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Consider a finite group G. The product of conjugacy classes can be defined in natural way just by multiplying the representatives and counting multiplicities (see e.g. MO 62088). So we get ring with a basis and structure constants are natural numbers. Similar to what one has for product of irreps. There are many analogies between conjugacy classes and irreps in particular see this article.

Tanaka-Krein duality states that group can be reconstructed from the tensor category of its representations which is semisimple for finite groups, and hence carries the same information as ring + basis of irreps.

Question: Can one reconstruct a group having (ring + basis) made of conjugacy classes ?

If not - what partial information (e.g. character table) one can get ?


Question: Is there any relation between this ring and ring of irreps of the same group ? or may be some other group ?

(Remark. For abelian group they are isomorphic.)

Question: Are there any further analogies between ring of irreps and conjugacy classes except mentioned in the paper cited above ?

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Both rings are lattices in the center of the group algebra $\mathbb C[G]$. So that's a pretty strong relation between them. There's also a duality between the two lattices, given by the character table / multiplication and the counit. I don't think the ring tells you the category structure, because it doesn't tell you the $\operatorname{Sym}$s and $\wedge$s. The ring depends only on the character table, which is not a complete group invariant. –  Will Sawin Oct 27 '12 at 20:49
    
Why only on char. table ? –  Alexander Chervov Oct 27 '12 at 20:55
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It is my understanding that the ring of conjugacy classes (with basis) carries exactly the same information as character table which in turn is the same information as character ring (with basis). On the other hand the tensor category of representations carries strictly more information: for example nonabelian groups of order 8 have the same character tables (and so the same character rings and conjugacy classes rings) but inequivalent tensor categories of representations (even if one ignores the commutativity constraint). –  Victor Ostrik Oct 27 '12 at 20:56
    
May be I misunderstanding TanakaKrein.... Still why you both so quickly say that "the same info as char table " is it obvious? –  Alexander Chervov Oct 27 '12 at 21:00
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Because you add representations by adding their characters, you can multiply representations by multiplying their characters, and you can check isomorphism of representations by checking equality of their characters. So the representation ring is the ring generated by the rows of the character table under entrywise addition and multiplication. –  Will Sawin Oct 27 '12 at 21:06

5 Answers 5

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The answer to your first question is negative. For a concrete example, you can show that the conjugacy class rings of the nonisomorphic groups $Q_8$ and $D_8$ are isomorphic, via an isomorphism that pairs off the bases as follows: $[1] \leftrightarrow [1]$, $[-1] \leftrightarrow [r^2]$, $[i] \leftrightarrow [r]$, $[j] \leftrightarrow [s]$ and $[k] \leftrightarrow [rs]$.

As to your question about the relationship between the conjugacy class ring and the character ring, there are lots of partial results that can be stated. Nonetheless, the answer to the question of when these two rings are isomorphic is completely known. This turns out to be the case if and only if the group is $p$-nilpotent with abelian Sylow $p$-subgroup. More generally, for arbitrary finite groups $G$ and $G'$, the following two conditions are equivalent.

  1. The character ring of $G$ is isomorphic to the conjugacy class ring of $G'$.

  2. $G$ and $G'$ are $p$-nilpotent groups with abelian Sylow $p$-subgroups. Moreover, if $g_1, \dots, g_l$ and $g_1',\ldots, g_{l'}'$ are complete sets of representatives for the conjugacy classes of $p'$-elements of $G$ and $G'$, resp., and if $D_i$ and $D_i'$ are Sylow $p$-subgroups of $C_G(g_i)$ and $C_{G'}(g_i')$, resp., then $l=l'$ and $D_i \cong D_i'$.

This is due to Saksonov, The ring of classes and the ring of characters of a finite group. Mat. Zametki 26 (1979), no. 1, 3–14, 156.

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Wow ! That is very interesting ! Thank you very much ! –  Alexander Chervov Oct 28 '12 at 7:45
    
mathnet.ru/php/… Russian pdf for free from mathnet.ru –  Alexander Chervov Oct 28 '12 at 12:49
    
springerlink.com/index/Q171286388N1262U.pdf English pdf also free –  Alexander Chervov Oct 28 '12 at 13:06

You can construct the ring of conjugacy classes from the character table in the following manner:

First note that the ring of conjugacy classes tensored with $\mathbb C$ can be identified with assignments of a number to each irrep, or column vectors, with entrywise addition and multiplication. So we have to find a basis. For each column vector of the form:

$\left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right)$

we know how to write it in terms of conjugacy classes: as the corresponding row in the character table, divided by the order of the group.

