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 - Posted by John Armstrong | Algebra, Representation Theory, Representations of Symmetric Groups

No comments yet.

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects

Subjects
Archives

January 2011

M T W T F S S

1 2

3 4 5 6 7 8 9

10 11 12 13 14 15 16

17 18 19 20 21 22 23

24 25 26 27 28 29 30

31

« Dec Feb »
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider