MANFRED EINSIEDLER AND SHIRALI KADYROV
Abstract. We study the relation between measure theoretic entropy and es- cape of mass for the case of a singular diagonal flow on the moduli space of three-dimensional unimodular lattices.
1. Introduction
Given a sequence of probability measures{µi}∞i=1on a homogeneous spaceX, it is natural to ask what we can say about weak∗limits of this sequence? Often one is interested in measures that are invariant under a transformationTacting onX, and in this case weak∗ limits are clearly also invariant under T. IfX is non-compact, maybe the next question to ask is whether any weak∗limit is a probability measure.
IfT acts onX = Γ\Gby a unipotent element whereGis a Lie group and Γ is a lat- tice, then it is known thatµis either the zero measure or a probability measure [12].
This fact relies on the quantitative non-divergences estimates for unipotents due to works of S. G. Dani [4] (further refined by G. A. Margulis and D. Kleinbock [9]).
On the other hand, ifT acts onX = SLd(Z)\SLd(R) by a diagonal element, then µ(X) can be any value in the interval [0,1] due to softness of Anosov-flows, see for instance [8]. However, as we will see there are constraints onµ(X) if we have additional information about the entropies hµi(T). This has been observed in [5]
for the action of the geodesic flow on SL2(Z)\SL2(R), see Theorem 1.2. In this paper we will generalize this theorem to the space SL3(Z)\SL3(R) with the action of a particular diagonal element.
We identify X = SLd(Z)\SLd(R) with the space of unimodular lattices in Rd, see§2.1. Using this identification we can define ford= 3 the height function ht(x) of a latticex∈X as follows.
Definition 1.1. For any 3-lattice x∈SL3(Z)\SL3(R) we define theheight ht(x) to be the inverse of the minimum of the length of the shortest nonzero vector inx and the smallest covolume of planes w.r.t. x.
Here, the length of a vector is given in terms of the Euclidean norm onRd. Also, if d= 2 then we consider the height ht(x) to be the inverse of the length of the shortest nonzero vector inx. Let
X≤M :={x∈X| ht(x)≤M} andX≥M :={x∈X|ht(x)≥M}.
By Mahler’s compactness criterion (see Theorem 2.3) X≤M is compact and any compact subset ofX is contained in someX≤M.
In [5], M. E., E. Lindenstrauss, Ph. Michel, and A. Venkatesh give the following theorem:
M.E. acknowledges the support of the NSF from the grant 0554373, and both authors acknowl- edge support by the SNF (200021-127145).
1
Theorem 1.2. Let X be the homogeneous space SL(2,Z)\SL(2,R), let T be the time-one-map for the geodesic flow, andµbe aT invariant probability measure on X. Then, there existsM0,such that
hµ(T)≤1 + log logM
logM −µ(X≥M) 2
for anyM ≥M0. In particular, for a sequence ofT-invariant probability measures µi with entropieshµi(T)≥c we have that any weak∗ limit µ has at least µ(X)≥ 2c−1 mass left.
Here,µis a weak∗ limit of the sequence {µi}∞i=1 if for some subsequenceik and for allf ∈Cc(X) we have
k→∞lim Z
X
f dµik → Z
X
f dµ.
The proof of Theorem 1.2 in [5] makes use of the geometry of the upper half planeH.
From now on we letX= SL3(Z)\SL3(R) and let α=
e1/2
e1/2 e−1
∈SL3(R).
We define the transformation T :X →X via T(x) =xα. We now state the main theorem of this paper.
Theorem 1.3. Let X and T be as defined above. Then there exists a function ϕ(M)(which is given explicitly), withϕ(M)→M→∞0,and M0 such that for any T-invariant probability measureµonX, and any M > M0, one has
hµ(T)≤3−µ(X≥M) +ϕ(M).
In this context we note that the maximal measure theoretic entropy, the entropy of T with respect to Haar measure on X, is 3. This follows e.g. from [10, Prop.
9.2]. We will see later thatϕ(M) =O(log loglogMM).
As a consequence of Theorem 1.3 we have:
Corollary 1.4. A sequence of T-invariant probability measures {µi}∞i=1 with en- tropy hµi(T) ≥c satisfies that any weak∗ limit µ has at leastµ(X)≥c−2 mass left.
This result is sharp in the following sense. For anyc∈(2,3) one can construct a sequence of probability measures µi with hµi(T) →c as i → ∞ such that any weak∗ limitµhas preciselyc−2 mass left, see [8].
Another interesting application of our method arises when we do not assume T-invariance of the measures we consider. In this case, instead of entropy consider- ation we assume that our measures have high dimension and study the behaviour of the measure under iterates ofT.
Let us consider the following subgroups ofG
(1.1) U+={g∈G:α−ngαn→1 asn→ −∞}, (1.2) U− ={g∈G:α−ngαn →1 asn→ ∞},
(1.3) C={g∈G:gα=αg}.
For any > 0, group H, andg ∈ H we write BH(g) for the -ball in H around g, see also§2.2. Throughout this paper we writeAB if there exits a constant c >0 such that A≤cB. If the constantc depends onM, then we writeAM B.
