Last time we introduced Poisson boundaries hoping to use them to distinguish between lattices of and , . More precisely, we followed Furstenberg to define and construct Poisson boundaries as a tool that would allow to prove the following statement:
Here, we recall that, very roughly speaking, these Poisson boundaries were certain “maximal” topological objects attached to locally compact groups with probability measures in such a way that the points in the boundary were almost sure limits of -random walks on and the probability measure was a -stationary measure giving the distribution at which -random walks hit the boundary.
For this second (final) post, we will discuss (below the fold) some examples of Poisson boundaries and, after that, we will sketch the proof of Theorem 1.
1. Some examples of Poisson boundaries
1.1. Abelian groups
The Poisson boundary of a locally compact Abelian group with respect to any probability measure is trivial.
Indeed, in terms of the characterization of Poisson boundaries via -harmonic functions, this amounts to say that bounded -harmonic functions on locally compact Abelian groups are constant, a fact proved by Choquet-Deny here.
More generally, there is a natural notion of random walk entropy (cf. Furman’s survey) allowing to characterize the triviality of Poisson boundary: more precisely, Kaimanovich-Vershik showed here that a discrete countable group equipped with a probability measure has trivial Poisson boundary if and only if .
Remark 2 In addition to these results, it is worth to mention that:
- Furstenberg showed that any probability measure on a locally compact non-amenable group whose support generates admits bounded non-constant -harmonic functions;
- Kaimanovich-Vershik and Rosenblatt showed that given a locally compact amenable group , there exists a (symmetric) probability measure with full support such that all bounded -harmonic functions on are constant.
See Furman’s survey for more comments on these results (as well as related ones).
1.2. Free group on 2 generators
Let be the free group on generators and equipped with the Bernoulli probability measure for . Last time, during the naive description of the Poisson boundary, we saw that the space
of infinite words on satisfy the non-cancellation condition for all was a natural candidate for a boundary of (as far as -random walks were concerned).
1.2.1. Computation of the Poisson boundary of
As the reader can imagine, equipped with an adequate measure is the Poisson boundary of . The proof of this fact goes along the following lines. Firstly, let us show that is a -space, i.e.,
Lemma 2 acts continuously on (for the topology of pointwise convergence).
Proof: The action of on is defined by continuity: given and , , let . In other words, if , then is the infinite word obtained by concatenation of and , and then collapsing if necessary: e.g., if , then
- , and , and
This proves the lemma.
Next, let us consider a stationary sequence of independent -valued random variables with distribution (so that ). We claim that converges to a point of with probability :
Lemma 3 with probability .
Proof: The basic idea to get convergence is simple: we write the product , we delete all pairs canceling out (i.e., ), and we continue to delete “canceling pairs” until we get a point in .
Of course, this procedure fails to produce a limit point only if the product keeps degenerating indefinitely.
However, this possibility occurs only with zero probability thanks to the transience of the random walk on the free group (i.e, the fact that this random walk wanders to with full probability; cf., e.g., this classical paper of Kesten).
Equivalently, letting denote the length of an element (i.e., the minimal length of products expressing in terms of the generators ), the transience of means that as with probability . In particular, with probability , for each , we can select the largest value of such that , so that the right multiplication of by will not change the first letters of the product , and, a fortiori, we get (with probability ) a well-defined limit word that we can denote without any ambiguity
This proves the lemma.
An immediate corollary of the proof of this lemma (and the definition of pointwise convergence) is:
Another way of phrasing this corollary is:
carries only one -stationary measure .
Indeed, from the limit point provided by the previous lemma, we can define a sequence of -valued random variables by setting and it is not hard to check that is a -process (as defined in the previous post). Therefore, if is a -stationary probability measure on , the convergence asserted in the corollary and Proposition 7 of the previous post says that is the distribution of a -process. Furthermore, Proposition 5 of the previous post asserts that this -process is exactly , that is, the unique -stationary measure on is the distribution of the -process .
In summary, denoting by the unique -stationary on (namely, the distribution of ), our discussion so far shows that is a boundary of . In fact, one can make this boundary entirely explicit because it is not hard to guess :
Lemma 5 Let be the probability measure on giving the maximal independence between the coordinates, i.e., takes one of the four values with equal probability, takes one of the three possible values with equal probability, etc., so that the -measure of a cylinder obtained by prescribing the first -entries is , i.e.,
Proof: From our previous discussion, it is sufficient to check that is -stationary, i.e., . This fact is not hard to check: we have just to verify that the -integral and the -integral of characteristic functions of cylinders coincide, i.e., we have to compute (and show the equality of) the -measure and the -measure of any given cylinder
In this direction, note that, by definition,
On the other hand, for , we have that is a cylinder of size (i.e., corresponding to the prescription of entries) unless where is a cylinder of size . In particular,
By plugging this into the previous equation, we deduce that
as desired. This proves the lemma.
