and just below the bottom of a column.

As an example, consider the partition $(5,4,4,2)$, with Ferrers diagram

$\displaystyle\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\end{array}$

We highlight the inner corners by shrinking them, and mark the outer corners with circles:

$\displaystyle\begin{array}{cccccc}\bullet&\bullet&\bullet&\bullet&\cdot&\circ\\\bullet&\bullet&\bullet&\bullet&\circ&\\\bullet&\bullet&\bullet&\cdot&&\\\bullet&\cdot&\circ&&&\\\circ&&&&&\end{array}$

That is, there are three ways we could remove a cell and still have the Ferrers diagram of a partition:

$\displaystyle\begin{array}{cccc}\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&&\\\bullet&\bullet&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&&&&\end{array}$

And there are four ways that we could add a cell and still have the Ferrers diagram of a partition:

$\displaystyle\begin{array}{cccccc}\bullet&\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&&\\\bullet&\bullet&\bullet&\bullet&&\\\bullet&\bullet&&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&&\end{array}\qquad\begin{array}{ccccc}\bullet&\bullet&\bullet&\bullet&\bullet\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&\bullet&\bullet&\\\bullet&\bullet&&&\\\bullet&&&&\end{array}$

If the first partition is $\lambda$, we write a generic partition that comes from removing a single inner corner by $\lambda^-$. Similarly, we write a generic partition that comes from adding a single outer corner by $\lambda^+$. In our case, if $\lambda=(5,4,4,2)$, then the three possible $\lambda^-$ partitions are $(4,4,4,2)$, $(5,4,3,2)$, and $(5,4,4,1)$, while the four possible $\lambda^+$ partitions are $(6,4,4,2)$, $(5,5,4,2)$, $(5,4,4,3)$, and $(5,4,4,2,1)$.

Now, as a quick use of this concept, think about how to fill a Ferrers diagram to make a standard Young tableau. It should be clear that since $n$ is the largest entry in the tableau, it must be in the rightmost cell of its row and the bottommost cell of its column in order for the tableau to be standard. Thus $n$ must occur in an inner corner. This means that we can describe any standard tableau by picking which inner corner contains $n$, removing that corner, and filling the rest with a standard tableau with $n-1$ entries. Thus, the number of standard $\lambda$-tableaux is the sum of all the standard $\lambda^-$-tableaux:

$\displaystyle f^\lambda=\sum\limits_{\lambda^-}f^{\lambda^-}$

January 26, 2011

## 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.

January 26, 2011