Posted by: matheuscmss | August 1, 2011

Conférence internationale Géométrie Ergodique (Orsay 2011) III

For my third post related to the “Conférence Internationale Géométrie Ergodique” (held at the Math. Department of Orsay – Univ. Paris 11 from 23th to 27th May), I’ll transcript below my notes for the lecture of Dmitry Kleinbock. The title of his talk was:

1. Ergodic Theory and Schmidt games

Definition 1 We say that {\alpha\in\mathbb{R}} is B.A. (badly approximable) if there exists {c>0} such that {|\alpha - p/q| > c/|q|^2} for all {p\in\mathbb{Z}}, {q\in\mathbb{Z}-\{0\}}. Equivalently, {\alpha\in\mathbb{R}} is B.A. if there exists {c>0} such that {\textrm{dist}(\alpha q,\mathbb{Z})>c/|q|} for all {q\in\mathbb{Z}-\{0\}}.

In the sequel, we denote by {BA} the set of badly approximable numbers. Among the basic properties of the set {BA} is the fact that it has zero Lebesgue measure. Nevertheless, we have the following results saying that {BA} is not so small:

For the proof of his result, W. Schmidt introduced the nowadays called “Schmidt games”. In the sequel, we will revisit the main features of this notion.

Remark 1 For the sake of the subsequent discussion, let us mention that the notion of “badly approximable” can be extended from real numbers to {m\times n}-matrices {A\in M_{m\times n}(\mathbb{R})} with real entries: we say that {A} is badly approximable (B.A.) if

\displaystyle \textrm{dist}(Aq,\mathbb{Z}^m)>\frac{c}{\|q\|^{n/m}}

for every {q\in\mathbb{Z}^n-\{0\}}. Here, {\textrm{dist}} is the Euclidean distance in {\mathbb{R}^m} and {\|.\|} is the Euclidean norm in {\mathbb{R}^n}. In what follows, we will also denote by BA the set of badly approximable matrices.

1.1. Schmidt games

We begin by explaining the rules of the game. Two players, say Alice and Bob (the common characters in physics and cryptography), are given two positive real numbers {\alpha} and {\beta} and a target set {S\subset\mathbb{R}^d}. In the first round, Bob chooses a ball {B_0\subset\mathbb{R}^d}, and then Alice chooses a ball {A_1\subset B_0} whose radius {\rho(A_1)} satisfies

\displaystyle \rho(A_1)=\alpha\rho(B_0).

In the second round, Bob chooses a ball {B_1\subset A_1} whose radius {\rho(B_1)} satisfies

\displaystyle \rho(B_1)=\beta\rho(A_1)

and then Alice chooses a ball {A_2\subset B_1} whose radius {\rho(A_2)} satisfies

\displaystyle \rho(A_2)=\alpha\rho(B_1)

The game continues indefinitely, so that Alice and Bob will produce a sequence of balls

\displaystyle B_0\supset A_1\supset B_1\supset A_2\supset\dots\supset B_{i-1}\supset A_i\supset B_i\supset\dots

whose radii verify

\displaystyle \rho(A_i)=\alpha\rho(B_{i-1})\quad \textrm{and} \quad \rho(B_i)=\beta\rho(A_i).

We say that Alice wins the game if

\displaystyle \bigcap\limits_{i=1}^{\infty} A_i\in S

i.e., the intersection of all {A_i}‘s belongs to the target set {S}. The figure below shows Bob’s choices of balls {B_i}‘s in blue, Alice’s choices of balls {A_i}‘s in red, and the target set {S} in black.

In this setting, we say that {S} is {(\alpha,\beta)}winning if Alice always has a winning strategy, i.e., she can always “hit” the target set independently of Bob’s choices of balls {B_i}‘s.

Also, we will say that {S} is {\alpha}winning if {S} is {(\alpha,\beta)}-winning for all {\beta} and sometimes we simply say that {S} is winning if {S} is {\alpha}-winning for some {\alpha}.

Below, we describe a few facts about winning sets:


  1. if {S} is {(\alpha,\beta)}-winning and {f:\mathbb{R}^d\rightarrow\mathbb{R}^d} is conformal, then {f(S)} is also {(\alpha,\beta)}-winning;
  2. if {\{S_i\}_{i\in\mathbb{N}}} is a countable collection of {\alpha}-winning sets, then {\bigcap\limits_{i=1}^{\infty}S_i} is also {\alpha}-winning;
  3. if {S} is winning, then {\textrm{dim}(S)=d}, i.e., winning sets have full dimension (here, {\textrm{dim}(S)} denotes the Hausdorff dimension of {S});
  4. {BA} is winning (as proved by Schmidt (1969)).

