# More conjugacy classes

I’m still thinking aloud (or the bloggerly equivalent thereof) about the topic of my last post, and I’m at this delightful stage of guessing there may well be interesting questions there, and yet not knowing too precisely which ones are easy, which are impossible, or even which are already hidden in the maze of MathSciNet under cleverly disguised search terms.

So consider the case of G=SL(2,Z) again, and assume given a subgroup H. In broadest terms, we’re trying to identify which conjugacy classes in G have representatives in H. We can’t exclude that all of them do; if that happens, we know that (1) H is of infinite index (see the first comment by D. Speyer to the earlier post); (2) but H surjects, by reduction modulo p, to SL(2,Fp) for every p. The latter condition implies in particular that H be Zariski-dense in SL(2) (otherwise, its reduction would be in G(Fp) for some proper algebraic subgroup, and this would be strictly contained in SL(2,Fp) if p is large enough). Nicely enough, such subgroups (especially when finitely generated) are currently the topic of much work in terms of spectral theory, expansion and the like (see for instance these recent preprints by Bourgain, Gamburd and Sarnak, and by Bourgain and Kantorovich).

The conjugacy classes of G have been classified for a long time (for instance, this is needed for the Selberg Trace Formula). The most interesting, or at least those I’m going to look at first, are the so-called hyperbolic ones, which are characterized by the fact that, for some (unique) a>1, they contain a representative which is conjugate in SL(2,R) to

$\left(\begin{array}{c} a\quad\quad 0\\ \ 0\quad\quad a^{-1}\end{array}\right),$

which acts as a dilation

$z\mapsto a^2 z$

on the Poincaré upper half-plane. A more direct characterization, in terms of an arbitrary representative g of the conjugacy class, is that

$|\mathrm{Tr}(g)|>2.$

So, for instance, we can take the conjugacy class of

$g=\left(\begin{array}{c} 2\quad\quad 3\\ \ 1\quad\quad 2\end{array}\right)$

In the case of a conjugacy class in G, the dilation a is a real quadratic integer (it is the largest eigenvalue of the matrix, and the determinant, which gives the constant term of the minimal polynomial, is 1). In the example above, we get

$a=2+\sqrt{3}=3.7305\ldots$

In SL(2,R), the dilation is the unique invariant of a hyperbolic conjugacy class (and visibly any a>1 occurs as a dilation). In G, things get a bit more arithmetic (which means more complicated, though the two words are maybe not quite synonyms). Essentially (I am here forgetting or glossing over some important semi-technical issues), for a given discriminant

$\Delta=\mathrm{Tr}(g)^2-4=(a+a^{-1})^2-4>0,$

there are only finitely many G-conjugacy classes, and the number of them is the class number of the associated real quadratic field. (Precise details are given in this old paper of Sarnak).

From my point of view of conjugacy classes, the following seems the obvious salient features:

(1) to have a chance to find a given hyperbolic conjugacy class in a subgroup H, a necessary condition is that H contains a matrix with a certain trace (up to sign; if we assume that minus the identity is in H, the sign ambiguity disappears); this condition, in turn, is obviously susceptible to local congruence obstructions — but we know that for a Zariski-dense (finitely generated) subgroup of G, all but finitely many of these congruence obstructions modulo primes will vanish by Strong Approximation.

(2) if we have a subgroup where all local obstructions disappear (for instance, all reductions modulo primes are surjective; not I don’t actually have an example of a proper subgroup of infinite index where this holds…), we are led to wonder whether all ideal classes associated with hyperbolic elements of G have representatives in H; this question is reminiscent of the representation problem for integers by ternary definite quadratic forms (where there are fairly simple necessary conditions for this to happen, and those are fairly classically also sufficient for an integer to be representation by some form in the same genus as the given one, which means by some form everywhere locally equivalent to it, while the representability by the given form holds for sufficiently large integers by much deeper work involving Fourier coefficients of half-integral modular forms — a very beautiful story, where crucial work is due to Iwaniec and Duke and Schulze-Pillot).