Definition 1.5. For a probability measureν onX we say thatν has dimension at least din the unstable direction if for any δ >0 there exists κ >0 such that for any ∈(0, κ)and for anyη∈(0, κ)we have
(1.4) ν(xBU+BηU−C)δ d−δ for any x∈X.
Note that the maximum value fordin the definition is 2 sinceU+ is two dimen- sional. The most interesting case of this definition concerns a measureν supported on a compact subset, sayx0BU1+, of an orbitx0U+ under the unstable subgroup.
In this case, (1.4) is equivalent toν(x0uBU+)d−δ for allu∈U+ (which is one of the inequalities of the notion of Ahlfors regularity of dimensiond−δ) and for anyδ >0. See [11, Chaps. 4-6] for more information on Ahlfors regularity.
Let us consider the following sequence of measuresµn defined by µn= 1
n
n−1
X
i=0
Ti∗ν
where Ti∗ν is the push-forward ofν under Ti. We have
Theorem 1.6. For a fixedd, let ν be a probability measure of dimension at least d in the unstable direction, and let µn be as above. Let µ be a weak∗ limit of the sequence (µn)n≥1. Thenµ(X)≥ 32(d−43). In other words, at least 32(d−43)of the mass is left.
In particular, if d = 2 then the limit µ is a probability measure. In this case with a minor additional assumption onν one in fact obtains the equidistribution result, that is, the limit measureµis the Haar measure [15].
Another application of Theorem 1.6 is that it gives the sharp upper bound for the Hausdorff dimension of singular pairs. The exact calculation of Hausdorff dimension of singular pairs was achieved in [2]. We say that r ∈ R2 is singular if for every δ >0 there existsN0>0 such that for anyN > N0 the inequality
kqr−pk< δ N1/2
admits an integer solution for p∈Z2 and forq ∈ Zwith 0 < q < N. From our results we obtain the precise upper bound for the Hausdorff dimension of the set of singular pairs; namely this dimension is at most 43. This gives an independent proof for this fact which was proved in [2]. Letx∈SL3(Z)\SL3(R). Then we say xis divergent ifTn(x) diverges in SL3(Z)\SL3(R). We recall (e.g. from [2]) thatr is singular if and only if
xr= SL3(Z)
1
1 r1 r2 1
is divergent. An equivalent formulation1 of the above Hausdorff dimension result (see [2]) is that the set of divergent points in SL3(Z)\SL3(R) has Hausdorff dimen- sion 8−23 = 43+ 6.
However, we can also strengthen this observation as follows. A weaker require- ment on points (giving rise to a larger set) would be divergence on average, which we define as follows. A pointxisdivergent on average(under T) if the sequence of measures
1 N
N−1
X
n=0
δTn(x)
converges to zero in the weak∗ topology, i.e. if the mass of the orbit — but not necessarily the orbit itself — escapes to infinity.
Corollary 1.7. The Hausdorff dimension of the set of points that are divergent on average is also 43+ 6.
We finally note that the nondivergence result [3, Theorem 3.3] is related to Theorem 1.6. In fact, [3, Theorem 3.3] implies that µ as in Theorem 1.6 is a probability measure if ν has the additional regularity property; namely if ν is assumed to be friendly. However, to our knowledge these additional assumptions make it impossible to derive e.g. Corollary 1.7.
The next section below has some basic definitions and facts. In§3, we charac- terize what it means for a trajectory of a lattice to be above heightM in some time interval. Using this we prove Theorem 1.3 in§4-5. Theorem 1.6 and its corollary are discussed in§6.
Acknowledgements: We would like to thank Jim Tseng for discussions and for pointing out the reference to [7]. We also thank the anonymous referee for his detailed report and his suggestions.
2. Preliminaries
2.1. The space of unimodular lattices. In this section we will give a brief introduction to the space of unimodular lattices inR3.
Definition 2.1. Λ ⊂R3 is a lattice if it is a discrete subgroup and the quotient R3/Λ is compact.
Note that this is equivalent to saying that Λ =hv1, v2, v3iZ wherev1, v2, v3 are linearly independent vectors overR.
Definition 2.2. A latticeΛ =hv1, v2, v3iZis said to beunimodular if it has covol- ume equal to1, where the covolumeis the absolute value of the determinant of the matrix with row vectorsv1, v2, v3.
We identify a point SL3(Z)g ∈X with the unimodular lattice in R3 generated by the row vectors ofg ∈G. We leave it as an exercise for the reader to convince himself that this correspondence is well defined and a bijection.
We now state Mahler’s compactness criterion which motivates the definition of the height function in the introduction.
1Roughly speaking the additional 6 dimensions corresponding toU−C,are not as important as the 2 directions in the unstable horospherical subgroupU+. The latter is parametrized by the unipotent matrix as in the definition ofxr.
Theorem 2.3(Mahler’s compactness criterion).A closed subsetK⊂X is compact if and only if there existsδ >0such that no lattice inK contains a nonzero vector of length less thatδ.
For the proof the reader can refer to [13, Corollary 10.9]. We now deduce Corol- lary 1.4 from Theorem 1.3.
