The Unapologetic Mathematician

Mathematics for the interested outsider

Permutation Representations from Partitions

Now that we have an action of S_n on the Young tabloids of shape \lambda\vdash n, we can consider the permutation representation M^\lambda that corresponds to it. Let’s consider a few examples.

First, let \lambda=(n). This is a pretty trivial “partition”, consisting of one piece of length n. The Ferrers diagram of \lambda looks like

\displaystyle\begin{array}{cccc}\bullet&\bullet&\cdots&\bullet\end{array}

Any Young tableau thus contains all n numbers on the single row, so they’re all row-equivalent. There is only one Young tabloid:

\displaystyle\begin{array}{cccc}\cline{1-4}1&2&\cdots&n\\\cline{1-4}\end{array}

We conclude that M^{(n)} is a one-dimensional vector space with the trivial action of S_n.

Next, let \lambda=(1^n) — another simple partition with n parts of length 1 each. The Ferrers diagram looks like

\displaystyle\begin{array}{c}\bullet\\\bullet\\\vdots\\\bullet\end{array}

Now in every Young tableau each number is on a different line, so no two tableaux are row-equivalent. They each give rise to their own Young tabloid, such as

\displaystyle\begin{array}{c}\cline{1-1}1\\\cline{1-1}2\\\cline{1-1}\vdots\\\cline{1-1}n\\\cline{1-1}\end{array}

These tabloids correspond to permutations; a generic one looks like

\displaystyle\begin{array}{c}\cline{1-1}\pi(1)\\\cline{1-1}\pi(2)\\\cline{1-1}\vdots\\\cline{1-1}\pi(n)\\\cline{1-1}\end{array}

The action of S_n on these tabloids is basically the same as left-multiplication on the underlying set S_n. And so we find the left regular representation.

Finally, consider the partition \lambda=(n-1,1). This time the Ferrers diagram looks like

\displaystyle\begin{array}{cccc}\bullet&\bullet&\cdots&\bullet\\\bullet&&&\end{array}

and a sample Young tabloid looks like

\displaystyle\begin{array}{cccc}\cline{1-4}1&2&\cdots&n-1\\\cline{1-4}n&&&\\\cline{1-1}\end{array}

Any Young tabloid of shape \lambda is uniquely specified by the single entry on the second row. Any permutation shuffles them around exactly like it does these entries, and so M^{(n-1,1)} is isomorphic to the defining representation.

December 14, 2010 Posted by | Algebra, Representation Theory, Representations of Symmetric Groups | 7 Comments

   

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