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## Bijection between irreducible representations and conjugacy classes of finite groups

Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?

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I do not see how it is natural for $S_n$, $n>3$? Surely, you can use either a partition or its dual partition for a representation. Is there a mathematical reason to choose one over another or is it purely historical? – Bugs Bunny Jul 23 at 9:37
@Bugs Bunny: maybe one can argue that there are two natural choices in this case? It all boils down to interpreting the word "natural", of course. – Vladimir Dotsenko Jul 23 at 13:14
@Bugs It's fairly natural. For any partition $\lambda$ of $n$, let $S(\lambda)$ be the subgroup $\prod S_{\lambda_i}$ of $S_n$. Let $H(\lambda)$ be the induction of the trivial rep from $S(\lambda)$ to $S_n$. The characters of the $H(\lambda)$ are upper triangular in the characters of the irreps; the conjugacy classes pairs upper triangularly against the $H(\lambda)$. The pairing is forced by wanting these matrices to be upper triangular rather than a permutation of upper triangular. I'm not sure that there is no other pairing which achieves this, but most of them won't work. – David Speyer Jul 23 at 13:30
Is there any other case in math where bijection exists, but there is no natural one ? – Alexander Chervov Jul 24 at 7:30

This is a different take on Steven Landsburg's answer. The short version is that conjugacy classes and irreducible representations should be thought of as being dual to each other.

Fix an algebraically closed field $k$ of characteristic not dividing the order of our finite group $G$. The group algebra $k[G]$ is a finite-dimensional Hopf algebra, so its dual is also a finite-dimensional Hopf algebra of the same dimension; it is the Hopf algebra of functions $G \to k$, which I will denote by $C(G)$. (The former is cocommutative but not commutative in general, while the latter is commutative but not cocommutative in general.) The dual pairing $$k[G] \times C(G) \to k$$

is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing $$Z(k[G]) \times C_{\text{cl}}(G) \to k$$

on the subalgebras fixed by conjugation; $Z(k[G])$ is the center of $k[G]$ and $C_{\text{cl}}(G)$ is the space of class functions $G \to k$. Now:

The maximal spectrum of $Z(k[G])$ can be canonically identified with the irreducible representations of $G$, and the maximal spectrum of $C_{\text{cl}}(G)$ can be canonically identified with the conjugacy classes of $G$.

The second identification should be clear; the first comes from considering the central character of an irreducible representation. Now, the pairing above is nondegenerate, so to every point of the maximal spectrum of $Z(k[G])$ we can canonically associate an element of $C_{\text{cl}}(G)$ (the corresponding irreducible character) and to every point of the maximal spectrum of $C_{\text{cl}}(G)$ we can canonically associate an element of $Z(k[G])$ (the corresponding sum over a conjugacy class divided by its size).

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@Qiaochu: This is a nice abstract way to re-focus rigorously the original somewhat fuzzy question. But I still feel unable to treat concrete cases involving finite groups of Lie type in any explicit way: e.g., given the family $\mathrm{SL}_2(\mathbb{F}_p)$, how to prescribe an actual bijection (uniformly for all odd primes) and pass to the quotient group by the center as well? Even for $S_n$ I'm reminded that the original Springer Correspondence assigned characters/partitions to cohomology degrees of the flag variety in one way which later got dualized/transposed. – Jim Humphreys Jul 24 at 20:47
I'm not sure I understand your question. I'm not claiming that a bijection between irreducible representations and conjugacy classes exist. What this argument does is exhibit irreducible representations and conjugacy classes as canonical bases of two vector spaces which are canonically dual to each other. But these bases are not dual to each other, so I don't get a bijection this way. – Qiaochu Yuan Jul 24 at 22:39
@Qiaochu: I guess I was expecting a more explicit answer to the original question, such as "no" (?). My concern beyond that is whether your interesting higher level viewpoint adds any concrete detail about actual finite groups of interest, starting with symmetric groups. – Jim Humphreys Aug 2 at 18:08
@Jim: yes, my answer is essentially "no." That is, I'm suggesting that there shouldn't be a nice bijection for arbitrary finite groups. I think one way to see this is to note that there is no reasonable sense in which this is true for infinite groups (e.g. $\text{SU}(2)$ has countably many irreducible representations but uncountably many conjugacy classes). The correct generalization of the duality above is that by Peter-Weyl, for a compact Hausdorff group $G$ the characters of irreducible representations are dense in the space of class functions. – Qiaochu Yuan Aug 2 at 18:35

In general there is no natural bijection between conjugacy classes and irreducible representations of a finite group. To see this think of abelian groups for example. The conjugacy classes are the elements of the group, while the irreducible representations are elements of the dual group. These are isomorphic, via the Fourier transform, but not canonically.

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 Nevetheless among all set-theoretic bijections one has subclass of "good" bijections which are isomorphisms of the groups G and and G^dual. – Alexander Chervov Sep 16 at 14:42

Let $k$ be an algebraically closed field whose characteristic is either zero or prime to the order of $G$.