Proof. We need to approximate 1X≤M by functions of compact support. So, let f ∈Cc(X) be such that
f(x) =
(1 forx∈X≤M 0 forx∈X≥(M+1)
and 0≤f(x)≤1 otherwise. Suchf exists by Urysohn’s Lemma. Hence, Z
f dµi≥ Z
1X≤Mdµi=µi(X≤M)≥c−2−ϕ(M) Letµbe a weak∗ limit, then we have
lim
ik→∞
Z
f dµk= Z
f dµ and hence we deduce that
Z
f dµ≥c−2−ϕ(M).
Now, by definition off we getR
f dµ≤µ(X<(M+1)). Thus, µ(X<(M+1))≥c−2−ϕ(M).
This is true for anyM ≥M0, so lettingM → ∞finally we have µ(X)≥c−2
which completes the proof.
2.2. Riemannian metric on X. Let G = SL3(R) and Γ = SL3(Z). We fix a left-invariant Riemannian metricdG(or simplyd) onGand for anyx1= Γg1, x2= Γg2∈X we define
dX(x1, x2) = inf
γ∈ΓdG(g1, γg2)
which gives a metricdX onX = Γ\G.For more information about the Riemannian metric, we refer to [14, Chp. 2].
For a given subgroupH of Gwe letBHr (g) :={h∈H|dG(h, g)< r}. It makes sense to abbreviate and writeBrH=BrH(1), where we write 1 for the identity inG.
Definition 2.4. We say that r > 0 is an injectivity radius of x∈ X if the map g7→xgfrom BrG→BXr (x)is an isometry.
Lemma 2.5. For anyx∈X there existsr >0 which is an injectivity radius ofx.
Note that since X≤M is compact, we can chooser > 0 which is an injectivity radius for every point in X≤M. In this case, r is called an injectivity radius of X≤M. We refer to Proposition 9.14 in [6] for a proof of these claims.
2.2.1. Operator norms. We endowR3with the standard euclidean metric, writing
|u|for the norm ofu∈R3. Rescaling the Riemannian metric if necessary we may assume that there exists someη0>0 such that|u−ug|<|u|dG(1, g) for anyu∈R3 andg∈BηG
0.
2.2.2. Metric on U+. We may identifyU+ withR2 using the parametrization (t1, t2)∈R2→
1
1 t1 t2 1
.
It will be convenient to work with the maximum norm on R2. We will write DηU+={
1
1 t1 t2 1
:|t1|,|t2|< η}for a ball inU+ of radiusη centred at the identity. Rescaling the maximum norm onR2if necessary we will assume that
DU+⊂BU+.
2.3. Entropy. Instead of giving here the formal definition of the ergodic theoretic entropyhµ(T) we will state only a well-known and important lemma that will enter our arguments later. We refer to [16,§4] for a complete definition.
Fix η > 0 small enough so that BηSL3(R) is an injective image under the expo- nential map of a neighborhood of 0 in the Lie algebra. Define a Bowen N-ball to be the translatexBN for somex∈X of
BN =
N
\
n=−N
α−nBηSL3(R)αn.
Roughly speaking the BowenN-ball xBN consists of all y near xwhich have the property that the trajectories from time −N to time N ofx andy areη-close to each other.
The following lemma gives an upper bound for entropy in terms of covers of Bowen balls.
Lemma 2.6. Let µ be a T-invariant probability measure on X. For any N ≥ 1 and >0 let BC(N, ) be the minimal number of Bowen N-balls needed to cover any subset of X of measure bigger that 1−. Then
hµ(T)≤lim
→0lim inf
N→∞
logBC(N, )
2N .
We omit the proof which is very similar to [5, Lemma 5.2] and goes back to [1].
3. Sets of labeled marked times
LetN, M > 0 be given. In this section we define for every x∈TN(X≤M) the set of labeled marked times. Each configuration of such markings will correspond to a particular element of a partition of X, and we will estimate the cardinality of this partition (which is desirable due to the link of entropy and the logarithmic growth of covers as in Lemma 2.6). This marking has the property that it will tell whether the lattice Tn(x) is above or below heightM, without having to knowx.
However, we do not want to consider all vectors (or planes) ofxthat become short at some point - it is likely that a partitioning ofX that uses all such vectors (or planes) will be too large to be of use.
Rather whenever there are two linearly independent primitive 1/M-short vectors, our strategy is to consider a plane inxthat contains both vectors. So, for a given latticexwe would like to associate a set of labeled marked times in [−N, N] which tells us when a vector or a plane is getting resp. stops being 1/M-short. Choosing the vectors and planes ofxcarefully in the following construction we obtain a family MN of sets of labeled marked times. This will give rise to a partition of X, which will be helpful in the main estimates given in§4.
3.1. Short lines and planes. Letu, v ∈R3 be linearly independent. We recall that the covolume of the two-dimensional latticeZu+Zvin the planeRu+Rvequals
|u∧v|. Here,u∧v= (u1, u3, u3)∧(v1, v2, v3) = (u2v3−u3v2, u3v1−u1v3, u1v2−u2v1).
Below,u, v∈R3 will always be such thatZu+Zv=x∩(Ru+Rv) for a latticex.
In this case we callRu+Rv rationalw.r.t. xand will call|u∧v|the covolume of the planeRu+Rvw.r.t. x. We sometimes writea planeP inxto mean the plane P =Ru+Rv rational w.r.t. x.
