How frequent are thin groups among Kontsevich-Zorich monodromies?
Instead of explaning the meaning of Sarnak’s question in general, we shall restrict ourselves to the case of Kontsevich-Zorich (KZ) monodromies associated to square-tiled surfaces.
More concretely, let be a square-tiled surface (also called origamis) of genus , i.e., is a finite branched covering which is unramified off and is the pullback of on . We have a natural representation
from the group of affine homeomorphisms of to the group of symplectic matrices of the subspace of integral homology classes of projecting to zero under . In this setting, the Kontsevich-Zorich monodromy (associated to the -orbit of in the moduli space of translation surfaces) is the image of , i.e.,
By following Sarnak’s terminology, we will say that is a thin group if is an infinite index subgroup of whose Zariski closure is
In the particular case of square-tiled surfaces, Sarnak’s question above is related to the following two problems:
- (a) find examples of square-tiled surfaces whose KZ monodromies are thin;
- (b) decide whether the “majority” of square-tiled surfaces in a given connected component of a stratum of the moduli spaces of unit area translation surfaces has thin KZ monodromy (here, “majority” could mean “all but finitely many” or “almost full probability as the number of squares/tiles grows”.)
Remark 1 While we will not give answers to items (a) and/or (b) in this post, we decided to write it down anyway with the hope that it might be of interest to some readers of this blog: in fact, by the end of this post, we will show the following conditional statement: if the group generated by the matrices
has infinite-index in , then a certain square-tiled surface of genus answers item (a) affirmatively.
Remark 2 Some “evidence” supporting a positive answer to item (b) is provided by this recent paper of Fuchs-Rivin where it is shown that two “randomly chosen” elements (in ) “tend” to generate thin groups.
1. Faithfulness and thinness
Let be a square-tiled surface of genus . For the sake of simplicity, suppose that has no non-trivial automorphisms. In this case, the group is naturally identified with a finite-index subgroup of called Veech group, and, therefore, we get a representation
The following proposition produces a certificate of thinness for .
Proof: If is virtually free, is faithful and has finite index in , then the group of higher rank would contain a lattice isomorphic to a free group.
However, this is impossible because higher rank linear groups do not contain lattices isomorphic to free groups: on one hand, a higher rank linear group satisfies Kazhdan’s property (T), so that its lattices also have this property (by Kazhdan’s theorem), and, on the other hand, a free group does not have Kazhdan’s property (T).
This proposition says that a positive answer to item (a) would follow from a positive answer to the following problem:
- (c) find a square-tiled surface of genus (without non-trivial automorphisms) such that the Veech group is virtually free, the image of is Zariski dense in and is faithful.
As we are going to see in the next section, it is not hard to construct explicit examples of square-tiled surfaces (of genus without non-trivial automorphisms) such that the Veech group is virtually free. In other words, a positive solution of (c) is somewhat related to a positive answer to the following problem of independent interest (for the experts on translation surfaces):
- (d) find a square-tiled surface of genus such that the monodromy representation is faithful.
Remark 3 If a square-tiled surface has a rational direction of homological dimension one (i.e., the waist curves of maximal cylinders of in a direction of rational slope span a one-dimensional subspace in absolute homology), then is not faithful: indeed, this happens because Dehn twists in such a direction act by the identity matrix on (see, e.g., Lemma 5.3 of this paper here), that is, they induce elements of . In particular, since any square-tiled surface of genus always have rational directions of one-cylinder decompositions (by the works of Hubert-Lelièvre and McMullen), it follows that the answer for the analog of item (d) in genus is always negative.
This remark suggests to try to answer items (c) and (d) by looking at square-tiled surfaces of genus with no one-cylinder rational directions. After a computer search (using Sage), Vincent Delecroix found a beautiful example of such a square-tiled surface whose description occupies the next section.
Remark 4 For any prescribed , one can construct infinite families of square-tiled surfaces without -cylinders rational directions (via the Hubert-Lelièvre-Kani invariant described in this paper here).
2. An origami without one-cylinder decompositions
Consider the square-tiled surface associated to the pair of permutations
The commutator is
so that is a genus square-tiled surface.
The -orbit of consists of four elements. Indeed, this fact can be checked as follows. We recall that:
- the generators and of act on pairs of permutations by the rules and ;
- the pairs of permutations and give rise to the same square-tiled surface.
Therefore, the -orbit of is where is given by the pair of permutations with
As it turns out, the -orbit of accounts for its entire -orbit because
where and .
In summary, the -orbit of can be depicted as follows.