Then the center of the group ring $kG$ has one basis in natural bijective correspondence with the set of irreducible representations of $G$ over $k$, and another basis in natural bijective correspondence with the conjugacy classes of $G$.

Namely:

1) $kG$ is semisimple (this is called Maschke's Theorem) and Artinian, so it is a direct sum of matrix rings over division rings, hence (because $k$ is algebraically closed) a direct sum of matrix rings over $k$. There is (up to isomorphism) one irreducible representation for each of these matrix rings. Those representations are therefore in natural one-one correspondence with the central idempotents that generate those matrix rings, and these form a basis for the center.

2) For each conjugacy class, we can form the sum of all elements in that conjugacy class. The resulting elements of $kG$ form a basis for the center.

This gives a (non-natural) bijection between irreducible representations and conjugacy classes, because there is a (non-natural) bijection between any two bases for a given finite-dimensional $k$-vector space. I do not see any way you can make this natural.

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 This only works when $k$ is algebraically closed, and the result itself also only works in that generality. For example, over $\mathbb{R}$, the cyclic group of order $3$ has two irreducible representations but still has three conjugacy classes. The error in the argument is that, for a division ring $\Delta$ occuring in the decomposition of $k[G]$, the contribution of $M_n(\Delta)$ to $Z(k[G])$ is $Z(\Delta)$, which may be larger than $k$. – David Speyer Jul 22 at 22:47 David Speyer: Point taken; I've inserted the words "algebraically closed" where I ought to have inserted them in the first place. – Steven Landsburg Jul 22 at 23:30 May I suggest that you then change "matrix rings over division rings" to "matrix rings over $k$"? Since an algebraically closed field has nonnontrivial finite division ring extensions. – David Speyer Jul 23 at 10:08 David Speyer: Done. Thanks. – Steven Landsburg Jul 23 at 13:00

I would suggest that no such general natural bijection has been found to date. I am not sure how one would prove" that such a natural bijection could not be found, notwithstanding Gjergi's answer. I take the view that the equality between numbers of irreducible characters (over an algebraically closed field of characteristic zero) and the number of conjugacy classes is most naturally obtained by counting the dimension of the center of the group algebra in two different categorical settings: from the group theoretic perspective, the natural distinguished basis for the group algebra (the group elements) makes it clear that the dimension of the center is the number of conjugacy classes. On the other hand, from a ring-theoretic perspective, the structure of semi-simple algebras makes it clear that the dimension of the center of the group algebra is the number of isomorphism types of simple modules, that is, the number of irreducible characters. Moving to prime characteristic (still over an algebraically closed field, now of characteristic $p$, say), it is rather more difficult to prove, as R. Brauer did, that the number of isomorphism types of simple modules is the number of conjugacy classes of group elements of order prime to $p.$ However, there are contemporary conjectures in modular representation theory which suggest that there may one day be a different explanation for this equality. In particular, Alperin's weight conjecture suggests counting the number of (isomorphism types of) absolutely irreducible modules in characteristic $p$ in quite a different way, but one which still degenerates to the usual "non-natural" count when the characteristic $p$ does not divide the group order, which is essentially the same as the characteristic zero case. No general conceptual explanation for the conjectural count of Alperin has been found to date, though a number of approaches have been suggested, including a 2-category perspective. But it is not impossible that such an explanation could one day be found, and such an explanation might shed light even on the "easy" characteristic zero situation.

Later edit: In view of some of the comments below on the action of the automorphism group on irreducible characters and on conjugacy classes (which is really an action of the outer automorphism group, since inner automorphisms act trivially in each case), I make some comments on (well-known) properties of these actions, which while not identical, have many compatible features.

Brauer's permutation lemma states that for any automorphism $a$ of the finite group $G,$ the number of $a$-stable complex irreducible characters of $G$ is the same as the number of $a$-stable conjugacy classses. Hence any subgroup of ${\rm Aut}(G)$ has the same number of orbits on irreducible characters as it does on conjugacy classes. The Glauberman correspondence goes further with a group of automorphisms $A$ of order coprime to $|G|$. In that case the $A$-actions on the irreducible characters of $G$ and on the conjugacy classes of $G$ are permutation isomorphic.

While the actions of a general subgroup of the automorphism group are not always as strongly compatible as in the coprime case, various conjectures from modular representation theory suggest that it might be possible to have more compatibilty when dealing with complexes of modules than with individual modules. As a matter of speculation, I have sometimes wondered whether there might be some analogue of Glauberman correspondence in the non-coprime situation for actions on suitable complexes, although I have no idea for a precise formulation at present. Since the dimension of the center of an algebra is invariant under derived equivalence, this is one reason why I do not dismiss the idea of a more subtle explanation for numerical equalities.

