1. If two nitely generated groups are quasi-isometric, are they bilipschitz equivalent, (see [5, p. 3])? Here the groups are assumed to be equipped wi

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1 Rectifying separated nets Dmitri Burago Bruce Kleiner y April 1, Introduction In our earlier paper [] (see also McMullen's paper [7]), we constructed examples of separated nets in the plane R which are not bilipschitz equivalent to the integer lattice Z. These examples gave a negative answer to a question raised by H. Furstenberg and M. Gromov. Furstenberg asked this question in connection with Kakutani equivalence for R - actions, [4]. Return times for a section of an R -action form a separated net, and to represent the returns of an R -action by a Z -action, one has to have a bilipschitz identication of the return times for each point with Z (depending measurably on the point). As was pointed out to us by A. Katok, one can use a standard construction of R -actions to represent our example as the set of return times for points from a set of positive measure, thus showing that not every section can be used (it is worth mentioning here that an old result of Katok [6] asserts that every R -action admits a section whose return times are bilipschitz equivalent to Z ). Gromov's motivation for the question came from large scale geometry, and the denition of quasi-isometries. Two metric spaces are quasi-isometric if they contain bilipschitz equivalent separated nets hence one would like to know if the choice of separated net matters, and if a given space can contain nets which are not bilipschitz equivalent. This question is particularly interesting for spaces with cocompact isometry groups. The counterexample in [] was based on a counterexample to another question which had been posed by J. Moser and M. Reimann in the 60's, namely whether every positive continuous function on the plane is locally the Jacobian of a bilipschitz homeomorphism. Using well-known properties of quasi-conformal homeomorphisms, one can actually show that any function that is not the Jacobian of a bilipschitz homeomorphism cannot be the Jacobian of a quasi-conformal homeomorphism either. The resolution of the original question suggested several intriguing problems, which we present below. These questions came from our discussions with C. Mc- Mullen and several other mathematicians. Supported by a Sloan Fellowship and NSF grant DMS y Supported by Sloan Fellowship and NSF grant DMS

2 1. If two nitely generated groups are quasi-isometric, are they bilipschitz equivalent, (see [5, p. 3])? Here the groups are assumed to be equipped with word metrics. Several special cases of this question are striking. Can G Z ever fail to be bilipschitz to G? More generally: is every nite index subgroup of G bilipschitz to G? By [1, 11] (see also [8, 1]) it is known that the answers are always armative when the group is nonamenable.. If G 1 and G are uniform lattices in the same connected Lie group, must they be bilipschitz equivalent? (Logically speaking, this is a subcase of Problem 1.) Again this is known in the nonamenable case, i.e. the case of nonsolvable Lie groups. Lattices with the same covolume are bilipschitz equivalent by an argument using Hall's Marriage Lemma (see Lemma 4.1). A similar argument shows that the the problem has an armative solution when the Lie group admits bilipschitz homeomorphisms with constant Jacobian t for every t > 0. For instance, graded nilpotent Lie groups and the isometry group of 3-dimensional Solv geometry admit bilipschitz automorphisms which scale volume by an arbitrary factor. 3. If one forms a separated net in the plane by placing a point in the center of each tile of a Penrose tiling, is the resulting net bilipschitz equivalent to Z? More generally, one can consider nets constructed as follows. Let P R n be a -plane with irrational slope, and let B R n be a bounded subset with nonempty interior. Take the set of points z Z n for which the intersection (z + B) \ P is nonempty, and project it orthogonally to P. When B has small diameter this example can also be described dynamically as the set of return times for a linear R -action on an n-torus to a section. 4. How can one characterize Jacobians of bilipschitz homeomorphisms R! R? Several authors have studied the prescribed Jacobian problem in other regularity classes [9, 3, 10, 13]. This question is already nontrivial if one restricts one's attention to nonconstant functions which are locally constant on the complement of a simple closed curve for instance it seems plausible that in the case of a closed snowake curve such a function is never the Jacobian of a bilipschitz homeomorphism. On the other hand it seems likely that a function that assumes one value on the subgraph of a continuous function R! R and another value elsewhere is always the Jacobian of a bilipschitz homeomorphism. One can reduce the n = 3 case of Problem 3 to a situation similar to this. In the remainder of this paper, we settle the n = 3 case of Problem 3 for planes with Diophantine slope. More precisely, let R satisfy the Diophantine condition j p q j > C q d (1.1) for some C > 0 d > and all p q Z recall that the set of such has full measure, and contains all algebraic irrationals. Now take P R 3 to be the graph of the linear function x + y. Let X P be the separated net obtained by projecting the set fz Z 3 : (z + B) \ P 6= g orthogonally to P. Theorem 1.. X is bilipschitz homeomorphic to Z. We prove Theorem 1. via a general criterion which shows that a separated net is