As the reader may suspect from the title of the talk, this number-theoretical discussion has a dynamical interpretation. Indeed, S. Dani (1985–1988; see e.g. here and here) showed that {A\in M_{m\times n}(\mathbb{R})} is B.A. if and only if {\{g_t\Lambda_A\}} is bounded in {SL_d({\mathbb R})/SL_d({\mathbb Z})}, {d=m+n}, for some appropriate choice of one-parameter subgroup {g_t} and lattice {\Lambda_A}. In particular, as a corollary of the fourth item, we deduce that:

Corollary 2 For some appropriate choices of one-parameter subgroups {g_t} in certain homogenous spaces {G/\Gamma}, the set of bounded orbits has full (Hausdorff) dimension.

Remark 2 It is possible to show that this corollary is true when {G} is a Lie group with real rank 1 (and {g_t} is the corresponding one-parameter Cartan subgroup).

In a similar vein, S. Dani also showed (1988) the following result:

Theorem 3 Let {R\in GL_d(\mathbb{R})\cap M_{d\times d}(\mathbb{Z})} be a semisimple matrix. Then, for any {z\in\mathbb{Q}^d/\mathbb{Z}^d}, the set

\displaystyle \left\{x\in\mathbb{T}^d:=\mathbb{R}^d/\mathbb{Z}^d:z\notin\overline{\{R^kx\textrm{ mod }1: k\in\mathbb{N}\}}\right\}

is winning (actually, {1/2}-winning), and, a fortiori, it has full Hausdorff dimension. In particular, the set

\displaystyle \left\{x\in\mathbb{T}^d:0\notin\overline{\{R^kx\textrm{ mod }1: k\in\mathbb{N}\}}\right\}

is winning (and has full Hausdorff dimension as well).

Below, we list some other closely related results:

Theorem 4 (Pollington, de Mathan (1980)) Let {\{r_k\}} be a sequence of lacunary positive real numbers (i.e., there exists {Q>1} such that {r_{k+1}\geq Q r_k} for all {k\in\mathbb{N}}). Then, the set

\displaystyle \left\{x\in\mathbb{R}:0\notin\overline{\{r_kx \textrm{ mod } 1: k\in\mathbb{N}\}}\right\}

has full Hausdorff dimension (i.e., its Hausdorff dimension is 1).

Theorem 5 (Urbanski (1991)) Let {f:X\rightarrow X} be a {C^2} expanding map (when {f} is not invertible) or Anosov diffeomorphism (when {f} is invertible). Then, for any {z\in X}, the set

\displaystyle \left\{x\in X: z\notin\overline{\{f^n(x):n\in\mathbb{N}\textrm{ or }\mathbb{Z}\}}\right\}

has full Hausdorff dimension (i.e., its Hausdorff dimension is {\textrm{dim}(X)}).

Theorem 6 (Kleinbock-Margulis (1996)) In the context of one-parameter subgroups {\{g_t\}} in homogenous spaces {G/\Gamma}, if {g_t} is partially hyperbolic and mixing, then the set

\displaystyle \left\{x\in G/\Gamma: \{g_tx:t\in\mathbb{R}\} \textrm{ is bounded}\right\}

has full Hausdorff dimension ({=\textrm{dim}(G/\Gamma)}).

Remark 3 The previous theorem was an answer to a conjecture of Margulis (1991): see Conjecture A of this paper by Margulis here.

Theorem 7 (Kleinbock (1998))In the same setting as before, for any {z\in G/\Gamma}, the set

\displaystyle \left\{x\in G/\Gamma: z\notin\overline{\{g_tx:t\in\mathbb{R}\}}\right\}

also has full Hausdorff dimension ({=\textrm{dim}(G/\Gamma)}).

Theorem 8 (Kleinbock (1999)) Let {\widetilde{BA}} be the set of pairs {(A,x)\in M_{m\times n}(\mathbb{R})\times \mathbb{R}^m} such that, for some constant {c=c(A,x)>0},

\displaystyle \textrm{dist}(Aq-x,\mathbb{Z}^m)>c/\|q\|^{n/m} \quad \forall\, q\in\mathbb{Z}^m-\{0\}.