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If I read the discussion at mathoverflow.net/questions/46900/… correctly, no bijection between the conjugacy classes and the irreducible representations respects the action of outer automorphisms. That seems like pretty convincing evidence to me. – Qiaochu Yuan Jul 22 at 21:53
mathoverflow.net/questions/21606/… seems to say the same thing. – Qiaochu Yuan Jul 22 at 21:54
@Qiaochu: It's a matter of opinion to some extent. That is some evidence, but I wouldn't consider it conclusive myself. – Geoff Robinson Jul 22 at 22:06
In my opinion, before I'd call a bijection natural, I'd want it to be invariant under isomorphisms of groups. That is, it should depend only on the group structure, not on what the specific elements of the group are. In particular, then, it should be invariant under arbitrary automorphisms, including outer ones. – Andreas Blass Jul 22 at 23:22
For what it's worth, even if the original straightforward version of the question has a somewhat negative answer (instead, duality...), as a general methodological attitude I endorse "not giving up tooo easily". Thus, I like Geoff R.'s points. Yet, dangit, there's that pesky factoid about (outer) automorphisms...?!? Ok, well, I myself "concede" that automorphisms' action on repns is natural in all ways that likely concern me (tho' I'm open to persuasion otherwise). So then the scope of that negative result is relevant. Still, "the orbit method"... and all that. Fun stuff... – paul garrett Jul 23 at 0:39
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Expanding slightly on the other answers:

To ask for a "natural" bijection is presumably to ask for a natural isomorphism between two functors from the category of finite groups to the category of sets. First, we have the contravariant functor $S$ that associates to each $G$ the set of isomorphism classes of irreducible representations. Then we have the covariant "functor" $T$ that associates to each $G$ the set of its conjugacy classes.

The first problem is that $T$ is not in fact functorial, because the image of a conjugacy class might not be a conjugacy class. So at the very least we should restrict to some subcategory on which $T$ is functorial, e.g. finite groups and surjective morphisms.

But the key problem still remains: There is no good way to define a natural transfomation between two functors of opposite variances. So when I said in my earlier answer that "I do not see any way you can make this natural" I might better have said "This is not a situation in which the notion of naturality makes sense".

All of this, of course, is really just an expansion of Gjergji's and Qiaochu's observations.

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 Representations can be induced (loosing irreducibleiity) so it also can thought as contr variant functor. There are many problems still exist but ..... – Alexander Chervov Sep 8 at 9:25 There is no hope for unique bijection in general but there are some "good" bijectionS in many examples although I do not know definitive characterization.... – Alexander Chervov Sep 8 at 9:32

Steven's and Gjergji answers points that there is no bijection, however possibly this idea should not be put into the rubbish completely.

Ideologically conjugacy classes and irreducible representations are somewhat dual to each other.

The other instances of this "duality" is Kirillov's orbit method - this is "infinitesimal version" of the duality: orbits in Lie algebra are infinitesimal versions of the conjugacy classes. But pay attention orbits are taken not in Lie algebra, but in the dual space g^. This again manifests that there irreps and conj. classes are dual to each other. However think of semi-simple Lie algebra - then g^ and g can be canonically identified...

Another instance is Langlands parametrization of the unitary irreducible representations of the real Lie group G. They are parametrized by conjugacy classes in Langlands dual group G^L. Again here are conjugacy classes in G^L, not in G itself. However for example GL=GL^L...

So it might be one should ask the question what are the groups such that conjugacy classes and irreps are in some natural bijection or something like this ?

PS

Here is some natural map conjugacy classes -> representations. But it does not maps to irreducible ones, and far from being bijection in general.

A colleague of mine suggested the following - take vector space of functions on a group which are equal to zero everywhere except given conjugacy class "C". We can act on these functions by $f \to g f g^{-1}$ - such action will preserve this class. So we get some representation. In the case of abelian group this gives trivial representation, however in general, it might be non-trivial. It always has trivial component - the function which is constant on "C".

I have not thought yet how this representation can be further decomposed, may be it is well-known ?

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 Take a finite abelian group $G$ and fix a non-degenerate pairing $G \times G \to \mathbb{C}^{\times}$. Unlike in the case of semisimple Lie algebras I do not see a canonical way to pick such a pairing. – Qiaochu Yuan Jul 22 at 19:21 @Qiaochu Yuan I agree. I did not pretend that there always exists some natural bijection. Just wanted to softly point out that "not giving up tooo easily", as Paul Garrett wrote in his comment above. E.g. Take G=Z/2Z in this case we may say there is natural bijection :) – Alexander Chervov Jul 23 at 7:39 @Geoff Sorry may be I misunderstanding however it seems my question was different. I do not take "class functions" (functions constant on conj. class (is it correct?)) but take NON constant on conj. class. Pay attention I am NOT acting by G in standadrd way - but act by conjugation f-> g f g^-1 hence non-constant functions are preserved by this action. This is a representation which clear have trivial one as a submodule (class function is trivial submodule). How this repr. is decomposed ? – Alexander Chervov Jul 24 at 9:50 E.g. take S_3 then conj. class corresponding to transpositions (just 3 elements) generaters standard 3D representation. We know it is trivial + 2D irreducible representation – Alexander Chervov Jul 24 at 9:54 OK Alexander. I did not read carefully enough. – Geoff Robinson Jul 24 at 10:31