As we mentioned in the first post of this series, our goal today is to discuss some elementary facts about and conditional measures. For this sake, we divide this post into two completely independent sections: in the next one we’ll exclusively talk about , and in the final section we’ll discuss a variant of the so-called Rokhlin’s disintegration theorem.

**1. Some facts about **

In this section, we’ll discuss 3 results where:

- we’ll study the action of on the Euclidean norms of non-collinear vectors (cf. Proposition 2 below),
- we’ll compute the Haar measure on a certain open subset of (cf. Proposition 3 below), and
- we’ll estimate the amount of time that the diagonal subgroup keeps the Euclidean norm of the vector below a given threshold (cf. Proposition 4 below).

During this entire section (and, actually, in the other posts of this series also), we will use the following notation:

- is the standard basis of and denotes the Euclidean norm of ;
- is the (positive) diagonal subgroup of ;
- is the rotation by angle ;
- is a lower triangular matrix.
- is a upper triangular matrix.

** 1.1. Euclidean norms under -action on **

The following lemma computing the Haar measure on is well-known, see, e.g., A. Knapp’s book (especially Chapter 5 on Iwasawa decomposition and integral formulas, and Equation (10.7) in Chapter 10).

Lemma 1The map from to is a diffeomorphism and the measure is sent to a Haar measure by this diffeomorphism.

Using this lemma, we will estimate the Haar measure of the set of elements of with bounded norm keeping two fixed non-collinear vectors inside a given annulus.

Proposition 2Let be vectors with . Then, the Haar measure of the set

is when is small. Here, the implied constant is uniform when is bounded away from .

Remark 1For the application we have in mind, it suffices to know that the Haar measure of is , but we prefer to state this proposition in this form in order to highlight the following phenomenon. A naive computation reveals that the Haar measure of the set of elements of with bounded norm keeping a given vector inside the annulus is . In particular, it would be natural to think that has Haar measure as we are trying to confine two non-collinear vectors inside the annulus . However, as we’ll see during the proof of the proposition, this intuitive picture is wrong because of certain singularities (critical points) of “square root” type develop (in other words, the events of confining two non-collinear vectors inside an annulus are not always independent). Fortunately, the singularity is mild and the estimate of the Haar measure of is sufficient for our future purposes.

*Proof:* The set is empty (for small ) unless , so we’ll assume that and satisfy these inequalities. Also, from the right-invariance of the Haar measure under , we can suppose that with .

Now, let us write and , and let us introduce the function

Clearly doesn’t depend on , so that we will write in the sequel.

We have . Thus, the Jacobian matrix of is

Since and , this matrix is invertible unless , i.e., and are collinear () or orthogonal ().

We will complete the proof of this proposition by considering the following cases:

**Suppose that with and .**Then, with when . Since , we*can’t*haveat the

*same time*if is small enough*depending only on*, i.e., . In other words, in this situation, the set is empty for small enough depending on .**Suppose that and are orthogonal, i.e., .**Then, . The condition determines an interval of ‘s of length : indeed, is equivalent to ; since , the interval has length . Next, we observe that, for a fixed in this interval, the condition determines an interval of ‘s of length (and*exactly*of order in the*worst case*): in fact, is equivalent to , that is, ; since (as we mentioned above) and , we get from that , and, thus, belongs to an interval of size . From Lemma 1, it follows that, in this situation, the Haar measure of is . Moreover, the same estimate is true if we relax the condition to in the definition of .**Assume that there exists such that and is orthogonal to .**In this case, the fact that has Haar measure follows from the previous item after performing a right-translation of by .**Assume that none of the assumptions hold.**Then, the Jacobian matrix of is invertible on the set and the norm of its inverse is uniformly bounded by a constant depending only on how far is from . In this case, from the inverse function theorem and Lemma 1, we deduce that the Haar measure of is where the implied constant depends only on .

This completes the proof of the proposition.

** 1.2. The decomposition and Haar measure on **

Let .

Proposition 3The map from is a diffeomorphism onto . Furthermore, up to a multiplicative constant, the restriction to of a Haar measure is equal to in the coordinates .