We also note that the action of T extends toV2
R2via (3.1) T(u∧v) = (u1e1/2, u2e1/2, u3e−1)∧(v1e1/2, v2e1/2, v3e−1)
= ((u2v3−u3v2)e−1/2,(u3v1−u1v3)e−1/2,(u1v2−u2v1)e1).
For a planeP =Ru+Rv as above, we sometimes write T(P) for T(u∧v). For a vectorv= (v1, v2, v3)∈R3 we let T(v) :=vα= (v1e1/2, v2e1/2, v3e−1).
Let >0 be given. Fixx∈X, a vectorvinxis-short at timenif|Tn(v)| ≤. Similarly for planeP ⊂R3we say that it is-short at timen(w.r.t. x) if Tn(P) is rational w.r.t. Tn(x) and its covolume is≤.
3.2. (Labeled) Marked Times. For a positive number N and a lattice x ∈ TN(X≤M) we explain which times will be marked in [−N, N] and how they are labeled. The following lemma which is special to SL3(Z)\SL3(R) is crucial.
Lemma 3.1 (Minkowski). Let 1, 2 ∈ (0,1) be given. If there are two linearly independent1-short and2-short vectors in a unimodular lattice inx, then there is a unique rational plane in xwith covolume less than1 which in fact is12-short.
If there are two different rational planes of covolumes1 and2 in a unimodular latticex, then there is a unique primitive vector of length less than1 which in fact is 12-short. In this case, the unique 12-short vector lies in the intersection of the two short planes.
The first part of the lemma follows quickly from the assumption that xis uni- modular. The second follows by considering the dual lattice tox. We will use these facts to mark and label certain times in an efficient manner so as to keep the total number of configurations as low as possible.
3.2.1. Some observations. Let us explain how we will use Lemma 3.1. Assume that we have the following situation: There are two linearly independent primitive vectorsu, v in a unimodular lattice such that
|u| ≤1/M and|T(v)| ≤1/M.
Letu= (u1, u2, u3). It is easy to see that
|T(u)|=|(e1/2u1, e1/2u2, e−1u3)| ≤ e1/2 M .
Assume M ≥ e1/2. From Lemma 3.1 we have that the plane containing both T(u),T(v) has covolume at most eM1/22 ≤ M1, and it is unique with this property.
The similar situation arises when we have two different planesP, P0 which are rational for a unimodular lattice such that
|P| ≤1/M and|T(P0)| ≤1/M
where | · | means the covolume. Assume M ≥e. One can see that |T(P)| ≤ Me. Thus, we conclude from Lemma 3.1 that there is a unique vector of length at most
e
M2 ≤ M1 contained in both planes T(P) and T(P0).
3.2.2. Marked times. LetVN,x ={i∈[−N, N] : Ti(x)6∈X≤M}. VN,x is a disjoint union of maximal intervals and letV = [a, b] be one them.
(a) eithera=−N (and so ht(Ta(x))≤M) ora >−Nand ht(Ta−1(x))< M, (b) eitherb=N or ht(Tb+1(x))< M, and
(c) ht(Tn(x))≥M for alln∈V.
We first show how one should inductively pick the marked times for this interval V:
We will successively choose vectors and planes inxand mark the time instances with particular labels when these vectors and planes get 1/M-short on V and when they become big again. At time a we know that there is either a unique plane or a unique vector getting 1/M-short. Here, uniqueness of either follows from Lemma 3.1. Moreover, we cannot have two 1/M-short vectors (1/M-short planes) as otherwise there is a 1/M2-short plane (or vector) which contradicts the assumption that V = [a, b] has a as a left endpoint. If we have both a unique 1/M-short plane and vector then we consider whichever stays 1/M-short longer (say with preference to vectors if again this gives no decision). Assume that we have a unique plane. The case where we start with a unique vector is similar. Mark abyp1 which is the time when the plane is getting 1/M-short, and also mark by p01the last time in [a, b] when the same plane is still 1/M-short. Ifp01=bwe stop marking. If not, then there is again by Lemma 3.1 a unique 1/M-short plane or vector atp01+1. If it is a 1/M-short plane then at timep01+1 we must have a unique 1/M-short vector by the discussions in §3.2.1. In either case, we have a unique 1/M-short vector at timep01+ 1. Let us mark byl1the instance in [a, p01+ 1] when this vector is getting 1/M-short. Also, mark byl01, the last time in [p01+ 1, b] for which this vector is still 1/M-short. Ifl01=bwe stop, otherwise at timel01+ 1 there must be a unique 1/M-short plane or vector. If it is a short vector then we know that there must be a unique plane of covolume at most 1/M by the discussions in
§3.2.1. So, in either case there is a unique 1/M-short plane at timel10+ 1. So, there is an instance in [a, l01+ 1] which we mark byp2 when for the first time this plane is 1/M-short. Also, mark by p02, the last instance of time in [l01+ 1, b], for which the plane is 1/M-short. Ifp02=b we stop here, otherwise we repeat the arguments above and keep marking the time instances inV byli, l0i, pj, p0j until we hit timeb.
Given a positive number N and a lattice x∈ TN(X≤M) we first consider the disjoint intervals Vi of maximum length with the property asV above. Now start labeling some elements of the sets Vi as explained earlier starting with V1 and continuing withV2 etc. always increasing the indices ofli, li0, pi, p0i.