(compare the definition of {\widetilde{BA}} with the definition of BA in Remark 1). Then, the set of {\widetilde{BA}} has full Hausdorff dimension.

It is worth to notice that, even though the statements of the last three results (by Kleinbock and Margulis) are similar in spirit to the results of S. Dani, their methods are different in that while Dani’s results are based on Schmidt games, there are no Schmidt games in the results by Kleinbock and Margulis!

In particular, a natural question is: by intersecting the two sets (of full Hausdorff dimension) provided by Kleinbock-Margulis (1996) and Kleinbock (1998), do we get a non-trivial (e.g., full Hausdorff dimension) set? Of course, if one can show that these sets are winning, their intersection would be winning as well and hence of full Hausdorff dimension. As it turns out, it was shown by Kleinbock-Weiss (2010) that these sets are actually winning (and thus their intersection of these sets has full Hausdorff dimension).

At this point, Kleinbock mentioned that during the period between 1998 and 2009 Schmidt games were a dormant subject, but he has the impression that Schmidt games are experiencing its “renaissance”.

1.2. Renaissance of Schmidt games

To quote a few papers of the renaissance period of Schmidt games, Kleinbock mentioned the articles by L. Fishman, D. Färm, C. McMullen, S. Kristensen, J. Tseng and collaborators (besides his 2010 paper with Weiss quoted above). Kleinbock encouraged the audience to look at the webpages of these authors (you can look at them by following the link in their names) and he quoted this paper by J. Tseng where it is proved that, in the case of the circle {X=S^1=\mathbb{R}/\mathbb{Z}}, given a {C^2} expanding map {f:X\rightarrow X}, a point {z\in X}, the set of points {\{x\in X: z\notin\overline{\{f^n(x):n\in\mathbb{N}\}}\}} is {\alpha}-winning for every {0<\alpha\leq 1/2} (so that this provides an extension of Urbanski’s Theorem 5 in dimension one).

Then, Kleinbock announced the main result of his talk:

Theorem 9 (Broderick, Fishman, Kleinbock) Let {\mathcal{R}=\{R_k\}_{k\in\mathbb{N}}\subset M_{m\times n}(\mathbb{R})} be a lacunary sequence of {m\times n} matrices (i.e., the sequence of real numbers {\{\|R_k\|\}_{k\in\mathbb{N}}} is lacunary [see Theorem 4 above]) and fix {z\in\mathbb{T}^n}. Then, the set

\displaystyle \{x\in\mathbb{R}^m: z\notin\overline{\{R_kx \textrm{ mod }\mathbb{Z}^m\}}\}

is winning.

In fact, this theorem is a particular case of a more general theorem (Theorem 1.3) of Broderick, Fishman, Kleinbock’s paper, but we will stick to this version in order to be able to give a rough sketch of proof of this result in the next (final) subsection.

For now, we will present some examples connecting this result with previous ones.

Example 1 When {m=n=1}, this Theorem allows to recover Theorem 4 of Pollington and de Mathan.

Example 2 In the case of square-matrices (i.e., {m=n=d}), by taking {R_k=R^k} with {R} a matrix whose spectral radius is {\rho(R)>1}, we see that this theorem allows to recover Dani’s theorem 3 without any semisimplicity or non-simplicity assumption!

For the third example, we need to recall the following fact: given a matrix {A\in M_{m\times n}({\mathbb R})}, there exists a lacunary sequence {\{y_k\}_{k\in{\mathbb N}}\subset {\mathbb Z}^m} (a subsequence of the “best approximants” of {A} in the sense of the works of Jarnick, Cassels and Bugeaud-Laurent) such that if {x\in{\mathbb R}^n} satisfies

\displaystyle (*) \quad \inf\limits_{k\in{\mathbb N}}\textrm{dist}(y_k\cdot x,{\mathbb Z})>0

then {(A,x)\in\widetilde{BA}} (see Theorem 8 above for the definition of this set). In particular, one gets the following application of Theorem 9:

Example 3 By taking {n=1} and {R_k=y_k\in{\mathbb R}^m} (where {y_k} is as above) in Theorem 4, we have that the set

\displaystyle \{x\in{\mathbb R}^m: (*)\, \inf\limits_{k\in{\mathbb N}}\textrm{dist}(y_k\cdot x,{\mathbb Z})>0\}

is winning. Thus, since we saw that the condition {(*)} for {x\in{\mathbb R}^m} implies that {(A,x)\in\widetilde{BA}}, one conclude that the set

\displaystyle \{x\in{\mathbb R}^m: (A,x)\in\widetilde{BA}\}

is also winning. This last fact was also derived by Einsiedler-Tseng (who gave an alternative proof of it).