*Proof:* The vertical basis vector is fixed by and it is mapped to under . Therefore, for , , , the matrix satisfies and , i.e., belongs to .

Conversely, given a vector with and , there exists a unique pair depending smoothly on with . Indeed, this follows from the facts that the matrix moves vectors along hyperbolas and, for , the matrix moves along the arc of unit circle containing between the hyperbolas and , see the figure below:

This proves the first assertion of the proposition because, after writing a vector with and uniquely as for , , if we are given a vector such that , then we can write by choosing the unique such that where is the horizontal basis vector of .

Now, let us write the restriction of a Haar measure to as where is a positive smooth density function. Since any Haar measure on is left and right invariant (i.e., is unimodular), we deduce that depends only on , i.e., . In order to compute , we will use the left-invariance of Haar measures under . More concretely, let us fix , and let us consider tiny open sets around the origin and their respective images under . In terms of matrices, this amounts to consider the equation:

for close to , and , , . For sake of simplicity, since is fixed, we will omit the dependence of the functions on in what follows. In this language, we have from the -invariance of Haar measures and the change of variables formula that where is the determinant of the Jacobian matrix at the origin . In particular, our task is reduced to show that . Keeping this goal in mind, let us notice that implies that , and in Equation (1). Thus, we have that

and

In other words, the Jacobian matrix has the form

at the origin , so that . Hence, it remains only to show that . In this direction, we apply both matrices in Equation (1) to the vertical basis vector to get the equations:

By multiplying these equations, we get the relation:

By taking the partial derivative with respect to , we deduce that

Therefore, since at the origin , we have that

Now, we can plug this information into Equation (3) to calculate . Indeed, by taking the partial derivative with respect to in (3), we get

Since at the origin , we obtain that

By combining this relation with (4) we conclude that

Of course, this proves the second assertion of the proposition.

Remark 2Note that the computations above give all entries of the Jacobian matrix at the origin but . Evidently, the knowledge of this particular entry was irrelevant for the computation of in the previous proposition, but the curious reader is invited to compute this entry along the following lines. By applying both matrices in (1) to the horizontal basis vector , one gets two relations:

and

By taking the partial derivative of the second relation above with respect to at the origin and by plugging the values and already computed, one deduces that

i.e.,

** 1.3. On the action of the diagonal subgroup **

Let be a given rotation. For , define:

Geometrically, is the interval of times such that, after applying to the unit vector , the norm of the vector stays below the “threshold” .

Proposition 4The set is empty if and only if . If , then, by writing with , one has that is an open interval of length .

*Proof:* By definition, if and only if , i.e.,

By performing the change of variables or depending on whether or , and by multiplying the previous inequality by , we get or . This inequality has a solution if and only if the discriminant of the corresponding second degree equation is positive, i.e.,

This proves the first assertion of the proposition.

If and we write with , then , so that . It follows that the solutions of

belongs to the open interval between the roots

of . By recalling the change of variables , we deduce that is an open interval where . In particular, the length of is

This completes the proof of the second assertion of the proposition.

Later in our discussion, we will combine Propositions 3 and 4 to derive some measure estimates of the sets of translation surfaces with short saddle-connections, and, for this sake, we will need the following equality:

Lemma 5

*Proof:* The change of variables gives

Since (as ) and for (by integration by parts), we deduce that

By Leibniz formula, we know that . By plugging this into Equations (6) and (5), we obtain the desired lemma.

Also for later use, we will need the following corollary of the proof of Proposition 4.

Corollary 6Given , there exists a constant such that, if

then

for all .

*Proof:* Recall that , so that

In particular, from the definition of , we have that for . By combining these two estimates, we deduce that

On the other hand, from the proof of Proposition 4, we know that, by writing ,

for every .

By hypothesis, (i.e., ), so that we conclude from the previous inequality that

By plugging this into our estimate of above, we deduce that

and thus the proof of the corollary is complete.

At this point we have all elementary facts about and its action on , and we will close this post with the following remark:

Remark 3Historically, the proof of Proposition 4 was the starting point of our solution with A. Avila and J.-C. Yoccoz of the Eskin-Kontsevich-Zorich regularity conjecture. Interestingly enough, after Artur, Jean-Christophe and I completed our article, A. Eskin pointed out to us that this lemma was previously known by G. Margulis and, indeed, they planned to include this fact in one of their joint articles but they ended up by not relying on it.

