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 is B.A. (badly approximable) if there exists such that for all , . Equivalently, is B.A. if there exists such that for all .
In the sequel, we denote by the set of badly approximable numbers. Among the basic properties of the set is the fact that it has zero Lebesgue measure. Nevertheless, we have the following results saying that is not so small:
- Cassels (1956): , for any choice of ;
- Davenport (1962): , for any choice of affine maps ;
- W. Schmidt (1966): , for any choice of .
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 -matrices with real entries: we say that is badly approximable (B.A.) if
for every . Here, is the Euclidean distance in and is the Euclidean norm in . 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 and and a target set . In the first round, Bob chooses a ball , and then Alice chooses a ball whose radius satisfies
In the second round, Bob chooses a ball whose radius satisfies
and then Alice chooses a ball whose radius satisfies
The game continues indefinitely, so that Alice and Bob will produce a sequence of balls
whose radii verify
We say that Alice wins the game if
i.e., the intersection of all ‘s belongs to the target set . The figure below shows Bob’s choices of balls ‘s in blue, Alice’s choices of balls ‘s in red, and the target set in black.
In this setting, we say that is –winning if Alice always has a winning strategy, i.e., she can always “hit” the target set independently of Bob’s choices of balls ‘s.
Also, we will say that is –winning if is -winning for all and sometimes we simply say that is winning if is -winning for some .
Below, we describe a few facts about winning sets:
- if is -winning and is conformal, then is also -winning;
- if is a countable collection of -winning sets, then is also -winning;
- if is winning, then , i.e., winning sets have full dimension (here, denotes the Hausdorff dimension of );
- 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 is B.A. if and only if is bounded in , , for some appropriate choice of one-parameter subgroup and lattice . In particular, as a corollary of the fourth item, we deduce that:
Corollary 2 For some appropriate choices of one-parameter subgroups in certain homogenous spaces , the set of bounded orbits has full (Hausdorff) dimension.
Remark 2 It is possible to show that this corollary is true when is a Lie group with real rank 1 (and is the corresponding one-parameter Cartan subgroup).
In a similar vein, S. Dani also showed (1988) the following result:
Theorem 3 Let be a semisimple matrix. Then, for any , the set
is winning (actually, -winning), and, a fortiori, it has full Hausdorff dimension. In particular, the set
is winning (and has full Hausdorff dimension as well).
Below, we list some other closely related results:
has full Hausdorff dimension (i.e., its Hausdorff dimension is 1).
Theorem 5 (Urbanski (1991)) Let be a expanding map (when is not invertible) or Anosov diffeomorphism (when is invertible). Then, for any , the set
has full Hausdorff dimension ().
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 , the set
also has full Hausdorff dimension ().
Theorem 8 (Kleinbock (1999)) Let be the set of pairs such that, for some constant ,
(compare the definition of with the definition of BA in Remark 1). Then, the set of 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 , given a expanding map , a point , the set of points is -winning for every (so that this provides an extension of Urbanski’s Theorem 5 in dimension one).
Then, Kleinbock announced the main result of his talk:
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 , this Theorem allows to recover Theorem 4 of Pollington and de Mathan.
Example 2 In the case of square-matrices (i.e., ), by taking with a matrix whose spectral radius is , 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 , there exists a lacunary sequence (a subsequence of the “best approximants” of in the sense of the works of Jarnick, Cassels and Bugeaud-Laurent) such that if satisfies
then (see Theorem 8 above for the definition of this set). In particular, one gets the following application of Theorem 9:
Example 3 By taking and (where is as above) in Theorem 4, we have that the set
is winning. Thus, since we saw that the condition for implies that , one conclude that the set
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 , we observe that
Hence, if verifies , one can deduce that, for some ,
Therefore, by choosing such that but not “very large”, one gets (from the previous estimate)
an inequality in the same spirit of the condition defining .
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 . Then, there exists such that, for every ball , for every , and for every hyperplanes in , there is a ball with
for at least of the hyperplanes .
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 ) is schematically presented in the figure below.
Here, the line represents a hyperplane , the white part (represent by two half-circles) is the set (where ), and the “dashed” part is the complement of inside the ball . Then, a direct computation with this choice of shows that the desired lemma follows (see, e.g., the (short) proof of Lemma 3.4 of Broderick, Fishman, Kleinbock paper for more details).
After this preparatory lemma, the sketch of proof of Theorem 9 goes as follows:
Proof: Recall that we wish to show that the set
is winning. To do so, fix , , and denote by the ball choosen by Bob in the first round of the Schmidt’s game. Let (so that, by hypothesis, for all ).
Of course, since is a matrix (linear operator), for all , the set is contained in a neighborhood of a proper affine subspace (actually a sort of ellipsoid by the singular values theorem). Also, given , , we have that , so that . In particular, since Alice is only able to choose balls of radius of the form at the th round of the game, it would be desirable to have, for every , that and, moreover, that these balls are avoiding all hyperplanes associated to certain ellipsoidal neighborhoods of , . 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 is the positive real number provided by Lemma 1 and we choose (once and for all) large enough so that
where (so that ). By lacunarity, it is not hard to see that, for all ,
Given with , we combine our observations that is contained in a neighborhood of radius of some hyperplane and two distinct points , verify
to see that any ball chosen by Bob from the round on can intersect at most one set of the form with (for sufficiently small, e.g., ). On the other hand, since for each there are at most indices satisfying the inequalities (for ) above, we deduce that the ball choosen by Bob at the round satisfies
Now, by using Lemma 1 times together with the fact that (by definition), Alice can choose (at the th round of the game) a ball of radius avoiding a neighborhood of all hyperplanes , .
In other words, it was shown that, for all , for all , and all with , one has
(for an appropriate choice of universal constant ). 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
is -winning for every and .