Just to see why condition {(*)} has something to do with {\widetilde{BA}}, we observe that

\displaystyle y_k\cdot x = y_k\cdot Aq - y_k(Aq-x) = q\cdot A^ty_k - y_k(Aq-x)

Hence, if {x} verifies {(*)}, one can deduce that, for some {\varepsilon>0},

\displaystyle \varepsilon<\textrm{dist}(y_k\cdot x,{\mathbb Z})\leq n\|q\|\textrm{dist}(A^ty_k,{\mathbb Z}^m)+m\|y_k\|\textrm{dist}(Aq-x,{\mathbb Z}^m)

Therefore, by choosing {y_k} such that {n\|q\|\textrm{dist}(A^ty_k,{\mathbb Z}^m)<\varepsilon/2} but {\|y_k\|} not “very large”, one gets (from the previous estimate)

\displaystyle \frac{\varepsilon}{2m\|y_k\|}<\textrm{dist}(Aq-x,{\mathbb Z}^m)

an inequality in the same spirit of the condition defining {\widetilde{BA}}.

Now, we will quickly describe a sketch of proof of Theorem 9.

1.3. A couple of words on the proof of Theorem 9

We begin by recalling the following lemma:

Lemma 1 (Schmidt, Moshchevitin) Let {0<\alpha<1/3}. Then, there exists {\varepsilon>0} such that, for every ball {B=B(x,\rho)\subset{\mathbb R}^d}, for every {N\in{\mathbb N}}, and for every {\mathcal{L}_1,\dots,\mathcal{L}_N} hyperplanes in {{\mathbb R}^d}, there is a ball {B'=B(x',\alpha\rho)\subset B} with

\displaystyle \textrm{dist}(B',\mathcal{L}_i)>\alpha\rho

for at least {\varepsilon N} of the hyperplanes {\mathcal{L}_i}.

In other words, this lemma says that, while playing Schmidt’s game, Alice can keep a safe distance from hyperplanes in a efficient way.

Proof: The idea of the proof (in particular, the choice of {\varepsilon}) is schematically presented in the figure below.

Here, the line represents a hyperplane {\mathcal{L}_i}, the white part (represent by two half-circles) is the set {A_i:=B(x,(1-\alpha)\rho)-V_{2\alpha}(\mathcal{L}_i)} (where {V_{\theta}(\mathcal{L}) = \{y:\textrm{dist}(y,\mathcal{L})\leq\theta\}}), and the “dashed” part is the complement of {A_i} inside the ball {B(x,\rho)}. Then, a direct computation with this choice of {\varepsilon} shows that the desired lemma follows (see, e.g., the (short) proof of Lemma 3.4 of Broderick, Fishman, Kleinbock paper for more details). \Box

After this preparatory lemma, the sketch of proof of Theorem 9 goes as follows:

Proof: Recall that we wish to show that the set

\displaystyle \{x:z\notin\overline{\{R_kx\textrm{ mod }{\mathbb Z}^m\}}\}

is winning. To do so, fix {0<\alpha<1/3}, {0<\beta<1}, {\rho>0} and denote by {B(x,\rho)=B_1} the ball choosen by Bob in the first round of the {(\alpha,\beta)} Schmidt’s game. Let {r_k=\|R_k\|} (so that, by hypothesis, {r_{k+1}/r_k\geq Q>1} for all {k\in{\mathbb N}}).