For any latticexas above we construct in this way a set of labeled marked times in [−N, N]. We denote this set by
N(x) =N[−N,N](x) = (L,L0,P,P0).
Here L = L(x),L0 = L0(x),P = P(x),P0 = P0(x) are subsets in [−N, N] that contain all the labeled marked timesli, l0i, pj, p0j forxrespectively. Finally, we let
MN ={N(x) : x∈TN(X≤M)}
be the family of all sets of labeled marked times on the interval [−N, N].
3.2.3. The Estimates.
Lemma 3.2(Noninclusion of marked intervals). Let(L,L0,P,P0)∈ MN be given.
For any qinL or in P there is nor inL or inP with q≤r≤r0 ≤q0.
Proof. We have four cases to consider. Let us start with the case thatr=pi, r0 =p0i and q =pj, q0 = p0j (where j > i as it is in our construction only possible for a later marked interval [q, q0] to contain an earlier one). However, by construction the plane Pi that is 1/M-short at that time we introduce the marked interval [pi, p0i] (which is either the beginning of the intervalV or is the time the earlier short vector stops to be short) is the unique short plane at that time. Hence, it is impossible to have the stated inclusion as the planePj (responsible for [pj, p0j]) would otherwise also be short at that time. The case of two lines is completely similar.
Consider now the case q = pj ∈ P and r = li ∈ L with pj ≤ li ≤ l0i ≤ p0j. If li =a (and so also li =pj = a) is the left end point of interval V = [a, b] in the construction, then we would have marked eitherli, l0i or pj, p0j but not both as we agreed to start by marking the end points of the longer interval (if there is a choice). Hence, we may assumeli > a and that times li, l0i have been introduced after consideration of a plane with marked timespk, p0k satisfyingli≤p0k+ 1≤l0i, in particular j 6= k. We now treat two cases depending on whether pk ≥ li or not. If pk ≥ li then pj ≤ pk ≤ p0k ≤ p0j which is impossible by the first case.
So, assume pk < li then we have two different planes that are 1/M-short at time li. This implies that the vector responsible for the interval [li, l0i] is 1/M2-short by Lemma 3.1. However, this shows that the same vector is also 1/M-short at time li−1 for M ≥ e, which contradicts the choice of li. The case of q =li ∈ L and
r=pj∈ P is similar.
We would like to know that the cardinality ofMN can be made small (important in Lemma 2.6) withM large. In other words, forM large we would like to say that limN→∞log #MN
2N can be made close to zero. The proof is based on the geometric facts in Lemma 3.1.
LetN = (L,L0,P,P0)∈ MN and letL={l1, l2, ..., lm}andP ={p1, p2, ..., pn} be as in the construction of marked times. It is clear from the construction that li0 < l0i+1 forl0i, l0i+1∈ L0. Thus from Lemma 3.2 we conclude thatli≤li+1. Hence we have L={l1 ≤l2 ≤... ≤lm}. Similarly, we must have P ={p1 ≤p2 ≤...≤ pn}. In fact, we have the following.
Lemma 3.3 (Separation of intervals). For any i = 1,2, ..., m−1 and for any j= 1,2, ..., n−1we have
li+1−li>blogMcandpj+1−pj >blogMc.
Also,
li+10 −li0>blogMcandp0j+1−p0j >blogMc.
Proof. For 1/M-short vectors in R3, considering their forward trajectories under the action of the diagonal flow (et/2, et/2, e−t), we would like to know the minimum possible amount of time needed for the vector to reach size≥1. Letv= (v1, v2, v3) be a vector of size≤1/M which is of size≥1 at timet≥0. We have
1≤v21et+v22et+v23e−2t≤(v21+v22+v32)et≤ et M2. So, we have
t≥logM2.
Hence, it takes more than 2blogMcsteps for the vector to reach size≥1. Similarly, for a vector v = (v1, v2, v3) of size ≥ 1, we calculate a lower bound for the time t≥0 when its trajectory reaches size ≤1/M. We have
1
M2 ≥v12et+v22et+v32e−2t≥(v21+v22+v32)e−2t≥e−2t.
So, we must have t≥logM and hence it takes at least t=blogMcsteps for the vector to have size≤1/M.
Now, assume that li+1−li ≤ blogMc. Let u, v be the vectors in x that are responsible for li, li+1 respectively. That is, u, v are 1/M-short at times li, li+1
respectively but not before. Then the above arguments imply that
|Tli(v)| ≤1 and|Tli+1(u)| ≤1
so the planeP containing bothuandv is 1/M-short at timesli andli+1. The covolume of Tn(P) w.r.t. Tn(x) isp
a1en+a2e−n/2 for some nonnegative a1 and a2. In particular, it is a concave function ofn and hence the plane P is 1/M-short in [li, li+1] (and soli, li+1are constructed using the sameV). From our construction we know thatl0i< li+10 . By Lemma 3.1 the same planePis 1/M2-short on [li, l0i]∩[li+1, l0i+1]. If this intersection is non-empty, thenP is alsoe/M2-short at time l0i+ 1. As M ≥ e this shows that it is the unique plane that is used to mark points, say pk, p0k, after marking li, l0i. If on the other handl0i < li+1, then we already know that P is also 1/M-short at time l0i+ 1 ∈ [li, li+1] and get the same conclusion as before. Therefore,pk≤li≤l0i≤p0k which is a contradiction to Lemma 3.2.