So we invert that matrix to find out how to write the conjugacy classes in terms of irreps.

Concretely, the ring of conjugacy classes is the ring generated by the columns of the inverse transpose of the character table times the order of the group under entrywise multiplication.

Thus, it gives no more information than the character table.

Edit: It also gives no less information than the character table. The reason is that one can reconstruct the center of the group algebra from this ring by tensoring with $\mathbb C$. One then has one minimal idempotent for each irreducible representation. We can find the minimal idempotents algebraically and compute how to write them in terms of the conjugacy classes, which tells us the character table.

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Thank you very much ! –  Alexander Chervov Oct 28 '12 at 7:49
    
@Will, honestly speaking I do not fully understand your answer, may be just bad day for me. What you mean by "First note that the ring of conjugacy classes tensored with C can be identified with assignments of a number to each irrep" –  Alexander Chervov Oct 28 '12 at 16:51
    
That ring is the center of the group algebra, which has a basis of idempotents, one for each irrep. Write an element in terms of this basis to identify the ring with the ring of functions $\\{irreps\\} \to \mathbb C$, aka column vectors, with entrywise addition and multiplication. –  Will Sawin Oct 28 '12 at 18:42
    
Functions from the set of irreducible representations to $\mathbb C$ –  Will Sawin Oct 28 '12 at 18:43
    
Ahhh, I misread - I thought you mean: conjuagacy class = irrep + number... –  Alexander Chervov Oct 28 '12 at 18:52

What Tannaka-Krein duality actually states is that a (say, finite) group $G$ can be reconstructed from the monoidal category $\text{Rep}(G)$ together with the forgetful functor $\text{Rep}(G) \to \text{Vect}$. This functor is absolutely crucial to the story; $G$ is reconstructed as its group of automorphisms (as a monoidal functor).

The data of $\text{Rep}(G)$ as a monoidal category alone is not enough to reconstruct $\text{Rep}(G)$. Its structure as a category is completely determined by the number of irreducible representations of $G$ (so it only knows the number of conjugacy classes of $G$) and the monoidal operation is completely determined by the multiplicities of tensor products. Both of these pieces of information can be read off of the character table, so $\text{Rep}(G)$ does not distinguish groups with the same character table. See Victor Ostrik's comments below.

If you don't want to use forgetful functors, you need more data. If I understand correctly, the Doplicher-Roberts reconstruction theorem asserts that you can recover $G$ from $\text{Rep}(G)$ as a "symmetric tensor $\ast$-category." See Müger's appendix for a thorough proof and discussion.


As for conjugacy classes, the "conjugacy class ring" is precisely the center of the group ring $\mathbb{Z}[G]$. The center of the group algebra $\mathbb{C}[G]$ has two distinguished choices of basis, one given by sums over conjugacy classes and one given by its primitive idempotents, which are in natural bijection with the irreducible representations of $G$.

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Actually the monoidal category $Rep(G)$ is not determined by the character ring of $G$. Namely, the character ring determines the category and the monoidal operation but it says nothing about the associativity isomorphisms. For example the groups $Q_8$ and $D_8$ have the same character rings but their representation categories are not monoidally equivalent. –  Victor Ostrik Oct 28 '12 at 0:20
    
@Victor: my mistake! That's interesting. How does one go about proving that two monoidal categories are not monoidally equivalent? And do you have an example of nonisomorphic groups with equivalent monoidal categories of representations? –  Qiaochu Yuan Oct 28 '12 at 1:12
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@Qiaochu: the fact that categories $Rep(Q_8)$ and $Rep(D_8)$ are not monoidally equivalent was originally established by Tambara and Yamagami via explicit classification of possible associativity constraints. Another approach is via counting: the category $Rep(Q_8)$ has a unique (up to isomorphism) monoidal functor to the vector spaces and the category $Rep(D_8)$ has 3 isomorphism classes of such functors. Examples of non-isomorphic groups with monoidally equivalent representation categories do exist, see e.g. paper "Isocategorical groups" by Etingof and Gelaki. –  Victor Ostrik Oct 28 '12 at 1:25
    