Of course, since {R_k} is a matrix (linear operator), for all {y\in z+{\mathbb Z}^m}, the set {R_k^{-1}(\textrm{small fixed neighborhood of }y)} is contained in a neighborhood of a proper affine subspace (actually a sort of ellipsoid by the singular values theorem). Also, given {y_1\neq y_2}, {y_1,y_2\in z+{\mathbb Z}^m}, we have that {\textrm{dist}(y_1,y_2)\geq 1}, so that {\textrm{dist}(R_k^{-1}(y_1),R_k^{-1}(y_2))\geq 1/r_k}. In particular, since Alice is only able to choose balls of radius of the form {\rho(\alpha\beta)^{M}} at the {M}th round of the game, it would be desirable to have, for every {k}, that {\rho(\alpha\beta)^{k-1}=1/2r_k} and, moreover, that these balls are avoiding all hyperplanes associated to certain ellipsoidal neighborhoods of {R_k^{-1}(y)}, {y\in z+{\mathbb Z}^m}. However, in general, this is too good to be true. Thus, we should aim at a more modest property (which is still sufficient to conclude the result). In the case at hand, we fix {\varepsilon} is the positive real number provided by Lemma 1 and we choose (once and for all) {N} large enough so that

\displaystyle (\alpha\beta)^{-l}\leq Q^N

where {l:=\lfloor\log_{1/(1-\varepsilon)}N\rfloor+1} (so that {(1-\varepsilon)^l N\sim 1}). By lacunarity, it is not hard to see that, for all {j\in{\mathbb N}},

\displaystyle \#\{k: \frac{1}{2\rho}(\alpha\beta)^{-(j-1)l}\leq r_k\leq \frac{1}{2\rho}(\alpha\beta)^{-jl}\}\leq N

Given {k} with {\frac{1}{2\rho}(\alpha\beta)^{-(j-1)l}\leq r_k\leq \frac{1}{2\rho}(\alpha\beta)^{-jl}}, we combine our observations that {R_k^{-1}(B(y,c))} is contained in a neighborhood {V_{c/r_k}(\mathcal{L})} of radius {c/r_k\leq 2\rho(\alpha\beta)^{(j-1)l}c} of some hyperplane and two distinct points {y_1\neq y_2}, {y_1,y_2\in z+{\mathbb Z}^m} verify

\displaystyle \textrm{dist}(R_k^{-1}(y_1),R_k^{-1}(y_2))\geq 1/r_k\geq 2\rho(\alpha\beta)^{jl}

to see that any ball chosen by Bob from the round {lj} on can intersect at most one set of the form {R_k^{-1}(B(y,c))} with {y\in z+{\mathbb Z}^m} (for {c} sufficiently small, e.g., {c=(\alpha\beta)^{2l}}). On the other hand, since for each {j} there are at most {N} indices {k} satisfying the inequalities (for {r_k}) above, we deduce that the ball {B_{lj}} choosen by Bob at the round {lj} satisfies

\displaystyle \bigcup\limits_{\frac{1}{2\rho}(\alpha\beta)^{-(j-1)l}\leq r_k\leq \frac{1}{2\rho}(\alpha\beta)^{-jl}}B_{lj}\cap R_k^{-1}(V_c(z+{\mathbb Z}^m))

\displaystyle\subset \bigcup\limits_{i=1}^N V_{2\rho(\alpha\beta)^{(j-1)l}c}(\mathcal{L}_i)\subset \bigcup\limits_{i=1}^N V_{2\rho(\alpha\beta)^{(j+1)l}}(\mathcal{L}_i)

Now, by using Lemma 1 {l} times together with the fact that {(1-\varepsilon)^lN<1} (by definition), Alice can choose (at the {l(j+1)}th round of the game) a ball {A_{l(j+1)}} of radius {2\rho(\alpha\beta)^{(j+1)l}} avoiding a {2\rho(\alpha\beta)^{(j+1)l}} neighborhood of all hyperplanes {\mathcal{L}_i}, {i=1,\dots, N}.

In other words, it was shown that, for all {j}, for all {x\in A_{(j+1)l}}, and all {k} with {\frac{1}{2\rho}(\alpha\beta)^{-(j-1)l}\leq r_k\leq \frac{1}{2\rho}(\alpha\beta)^{-jl}}, one has

\displaystyle \textrm{dist}(R_kx,z+{\mathbb Z}^m)>c

(for an appropriate choice of universal constant {c>0}). In fact, this is the modest property announced in the beginning of the argument. As the reader can check, this property suffices to conclude that

\displaystyle \{x:z\notin\overline{\{R_kx\textrm{ mod }{\mathbb Z}^m\}}\}

is {(\alpha,\beta)}-winning for every {0<\alpha<1/3} and {0<\beta<1}. \Box

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