At this point, it remains only to check that is the Poisson boundary, i.e., all boundaries of are equivariant images of and the Poisson formula for -harmonic functions.
We will skip the proof of this last fact because it would lead us far from the scope of this post. Instead, we refer to Dynkin-Maljutov paper where it is shown that is the Martin boundary (and, a fortiori, the Poisson boundary).
1.2.2. A law of large numbers for
Using our knowledge of the Poisson boundary of , let us sketch a proof of the fact that
as with probability (where is the length of ).
Firstly, let us observe that is absolutely continuous with respect to for each . Indeed, since is -stationary, i.e.,
our claim is true for the generators of , and, a fortiori, it is also true for all .
Actually, the Radon-Nikodym density is not hard to compute: if is the limit of the sequence , then one can check by induction on that
Consider now the quantity
For each , we know that . From this, it is possible to prove (after some work [related to the fact that we want to let and vary…]) that
By Birkhoff’s ergodic theorem, the time-average on the right-hand side of this equality converges to the spatial-average with probability , i.e.,
Using (1), we can compute this integral as follows. By definition of ,
Now, for each and , we have that unless in which case . In particular, from (1), we get that
for each . By plugging this into the previous equation, we obtain that
i.e., as with probability .
1.2.3. A random walk on the free group on generators
The arguments of the previous subsubsections also apply to the free groups on generators equipped with the Bernoulli probability assigning equal probabilities to the singletons consisting of the generators and their inverses.
However, the simple-minded extension of this discussion to the free group on countably many generators fails because we do not have a Bernoulli measure.
Nevertheless, it is possible to equip with some probability measure such that the Poisson boundary of coincides with the Poisson boundary of .
In fact, this is so because “behaves like a lattice” of . More concretely, the important fact about is that it is isomorphic to the commutator subgroup of . Using this fact, we can formalize the idea that is a lattice of through the following lemma:
Lemma 6 (or, more precisely, the commutator subgroup ) is a recurrence set for the random walk on in the sense that meets infinitely often with probability .
Proof: The quotient group is the free Abelian group on generators, i.e., . Now, by projecting to the random walk on , we obtain the simple random walk on . In this notation, the assertion that is a recurrence set of corresponds to the well-known fact that the simple random walk on returns infinitely often to the origin.
From this lemma, we can construct (at least heuristically) a probability measure on whose Poisson boundary is the same of (namely, ). In fact, letting be the random walk on , we know that (with probability ) there is a subsequence hitting . On the other hand, converges to a limit point . In particular, it follows that the boundary points are also limits of the -valued random variables . Thus, if one can show that are independent random variables with a fixed distribution , then has Poisson boundary . Here, the keyword is the strong Markov property of : more precisely, we notice that from to we multiply (to the right) ; since and belong to , we have that and, in particular, this suggests that is the distribution of the position of at the first time that enters ; of course, to formalize this, we need to know that the the entrance times of on are themselves random variables (or rather that are still independent random variables) and this is precisely a consequence of the strong Markov property of .
Alternatively, one can show that has Poisson boundary by working with harmonic functions. More precisely, our task consists into showing that the restrictions to of -harmonic functions on are -harmonic and all -harmonic functions on arise in this way (by restriction of -harmonic functions on ). In this direction, one recalls that, for each , the quantity is the probability that takes value at the first time it enters , and then one shows the desired facts about -harmonic functions versus -harmonic functions with the aid of the following abstract lemma:
Lemma 7 Let be a discrete, a probability measure on and a stationary sequence of independent random variables with distribution . Suppose that is a recurrence set of and let be a -harmonic function. Then,
where is the distribution of the first point of hit by .
We refer to pages 31 and 32 of Furstenberg’s survey for a (short) proof of this lemma and its application to the computation of the Poisson boundary of .
Let , , and denote by the (maximal compact) subgroup of orthogonal matrices in and the (Borel or minimal parabolic) subgroup upper triangular matrices in . From the Gram-Schmidt orthogonalization process, we know that .
Definition 8 A probability measure on is called spherical if
- is absolutely continuous with respect to Haar measure on and
- is -bi-invariant, i.e., for all , or, equivalently, for all .
Since acts transitively on , there exists an unique -invariant measure on that we will denote (philosophically, this is comparable to the fact that the unique translation-invariant measure on the real line is Lebesgue). Now, given any probability measure on and any spherical measure on , note that is a -invariant probability measure on , and, a fortiori, . In particular, is the unique -stationary measure on .
In this context, Furstenberg proved the following theorem:
Theorem 9 (Furstenberg) is the Poisson boundary of whenever is a spherical measure.
We will not comment on the proof of this theorem here: instead, we refer the reader to the original article of Furstenberg for more details (and, in fact, a more general result valid for all semi-simple Lie groups ).
An interesting consequence of this theorem is the following fact:
Corollary 10 The class of -harmonic functions on is the same for all spherical measures .
Proof: By the definition of the Poisson boundary, the fact (ensured by Furstenberg’s theorem) that is the Poisson boundary of means that we have a Poisson formula
giving all -harmonic functions on as integrals of bounded measurable functions on . Since the right-hand side of this equation does not depend on (only on ), the corollary follows.
From this corollary, it is natural to define a harmonic function on as a -harmonic function with respect to any spherical measure .
For later use, let us describe more geometrically the Poisson boundaries and (resp.) of and (resp.). In fact, we already hinted that, in general, is the complete flag variety of , but we will discuss the particular cases of and because our plan is to show Theorem 1 in the context of and only.
1.3.1. Poisson boundary of
Consider the usual action of on and the corresponding induced action on the projective space of directions/lines in associated to non-zero vectors .
By definition, the subgroup of upper triangular matrices in is the stabilizer of the direction generated by the vector . In particular, since acts transitively on , we deduce that .
Let us now try to understand how is attached to in terms of the measure topology described in the previous post.
By definition, an element is close to whenever the measure is close to .
On the other hand, a “large” element
i.e., a matrix with large operator norm has at least one large of the column vector, either or , and, if both column vectors are large, then their directions and are close by unimodularity of (indeed, if they are both large and they form a rectangle of area , then their angle must be small). In particular, we conclude that, for large elements , most directions are close to the larger of , except when is approximately orthogonal to , (where is the transpose of ).
In summary, we just proved the following lemma:
An interesting consequence of this lemma concerns the action of large elements of on non-atomic measures on .
Indeed, given an arbitrary non-atomic probability measure on , we have that for each there exists such that any interval of size has -measure (because any point has zero -measure by assumption).
In particular, by combining this information with the previous lemma, we deduce that any large element of moves most of the mass of to a small interval. More precisely, denoting by and the intervals of sizes provided by the lemma, we have that , i.e., . Therefore, (since ), i.e., most of the mass of is concentrated on . In other words, we proved the following crucial lemma:
An alternative way of phrasing this lemma is: as , it approaches a definite point of . In particular, is compact, that is, we get a compactification of by attaching ! (This result is comparable to Corollary 4 in the context of free groups.) As we will see below, this compacteness property is no longer true for (i.e., in the context of ), and this is one of the main differences between the Poisson boundaries of and . As the reader might imagine, this observation will be important for the proof of Theorem 1 (that is, at the moment of distinguishing between the lattices of and ).
Note that, from this lemma, we can show directly that is a boundary of whenever is spherical (without appealing to Furstenberg’s theorem above). Indeed, denoting by the random walk with respect to , we have that, by definition, is a boundary of if and only if converges to a Dirac mass with probability . Since is a compact space, we know that converges to some measure with probability (cf. Corollary 11 of the previous post) and the previous lemma says that this measure is a Dirac mass unless the products stay bounded (with positive probability). Now, a random product of elements in a topological group remains bounded (with positive probability) only if the distribution is supported on a compact subgroup (by Birkhoff’s ergodic theorem and the fact that a compact semigroup of a group is a subgroup), but this is not the case as is spherical (and thus absolutely continuous with respect to Haar).
Finally, once we know that is a boundary of , i.e., converges to a Dirac mass where is a -process on , we can obtain the following interesting consequences. Firstly, , i.e., the random walk is transient. Moreover, the direction of the larger of the column vectors of converges with probability to a direction , and, in fact, it can be shown that both column vectors become large. In particular, by unimodularity, it follows that, with probability , both column vectors of converge to a random point (direction) of .
1.3.2. Poisson boundary of
By Furstenberg’s theorem, is the space underlying the Poisson boundary of equipped with any spherical measure . Here, we recall that is the subgroup of upper triangular matrices in .
We proceed as before, i.e., let us consider the usual action of on and the induced action on the space of pairs of points in the projective space corresponding to orthogonal directions and in .
Note that acts naturally on via (our choice of is natural because the orthogonality condition is preserved and is an automorphism of [in particular, our matrices are multiplied in the “correct order”]). Moreover, this action is transitive, so that is the quotient of by the stabilizer of a point of . For sake of concreteness, let us consider the point (where are the vectors of the canonical basis of ) and let us determine its stabilizer in . By definition, if , say
stabilizes , then and , i.e.,
that is, is upper triangular. In other words, is the stabilizer of and, thus, .
Next, let us again try to understand how is attached to in terms of the measure topology. Once more, by performing the random walk with distribution , one can check that the column vectors of converge to a limit vector in . In terms of the boundary , we can recover by letting be the point (-process) of obtained as the almost sure limit of .
Of course, in this description, the role of remains somewhat mysterious, and so let us try to uncover it now. By definition, is the limit point of the column vectors of , i.e., the random walk obtained from by applying the automorphism . On the other hand, if are the column vectors of , then the vectors , and are the column vectors of (where is the vector product of ). In particular, the fact that these vectors converge to means that the perpendicular directions to the 3 planes passing through the vectors converge to , i.e., all of these 3 planes converge to the plane perpendicular to .
Finally, let us make more precise the observation from the previous subsubsection that the boundary behaviors of elements in and are different as this is one of the main points in the proof of Theorem 1.
Once more, let us point out that Lemma 12 above says that any element approaches the boundary (i.e., becomes large) by getting close to a specific point . On the other hand, this is no longer true for elements : for instance, the sequence
goes to infinity without getting close to any specific point of because its larger column vectors and do not converge together to a single direction in . Instead, it is not hard to convince oneself (by playing with the definitions) that the sequence of measures converges to a measure supported on the great circle . In general, a great circle is a fiber of one of the two natural fibrations and it is possible to show that the measures can approach measures supported on any given great circle.
Anyhow, the basic point is that is compact but is not compact. For sake of completeness, let us point out what are the measure that we must add to to get a compactification of . It can be proven that any convergent sequence tends to a Dirac mass, a circle measure (i.e., a measure supported on a great circle) or an absolutely continuous measure with respect to (this last case occurring only if is bounded in ). Moreover, in the case of getting a circle measure as a limit, we get a very specific object: by identifying the great circle with and denoting by the Lebesgue measure, one has that all circle measure have the form where (and acts on its boundary ).
1.3.3. Harmonic functions on
Closing this quick discussion on the Poisson boundaries of , let us quickly comment on the relationship between -harmonic functions on and classical harmonic functions on Poincaré’s disk.
Let be a spherical measure on . For the sake of simplicity of notation, we will say that a random walk with respect to a spherical measure is a Brownian motion. By definition of sphericity of , the transition from to has the same probability as the transition from to , and from to for each . In particular, the Brownian motion can be transferred to a Brownian motion on the symmetric space of .
Now, we recall that is Poincaré’s disk /hyperbolic upper-half plane : more concretely, by letting act on the upper-half plane by Möebius transformations, i.e., isometries of equipped with the hyperbolic metric,
for , we see that is the stabilizer of . In particular, is naturally identified with via , and, hence, we can also think of as the Poincaré’s disk after considering the fractional linear transformation sending to in such a way that is sent to the origin and is sent to .
In summary, we can think of the Brownian motion as a Brownian motion on Poincaré’s disk. Here, it is worth to point out that the transitions of are not given by group multiplication as acts on the left and the ‘s are multiplied from the right. Finally, if we transfer the measure topology on to , we get the usual Euclidean topology on the closed unit disk . Indeed, suppose that with . Then, transfers most of the mass of to a point of close to . In particular, if , then , that is, converges to , and, a fortiori, the Euclidean topology on is the topology we get after transferring the measure topology.
Finally, note that the value of a harmonic function (with respect to any spherical measure) depends only on the coset (by sphericity and the mean value property of harmonic functions). Thus, induces a function on . By Poisson’s formula (in the definition of Poisson boundary), we have that
On the other hand, by computing the density (using that acts via Möebius transformations and is the Lebesgue measure), we can show that
where is the classical Poisson kernel on the unit disc. In other words, by letting , we obtain that
and, hence, the function is harmonic in the classical sense. In summary, the two notions of `harmonic’ are the same. Probabilistically speaking, the formula above says that the value is obtained by integrating the boundary values with respect to the `hitting measure on the boundary starting at ‘ (as is the distribution of the limit of [because is the distribution of the limit of and by invariance of the Brownian motion under the group]).
1.4. Mapping class group and Teichmüller space
As it is “customary”, the mapping class groups and Teichmüller spaces are very close to lattices in Lie groups and homogenous spaces. In particular, this partly motivates these two articles of Kaimanovich and Masur about the Poisson boundary of the mapping class group and Teichmüller spaces, where it is shown that it is the Thurston compactification (via projective measure foliations) equipped with a natural harmonic measure. Of course, it is out of the scope of this post to comment on this subject and we refer the curious reader to (very well-written) papers of Kaimanovich-Masur.
2. Poisson boundary of lattices of
After this series of examples of Poisson boundaries, let us come back to the proof of Theorem 1. At this stage, we know that equipped with any spherical measure has Poisson boundary and now we want to distinguish between lattices of .
As we mentioned in the previous post, the basic idea is that a `nice’ random walk in a lattice of should see the whole boundary of . In fact, this statement should be compared with the results in Subsubsection 1.2.3 above where we saw that an adequate random walk in the free group in generators sees the whole boundary of the free group in two generators because behaved like a lattice in , or, more accurately, it was a recurrence set for the symmetric random walk in .
However, this heuristic for free groups can not be applied ipsis-literis to lattices of because a countable set can not be a recurrence set for a Brownian motion in . Nevertheless, one still has the following result:
Theorem 13 (Furstenberg) If is a lattice of , then there exists a probability measure on such that the Poisson boundary of coincides with the Poisson boundary of (with respect to any spherical measure).
In order to simplify the exposition, we will restrict our attention to the low dimensional cases. More concretely, we will sketch the construction of in the case of cocompact lattices in and we will show that has as a boundary. However, we will not enter into the details of showing that is the Poisson boundary of : instead, we refer the reader to Furstenberg’s survey for a proof in the case of (i.e., ) and his original article for the general case.
2.1. Construction of in the case
Consider again the symmetric space associated to . This space has a natural -invariant metric and, using this metric, we have a function
measuring the distance to the origin. Note that, in the particular case , the function has a very simple expression:
- (a) , i.e., for all ,
- (b) is -stationary, i.e., ,
- (c) the restriction of the function to is -integrable, i.e.,
Remark 3 The condition (b) above implies that the restriction to of an arbitrary harmonic function on is -harmonic. In particular, this means that a harmonic function on satisfies plenty (at least one per cocompact lattice of ) of discrete mean-value equalities.
For the proof of this proposition, we will focus on the construction of a measure satisfying item (b) and then we will adjust it to satisfy items (a) and (c). Also, we will discuss only the case of cocompact lattices in .
In this direction, let us re-interpret item (b) in terms of the Brownian motion on Poincaré’s disk (that is, the symmetric space of ). As we saw above, is the hitting distribution on of the Brownian motion starting at .
In particular, the stationarity condition in item (b) says that the hitting probability on starting at the origin is a convex linear combination of the hitting probabilities on starting at the points (for ). This hints how we must show item (b): the measure will correspond to the weights used to write is a linear convex combination of , . Keeping this goal in mind, it is clear that the following lemma will help us with our task:
Lemma 15 Let be a cocompact lattice of and denote the hitting probability on of a Brownian motion starting at . Then, there are two constants and such that, for any , one has
where , the points and are within a hyperbolic distance of , and is a non-negative measure of total mass .
Proof: Since is cocompact, it has a compact fundamental domain. In particular, we can find a large constant such that the hyperbolic ball of radius around any contains in its interior at least one point of the form with .
For sake of simplicity of the exposition, during this proof, we will replace the discrete-time Brownian motion by its continuous-time analog for a technical reason that we discuss now.
For each and in the interior of , the hitting measure on starting at can be computed by noticing that a continuous-time Brownian motion emanating from must hit the circle before heading towards . Of course, the same is not true for the discrete-time Brownian motion (as we can jump across ), but we could have overcome this small difficulty by considering an annulus around . However, we will stick to the continuous-time Brownian motion to simplify matters. Anyhow, by combining this observation with the strong Markov property of the Brownian motion, one has that
where is the hitting distribution of for a Brownian motion starting at (i.e., for each interval , the measure is the probability that a Brownian motion starting at first hits at a point in .
Next, let us consider the points of the form , , inside (the interior of) , choose positive numbers and consider the measure
At this point, we observe that the measures are absolutely continuous with respect to the Lebesgue measure on whenever belongs to the interior of (as our Brownian motion is guided by a spherical measure [by definition]). Therefore, by taking small (depending on how close is to the circle ), we can ensure that the measure
appearing in the right-hand side of (2) is positive. Furthermore, note that this scenario is –invariant: if we replace by for , the circle is replaced by and the elements are replaced by , but we can keep the same values of .
In other words, the values of (making the measure in (3) positive), and, a fortiori, the quantity , depends only on the coset where satisfies . Therefore, by the compactness of , we can find some such that, for all , the values of (making (3) positive) can be chosen so that
In particular, it follows that is a positive measure of total mass .
In summary, we managed to write
where is a positive measure of total mass , as desired.
By taking in this lemma, we know that one can write as a convex linear combination of a “main contribution” coming from ‘s (where the distance from ‘s to are ) and a “boundary contribution” coming from an integral of with respect to a measure of total mass .
From this point, the idea to construct a probability measure on satisfying item (b) of Proposition 14 is very simple: we repeatedly apply the lemma to the ‘s appearing in the “boundary contribution” in order to push it away to infinity; here, the convergence of this procedure is ensured by the fact that the boundary measure loses a definite factor (of ) of its mass at each step.
By applying the lemma to , we also have
where only for , and is supported in and it has total mass .
By combining these equations, we deduce that
where only for , and is a positive measure supported on whose total mass is
In particular, by setting , we find that
that is, the stationarity condition of item (b) of Proposition 14 is proved.
Now, we claim that item (c) of Proposition 14 is satisfied by . Indeed, by construction, for the elements with , the quantity comes from the -combination of the contributions of the measures in the right-hand side of (4). Since the measure has total mass , we deduce that
In particular, given that any element with appears times in the sum above, we conclude that
that is, the -integrability condition on in item (c) of Proposition 14 is verified.
Finally, the full support condition in item (a) of Proposition 14 might not be true for the probability measure constructed above. However, it is not hard to fulfill this condition by slightly changing the construction above. Indeed, it suffices to add all points at distance from in the th step of the construction of and then assign to them some tiny but positive probabilities so that the measure in the right-hand side of (4) is still positive. In this way, we are sure that in the end of the construction of , all ‘s in were assigned some non-trivial mass.
This completes the sketch of the proof of Proposition 14 in the case (i.e., cocompact lattices in ).
After constructing , let us show that the Poisson boundary of equipped with a spherical measure is a boundary of .
2.2. is the Poisson boundary of
A reasonably detailed proof that is the Poisson boundary of is somewhat lengthy because the verification of the maximality property (i.e., any boundary of is an equivariant image of ) needs a certain amount of computation (in fact, we might come to this point later in a future post, but, for now, let us skip this point). In particular, we will content ourselves with checking only that is a boundary of in the cases and .
As it turns out, the fact that is a boundary of (in the case ) is not hard. By definition, we have to show that a stationary sequence of independent random variables with distribution has the property that converges to a Dirac mass with probability . On the other hand, by Corollary 11 of the previous post and the compactness of , we know that converges to some measure with probability , and, by Lemma 12, this limit measure is a Dirac mass if the elements are unbounded. Now, it is clear that these elements are unbounded because has distribution and, by construction (cf. item (a) of Proposition 14), is fully supported on a lattice of (and, thus, its support is not confined in a compact subgroup).
Next, let us show that is a boundary of (in the case ). In this direction, we will need the following lemma playing the role of an analog of Lemma 12 in the context of :
- (i) has a rich support: the support of is not confined to a compact or reducible subgroup of ;
- (ii) the norm function is –-integrable: ;
- (iii) is -stationary:
Then, for any stationary sequence of independent random variables with distribution , the sequence of measures converges to a Dirac mass on with probability .
Before proving this lemma, let us see how it allows to prove that is a boundary of for the measure constructed in Proposition 14, thus completing our sketch of proof of Furstenberg’s theorem 13. By definition of boundary, it suffices to check that the measure provided by Proposition 14 fits the assumptions of Lemma 16. Now, by item (a) of Proposition 14, is fully supported on the lattice of . Since a lattice of is Zariski dense (by Borel’s density theorem), is not a compact subgroup nor reducible subgroup of , that is, satisfies item (i) of the lemma above. Next, we notice that a computation shows that the distance function to origin in the symmetric space satisfies and . In particular, the integrability condition in item (c) of Proposition 14 implies that satisfies item (ii) of the lemma above. Finally, the item (b) of Proposition 14 is precisely the stationarity condition in item (iii) of the lemma above.
So, let us complete the discussion in this section by sketching the proof of Lemma 16.
Proof: By the compactness of (and Corollary 11 of the previous post), we know that converges to some measure with probability . This allows us to consider the sequence
of measure-valued random variables.
Our task consists into showing that are Dirac masses. Keeping this goal in mind, note that and is independent of for . Also, let us observe that we can extend the sequence can be extended to non-positive indices by relabeling as and shifting the remaining variables. Here, by stationarity of (by definition) and (by item (iii)), the variables with positive indices are probabilistically isomorphic to the original sequence (before shifting). In other words, we can assume that our sequence to is defined for all integer indices . (In terms of Dynamical Systems, this is analog to taking the natural extension , of the unilateral shift , ). In any case, we can write where is a stationary sequence of independent random variables and is independent of ‘s.
