Today we will complete the description of the solution of Eskin-Kontsevich-Zorich regularity conjecture, that is, we will prove that, for any -invariant probability measure on a connected component of a stratum of the moduli space of translation surfaces of genus , the -measure of the set of with two non-parallel saddle-connections of lengths is
For this sake, let us recall that, in the previous post of this series, we considered an arbitrarily fixed level of the systole function and we introduced a subset consisting of such that all its non-vertical saddle-connections have length . Then, we defined the set
and we studied the -measure of the subsets
of translation surfaces with systole that are “accessible” by and movements from . In this setting, the results of the previous post of this series can be summarized as follows:
- , where is a finite measure on ;
- for all ;
- is a density measure in the sense that
From this point, we will divide this final post into two sections. In the first one, we will formalize the idea that occupies most of for adequate choices of , so that the proof of Eskin-Kontsevich-Zorich regularity conjecture will be reduced to the computation of the -measure of . Then, in the final section, we will show that the set of such that has small -measure for large because these ‘s have a pair of non-parallel saddle-connections with a small angle (and is ).
1. A “perfect cancellation” result
Let . From the fact that imply that the level sets of the systole function have zero -measure, and, thus, is a continuous non-increasing function.
Furthermore, the fact that is a density measure implies that has a left derivative
equal to .
Moreover, the Siegel-Veech formula says that for all , so that the previous estimate imply that
In particular, the left derivative is bounded.
From this, it follows that is absolutely continuous: indeed, if the left derivative of a continuous function is bounded by on an interval , then because, for any , one can check that the supremum of the such that is (in view of the assumption that for all ).
At this point, we are ready to “improve” the Siegel-Veech formula by showing that there exists a constant such that
Since is absolutely continuous, it is the integral of its left-derivative. It follows that
for all , and, hence,
On the other hand, from (1), we know that
for all . Therefore, by taking , we obtain that
In summary, we showed that
Once we know that , we can show the following “perfect cancellation” equality:
In fact, it is obvious that , and, for the converse inequality, we can use that is the integral of its left derivative:
We call the equality
a “perfect cancellation” by the following reason. Recall that (where and ). So, by taking a sequence such that as , and by considering the sets , we conclude that
Since , we conclude that occupies most of in the sense that
In particular, our task of showing that
is reduced to show that, for each , and all sufficiently large (depending on ),
The proof of this last assertion will occupy the next (last) section of this post.
2. End of proof of EKZ regularity conjecture
Let us fix once and for all one of the ‘s, , as above, and let us consider the set . By definition, they consist of of the form with and possessing two non-parallel saddle-connections of lengths .
Without loss of generality, we can assume that the lengths of these two saddle-connections are comparable within a large multiplicative factor . Indeed, if is very large and the ratio of these saddle-connections is larger than , then the systole of is actually and, by the Siegel-Veech formula, the -measure of the set
is for large enough.
Next, we consider the set of giving birth to with a pair of saddle-connections of lengths comparable within a large multiplicative factor from the systole.
We affirm that, for large enough, the angle is small, that is, for each and , there are and such that if with and , then:
- or has a pair of non-parallel saddle-connections of lengths making an angle .
Assuming momentarily this claim, let us complete the proof of EKZ regularity conjecture. In this direction, let us consider the first item above and let us show that the -measure of this event is if small enough. Recall that . Therefore, given , for all , the -measure of the subset of translation surfaces in with is
Since this quantity is for small enough, it suffices to estimate the -measure of the set of with where denotes the event described in the second item of our claim. Here, we observe that, by discreteness of the set of holonomy vectors of saddle-connections, is for small enough. Therefore, the -measure of the set of with is
so that the proof of EKZ regularity conjecture is complete.
Now, let us prove our affirmation. Since , has a saddle-connection in the direction of . Now, let us try to figure out what non-vertical vectors can represent (the holonomy vector of) a saddle-connection on such that the saddle-connection of represented by has length comparable to within the multiplicative factor .
Suppose that the first item of our claim is violated, i.e., . In this situation, as we saw in Subsection “On the action of the diagonal subgroup ” of the second post of this series, given , there exists a constant such that if , then for all . It follows that, if , then
In other words, if we start with a saddle-connection of whose holonomy vector satisfies , then is not comparable to within the multiplicative factor . Equivalently, if is the holonomy vector of a saddle-connections of leading to saddle-connections of whose lengths are between and , then .
At this point, our task consists into checking that if with is the holonomy vector of a saddle-connections of leading to saddle-connections of whose lengths are between and , then the angle between and is if is large enough and . However, this last property is not hard to get: if and the angle between and is , then
for some . Since expands horizontal vectors, one can check that
for and large enough depending on , e.g., and . In other terms, if with , then a saddle-connection of () of length non-parallel to a length-minimizing saddle-connection of must come from a saddle-connection of of length making an angle with the vertical direction.
In summary, we showed that if the first item of our claim is violated, then the second item holds. This proves the claim and, a fortiori, the proof of EKZ regularity conjecture.