Since the cylinder decompositions of in any rational direction described by the horizontal cylinder decomposition of some element of , we have that all cylinder decompositions in rational directions of have exactly three cylinders.
Remark 6 It follows from this discussion that has a single cusp (i.e., single -orbit in ). Also, the homological dimension of in the sense of Forni is three. Thus, by the results in this paper of Forni, the Lyapunov spectrum of the Kontsevich-Zorich cocycle over with respect to the Haar measure has the form
Moreover, the Eskin-Kontsevich-Zorich formula for the sum of non-negative Lyapunov exponents of the Kontsevich-Zorich cocycle imply that , i.e.,
Some numerical experiments (with Sage) indicate that and
3. The affine homeomorphisms of
The group of affine homeomorphisms of is the stabilizer of in the moduli space of translation surfaces.
It is not hard to see that the subgroup of automorphisms of is trivial. It follows that the elements of are determined by their linear parts in , that is, the natural map
is injective. Hence, is isomorphic to its image under this map.
The group is the finite-index subgroup of consisting of all elements of stabilizing : in the literature, is called the Veech group of .
We saw in the previous section that . Thus, is an index four subgroup of . Furthermore, is a congruence subgroup of level , and the Teichmüller curve has genus zero. Thus, is generated by elliptic and parabolic elements: indeed, one can check that is generated by the following two elliptic matrices
of orders .
The group structure of is provided by the following lemma:
Proof: Consider the twelve cones defined by the following properties:
- for each ;
- each , , consists of the convex combinations of positive multiples of the vectors and , where , , , , , and .
A simple calculation shows that
- for each ;
- , , , , , and .
It follows that and play ping-pong with the tables
in the sense that and are disjoint subsets of such that
- , ;
- , .
By the ping-pong lemma, we conclude that .
Remark 7 The construction of the cones above is similar to the cones studied in Section 2 of this paper of Brav-Thomas.
4. The Kontsevich-Zorich monodromy of
4.1. The relative homology groups of ,
Given , , let us denote by , resp., the relative cycles on consisting of the bottommost horizontal and leftmost vertical sides of the square numbered .
Note that each square of gives a relation , that is,
- , , , , , , , , ;
- , , , , , , , , ;
- , , , , , , , , ;
- , , , , , , , , .
4.2. The action of on the relative homology groups
The matrix takes to , and it acts on the corresponding relative homology groups by the matrix such that
Similarly, the matrix takes to , and it acts on the corresponding relative homology groups by the matrix such that
Finally, exchange and , resp. and , and it acts on the corresponding relative homology groups by the matrices and such that
4.3. The absolute homology groups of ,
Note that this basis is adapted to the decomposition in the sense that this decomposition corresponds to the partition where and , i.e.,
Moreover, it is worth to point out that, in the basis , the matrix of the restriction to of the intersection form is
4.4. The action of on the absolute homology group
The matrices of , and with respect to the basis are
This allows us to compute the images and of the generators and of under the KZ cocycle . Indeed,
For later use, we observe that these formulas give that the non-tautological subrepresentation of takes values
(with respect to the basis of ) at the two generators and of . Moreover, if we denote by , , then the characteristic polynomials and of the matrices and are
5. Zariski density of in
The matrices and are Galois-pinching, i.e., all roots of their characteristic polynomials and are real and simple, and the Galois groups of and have order (that is, the largest possible for symplectic matrices). Furthermore, the splitting fields of their characteristic polynomials are disjoint.
Indeed, these facts follow from the analysis of the following discriminants
related to the quadratic subfields of the splitting fields of and : see Proposition 6.14, Remark 6.15 and Proposition 6.16 in this paper here for more details.
Remark 8 By the main result in this paper here, the Zariski-density of in implies that the Lyapunov spectrum in Remark 6 is simple, i.e.,
6. Non-faithfulness of the representation
Once we have Proposition 3 in our toolkit, it is natural to investigate the thinness of . Here, it is tempting to try to use the thinness certificate discussed in Proposition 1 from Section 1 in order to give a positive (partial) answer to Sarnak’s question.
By definition, the faithfulness of would amount to show that has the same group structure of the Veech group (described in Lemma 2), i.e., .
After some numerical experiments with all non-trivial words of length on and and some non-trivial words of length on and , I thought that could be faithful.
As it turns out, after I asked about the faithfulness of on MathOverflow, Stefan Kohl noticed that is not faithful because (a computer-assisted calculation shows that) the kernel of contains a certain non-trivial word of length in and .
Proposition 4 (S. Kohl) The representation is not faithful because
Remark 9 The matrices and appearing in my MathOverflow question generate a conjugate of : indeed,