The proof of the remaining three cases are very similar to the arguments above
and are left to the reader.
Let us consider the marked points of Lin a subinterval of lengthblogMcthen there could be at most 1 of them. Varying x while restricting ourselves to this interval of lengthblogMcwe see that the number of possibilities to set the marked points in this interval is no more thanblogMc+ 1. ForM large, sayM ≥e4, we have
=blogMc+ 1≤ blogMc1.25. Therefore, there are
≤ blogMc1.25(bblog2NMcc+1)M e2.5NbloglogblogMcMc
possible ways of choosing labeled marked points forLin [−N, N]. The same is true forL0,P,P0. Thus we have shown the following.
Lemma 3.4 (Estimate ofMN). ForM ≥e4 we have
#MN M e10NbloglogblogMcMc.
3.3. Configurations. Before we end this section, we need to point out another technical detail. For our purposes, we want to study a partition element in X≤M corresponding to a particular set of labeled marked times. SinceX≤M is compact, it is sufficient for us to study anη-neighborhood of somex0in this partition. These are the close-by lattices which have the same set of labeled marked times. We shall see that the fact thatN(x) =N(x0), forxinx0BηG, gives rise to restrictions on the position ofx with respect tox0 (see §4.1). However, just knowing that N(x0) = N(x) will not be sufficient for the later argument. Hence, we need to calculate how many possible ways (in terms of vectors and planes) we can have the same labeled marked times. For this purpose, we consider the following configurations.
3.3.1. Vectors. Let l be a marked time in the first componentL of the marking N(x0). Letv0 be the vector inx0 that is responsible forl in the construction of marked times forx0. Lety = Tl−1(x) be in Tl−1(x0)BηSL3(R) withN(x) =N(x0) andv in xthat is responsible forl in the construction of marked times for x. Let v0∈x0be such that Tl−1(v0)g= Tl−1(v) for someg∈BηSL3(R)withy= Tl−1(x0)g.
We want to know how many choices forv0 are realized by the various choices ofx as above.
Lemma 3.5. Let N(x0) be given. Also, let l ∈ L = L(x0) and v0 ∈ x0 be the vector which is responsible forl. There are two possibilities:
(1) If l is the end point of a maximal interval V in Vx0, then for any x with N(x) = N(x0) and Tl−1(x) = Tl−1(x0)g, with g ∈ BGη, the vec- tor±v0αl−1gα−(l−1)is responsible forl inL(x).
(2) If not, then there arep, p0 inP(x0),P0(x0)respectively, withp≤l−1≤p0, and a setW ⊂x0, of sizemin{ep0−l, e(l−p)/2}, such that ifxis a lattice such that N(x) =N(x0), andTl−1(x) = Tl−1(x0)g, with g ∈BηG, then for somew∈W,wαl−1gα−(l−1), is the vector responsible for l inL(x).
Proof. To simplify the notation below we set w0 = Tl−1(v0) ∈ Tl−1(x0), w = Tl−1(v)∈y, andw0= Tl−1(v0) =wg∈Tl−1(x0).
We have
1
M ≤ |w| ≤ e M, and so
|w0| ≤ |w0−w|+|w|
≤ |w|d(g−1,1) +|w|
≤e(1 +η)/M.
Also,
|w0| ≥ |w| − |w−w0|
≥(1−η)/M.
Together
(3.2) 1−η
M ≤ |w0| ≤ e(1 +η)
M .
Assume first that l = a is the left end point of the interval V = [a, b] in the construction of marked times. In this case, w0 and w0 lie in the same line inR3. Otherwise, if they were linearly independent then the plane containing both would be e2(1 +η)/M2-short by Lemma 3.1. ForM ≥3e2 this is a contradiction to the assumption thatl =a. Since we only consider primitive vectors we only have the choice ofw0=±w0.
Now, assume thatl is not the left end point of the intervalV. Then, there is a planeP inx0 responsible forp, p0 withp≤l−1≤p0 such that
|Tp−1(P)| ≥1/M and|Tp0+1(P)|>1/M
|Tk(P)| ≤1/M fork∈[p, p0].
Let us calculate how many possibilities there are forw0∈Tl−1(x0). By (3.2)w0 is in the plane Tl−1(P) of covolume<1 w.r.t. Tl−1(x0) since Tl−1(x0) is unimodular.
Since
1
M <|Tp0+1(P)|and 1
M ≤ |Tp−1(P)|, we get
max
(e−(p0−l+2)
M ,e−(l−p)/2 M
)
≤ |Tl−1(P)|
(see§3.1 for the action of T on planes). We note that the ball of radiusrcontains at most max{rA2,1} primitive vectors of a lattice in R2 of covolume A. This follows since in the case of rbeing smaller than the second successive minima we have at most 2 primitive vectors, and ifr is bigger, then area considerations give rA2 many lattice points in the r-ball.
We apply this for A=|Tl−1(P)| ≥max n
e−(p0 −l+2)
M ,e−(l−p)/2M o
andr= (1+η)eM where
r2
A = (1 +η)2e2/M2 maxn
e−(p0 −l+2)
M ,e−(l−p)/2M o min{e(p0−l), e(l−p)/2},
which proves the lemma.
