In this previous post, we discussed some results of Furstenberg on the Poisson boundaries of lattices of (mostly in the typical low-dimensional cases and/or ). In particular, we saw that it is important to know the Poisson boundary of such lattices in order to be able to distinguish between them.
More precisely, using the notations of this post (as well as of its companion), we mentioned that a lattice of can be equipped with a probability measure such that the Poisson boundary of coincides with the Poisson boundary of equipped with any spherical measure (cf. Theorem 13 of this post). Then, we sketched the construction of the probability measure in the case of a cocompact lattice of , and, after that, we outlined the proof that is a boundary of in the cases and .
However, we skipped a proof of the fact that is the Poisson boundary of by postponing it possibly to another post. Today our plan is to come back to this point by showing that is the Poisson boundary of .
More concretely, we will show the following statement due to Furstenberg. Let be a cocompact lattice of . As we saw in this previous post (cf. Proposition 14), one can construct a probability measure on such that
- (a) has full support: for all ,
- (b) is –stationary: ,
- (c) the –norm function is –integrable: .
Here, we recall (for the sake of convenience of the reader) that: is the “complete flag variety” of or, equivalently, where is the subgroup of upper-triangular matrices, is the Lebesgue (probability) measure and
where acts on Poincaré’s disk via Möebius transformations (as usual) and denotes the hyperbolic distance on Poincaré’s disk .
Then, the result of Furstenberg that we want to show today is:
As we already mentioned above, we know that is a boundary of (cf. Subsection 2.2 of this post).
Thus, if we denote by the Poisson boundary of (an object constructed in Section 4 of this post), then, by the maximality of the Poisson boundary, is an equivariant image of under some equivariant map .
Our goal consists into showing that is an isomorphism, and, for this sake, it suffices to show that we can recover all bounded measurable functions of from the corresponding functions on via , i.e., the proof of Theorem 1 is reduced to prove that:
In this direction, it is technically helpful to replace by and consider the subspace
In fact, since is a closed subspace of the Hilbert space , we have an orthogonal projection and our task of proving Proposition 2 is equivalent to show that is the identity map .
Now, the basic strategy to show that is to prove that, for each , the functions and induce the same -harmonic function on (via Poisson formula). Indeed, since is the Poisson boundary of , we have (by definition) that the Poisson formula associates an unique -harmonic function
on to each . Hence, if and are associated to the same -harmonic function on , then . In other words, we reduced the proof of Proposition 2 to the following statement:
In other to show this proposition, we rewrite the -harmonic function associated to in terms of the -inner product as follows:
In particular, if for all , then
Equivalently, we just showed that and induce the same -harmonic function if for all , that is, the proof of Proposition 3 will be complete once we prove that:
As it turns out, the functions admit a nice characterization in terms of Jensen’s inequality. More concretely, since consists of all functions in which are measurable with respect to the field of sets (with measurable), one can show that the projection enjoys a “Jensen’s inequality property”:
As the reader might suspect, we intend to use Jensen’s inequality to produce an equality characterizing whether . For this, we will compute for .
In fact, it is not hard to guess who must be: since is an equivariant map sending to , it is not surprising that . Now, let us formalize this naive guess as follows. Recall that, by definition, is the (unique) function in such that
for each , i.e., with ). We rewrite this identity as
For , this identity becomes
Observe that the right-hand side of this equality is the -harmonic function of induced by . On the other hand, since is an equivariant map between the Poisson boundary and the boundary , we have that the functions and induce the same -harmonic function, i.e.,
By putting the previous two equalities together, we get that
Next, we recall that sends to (i.e., ). Therefore, if we denote , we obtain that the right-hand side of the previous equality becomes
By combining the last two equalities above, we deduce that
Since this identity holds for an arbitrary function , we conclude that
as it was claimed (or rather guessed).
