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For a group G, denote by c(G) the number of conjugacy classes in G.

If $S_n$ denotes the symmetric group on n letters, then:

$$\lim_{n \to \infty} \frac{\log \log c(S_n)}{\log \log |S_n|} = \frac{1}{2}$$

[NOTE: Assume all logs are to the same base. It doesn't matter what the base is, as long as we use the same one in the numerator and denominator.]

[Proof that I know of: $c(S_n)$ equals the number of unordered integer partitions of n, and there exist asymptotic formulas for that which show it to be exponential in $\sqrt{n}$. $|S_n|$ is exponential in $n \log n$. Taking double logs, we get roughly $(1/2)\log n$ and $\log n + \log(\log n)$ respectively, and the quotient goes to $1/2$.]

Obviously, the result also holds for alternating groups instead of symmetric groups.

Similarly, if $GL(n,q)$ denotes the general linear group of degree n over a field of q elements, then for fixed q:

$$\lim_{n \to \infty} \frac{\log \log c(GL(n,q))}{\log \log |GL(n,q)|} = \frac{1}{2}$$

[Proof that I know of: the number of conjugacy classes can be expressed as a polynomial in q of degree n, and the group order is a polynomial in $q$ of degree $n^2$, so we get the result.]

The result holds if we replace the general linear group by the special linear group, projective general linear group, or projective special linear group (which is the Chevalley group of type A).

My question:

  1. Does the result also hold for the groups in the other three infinite Chevalley families B, C, D, and in the other infinite families of simple groups?
  2. If the answer to 1 is yes, is there some deeper reason why the result holds both in the symmetric/alternating group case and in the linear groups case? I imagine that the linear groups cases can be tied together using some general Lie group or algebraic group principles. If so, what are they? How do the alternating groups fit into the picture? May be something to do with Iwahori-Hecke algebras or the field of one element? All explanations are welcome.
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Certainly in terms of the number of elements, $|S_n| = |GL(n,1)|$, and this can be traced to the Bruhat decomposition. I don't know why the same should hold for the number of conjugacy classes, in particular why it is (as you assert) a polynomial, but I presume it does. –  Allen Knutson Mar 25 '12 at 2:49
    
Outside the classical series, the only infinite families of simple groups that have a rank parameter are Steinberg's unitary and orthogonal series. –  S. Carnahan Mar 25 '12 at 5:08
    
Allen Knutson, the GL(n,q) number of conjugacy classes is a polynomial in q which you can work out by explicitly computing the number of conjugacy classes of each type; here's the GL(3,q) case: groupprops.subwiki.org/wiki/… For SL, PGL, PSL, we don't get polynomials but PORC functions; the statement for SL, for instance, is here: groupprops.subwiki.org/wiki/… –  Vipul Naik Mar 25 '12 at 7:12
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Allen: Regarding your original assertion that |S_n| = |GL(n,1)|, the naive interpretation of the polynomial formula for the order of GL(n,1) does not give the order of S_n (that naive formula would give an order of 0, because q - 1 divides the polynomial). So could you elaborate what you mean? Also, the term quasipolynomial is over-used; I think some people use it for linear combinations of positive power functions, others use it for PORC functions, others use it for functions that have polynomial-like growth. –  Vipul Naik Mar 25 '12 at 20:09
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@Vipul: the correct version replaces the cardinality of $\text{GL}_n$ with the cardinality of $\text{GL}_n/B$ for $B$ a Borel. I guess it should really be intepreted as a statement about flag varieties: over $\mathbb{F}_q$ one considers maximal chains in the poset of subspaces of $\mathbb{F}_q^n$ and over $\mathbb{F}_1$ one considers maximal chains in the poset of subsets of $[n]$. –  Qiaochu Yuan Mar 27 '12 at 17:37

1 Answer 1

(1) is true, and follows from a paper of Liebeck and Shalev ("Character degrees and random walks in finite groups of Lie type" on Liebeck's webpage). They count representations according to their degrees rather than conjugacy classes, but it follows from their results that if $\Phi$ is a (reduced, irreducible) root system and G is a corresponding algebraic group (possibly twisted), then the number of conjugacy classes of $G(F_q)$ is roughly $q^{rk \Phi}$ (the lower bound is easy, just take conjugacy classes of elements in the torus). The limit of $$ \frac{\log \log q ^{rk \Phi}}{\log \log |G(\mathbb{F}_q)|}$$ as the rank of $\Phi$ tends to infinity is 1/2 also for types B,C, and D.

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"The torus" usually means a split torus, but I don't think that has enough classes if the rank is large compared to $q$. But a maximal anisotropic torus seems to work. –  Ben Wieland Apr 4 '12 at 2:32

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