3.3.2. Planes. Let pbe a marked time in the third component P of the marking N(x0). LetP0be a plane inTp−1(x0) that is responsible forpin the construction of marked times forx0. Lety= Tp−1(x) be in Tp−1(x0)BSLη 3(R) withN(x) =N(x0) andP inxthat is responsible forpin the construction of marked times forx. Let P0 be a plane that is rational w.r.t. x0 such that Tp−1(P0)g = Tp−1(P) for some g ∈BηSL3(R) with y = Tp−1(x0)g. We want to know how many choices for P0 are realized by the various choices ofxas above. We have two cases.
Lemma 3.6. Let N(x0) be given. Also, let p∈ P =P(x0)and P0 in x0 be the plane which is responsible for p. There are two possibilities:
(1) If p is the end point of a maximal interval V in Vx0, then for any x with N(x) = N(x0) and Tp−1(x) = Tp−1(x0)g, with g ∈ BηG, the plane P0αp−1gα−(p−1) is responsible forpin P(x).
(2) If not, then there arel, l0 inL(x0),L0(x0)respectively, withl≤p−1≤l0, and a set of planes W ⊂x0, of size min{e(l0−p)/2, ep−l}, such that if x is a lattice such that N(x) = N(x0), andTp−1(x) = Tp−1(x0)g, with
g∈BGη, then for some P ∈W,P αp−1gα−(p−1), is the plane responsible forpinP(x).
We will not prove the lemma since a similar argument to that giving Lemma 3.5 gives this lemma.
4. Main Proposition and Restrictions
Fix a height M ≥ 1. Let N ≥ 1 and consider N = N(x0) ∈ MN. Let V =Vx0 ⊂[−N, N] be as before so that for any n∈[−N, N], n∈Vx0 if and only if there is a 1/M-short plane or a 1/M-short vector at timen. Define the set
Z≤M(N) :={x∈TN(X≤M)| N(x) =N }.
Now, we state the main proposition.
Proposition 4.1. There exists a constantc0>0, independent ofM, such that the setZ(N)can be covered byM e6N−|V|c
18N blogMc
0 BowenN-balls.
In the proof of Theorem 1.3 we will consider
N→∞lim
log #Z(N)
2N .
Thus, in this limit, the term arising from c
18N blogMc
0 can be made small for M large sincec0 does not depend on M. So, our main consideration is thee6N−|V| factor.
On the other hand, it is easy to see that the setZ(N) can be covered bye6Nmany BowenN-balls. But this does not give any meaningful conclusion. Therefore,e−|V| is the factor appearing in Proposition 4.1 that leads to the conclusion of Theorem 1.3.
In proving Proposition 4.1, we will make use of the lemmas below which give the restrictions needed in order to get the drop in the number of BowenN-balls to cover the setZ(N).
4.1. Restrictions of perturbations.
4.1.1. Perturbations of vectors. Letv= (v1, v2, v3) be a vector inR3.
Lemma 4.2. For a vector v of size≥1/M, if its trajectory under the action ofT stays1/M-short in the time interval[1, S] then we must have v21v+v222
3
<2e−S. Proof. We will prove a slightly stronger statement. For this letλ1≥1 andλ2≤1 and assume that
λ1(v12+v22+v23)≥ 1
M2 ≥λ2(v12eS+v22eS+v32e−2S).
This simplifies to
λ1v32>(v12+v22)(λ2eS−λ1).
Assumingλ1, λ2 are close to 1, we must havev36= 0 and v12+v22
v23 ≤ λ1 λ2eS−λ1
.
Assuming again thatλ1, λ2are close to 1 the last expression is bounded by 2e−S.
We would like to get restrictions for the vectors which are close to the vectorvand whose trajectories behave asv on the time interval [0, S] . So, let u= (u1, u2, u3) be a vector inR3 withu=vgfor some g∈BSLη 3(R) such that|u| ≥1/M and that its forward trajectory stays 1/M-short in the time interval [1, S].
Let us first assumeg=
1
1
−t1 −t2 1
∈BηU+ so that
u1 u2 u3
= v1 v2 v3
1
1
−t1 −t2 1
.
From Lemma 4.2 we know that u21u+u2 22 3
<2e−S.So, (v1−v3t1)2+ (v2−v3t2)2
v23 <2e−S.
We are interested in possible restrictions ontj’s since they belong to the unstable horospherical subgroup of SL3(R) under conjugation by α= diag(e1/2, e1/2, e−1).
Simplifying the left hand side, we obtain (v1
v3 −t1)2+ (v2
v3 −t2)2<2e−S. We also know vv212
3
+vv222 3
<2e−S.Together with the triangular inequality, we get t21+t22<(
√ 2e−S+
√
2e−S)2= 8e−S. In general, we have
g=
1
1
−t1 −t2 1
a11 a12 a13 a21 a22 a23 0 0 a33
∈BηSL3(R).
In this case, we still claim that
t21+t22<8e−S. Let
w= w1 w2 w3
= v1 v2 v3
1
1
−t1 −t2 1
so that
(4.1) u=vg=w
a11 a12 a13
a21 a22 a23 0 0 a33
.
