# What is natural about the well-known bijection between conjugacy classes and irreps of a symmetric group?

Symmetric groups possess a well-known bijection between conjugacy classes and irreducible representations. More precisely, both sets are indexed by Young diagrams.

Question: To what extent is this bijection natural?

This question is somewhat informal, however I hope it can be left in such a form. One of the ways to make it formal is to ask about the properties which distinguish the bijection above among all bijections between the two sets. E.g. properties like: identity class is mapped to trivial irreps...

Background:

It is very standard material that for any finite group the number of conjugacy classes and irreps is the same, however it is striking that no natural bijection between the two sets is known (and may not exist at all).

MathOverflow has already several discussions around the subject, however they are different from the present question. A few of them are listed below.

Bijection between irreducible representations and conjugacy classes of finite groups discusses bijections for general groups.

The symmetric groups are not the only groups with a "natural" bijection. Some examples are collected here: Examples of finite groups with “good” bijection(s) between conjugacy classes and irreducible representations?

Related questions: G/[G,G], irreps and conjugacy classes, Duality between conjugacy classes and irreducible characters for finite monoids?, among others.

Conjugacy classes and irreps have many similar properties. However, in general, they do not fit each other exactly: Action of Out(G) on both sets, "reality properties" (Are there “real” vs. “quaternionic” conjugacy classes in finite groups?, If all real conjugacy classes are strongly real, then all real irreps are “strongly real”(symmetric), true ?).

• IIRC there is a method to directly associate to each conjugacy class in a Weyl group a corresponding irreducible representation. I don't remember the details though. Here is another related question: mathoverflow.net/questions/46900/… Jul 5 '14 at 17:15
• Since you already found the other threads, maybe you should explain why you're not satisfied with the answers given there. Jul 5 '14 at 17:16
• @QiaochuYuan What is IIRC ? JohannesHahn Other question are not about that. Jul 5 '14 at 17:20
• "If I recall correctly." Jul 5 '14 at 17:29
• @JohannesHahn The other threads ask different questions. 1) Is there any "bijection" for GENERAL finite groups 2) What are OTHER (not S_n) groups which have some "well-known"/"natural" bijection. So on. Jul 6 '14 at 9:51

## 2 Answers

Here's a way to define the usual specific bijection between Young diagrams and irreducible representations representations of the symmetric group. Since the bijection between Young diagrams (or, equivalently, partitions) and conjugacy classes is obvious, this will give an answer to your question, provided you can consider the following to be natural.

For any partition $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k$ of $n$ consider first the obvious action of the symmetric group $S_n$ on the set of partitions $(A_1,A_2,\dots,A_k)$ of the set $\{1,2,\dots,n\}$ such that each set $A_i$ has size $\lambda_i$. The corresponding linear action of $S_n$ (on the vector space of formal linear combinations of such partitions) is, in general, reducible. Write it as a sum of irreducible representations. So at this point we've associated to each partition of $n$ a list of irreducible representations. Now apply the following process to these lists.

If a partition has only one irreducible representation in its list, marry that irreducible representation to that partition. (A priori, there could be a danger of bigamy here, but see below.) Delete that irreducible representation from any other lists in which it occurs.

Because of the deletions, it is possible that some other partitions that previously had lists of more than one irreducible representation now have only one (or none, but again see below). If a partition now has exactly one irreducible representation remaining on its list, then marry them and remove that irreducible representation from any other lists in which it occurs. Repeat the process of marriages and deletions until there are no remaining unmarried parttions with exactly one irreducible representation in their lists.

A (nontrivial, as far as I can tell) theorem is that, at the end of the process, the marriages constitute a bijection (in fact the standard bijection) between partitions and irreducible representations of $S_n$. In particular, bigamy never occurs; no partition's list ever becomes empty; and every partition eventually gets married.

(The order in which partitions get married is, if I remember correctly, an extension of the dominance order to a linear pre-order.)

• Is it just a coincidence that the terminology you're using recalls the marriage problem, or is it because the entire situation can be re-interpreted in terms of it, and then the canonical matching of irreducible reps with conjugacy classes is just given by (one of) the usual algorithms that solve the marriage problem? Jul 12 '14 at 18:29
• @Niccolo' My use of marriage terminology was not intended to suggest any connection to the marriage theorem. I'm not aware of any connection, but of course that doesn't imply that no connection exists. Jul 13 '14 at 1:49

Here is how Frobenius set up the bijection when he computed the irreducible characters of $S_n$, but I don't know whether you would consider this "natural." Let $\phi_\lambda$ denote the action of $S_n$ on the left cosets of the Young subgroup $S_\lambda = S_{\lambda_1}\times S_{\lambda_2}\times\cdots$. Equivalently, $\phi_\lambda$ is the action of $S_n$ by permuting coordinates on all $n$-tuples with $\lambda_i$ $i$'s for all $i\geq 1$. Then there is a unique irreducible character $\chi^\lambda$ appearing in $\phi^\lambda$ (necessarily with multiplicity one) that does not appear in any $\phi_\mu$ with $\mu>\lambda$ (dominance order, as defined in http://en.wikipedia.org/wiki/Dominance_order).

• This seems to be rephrasing of Andreas' answer. Jan 12 '17 at 22:16