3 bilipschitz to Z provided its density in large subsets approaches a limiting value rapidly enough: Theorem 1.3. Suppose Y R is a separated net. For > 0 and each measurable subset U R, dene e (U) to be the density deviation max juj #(U \ Y ) #(U \ Y ) juj : (1.4) Then dene E : N! R by letting E (k) be the supremum of the quantities e (U), where U ranges over all squares of the form [i i + k] [j j + k] for i j Z. If there exists a > 0 such that the product Q m E ( m ) converges, then Y is bilipschitz to Z. Remark 1.5. The technique from [] can be used to produce separated nets which have uniform asymptotic density (i.e. lim k!1 E(k) = 1), but which are not bilipschitz to Z. In outline, the proof of Theorem 1.3 goes as follows. First we associate a Voronoi type tiling with the net, and then introduce a function u whose value on each tile is the reciprocal of the area of the tile. We then use Hall's Marriage Lemma to show that the net is bilipschitz to Z if the function u is the Jacobian of a bilipschitz homeomorphism R! R. To construct such a homeomorphism, we let S i be the image under scaling by i of the usual tiling of R by unit squares, and then we let u i : R! R be a function whose value of each square of S i is equal to the average of u over that square. We dene a sequence of bilipschitz homeomorphisms i : R! R such that Jac( i ) = u u i, and we show that i subconverges Q to a bilipschitz homeomorphism by showing the the innite product bilip( i i 1 ) i 1 converges. The bilipschitz homeomorphism has Jacobian u, and we get the desired homeomorphism by composing with the scaling x 7! p x. The paper is organized as follows. In section we show in Corollary. that nets in Theorem 1. satisfy the hypotheses of Theorem 1.3. In section 3 we give a sucient condition for a function u : R! R to be the Jacobian of a bilipschitz homeomorphism R! R. In section 4 we use the main result from section 3 to prove Theorem 1.3. Acknowledgements. We would like to thank the anonymous referee for correcting many inaccuracies in the original version of the paper. We are grateful to C. McMullen and G. Margulis for stimulating discussions.. Density estimates for the net X Our objective in this section is to estimate the deviation of the density of X in squares from the asymptotic density of X in R. Note that X is bilipschitz to its vertical projection X to the xy-plane. It is easy to see that in the 3 dimensional situation it is enough to consider the case when B is a ball indeed, the set of points that we project to P to obtain X is the intersection of Z 3 with a slab W parallel to P. We will assume that the radius of B is small enough 3

