jump to navigation

Representations of the symmetric group September 20, 2009

Posted by Akhil Mathew in algebra, combinatorics, representation theory.
Tags: , , ,

I’ve now decided on future plans for my posts. I’m going to alternate between number theory posts and posts on other subjects, since I lack the focus have too many interests to want to spend all my blogging time on one area.

For today, I’m going to take a break from number theory and go back to representation theory a bit, specifically the symmetric group. I’m posting about it because I don’t understand it as well as I would like. Of course, there are numerous other sources out there—see for instance these lecture notes, Fulton and Harris’s textbook, Sagan’s textbook, etc.  Qiaochu Yuan has been posting on symmetric functions and may be heading for this area too, though if he does I’ll try to avoid overlapping with him; I think we have different aims anyway, so this should not be hard. 

Representation Theory of the Symmetric Group  

I summarized the basic general facts about group representation theory here, which we we will need here. For more complete details, see the references there to other mathematics blogs and this post .

The simple representations of the symmetric group {S_n} are indexed by the conjugacy classes of {S_n}, which correspond to the partitions of {n}. It turns out that we can explicitly construct a simple {L_{\pi}} associated to each {\pi}; this is the goal of today’s post. There are various equivalent ways of doing so; following Fulton and Harris and the arXived notes, I’ll use the Young projectors.

So, first of all, given the partition {\pi: n_1 + \dots + n_k = n}, we can assume {n_1 \geq n_2 \geq \dots}, and construct a Young diagram with {n_1} boxes in the first row, {n_2}, in the second, etc:

A Young diagram

Now insert the numbers from {1} to {n} in the boxes, in some order; the result is a Young tableau. It is clear that {S_n} acts on the set of Young tableaux. The Young projectors are constructed from a given tableau using the row and column stabilizer subgroups {P_{\pi}, Q_{\pi}} of {\tau}. Define

\displaystyle a_\pi = \sum_{\sigma \in P_{\pi}} \sigma , \quad b_{\pi} = \sum_{\sigma \in Q_{\pi}} (-1)^{\mathrm{sgn} \sigma} \sigma \in \mathbb{Q}[S_n].

The {S_n}-module {\mathbb{C}[S_n] a_{\pi}} can be described combinatorially as follows: say that two Young tableaux are equivalent if they have the same numbers in each row, and call an equivalence class a tabloid. Then {S_n} acts on the set of tabloids; the corresponding permutation representation is isomorphic to {\mathbb{C}[S_n] a_{\pi}}. Alternatively, this is

\displaystyle \mathrm{Ind}_{S_{n_1} \times \dots \times S_{n_k}}^{S_n} (1 \otimes \dots \otimes 1).

Finally, the Specht module is the submodule {L_{\pi} = \mathbb{C}[S_n] a_{\pi} b_{\pi}}. Up to isomorphism, this does not depend on the choice of tableau, since changing the tableau simply conjugates {P_{\pi}} and {Q_{\pi}}.  

Theorem 1 The Specht modules are non-isomorphic and simple. Every simple representation of {S_n} is isomorphic to one of them.  

Anyway, for now I’m going to leave the theorem as is without proof; in the next post I’ll prove it.   After that I will discuss how one computes the dimensions of these objects—there is an elegant “hook length formula” that gives the dimension from a glance at the Young diagram, and a Frobenius formula to compute the characters.

About these ads


1. Steven Sam - September 20, 2009

There’s an alternative approach which is pretty much the same thing as you mentioned in characteristic 0, but has the advantage of saying something in characteristic p.

We can build a representation corresponding to some partition \lambda of m of the general linear group using Schur functors over any base ring. In a characteristic 0 field this is irreducible, and the all 1′s weight space is an irrep for the mth symmetric group. In characteristic p, this need not be irreducible, but we can look at the submodule generated by the element which is a highest weight vector in the characteristic 0 setting. It turns out this is irreducible also, and the all 1′s weight space (when it exists) is an irrep for the symmetric group. In fact, all of them come about this way, and we can say precisely when the weight space exists (it’s when \lambda does not contain p columns of the same length, but I guess this depends on how you index representations, also).

