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 1 The 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 2 Let
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 1 For 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 have
at 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 3 The 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 2 Note 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 4 The 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 6 Given
, 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 3 Historically, 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 4 In 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 8 If
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
.
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