4 that W does not contain a vertical interval of length 1. The general case can be reduced to this one by splitting the slab W into a union of thin slabs, and observing that our density estimates are additive 1. Observe that Z := W \ Z 3 can be represented as the set of points (x y z) Z 3 satisfying jz x yj < for an appropriate. Since W does not contain vertical intervals of length 1, it follows that < 1, Z projects to the xy-plane injectively, and hence X is bilipschitz homeomorphic to the vertical projection of Z to the xyplane, by a bijection which displaces points by a distance at most 1. We denote the projection of Z to the xy-plane by Z. Since there is a bounded displacement bijection between Z and X, the reader may verify that it suces to obtain density estimates for Z the discrepancy between the densities in large squares is due to boundary eects which do not aect the convergence of the product in the statement of Theorem 1.3. Pick constants r < 1=. Set S := fi Z j d(r + i Z) < g: Proposition.1. There are constants C 0 and c > 0 such that for any j and k, j 1 k #(S \ [j j + k)) j C 0 k c Let =, and for any measurable U R, let e(u) be as in the statement of Theorem 1.3, with Y = Z. Corollary.. For all j 1 j Z, k N, if U is a square of the form [j 1 j 1 + k] [j j + k], then je (U) 1j C 1 Q for some constant C k c 1. It follows that the product E i ( i ) converges, where E is dened as in the Theorem 1.3. The corollary can be deduced from the proposition by breaking the square U into rows. The remainder of this section is devoted to the proof of Proposition.1. Let pn denote the convergents of the irrational number. Lemma.3. For any n k, j#(s \ [k k + 1]) j 3. Proof. This is an easy quantative renement of a standard argument showing that the sequence fi mod 1g is equidistributed in the circle S = R=Z. We will assume that pn > 0, the other case is absolutely analogous. Consider a nite sequence x i = (i + k 1) mod 1 S, i = 1 : : :. Then #(S \ [k k + 1]) is nothing but the number of x i 's that belong to the interval of length centered at r mod 1 in S. Consider another sequence y i = (k 1) + i pn mod 1, i = 1 : : :. Note that 0 < x i y i < ( pn ) (where the rst inequality means that y n preceeds x n with respect to the natural orientation of S.) This means that there are no other x j on the segment between y i and x i. Indeed, otherwise j(i j) mj < p n for some integer m, and this contradicts to the fact that pn is the best approximation for by 1 Finicky readers may note that the discrepancy between the density estimate for closed slabs and for open slabs is negligible. 4

5 rationals with denominators not exceeding. Hence the points x i and y i alternate in S. This means that there is exactly one member of the sequence fx i g between any two neighboring (with respect to their positions in S, as opposed to their indices) points y i 0 and y i 00. It is clear that the points y 1 y : : : y qn are equispaced in S, and hence the number of these points in any interval of length diers from by no more than 1. Since x i 's and y i 's alternate, the number of x i 's in any interval diers from the number of y i 's in the same interval by no more than two. This completes the proof of the lemma. We now return to the proof of Proposition.1. Recall that for all k 0. Combining this with (1.1), we get where h := d 1 > 1. j p k q k j < 1 q k q k+1 (.4) q k+1 < Cq d 1 k = Cq h k (.5) If l Z and l < +1, then from (.5) we get > ( l ) 1 h if moreover q C n jl, we may divide any interval of the form [j j + l) into l < C 1 h l 1 1 h intervals of length, and apply Lemma.3 to each of these, getting j#(s \ [j j + l)) lj < C 1 h 1 l 1 1 h = C l h 1 (.6) where C = C 1 h 1, h 1 = 1 1 (0 1). Given any interval [j j + k), let q h n be the largest denominator k, and set a := [ k ]. Then a > k, and we may apply (.6) to get j#(s \ [j j + a )) (a )j < C (a ) h 1 < C k h 1 : Repeating this estimate inductively to the leftover interval [j + a k), we get that j#(s \ [j j + k)) kj < C (k h 1 + ( k )h 1 + : : : ) = C 3 k h 1 where C 3 := C : 1 1 h 1 3. A sucient condition for a function to be a Jacobian The main result of this section is: Proposition 3.1. Let u : R! R be a positive function which is constant on each open unit square with vertices in Z, and let > 0 be given. For any square S in R, let e(s) be the quantity 5

6 max ( 1 jsj RS u R 1 jsj S u where jsj denotes the area of the square S. Dene an \error" function E : N! R by letting E(k) be the supremum of e() over the collection of k k squares of the form [i i + k] [j j + k], where i j Z. If the product Y E( i ) i ) converges, then there is a bilipschitz homeomorphism : R! R a.e. with Jac() = u The proof of Proposition 3.1 is based on the following: Proposition 3.. There is a constant C 1 with the following property. Let T := [0 ] [0 ] R, and let u : T! (0 1) be a function which is constant in each square (i i + 1) (j j + 1), i j = 0 1. Then there is a bilipschitz homeomorphism : T! T which pointwise, so that Jac() = u a.e., where 1 = 1 jt j and BiLip() C1. To prove Proposition 3., we will need two lemmas. Let k k 1 denote the l 1 norm on R, so k(x y)k 1 := max(jxj jyj). Let S r denote the square fx R j kxk 1 rg. Lemma 3.3. Suppose u R 1 : S r! R R and u : S r! R are a continuous positive functions of k k 1, and S r u 1 = S r u. Then there is a bilipschitz homeomorphism : S r! S r which r pointwise, so that Jac() = u 1 u a.e., and bilip() where C is independent of u and r. 1 1 C C Proof. We rst treat the case where u 1 and R S r u 1 = R S r u = js r j. Set u := u 1. Dene f : [0 r]! [0 r] by f(t) := 1 Z S t u 1 so f(0) = 0 and f(r) = r. Now dene : S r! S r by x (x) = f(kxk 1 ) kxk 1 when kxk 1 6= 0, and (0) = 0. The map is dierentiable on S r n f(x y) j jxj = jyjg, and calculation shows that kdk(x) k 1 max f(kxk1 ) f 0 (kxk 1 ) kxk 1 R T u, 6

