The Unapologetic Mathematician

Mathematics for the interested outsider

The Road Forward

We’ve been talking a lot about the general theory of finite group representations. But our goal is to talk about symmetric groups in particular. Now, we’ve seen that the character table of a finite group is square, meaning there are as many irreducible representations of a group G as there are conjugacy classes K\subseteq G. But we’ve also noted that there’s no reason to believe that these have any sort of natural correspondence.

But for the symmetric group S_n, there’s something we can say. We know that conjugacy classes in symmetric groups correspond to cycle types. Cycles correspond to integer partitions of n. And from a partition we will build a representation.

For a first step, let n=\lambda_1+\lambda_2+\dots+\lambda_k be a partition, with \lambda_1\geq\lambda_2\geq\dots\geq\lambda_k. We can use this to come up with a subgroup S_\lambda\subseteq S_n. Given a set X we will write S_X for the group of permutations of that set. For example S_{\{1,\dots,\lambda_1\}} permutes the first \lambda_1 positive integers, and S_{\{\lambda_1+1,\dots,\lambda_2\}} permutes the next \lambda_2 of them. We can put a bunch of these groups together to build

\displaystyle S_\lambda=S_{\{1,\dots,\lambda_1\}}\times S_{\{\lambda_1+1,\dots,\lambda_2\}}\times\dots\times S_{\{\lambda_{k-1}+1,\dots,\lambda_k\}}

Elements of S_\lambda permute the same set as S_n, and so S_\lambda\subseteq S_n, but only in certain discrete chunks. Numbers in each block can be shuffled arbitrarily among each other, but the different blocks are never mixed. Really, all that matters is that the chunks have sizes \lambda_1 through \lambda_k, but choosing them like this is a nicely concrete way to do it.

So, now we can define the S_n-module M^\lambda=1\!\!\uparrow_{S_\lambda}^{S_n} by inducing the trivial representation from the subgroup S_\lambda to all of S_n. Now, the M^\lambda are not all irreducible, but we will see how to identify a particular irreducible submodule S^\lambda\subseteq M^\lambda of each one, and the S^\lambda will all be distinct. Since they correspond to partitions \lambda, there are exactly as many of them as there are conjugacy classes in S_n, and so they must be all the irreducible S_n-modules, up to isomorphism.

December 7, 2010 - Posted by | Algebra, Group theory, Representation Theory, Representations of Symmetric Groups

1 Comment »

  1. […] It doesn’t really matter which we pick; any two tableaux in the same orbit — and they’re all in the same single orbit — have isomorphic stabilizers. But like we mentioned last time the usual choice lists the numbers from to on the first row, from to on the second row, and so on. We write for the stabilizer of this choice, and this is the subgroup of we will use. Notice that this is exactly the same subgroup we described earlier. […]

    Pingback by Dimensions of Young Tabloid Modules « The Unapologetic Mathematician | December 16, 2010 | Reply


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