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


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


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


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


These tabloids correspond to permutations; a generic one looks like


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


and a sample Young tabloid looks like


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


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