From this computation and Jensen’s inequality (1), we get the following lemma:
Proof: By setting , we see that the left-hand side of (2) is
while our computation of above reveals that the right-hand side of (2) is
It follows that the desired lemma is a consequence of Jensen’s inequality (1).
In summary, our task now becomes to prove that:
Proposition 6 For all , the inequality (3) above holds, i.e.,
The basic idea to prove this proposition is the following. The quantities and can be interpreted as spatial averages. In particular, the ergodic theorem will tell us that and drive the Birkhoff sums of the observables and along almost every sample of random walk in .
Now, assuming by contradiction that , we will see that the Birkhoff sums of are very well controlled by the Birkhoff sums of (with some “margin” coming from the strict inequality ). Using this and the fact that the density can be explicitly computed, we will be able to solve a counting problem to show that:
On the other hand, using the properties of -harmonic functions, we will show the following general fact about recurrence sets of :
Of course, by putting together Propositions 7 and 8, we deduce the validity of Proposition 6. Hence, it remains only to prove Propositions 7 and 8. In order to organize the discussion, we will show them in separate sections, namely, the next section will concern Proposition 7 while the final section of this post will concern Proposition 8.
2. Proof of Proposition 7
As we already mentioned above, the first step in the proof of this proposition is to observe that and are spatial averages, so that the ergodic theorem says that one can express them in terms of temporal averages along typical “orbits” (samples of random walk).
More precisely, let be a stationary sequence of independent random variables with distribution and consider the -process on . For technical reasons (that will become clear in a moment), we will think of as moving forward in time (rather than backward), i.e., the -process satisfies
with independent of (instead of and independent of ). Note that by setting
we get a -process on (because is an equivariant map from to ).
2.1. Interpretation of as a Birkhoff sum
In this language, we can convert the spatial average of the observable
in a Birkhoff average as follows. Let us consider the random walk on obtained by left-multiplication. Then,
Of course, in order to justify the application of the ergodic theorem, we need to check the (absolute) integrability of the corresponding observable, that is, we need to show that the following expectation
can be controlled as follows. By letting act on the Poincaré’s disk via
where and . We have that
A simple calculation using this expression and the fact that reveals that
Therefore, from the -integrability of , cf. item (c) above, we deduce that
In summary, the validity of (4) essentially follows from the -integrability condition on in item (c).
2.2. Interpretation of as a Birkhoff sum
Similarly to the case of , we want to convert into Birkhoff average. Again, let us consider the random walk on , and let us write
might be not true. Indeed, we have no prior information on the relationship between and , so that we can not use item (c) to get the integrability (contrary to the case of the Lebesgue measure where could be computed explicitly). Fortunately, it is not hard to overcome this little technical difficulty: as it turns out, the ergodic theorem also applies to observables that are bounded only on one side by a -integrable function; in particular, we can apply the ergodic theorem to because
2.3. Construction of a “weird” recurrence set when
During this subsection, let us assume that . Recall that the plan is to show that the Birkhoff sums of of are very well controlled by the Birkhoff sums of .
Using the properties of the Radon-Nikodym derivative (e.g., ), we can rewrite the numerator in the left-hand side of this equation as:
From this and (9) we deduce that
with probability . Since ‘s are independent of , we conclude from (11) that, for almost every , one has
for almost all random paths .
In particular, we can fix two distinct values and of so that (11) holds for almost every random path. For , let us consider the random variables
We are interested in the properties of (as is the random walk on ) but (11) provides information only about . Fortunately, and have the same distribution, so that all probabilistic statements about are also true for . In particular, for each , the probabilities of the events
go to for because the probabilities of the events
go to for in view of the fact that (11) implies
with probability (for ).
Therefore, if we choose a sequence going very fast to infinity as so that the sum of the probabilities of the events
is finite (for ), then we can use the Borel-Cantelli lemma to obtain that
with probability . In particular, it follows that the set
is a recurrence set for the random walk (i.e., this random walk visits infinitely often with probability ).