This stuff is in the book _The Representation Theory of the Symmetric Groups_ by Gordon James (but I haven’t read it).

Akhil Mathew - September 20, 2009

Thanks for the comment!

I had a question: Schur functors are basically just multiplication by the up-to-a-scalar-idempotent c_{\pi} := a_{\pi} b_{\pi}, right? How do you define them in characteristic p, though, if c_{\pi}^2 = 0 in that case?

I’ve looked at Gordon’s book though I did not get that far into it.

Steven Sam - September 20, 2009

You can’t use the Young idempotent construction to get characteristic free Schur functors. There are two approaches (one usually called Schur functor, the second called Weyl functor, see the paper “Schur functors and Schur complexes” by Akin, Buchsbaum, Weyman, or chapter 2 of the book _Cohomology of Vector Bundles and Syzygies_ by Jerzy Weyman for more details):

For the Schur functor, start with your partition \lambda with m parts, and let \mu be its transpose (say it has n parts). Pick a module V. We have a map

\bigwedge^{\mu_1} V \otimes \cdots \otimes \bigwedge^{\mu_n} V \to Sym^{\lambda_1} V \otimes \cdots \otimes \cdots Sym^{\lambda_m} V

which is obtained by doing antisymmetrization of the exterior powers (which are thought of as “columns”), then projecting to the symmetric powers (along the “rows”). I’m not being precise but I think you know what I mean. Conversely, we can do the dual construction which goes from divided powers (cannot use symmetric powers) to exterior powers.

None of the maps involved use coefficients which aren’t integers, so it is well-defined for any ring. Furthermore, it can be shown that they have bases corresponding to semistandard Young tableaux is M was free to begin with (but this will not be irreducible for positive characteristic in general).

Akhil Mathew - September 23, 2009

Thanks for the reference and explanation.

(And sorry about the delayed response-I’ve fallen somewhat behind on this blog due to other commitments.)

2. Qiaochu Yuan - September 21, 2009

Perfect – I wasn’t planning on describing the Specht modules but I do have some stuff to say about Young’s lattice, so this post complements the direction I’m going in.

3. Harrison - September 24, 2009

On a tangentially related note, it’s good to know the relationship between the theory of partitions and the representation theory of the symmetric group even for basic problem-solving. From the ’97 Putnam:

Let N_n denote the number of ordered n-tuples (a_1, a_2, \ldots, a_n) satisfying 1/a_1 + \ldots + 1/a_n. Is N_{10} even or odd?

What’s interesting about this problem is that, even though it’s not immediately apparent that it’s related to the representation theory of S_n (and, indeed, I don’t think that it can even be phrased entirely in representation-theoretic language), you have to use some fairly nontrivial such machinery to have a hope of solving it.

Harrison - September 24, 2009

Ah, as usual I screwed up a formula — that should be 1/a_1 + \ldots + 1/a_n = 1.

Akhil Mathew - September 25, 2009

Hm, interesting, but how does the representation theory of S_n get involved though? S_{10} acts on the set N_{10}, but it’s not a vector space or anything like that. Do you use an orbit-counting lemma?

Harrison - September 27, 2009

Yeah, it’s just an orbit-counting problem, but you have to do a good bit of work to solve it.

harrison - September 27, 2009

To clarify, everything seems weaker than the representation theory, but not much weaker — in particular, if you work the problem in the partition-theory domain instead of dealing with subgroups and stabilizers directly, you see a bunch of the relevant ideas pop out.

4. ulimited hosting - August 5, 2013

Wow, this piece of writing is nice, my sister is analyzing such things, thus I am going to tell her.|

5. http://www.spunka.se/omspunks.asp?p=5 - November 25, 2013

As was the situation in Qubec, a brief line operator (Southern Ontario Railway), owned by an american keeping (Rail The usa), was involved. The TSB report around the incident states that : The accident occurred once the crew left the teach (carrying gasoline) unattended on a 1 % grade, with no the teach getting properly secured. Does that ring any bell?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


Get every new post delivered to your Inbox.

Join 39 other followers

%d bloggers like this: