Abstract
We construct finitely supported symmetric probability measures on \(S{L}_{2}(\mathbb{R})\) for which the Furstenberg measure on \({\mathbb{P}}_{1}(\mathbb{R})\) has a smooth density.
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1 Introduction
In this note, we give explicit examples of finitely supported symmetric probability measures ν on \(S{L}_{2}(\mathbb{R})\) for which the corresponding Furstenberg measure μ on \({\mathbb{P}}_{1}(\mathbb{R})\) is absolutely continuous wrt to Haar measure dθ, and moreover \(\frac{d\mu } {d\theta }\) is of class C r, with r any given positive integer. Probabilistic constructions of finitely supported (non-symmetric measures ν on \(S{L}_{2}(\mathbb{R})\) with absolutely continuous Furstenberg measure appear in the paper [1], setting (in the negative) a conjecture from [4]. The construction in [1] may be viewed as a non-commutative analogue of the theory of random Bernoulli convolutions and uses methods from [5, 6].
It is not clear if this technique may produce Furstenberg measures with say C 1-density. Our method also addresses the issue of obtaining a symmetric ν (raised in [4]), which seems problematic with the [1] technique.
Our starting point is a construction from [2] of certain Hecke operators on \(S{L}_{2}(\mathbb{R})\) whose projective action exhibits a spectral gap. The mathematics underlying [2] is closely related to the paper [3] and makes essential use of results and techniques from arithmetic combinatorics. In particular, it should be pointed out that the spectral gap is not achieved by exploiting hyperbolicity, at least not in the usual way. Our measure ν has in fact a Lyapounov exponent that can be made arbitrary small, while the spectral gap (in an appropriate restricted sense) remains uniformly controlled (the size of supp ν becomes larger of course).
We believe that similar constructions are possible also in the \(S{L}_{d}(\mathbb{R})\)-setting, for d > 2 (cf. [4]). In fact, such Hecke operators can be produced using the construction from Lemmas 1 and 2 below in \(S{L}_{2}(\mathbb{R})\) and considering a suitable family of \(S{L}_{2}(\mathbb{R})\)-embeddings in SL d . We do not present the details here.
2 Preliminaries
We recall Lemmas 2.1 and 2.2 from [2].
Lemma 1.
Given ε > 0, there is \(Q \in {\mathbb{Z}}_{+}\) and \(\mathcal{G}\subset S{L}_{2}(\mathbb{R}) \cap \left ( \frac{1} {Q}Ma{t}_{2}(\mathbb{Z})\right )\) with the following properties
Here c 1, c 2 are constants independent of ε.
Define the probability measure ν on \(S{L}_{2}(\mathbb{R})\) as
Denote also P δ, δ > 0, an approximate identity on \(S{L}_{2}(\mathbb{R})\). For instance, one may take \({P}_{\delta } = \frac{{1}_{{B}_{\delta }(1)}} {\vert {B}_{\delta }(1)\vert }\) where B δ(1) is the ball of radius δ around 1 in \(S{L}_{2}(\mathbb{R})\).
Lemma 2.
Fix τ > 0. Then we have
provided
and assuming δ small enough (depending on Q and τ).
3 Furstenberg Measure
Denote for \(g \in S{L}_{2}(\mathbb{R})\) by τ g the action on \({P}_{1}(\mathbb{R})\) that we identify with the circle \(\mathbb{R}/\mathbb{Z} = \mathbb{T}\). Thus if \(g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right ),ad-bc = 1\), then
Assume μ on \({P}_{1}(\mathbb{R})\) is ν-stationary, i.e.
4 A Restricted Spectral Gap
Take \(\mathcal{G}\) as in Lemma 1 and \(\nu = \frac{1} {2r}{ \sum \nolimits }_{g\in \mathcal{G}}({\delta }_{g} + {\delta }_{g-1})\) with \(r = \vert \mathcal{G}\vert \).
Lemma 3.
There is some constant K > 0 (depending on ν), such that if \(f \in {L}^{2}(\mathbb{T})\) satisfies
then
Proof.
Define \({\rho }_{g}f = {({\tau }_{g}^{\prime})}^{1/2}(f \circ {\tau }_{g})\), hence ρ is the projective representation. Since \(\Vert 1 - g\Vert < \epsilon \), | τ g ′ − 1 | ≲ ε and (11) will follow from
Assume (12) fails. By almost orthogonality, there is \(f \in {L}^{2}(\mathbb{T})\) such that
Let ℓ < k to be specified. From (15), since ν is symmetric,
and hence
Take \(\delta = 1{0}^{-k}\). Recalling (13), straightforward approximation permits us to replace in (16) the discrete measure ν(ℓ) by ν(ℓ) ∗ P δ, where P δ(δ > 0) denotes the approximate identity on \(S{L}_{2}(\mathbb{R})\). Hence (17) becomes
Fix a small constant τ > 0 and apply Lemma 2. This gives
such that
Note that supp ν(ℓ) is contained in a ball of radius at most (1 + ε)ℓ, by (4).
Introduce a smooth function 0 ≤ ω ≤ 1 on \(\mathbb{R},\omega = 1\) on \([-{(1 + \epsilon )}^{4\mathcal{l}},{(1 + \epsilon )}^{4\mathcal{l}}]\) and ω = 0 outside \([-2{(1 + \epsilon )}^{4\mathcal{l}},2{(1 + \epsilon )}^{4\mathcal{l}}]\).
Let \({\omega }_{1}(g) = \omega ({a}^{2} + {b}^{2} + {c}^{2} + {d}^{2})\) for \(g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right )\).
From (20), the first term of (18) is bounded by
Note also that by assuming ε a sufficiently small constant, we can ensure that ℓ ≪ k and 2 − k < c 2ℓ. Thus
and applying Cauchy-Schwarz
Fix x ≠ y and consider the inner integral. If we restrict \(g \in S{L}_{2}(\mathbb{R})\) s.t. τ g x = θ (fixed), there is still an averaging in ψ = τ g y that can be exploited together with (13). By rotations, we may assume \(x = \theta = 0\). Write \(g = \left (\begin{array}{ll} a&b\\ c &d \end{array} \right ) \in S{L}_{2}(\mathbb{R})\), \(dg = \frac{dadbdc} {a}\) on the chart a ≠ 0. Since
the condition τ g 0 = 0 means c = 0 and thus
Hence, fixing a
Also
implying
Substituting (24), (26) in (23) gives for the inner integral the bound
The weight function restricts a to \({(1 + \epsilon )}^{-2\mathcal{l}} \lesssim \vert a\vert \lesssim {(1 + \epsilon )}^{2\mathcal{l}}\) and clearly
If we restrict \(\vert \sin (x - y)\vert > {2}^{-\frac{k} {10} }\), Assumption (13) gives a bound at most \({2}^{-k}\Vert {f\Vert }_{1}\) for the ψ-integral in (27). Indeed, if β is a smooth function vanishing on a neighborhood of 0 and | n | ∼ 2k, partial integration implies that for any given A > 0
Thus
The contribution to (23) is at most
Next we consider, the contribution of \(\vert \sin (x - y)\vert \leq {2}^{-\frac{k} {10} }\) to (23).
First, from (25), we have that
By Cauchy-Schwarz, the inner integral in (23) is at most
Hence, we obtain
and hence, by (19)
Taking (in order) τ and ε small enough, a contradiction follows.
This proves Lemma 3.
5 Absolute Continuity of the Furstenberg Measure and Smoothness of the Density
Our aim is to establish the following.
Theorem.
Let μ be the stationary measure introduced in (9). Given \(r \in {\mathbb{Z}}_{+}\) and taking ε in Lemma 1 small enough will ensure that \(\frac{d\mu } {d\theta } \in {C}^{r}\).
This will be an immediate consequence of
Lemma 4.
Let k > k(ε) be sufficiently large and \(f \in {L}^{\infty }(\mathbb{T}),\vert f\vert \leq 1\) such that \(\mathrm{supp}\hat{f} \subset [{2}^{k-1},{2}^{k}]\) . Then
where \({C}_{\epsilon } \rightarrow ^{\epsilon \rightarrow 0}\infty \).
Proof.
Clearly, for any \(\mathcal{l} \in {\mathbb{Z}}_{+}\)
We will iterate Lemma 3 and let K = K(ε) satisfy (10), (11).
We assume 2k > 10K 10. For m < ℓ and | n | < K, we evaluate \(\vert \widehat{{F}_{m}}(n)\vert \), denoting
Clearly \(\vert \widehat{{F}_{m}}(n)\vert {\leq \max }_{g\in \mathrm{supp}{\nu }^{(m)}}\vert {(f \circ {\tau }_{g})}^{\wedge }(n)\vert \) and by assumption on \(\mathrm{supp}\hat{f}\)
. Performing a change of variables gives
by partial integration and our assumptions. It follows from (36) that if ℓ satisfies
then for m < ℓ and k > k(r)
(with r a fixed large integer).
Next, decompose
Hence, by (38)
Estimate using (39) and Lemma 3
Iteration of (40) implies by (37)
Also
and interpolation between (41), (42) implies for r (resp. ε) large (resp. small) enough
provided k > k(ε, r).
In view of (34), this proves (33).
Remark.
For ν finitely supported (with positive Lyapounov exponent), one cannot obtain a Furstenberg measure μ that equals Haar measure on \({\mathbb{P}}_{1}(\mathbb{R}) \simeq \mathbb{T}\). Indeed, otherwise for any f on \(\mathbb{T}\), we would have
For \(g \in S{L}_{2}(\mathbb{R})\),
for some \(z \in D =\{ z \in \mathbb{C};\vert z\vert < 1\}\), with \({P}_{z}(\theta ) = \frac{1-\vert z{\vert }^{2}} {\vert 1-\bar{z}{e}^{i\theta }{\vert }^{2}}\) the Poisson kernel.
From (44), (45), taking \(\nu ={ \sum \nolimits }_{j=1}^{r}{c}_{j}{\delta }_{{g}_{j}},{c}_{j} > 0\) and ∑c j = 1 and {z j } the corresponding points in D, we get
This easily implies that \({z}_{1} = \cdots = {z}_{r} = 0\). But then each g j has unimodular spectrum and ν vanishing Lyapounov exponent.
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Acknowledgements
The author is grateful to C. McMullen and P. Varju for several related discussions. Research was partially supported by NSF grants DMS-0808042 and DMS-0835373
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Bourgain, J. (2012). Finitely Supported Measures on \(S{L}_{2}(\mathbb{R})\) Which are Absolutely Continuous at Infinity. In: Klartag, B., Mendelson, S., Milman, V. (eds) Geometric Aspects of Functional Analysis. Lecture Notes in Mathematics, vol 2050. Springer, Berlin, Heidelberg. https://doi.org/10.1007/978-3-642-29849-3_7
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