**2. Rokhlin’s disintegration theorem **

Very roughly speaking, given a probability measure on a space and a partition of , Rokhlin’s disintegration theorem concerns the problem of writing/decomposing as a “superposition” of (conditional) probability measures supported on the elements of the given partition. In other words, Rokhlin’s theorem addresses the question of disintegrating into supported on

Of course, such a decomposition might not exist in general, but Rokhlin’s disintegration theorem provides fairly general conditions ensuring the existence and uniqueness of disintegrations/conditional measures.

Theorem 7 (Rokhlin’s disintegration theorem)Let be a Lebesgue space and let be a measurable partition of , i.e., there is a sequence of finite partitions of with the following properties:

- denoting by denotes the element of a given partition containing , one has that for all and ;
- the sequence is monotonic: for all and , ;
- is the limit of : for all , .

Then, there exists a system of conditional measures for , i.e.,

- for all , (i.e., is supported on );
- for all , (i.e., is constant on elements of the partition );
- for any , the function is measurable and
(i.e., disintegrates ).

Moreover, the system of conditional measures is essentially unique: if is another system of conditional measures, then for -almost every .

Remark 4In some sense, Rokhlin’s disintegration theorem is a sort of martingale convergence theorem: indeed, for the finite partitions , it is easy to disintegrate by letting be the normalized restriction of to , and one has to work to show that the desired ‘s are the limits of ‘s. This point of view is nicely explained in these notes of M. Viana here (where a proof of a particular case of Rokhlin’s disintegration theorem is proved).

For our future applications, the probability measure will leave on a space with a partition given by pieces of a natural action of a Lie group and will be “invariant” under this Lie group. In this context, Rokhlin’s disintegration theorem alone will not suffice for our purpose because we will want to know that the conditional measures are essentially pieces of Haar measures. Here, it is worth to point out that similar settings were already considered by other authors, but the exact statement (below) needed in our applications in the next posts of this series doesn’t seem to appear in the literature. So, we will close today’s post with the following variant of Rokhlin’s disintegration theorem.

Let be a Polish space and let be a Lie group. Denote by a left-invariant Haar measure on and a left-invariant metric on .

Let act on by and denote by the canonical projection. Let be an open subset and let be a probability measure on .

Note that comes with a natural *measurable* partition . Thus, by Rokhlin’s disintegration theorem, has a system of conditional measures that we can think as a probability measure on the fiber of over .

We will assume that is *invariant*, that is, for all measurable and all such that , we have . In this context, we can assert that the conditional measures are pieces of Haar (-)measure on .

Proposition 8If is invariant, then, for -almost every , has finite (Haar) -measure and

*Proof:* For -a.e. , the conditional measure is “invariant”. Intuitively, this follows from the uniqueness of the system of conditional measures and the invariance of , but one has to be take a little bit of care about the meaning of “invariant” for .

For our purposes, we will consider the following notion. Let be a *countable* dense subset of and let be the *countable* set of balls of the left-invariant metric on centered at points in with positive rational radius.

Since and are countable, we can use the uniqueness of conditional measures and the invariance of (plus the fact that countable unions of sets of zero measure have zero measure) to deduce that, for -almost every , if and satisfy and , then , i.e., is “invariant” if we test the invariance properties *exclusively* with elements and of the countable (dense) sets and .

In any case, the fact that and are dense are enough to deduce that ‘s are Haar measures from the “weak” form of invariance. Indeed, notice that it suffices to show that ‘s are absolutely continuous with respect to because the “weak” invariance property is good enough to ensure that the density is constant. Now, once we reduced our task to the absolutely continuity of , we can proceed as follows. Given , with radius , by the nice (homogeneity) properties of the Haar measure on a Lie group, we can find an integer (with independent of ) and some elements such that the balls are mutually disjoint and contained in . By the “weak” invariance of , we obtain that

so that , and, hence, is absolutely continuous with respect to .

## Leave a Reply