At this point, let us recall the discussion of Subsubsection 1.3.2 on the Poisson boundary of . In this subsubsection we saw that there are only three possibilities for any limit of the measures , such as : it is either a Dirac mass, a circle measure or an absolutely continuous (w.r.t. ) measure, the latter case occurring only when stays bounded in .
By ergodicity (of the shift dynamics underlying the sequence ), we have that only one of these possibilities for can occur with positive probability.
Now, can not be absolutely continuous w.r.t. because this would mean that is bounded (with positive probability) and, a fortiori, is confined to a compact subgroup of , a contradiction with our assumption in item (i) about the distribution of ‘s.
Therefore, our task is reduced to show that ‘s are not circle measures with probability . For sake of concreteness, let us fix by assuming that is a circle measure supported on our “preferred” circle . In order to show that are Dirac masses, it suffices to check that the angles between the column vectors of the matrices converge to as . In other terms, denoting by , , the column vectors of , and by noticing that , , play symmetric roles, we want to check that
The idea to show this is based on the simplicity of the Lyapunov spectra of random products of matrices in with a distribution law that is not confined to compact or reducible subgroups. More concretely, we will show that the column vectors and have a definite exponential growth
with (the top Lyapunov exponent) and, similarly, the column vector of the matrix has a definite exponential growth
where (the sum of the two largest Lyapunov exponents) satisfies (i.e., the top Lyapunov exponent is simple, i.e., it does not coincide with the second largest Lyapunov exponent). Of course, if we show these properties, then converges exponentially to as (since ).
Let us briefly sketch the proof of these exponential growth properties. We write as a “Birkhoff sum”
where for and . As it turns out, the sequence is not stationary, but it is almost stationary, so that Birkhoff’s ergodic theorem says that the time averages converge to the spatial average (with probability ):
where is the rotation-invariant distribution of ‘s. Logically, this application of the ergodic theorem is valid only if we check that the observable is integrable. However, this is not hard to verify: by definition, , so that the desired integrability is a consequence of the integrability requirement in item (ii). A similar argument shows that
where . So, it remains only to check the simplicity condition . For this sake, we combine the two integrals defining and , and we transfer it from to . In this way, we obtain:
On the other hand, by definition, runs over orthogonal pairs of directions in . Thus, since is not confined to an orthogonal subgroup of , we have that the integral in the right-hand side of the equation above is strictly positive, i.e., .
In summary, we showed that, for each fixed , the sequence converges to a Dirac mass with probability . Since are independent of , this actually proves that converges to Dirac masses independently of . Finally, since the sequence is stationary, we conclude that the were Dirac masses to begin with.
This completes the proof of Lemma 16.
Closing this post, we will use the fact that is the Poisson boundary of (where is the probability measure constructed in Proposition 14) to show Furstenberg’s theorem 1 that the lattices of are “distinct” from the lattices of .
3. End of the proof of Furstenberg’s theorem 1
In this section, we will show the following statement (providing a slightly stronger version of Theorem 1) in the case :
The proof of this theorem proceeds by contradiction. Suppose that is isomorphic to a cocompact lattice of and also to a subgroup of .
By Theorem 14, we can equip with a fully supported probability measure such that has Poisson boundary .
Let us think now of as a subgroup of . We claim that can not be confined to a compact subgroup of : indeed, if this were the case, would be Abelian; however, we saw that Abelian groups have trivial Poisson boundary, while .
Next, let us observe that admits some -stationary probability measure , i.e., . Indeed, this is a consequence of the Krylov-Bogolyubov argument: we fix an arbitrary probability measure on the compact space , and we extract a -stationary probability as a limit of some convergent subsequence of the sequence
We affirm that is a boundary of . In fact, given a stationary sequence of independent random variables with distribution , we know that the elements are unbounded as is a non-compact subgroup of (as we just saw) and is fully supported on . By Lemma 12, it follows that converges to a Dirac mass, and so, by Proposition 7 of the previous post, we deduce that is a boundary of .
By definition of Poisson boundary, the facts that has Poisson boundary and is a boundary of imply that is an equivariant image of under some equivariant map . We will prove that this is not possible (thus completing today’s post). The basic idea is that the sole way of going to infinity in is to approach , i.e., converges to a Dirac mass as in , but, on the other hand, we can go to infinity in without approaching , i.e., we can let in in such a way that converges to a circle measure. Then, in the end of the day, these distinct (and incompatible) boundary behaviors (Dirac mass versus circle measure) will lead to the desired contradiction.