Now, if , we can take and such that
In other words, the density is very well-controlled by with a “margin” coming from the assumption that .
From this nice control our plan is to prove that the recurrence set has the “weird” property referred to in Proposition 7, i.e., we will show that
Keeping this goal in mind, given , let us denote by
By (12), we can bound the quantity as follows:
and we observe that , we can estimate the right-hand side of (13) as
Since was chosen so that (assuming ), we have that the right-hand side of this estimate is convergent if we can show that grows linearly (at most), i.e., the proof of Proposition 7 is complete once we can handle the counting problem of showing that
We will exploit the explicit nature of the densities in order to show this (counting) lemma. More precisely, given , recall that
if acts on Poincaré’s disk as with and .
Since and are distinct, the complex number can’t be close to both of them at the same time. Using this information, the reader can see that
for some constants and . Equivalently, since , one has
for some constants and .
In particular, Lemma 9 is equivalent to show that where
Actually, since the subset of elements with is finite (namely, it is the intersection of the lattice with the compact subgroup stabilizing ), we can convert the counting problem
for elements of into the following geometrical counting problem about points :
Now, this geometrical counting problem is not hard to solve, at least when is cocompact.
Indeed, let us consider first a large compact subset of containing a fundamental domain of about the origin . Then, by definition, the -translates of cover and, hence,
where is an appropriate constant (depending on ) and is the area of the hyperbolic disk of radius centered at .
Next, let us consider a small compact ball of around so that it is disjoint from its -translates. Then, we have that
where is an appropriate constant (depending on ).
In summary, there are two constants and such that
On the other hand, the area of the hyperbolic disk of radius centered at is not hard to compute:
where is the Euclidean radius of , i.e., . From this expression we see that
so that this ends the proof of Lemma 9.
This completes the proof of Proposition 7.
3. Proof of Proposition 8
Closing this post, let us show that the properties of -harmonic functions do not allow the existence of the “weird” recurrence sets constructed in Proposition 7. For this sake, let us suppose by contradiction that is a recurrence subset of such that
Next, let us observe the following facts. Firstly, since is -stationary and is fully supported on (cf. item (a) above), we have that is absolutely continuous with respect to and the density is bounded because
so that . Secondly, from the previous identity, we see that
so that, for almost every , the function
In particular, our plan is to use the mean value property of -harmonic functions to express the values of in terms of its values in in order to eventually contradict (14).
For this sake, let us show the following elementary abstract lemma about the mean value property of bounded -harmonic functions with respect to recurrence sets:
Lemma 10 Let be a discrete group with a probability measure and denote by a stationary sequence of independent random variables with distribution . If is a recurrence set of the random walk and is a bounded -harmonic function on , then the following mean value property with respect to holds:
where is the distribution of the first point of hit by .
Proof: We start with the usual mean value property
Now, for each term we can independently decide whether we want to use again the mean value relation to express as a convex combination of or not. Since our ultimate goal is to write as a convex combination of the values of on the recurrence set , we will take our decision as follows: if , we leave alone, and, otherwise, we apply the mean value relation.
After steps of this procedure, we have
where “something” is a combined weight of contributions coming from the values of on points outside that were reached by the random walk after steps.
Because is a recurrence set, the random walk reaches with probability . Therefore, since the function is bounded, we can pass to the limit as in the identity above to get the desired equality
This proves the lemma.
Coming back to the context of Proposition 8, we observe that this lemma does not apply directly to the -harmonic density function
because it might be unbounded.
Nevertheless, by revisiting the argument of the proof of the lemma above, one can easily check that, for an unbounded -harmonic (integrable) function , one has the mean value inequality
(but possibly not the mean value equality ) where is the probability that the first point of hit by the random walk is .
In any event, using this mean value inequality with we deduce that
for almost every .
In particular, we conclude that
Thus, in view of (14), we obtain that
that is, the total probability that the random walk hits is strictly smaller than , a contradiction with the fact that is a recurrence set of the random walk!