7 and kxk1 k[d(x)] 1 k k 1 max f(kxk 1 ) 1 f 0 (kxk 1 ) where k 1 is independent of u. These quantities are bounded by k independent of u. Hence where k is When which implies that is close to 1, then one gets that bilip() k : (3.4) k(d I)k k 3 1 k4 k[d] 1 k where k 3 and k 4 are independent of u. Combining this with (3.4) we get that when C is suciently large. In the general case, we set bilip() C (3.5) u i := u ijs r j RS r u i R so S r u i = js r j. Applying the special case above, we obtain bilipschitz homeomorphisms 1 : S r! S r and : S r! S r with Jac( i ) = u i a.e., whose bilipschitz constants satisfy (3.5) (with u replaced by u i ). Then := 1 1 : S r! S r has Jacobian u 1 u a.e. and bilip() C C 1 1 Lemma 3.6. We use the notation A(r 1 r ) for the annulus B(0 r ) B(0 r 1 ). Pick 0 < a < b, and set A := A(a b). If u 1 : A! R and u : A! R are positive Lipschitz functions with Z Z u 1 = u = jaj then there is a bilipschitz homeomorphism : A! A with A A and Jac() = u 1 u a:e: 7

8 C3 C3 1 bilip() (1 + Lip(u 1 )) (1 + Lip(u )) (3.7) 1 for C 3 = C 3 ( b a ) moreover, when u = u then can be chosen to pointwise. Proof. We rst assume that u 1, and set u := u 1. Using polar coordinates, we dene I : [a b] [0 ]! R by I(r ) := Z 0 u(r )d : Then dene 1 : A! A to be the polar coordinate shear I(r ) 1 (r ) := r : I(r ) The map 1 clearly denes a bilipschitz homeomorphism of A and Jac( 1 )(r ) = u(r ) I(r ) for a.e. (r ). Now dene f : [a b]! [a b] by the relation Z A(at) u(r ) = Area(A(a f(t))) and : A! A by (r ) = (f(r) ). Then satises Jac( )(r ) = I(r) a.e. (r ). So we can set = 1, and for a.e. (r ), for We now estimate bilip(). Jac()(r ) = [Jac( )( 1 (r )] [Jac( 1 )(r )] I(r ) = u(r ) = u(r ): I(r ) One gets bilip( 1 ) k 1 max( (1 + Lip(u))) where k 1 is independent of u. 1, then at points of dierentiability one has When which gives k (D 1 I) k k 1 + Lip(u) k[d 1 ] 1 k (1 + Lip(u)) k3 where k and k 3 are independent of u. It follows that bilip( 1 ) (1 + Lip(u)) k4 8