It's easy to see if you add in the symmetry isomorphisms, because then you can get syms and wedges, which tell you which elements are powers of which other elements. –  Will Sawin Oct 28 '12 at 3:13
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Joyal and Street's excellent 1991 paper on Tannakian reconstruction discusses some of these variations. The Doplicher–Roberts version is trying to reconstruct a real group (or monoid), whereas the original Deligne version reconstructs an algebraic group over C. From the algebraic perspective, the difference is the data of a Gal(R)=Z/2 action, which is encoded by the star-structure. –  Theo Johnson-Freyd Oct 28 '12 at 18:27

As suggested in the comments, it has been known almost since the beginning of the representation theory of finite groups that knowledge of the character table is equivalent to the knowledge of the "class algebra constants", which are the structure constants telling you the multiplicity with which the class sum $C_{z}$ occurs in the product of class sums $C_{x}C_{y}.$ A formula of Burnside tells us that this multiplicity is $\frac{|G|}{|C_{G}(x)||C_{G}(y)|} \sum_{\chi} \frac{\chi(x)\chi(y) \chi (z^{-1})}{\chi(1)}$, where $\chi$ runs through the comples irreducible characters of $G.$ In the other direction, the centre of the complex group algebra has a basis consisting of primitive idempotents. There are several standard ways to recover this basis given the explicit knowledge of multiplication in the algebra, which is given by the class algebra constants. These primitive idempotents are in bijection with the irreducible characters, and the primitive idempotent $e_{\chi}$ corresponding to the irreducible character $\chi$ can be written as $\frac{\chi(1)}{|G|}\sum_{y} \chi(y^{-1}) C_{y}$, where $y$ runs through a set of representatives for the conjugacy classes of $G$ and $C_{y}$ denotes the class sum of $y.$ It is also worth noting that the class sum $C_{y}$ may be expressed as $\sum_{\chi} \frac{[G:C_{G}(y)]\chi(y)}{\chi(1)} e_{\chi},$ where $\chi$ again runs through the irreducible characters.

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Thank you very much ! –  Alexander Chervov Oct 28 '12 at 11:51
    
Is "it" also equivalent to the knowledge of (ring+basis) of irreps ? –  Alexander Chervov Oct 28 '12 at 12:08
    
Well, yes, because once you know the character table, you know how to decompose $\chi_{i}\chi_{j}$ for irreducible characters $\chi_{i}$ and $\chi_{j}$. –  Geoff Robinson Oct 28 '12 at 12:30
    
This is one direction: from c-table->ring, and from ring to c-table is it also possible ? –  Alexander Chervov Oct 28 '12 at 12:44
    
I was assuming you were starting from the natural basis for the character ring. Without that, I think you may need the duality operation, given by complex conjugation in general. You then recognise the irreducible characters because you know the identity, and you can recognise the irreducible characters, given the integral character ring, because these are the elements $\theta$ for which $\theta(1) >0$ and the identity shows up with multiplicity $1$ in $\theta \overline{\theta}.$ –  Geoff Robinson Oct 28 '12 at 13:45

One nice fact, I think, is that the formula of Burnside that Geoff Robinson gives above,

$$ N^{C_z}_{C_x C_y} = \frac{|G|}{|C_G(x)| |C_G(y)|} \frac{\chi(x) \chi(y) \chi(z^{-1})}{\chi(1)} $$

can be understood nicely from a geometric / topological quantum field theory perspective. I think it is precisely the "Verlinde formula" for the modular category Rep(/\G), the representation category of the Drinfeld double of C[G]. The Verlinde formula says that in a general modular category, we have

$$ N^k_{ij} = \sum_r \frac{s_{ir} s_{jr} s_{k^* r}}{s_{0 r}} $$

where $s_{ij}$ is the S-matrix.

More concretely, this is to say that it has a natural interpretation in terms of G-bundles on the torus. This perspective also comes out in the appendix of Zagier to "Graphs on surfaces and their applications" by Lando and Zvonkin.

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May be Koornwinder's paper arxiv.org/abs/math/9904029 justifies your suggestion ? ... Added: Hm, but after writing this, I have look on a that paper one more time, and it does not seem to mention Burnside's formula... –  Alexander Chervov Dec 31 '13 at 12:35

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