Bijection between irreducible representations and conjugacy classes of finite groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T19:57:09Z http://mathoverflow.net/feeds/question/102879 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro Bijection between irreducible representations and conjugacy classes of finite groups Dan 2012-07-22T16:52:05Z 2012-07-24T10:28:05Z <p>Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?</p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102883#102883 Answer by Gjergji Zaimi for Bijection between irreducible representations and conjugacy classes of finite groups Gjergji Zaimi 2012-07-22T17:32:17Z 2012-07-22T17:32:17Z <p>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.</p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102884#102884 Answer by Steven Landsburg for Bijection between irreducible representations and conjugacy classes of finite groups Steven Landsburg 2012-07-22T17:34:04Z 2012-07-23T13:00:03Z <p>Let $k$ be an algebraically closed field whose characteristic is either zero or prime to the order of $G$.</p> <p>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 $G$. </p> <p>Namely: </p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102890#102890 Answer by Alexander Chervov for Bijection between irreducible representations and conjugacy classes of finite groups Alexander Chervov 2012-07-22T19:06:37Z 2012-07-23T10:52:23Z <p>Steven's and Gjergji answers points that there is no bijection, however possibly this idea should not be put into the rubbish completely.</p> <p>Ideologically conjugacy classes and irreducible representations are somewhat <strong>dual</strong> to each other.</p> <p>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^<em>. This again manifests that there irreps and conj. classes are dual to each other. However think of semi-simple Lie algebra - then g^</em> and g can be canonically identified... </p> <p>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...</p> <p>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 ? </p> <p>PS</p> <p><strong>Here is some natural map conjugacy classes -> representations</strong>. But it does not maps to irreducible ones, and far from being bijection in general.</p> <p>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".</p> <p>I have not thought yet how this representation can be further decomposed, may be it is well-known ?</p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102894#102894 Answer by Qiaochu Yuan for Bijection between irreducible representations and conjugacy classes of finite groups Qiaochu Yuan 2012-07-22T19:35:24Z 2012-07-22T19:35:24Z <p>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. </p> <p>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 $$k[G] \times C(G) \to k$$</p> <p>is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing $$Z(k[G]) \times C_{\text{cl}}(G) \to k$$</p> <p>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:</p> <blockquote> <p>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 $G$.</p> </blockquote> <p>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). </p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102896#102896 Answer by Geoff Robinson for Bijection between irreducible representations and conjugacy classes of finite groups Geoff Robinson 2012-07-22T20:09:49Z 2012-07-24T10:28:05Z <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gro/102948#102948 Answer by Steven Landsburg for Bijection between irreducible representations and conjugacy classes of finite groups Steven Landsburg 2012-07-23T17:11:52Z 2012-07-23T18:28:54Z <p>Expanding slightly on the other answers:</p> <p>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.</p> <p>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 $T$ <b>is</b> functorial, e.g. finite groups and <b>surjective</b> morphisms.</p> <p>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".</p> <p>All of this, of course, is really just an expansion of Gjergji's and Qiaochu's observations.</p>