9 where k 4 is independent of u. One easily gets that where k 5 is independent of u, and when 1 then which implies bilip( ) < k 5 (3.8) k(d I)k k 6 1 k[d ] 1 k k7 (3.9) in this case. Combining (3.8) with (3.9) we get k8 bilip( ) (3.10) where k 8 is independent of u. Therefore bilip() bilip( 1 )bilip( ) (1 + Lip(u)) k4k8 : (3.11) We now return to the general case when u 6 1. Applying the special case above, we get homeomorpisms 1 : A! A and : A! A with Jac( i ) = u i almost everywhere, whose bilipschitz constants are controlled as in (3.11). Then the composition := 1 1 has Jacobian u 1 u a.e., and satises (3.7) with C 3 = k 4 k 8. It remains only to observe that i is determined by u i and therefore will pointwise when u 1 = u Proof of Proposition 3.. We will produce as a composition of homeomorphisms 1 3 which are constructed by applying Lemmas 3.3 and 3.6. We rst introduce some notation. Let T ij := (i i + 1) (j j + 1) for i j = 0 1. Let S ij T ij be the square with side length p 1 with the same center as T ij note that S ij is the largest square concentric with T ij which is contained in the ball B((1 1) 1) T. Let S 0 be the square of side length p with center at (1 1) so S 0 is contained in B((1 1) 1) and contains the squares S ij. Let A T be the annulus centered at (1 1) with radii 1 p and 1 note that A contains the squares S ij. Finally, let S be the square of side length p with center (1 1) the \hole" B((1 1) 1 p ) of A is inscribed in S. R We may assume that ess inf u =, ess sup u =, and that u = jt j. T Set v 1 := u, and let m :=. Let v : T! R be a continuous positive function satisfying a. R T ij v = R T ij v 1 for i j = 0 1 b. v m on the complement of [S ij 9

10 c. The restriction of v to S ij is constant on the boundary of each square concentric with S ij and d. max v min v k1 and 1 + Lip(v ) k1 where k 1 is independent of u. Now let v 3 : T! R be a continuous positive function satisfying e. R T v 3 = jt j f. v 3 m on S and on the complement of S 0 g. v 3 is constant on the boundary of each square concentric with T k max v h. 3 min v 3 where k is independent of u. Finally, set v 4 1. and 1 + Lip(v 3 ) k We now apply Lemma 3.3 to the restrictions v 1 j Tij and v j Tij and obtain a homeomorphism 1 : T! T with Jac( 1 ) = v 1 v 1 a.e. and C k 1 bilip( 1 ) : Since v (x) = v 3 (x) = m 1 for m T A, it follows that R A v = R A v 3 jaj. Hence if we set v := v jaj RA and v v 3 := v 3jAj RA, then Lip(v ) Lip(v v 3 ) and Lip(v 3 ) Lip(v 3 ). We may therefore apply Lemma 3.6 with u 1 = v and u = v 3 to get : T! T with Jac( ) = v v 3 = v v 3 a.e. and bilip( ) 4k C 3 : Finally, we apply Lemma 3.3 with u 1 = v 3 and u = v 4 to get 3 : T! T with Jac( 3 ) = v 3 v 4 3 a.e. and bilip( 3 ) k C : Then := 3 1 satises the requirements of Proposition 3. where C 1 := C k 1 + 4k C 3 + k C. Proof of Proposition 3.1. We may assume without loss of generality that = 1, since we may postcompose a bilipschitz homeomorphism 0 : R! R whose Jacobian satises Jac( 0 ) = u with the map x 7! p x, to get : R! R with Jac() = u. For each nonnegative integer q, let u q : R! R be a function whose value on each open square of the form (m q (m + 1) q ) (n q (n + 1) q ), m n Z, is equal to the average of u over that square. For each q N, we apply Proposition 3. to each of the above squares, to obtain a bilipschitz homeomorphism q : R! R with Jac( q ) = v q := u q 1 u q q a.e. and bilip( q ) max vq min v q C1 E( q 1 ) E( q ) C 1 : Then q := q : : : 1 : R! R will have Jacobian 10