In order to get Dirac mass behavior in the context of , our plan is to apply Lemma 12. But, before doing so, we need to know that is not atomic, a property that we claim to be true. Indeed, suppose to the contrary that for some point , and denote by , a set of -measure . Consider the translates of under . On one hand, if intersects in a subset of positive measure, then their images and under in intersect, and, thus, by equivariance of and the fact that is a singleton, , and, a fortiori, . In particular, the property that whenever intersects with positive measure implies that does not act ergodically on (because the ‘s, , do not get mixed together). However, this is a contradiction with Moore’s ergodicity theorem (implying that a lattice of acts ergodically on ). This shows that is non-atomic.
In particular, given and two disjoint closed subsets of , if is any sequence going to , then
Indeed, this is so because Lemma 11 (and the proof of Lemma 12) says that, for each , the measure concentrates at least of its mass in an interval of length for all sufficiently large. In particular, for any smaller than the distance separating the disjoint closed sets and (i.e., ), we obtain that either or for sufficiently large, as desired.
We can rephrase the “concentration property” of the last paragraph in terms of -harmonic functions as follows. Let , , be measurable functions supported on , , and consider the associated -harmonic functions
Now, let us “transfer” this picture to the context, i.e., let us think of the -harmonic , , as defined on . By the Poisson formula (and the fact that ), we can represent the -harmonic functions , , as
In what follows, we will try to understand the boundary behavior of ‘s, ultimately to contradict the concentration property (6). In this direction, let us first “transfer” the concentration property (6) from to as follows. Observe that is a cocompact lattice of , so that we can select a bounded fundamental domain , i.e., a bounded set such that any element of has the form with and . Now, given , let us compare the values of and one of its “neighbors” , . Here, we observe that
In particular, since the right-hand side is bounded for in the bounded set , we conclude that the ratio between
is uniformly bounded away from and for and . Therefore, since the values of a positive -harmonic function at and can be written as
we deduce that the values and are uniformly comparable for and , that is, there exists a constant such that
for all and . Hence, given , there exists such that
because for some , . Furthermore, when letting , we have that (as and is bounded). Thus, by combining this with the “concentration property” (6), we get the following concentration property in the context of :
Let us now try to contradict the “transferred concentration property” (8) by analyzing the values when without approaching (something that is not possible in !). More concretely, let us consider the sequences
We want to investigate the “boundary” values of the harmonic functions and along the sequences and (and some adequate translates). For this, we need an analog of Fatou’s theorem saying that harmonic functions have boundary values along almost all radial direction. In our context, the analog of Fatou’s theorem goes as follows.
The limits of the sequences and are circle measures and supported on the great circles
Also, it is possible to check that all circle measures have the form or for in the orthogonal subgroup of . Now, we affirm that, given a bounded measurable function on , the integrals
are defined for almost every . Indeed, we recall that acts transitively on , so that where is a finite group, i.e., is a finite cover of . The great circles and correspond to two -parameter subgroups and of (as any great circle of passing through the identity coset). Also, the great circles of are just the cosets and , , modulo . In particular, given a bounded measurable function on , we can lift it to a bounded measurable function on and then define
Anyhow, in this setting, Fatou’s theorem implies that
(at least) in measure as .
Now let us come back to the harmonic functions and constructed above satisfying the concentration property (8). Denoting by and the “boundary values” of (along and ), we see that (10) and the concentration property (8) imply that
We will show that this is not possible by choosing the disjoint closed sets and of leading to and in a careful way and by using the fact that the -stationary measure on is non-atomic.
More concretely, since is not atomic, we can choose a compact subset of with that we fix once and for all. Set and construct from a harmonic function as above (cf. (5)) and let the associated function in (7). Denote by and the boundary values of along sequences and .
Now, we consider an increasing sequence of compact subsets exhausting . Again, set , construct the corresponding harmonic functions (cf. (5)), and let and be the boundary values of along sequences and . By construction, since is an increasing sequence, the functions and form two sequences of non-negative functions uniformly bounded by that do not decrease as increases. It follows that we can extract limits and . Moreover, from the definition of and , we get
Also, since exhausts , i.e., , we obtain from (9) that
Furthermore, the concentration property (11) implies that
for almost every .
Finally, since the harmonic functions have nice integral representations as in (7), i.e.,
where and are circle measures in our preferred circles and .
for almost every . By (15), this means that the functions and are constant ( or ) on each great circle (up to some neglectable expectional set of zero measure).
In particular, after lifting the function on to a function on , we have that for (where and are the lifts of the great circles and supporting and ), that is,
Note that is a subgroup of . It follows that : indeed, it is a general fact that the group generated by two distinct -parameter subgroups of (such as and ) is the whole . This shows that , and, a fortiori, is the constant function or .
a contradiction with the fact that or by our choice of with .