As before, hopefully more to come…

### Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

## 5 thoughts on “More conjugacy classes”

1. A few thoughts.

1. A topologist friend (who can comment here himself if he wants) says that if you choose a random element from each conjugacy class in SL_2(Z), you will probably generate an infinite-index subgroup, which obviously has a representative of each conjugacy class (but which is very unlikely to be finitely generated….) He though the problem of finding a finitely generated H sounded harder.

2. If you choose TWO random elements in SL_2(Z), I think it’s very unlikely they generate SL_2(Z), but the probability they generate SL_2(F_p) should be at least 1-1/p or so (there is an old paper of Kantor and Lubotzky about this.) I wonder if any counting argument (say, letting the two elements be in some big wordlength ball in SL_2(Z)) can show “by pure thought” that you can choose two elements which generate SL_2(F_p) for every p but which generate an infinite-index subgroup of SL_2(Z)?

3. More general version of your question: are there infinite-index subgroups of the mapping class group Gamma_{g,n} intersecting every conjugacy class nontrivially?

2. Regarding Jordan’s question 3: One way to find infinite index subgroups of a lattice in SL_2(R) or SL_2(C) intersecting each conjugacy class is to pick a representative from each class and do the following. Conjugate the second one by a high power of some hyperbolic element until the first two generate a free product (by a “ping-pong” argument). Conjugate the third one by a new hyperbolic element until the first three generate a free product. And so on. If you’re careful, you can guarantee that the resulting subgroup is a free product of the cyclic groups generated by representatives of the conjugacy classes. This won’t be finite index as the free product decomposition has too many factors (in the SL_2(C) case, you could use the fact that a lattice is virtually freely indecomposable).

This ping pong argument can be made in the mapping class group by considering the action on the sphere of projective measured laminations. So, the answer to Jordan’s 3 should be yes.

3. Thanks for the comments!
Concerning 2. in Jordan’s comment, I had actually checked a while ago that two random elements of SL(2,Z), obtained by two independent random walks of same length, will generate a subgroup that surjects on SL(2,Z/pZ) for some small p with large probability (using sieve and the Kantor-Lubotzky result; see here). I haven’t thought about the other part of (2.), but that seems an interesting possibility…

4. Anonymous says:

Regarding Jordan’s comment 2,

The elements (1 N)(0 1) and (1 0)(N 1) generate a group of infinite index in G = PSL_2(Z) (if N is at least 3) which surjects on PSL_2(F_p) for all p not dividing N. To promote this to an example of an infinite index subgroup surjecting onto PSL_2(F_p) for all p, one can proceed as follows (there may be a slicker way).
Start with a surjection of G–>>M onto some simple group M, which we suppose is not a congruence quotient. (Simple is not really necessary, it’s just an easy way to ensure that the kernel surjects onto PSL_2(F_p) for all p.) Let K denote the kernel. Suppose that M is “big enough” such that K has no elements of finite order, and so K is free. Let M have order N, and suppose that N has d distinct prime divisors. We may find d pairs of elements in K whose images generate PSL_2(F_p) for all p dividing N. Let H be the subgroup of K generated by these 2d elements as well as the elements (1 N)(0 1) and (1 0)(N 1). H is free of rank at most 2d+2. If H had finite index inside K, then K would also have rank at most 2d+2. On the other hand, using Riemann-Hurwitz for X(H)–>X(Gamma) (and the fact that the ramification at i and rho is bounded), H is generated by very close to |M|/6 elements. So as long as M is big enough, |M|/6 > 2d+2 and H has infinite index.
It’s a little hard by “pure thought” to come up with non-congruence quotients of G, but an easy example to verify is to take M = A_9, which is generated by (1,2,3)(4,5,6)(7,8,9) of order three and (1,4)(2,5)(6,7)(8,9) of order two, and is hence a quotient of G = . The required inequality in this case is 30241 > 10.

5. Thanks a lot for this explanation!
I’ve started doing a bit of numerics on the conjugacy classes in SL(2,Z) of elements of the group generated by (1 3;0 1) and (1 0;3 1) to try to see it it seems that “most” (permitted) classes appear. I’ll report on this soon.