11 and u u q q = u u q bilip( q ) " Y q 4C1 " Y 1 E( )# i i=1 i=1 E( i )# 4C1 < 1: Hence bilip( q ) is uniformly bounded and we may apply Arzela-Ascoli to get the desired bilipschitz homeomorphism : R! R. 4. Proof of Theorem 1.3 For j N, let S j be the tiling of R produced by rescaling the standard unit tiling by the factor 1. Pick i large enough that no tile of S j i contains more than one element of Y. Now form a tiling T = ft y g yy whose tiles are indexed by Y, where T y is a union of tiles from S i which are closer to y than to any other point of Y. The tiles of T have uniformly bounded diameters and inradii, and y is an interior point of T y for every y Y. Let u : R! R be a function that takes the value 1 jt yj on the interior of T y, for all y Y. Observe that if U R R is a square of side length k, then u U agrees with #(Y \ U) to within an error c 1 k. Hence our assumption on Y implies that u satises the hypotheses of Proposition 3.1. Therefore by Proposition 3.1, u is the Jacobian of a bilipschitz homeomorphism. The following lemma then completes the proof of Theorem 1.3. Lemma 4.1. Let T = ft i g ii be a tiling of the plane by tiles with uniformly bounded diameters and inradii. Let u : R! R be the function such that u(x) = 1 jt i j for a.e. x T i, and suppose u = Jac() for some bilipschitz homeomorphism : R! R. Let X = fx i g ii be a separated net where x i lies in the interior of T i for each i I. Then X is bilipschitz homeomorphic to Z. Proof. This is an application of Hall's Marriage Lemma, which has been used for similar problems by several authors (see [7, 1]). Consider a bi-partite graph whose set of vertices is Z S X, and there is an edge between (z 1 z ) Z and x i X i the intersection of (T i ) with the square [z 1 0:5 z 1 +0:5][z 0:5 z +0:5] is non-empty. Notice that the area of each set (T j ) is 1. Hence, for every k, any collection of k points in Z is connected with at least k elements in X, because otherwise k squares of the form [z 1 0:5 z 1 + 0:5[[z 0:5 z + 0:5] would be entirely covered by less than k tiles of the form (T j ), and this is impossible since the total area of the tiles is at most k 1. By the same reason, for every k, any collection of k points of X is connected with at least k points in Z. Hence by Hall's Marriage Lemma, our graph contains a bijection F : X! Z. Notice that (X) is also a separated net, which is bilipschitz equivalent to X. To complete the argument, it is enough to show that (X) in its turn is bilipschitz equivalent to Z. To see this consider the bijection F 1 : (X)! Z. If F 1 ((x i )) = (z 1 z ), then (by denitions of F and ), the distance dist((x i ) (z 1 z )) is bounded by p 0:5+BiLip() sup diam(t i ). Now it remains to notice that a bijection 11

12 between two separated nets that moves every point by a uniformly bounded distance is a bilipschitz equivalence. References [1] O. V. Bogopolsky. Preprint, [] D. Burago and B. Kleiner, Separated nets in Euclidean space and Jacobians of bi-lipschitz maps, Geom. Funct. Anal., 8 (1998), pp. 73{8. [3] B. Dacorogna and J. Moser, On a partial dierential equation involving the Jacobian determinant, Ann. Inst. H. Poincare Anal. Non Lineaire, 7 (1990), pp. 1{6. [4] H. Furstenberg. Private communication, [5] M. Gromov, Asymptotic invariants of innite groups, in Geometric group theory, Vol. (Sussex, 1991), Cambridge Univ. Press, Cambridge, 1993, pp. 1{95. [6] A. B. Katok, The special representation theorem for multi-dimensional group actions, in Dynamical systems, Vol. I Warsaw, Soc. Math. France, Paris, 1977, pp. 117{140. Asterisque, No. 49. [7] C. T. McMullen, Lipschitz maps and nets in Euclidean space, Geom. Funct. Anal., 8 (1998), pp. 304{314. [8] P. Papasoglu, Homogeneous trees are bi-lipschitz equivalent, Geom. Dedicata, 54 (1995), pp. 301{306. [9] H. M. Reimann, Harmonische Funktionen und Jacobi-Determinanten von Diffeomorphismen, Comment. Math. Helv., 47 (197), pp. 397{408. [10] T. Riviere and D. Ye, Resolutions of the prescribed volume form equation, NoDEA Nonlinear Dierential Equations Appl., 3 (1996), pp. 33{369. [11] O. Schramm. Private communication, [1] K. Whyte, Amenability, bi-lipschitz equivalence, and the von Neumann conjecture, Duke Math. J., 99 (1999), pp. 93{11. [13] D. Ye, Prescribing the Jacobian determinant in Sobolev spaces, Ann. Inst. H. Poincare Anal. Non Lineaire, 11 (1994), pp. 75{96. 1

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