We observe
TS(u) = TS(w)
a11 a12 a13e−3S/2 a21 a22 a23e−3S/2
0 0 a33
,
so that TS(u)∈TS(w)BηSL3(R)and|TS(u)−TS(w)|< η|TS(u)|by the discussion in§2.2.1. Hence,|TS(u)|<1/M implies
(4.2) |TS(w)| ≤ |TS(u)|+|TS(u)−TS(w)|<1 +η M .
On the other hand, sinceg∈BηSL3(R)we have
(4.3) |w| ≥ |u| − |u−w|>1−η M Combining (4.2) and (4.3) we get
|w|
1−η > 1
M > |TS(w)|
1 +η .
Now, the proof of Lemma 4.2, for sufficiently smallη >0 implies w21+w22
w23 <2e−S.
Hence, we are in the previous case withureplaced byw. So, we havet21+t22<8e−S which proves the claim. We have shown the following.
Lemma 4.3. There exists a sufficiently smallη >0, such that for any M, S >0 the following holds. Letv, ube vectors inR3with sizes≥1/M whose trajectories in [1, S] stay1/M-short. Assume thatu=vgwith g∈BηSL3(R) and that the notation is as in (4.1). Then
t21+t22≤8e−S.
Lemma 4.4. Let η >0 be given. For any S, S0 >0, let us divide[−2η,2η]2 into small squares of side length 12ηe−3S0/2. Then there exists a constantc >0such that there aremax{1, e3S0−S}small squares that intersect with the ballt21+t22≤8e−S on[−2η,2η]2.
Proof. Note thatt21+t22≤8e−Sdefines a ball with diameter 2√
8e−S/2. If12ηe−3S0/2≥ 2√
8e−S/2then there are 4 squares that intersects the ball. Otherwise (which makes 3S0−Sbounded below), there can be at most (e(e−3S−S/20/2)2)2 =e3S0−S small squares
that intersect with the given ball.
What Lemma 4.3 and Lemma 4.4 say is the following:
Consider a neighborhoodO =x0Bη/2U+Bη/2U−Cofx0inX where as beforeU+, U−, and C are the unstable, stable, and centralizer subgroups of SL3(R) with respect toα, respectively. If we partition the square with side length 2ηin Bη/2U+ into small squares with side lengthsηe−3S0/2, then we have dηe−3S2η0/2e2 de3S0/2e2 many elements in this partition. Now, assume that there is a vectorv∈x0with|v| ≥1/M that stays 1/M-short in [1, S] and consider the set of latticesx=x0ginO with the property that the vector w=vg in xbehaves as v in [0, S]. Then the above two lemmas say that this set is contained in≤c0e3S0−S many partition elements (small squares). Hence, in the proof of Proposition 4.1, instead of ≤ c0de3S0/2e2 many Bowen balls we will only consider≤c0e3S0−S many of them and this (together with the case below) will give us the drop in the exponent as appeared in Proposition 4.1.
4.1.2. Perturbations of planes. Assume that for a latticex∈X there is a rational planeP w.r.t. xwith
|P| ≥1/M and|Tk(P)| ≤1/M fork∈[1, S].
Letu, v be generators ofP with|P|=|u∧v|. So we have
|u∧v| ≥1/M ≥ |TS(u∧v)|.
Thus, substitutinga=u2v3−u3v2, b=u3v1−u1v3, c=u1v2−u2v1 (cf. 3.1) we obtain
a2+b2+c2≥a2e−S+b2e−S+c2e2S, which gives
c2
a2+b2 ≤ 1−e−S
e2S−1 =e−2S 1−e−S
1−e−2S =e−2S 1
1 +e−S < e−2S. Assumex0=xgfor some g∈BηSL3(R). For now, let us assume that
g=
1
1 t1 t2 1
.
Letu0, v0∈x0 be such that u0
v0
=
u01 u02 u03 v10 v20 v30
=
u1 u2 u3
v1 v2 v3
1
1 t1 t2 1
=
u1+t1u3 u2+t2u3 u3 v1+t1v3 v2+t2v3 v3
.
We leta0=u02v30−u03v02= (u2+t2u3)v3−u3(v2+t2v3) and hencea0=a. Similarly, b0 =u03v10 −u01v03=b and let
c0 =u01v20 −u02v01=
(u1+t1u3)(v2+t2v3)−(u2+t2u3)(v1+t1v3) =c−at1−bt2. Now, assume that
|u0∧v0| ≥1/M and|Tk(u0∧v0)| ≤1/M fork∈[1, S]
which by the above implies c02
a02+b02 = (c−at1−bt2)2
a2+b2 < e−2S.
For a generalg∈BSLη 3(R) we would like to obtain a similar equation. Let us write g as
(4.4) g=
1
1 t1 t2 1
g11 g12 g13
g21 g22 g23
0 0 g33
.
Then we have
Tl(x0) = Tl(xg) = Tl
x
1
1 t1 t2 1
g11 g12 g13e−32l g21 g22 g23e−32l
0 0 g33
.
Hence the forward trajectories of x0 and x
1
1 t1 t2 1
stay η close. Thus, we have
(c−at1−bt2)2
a2+b2 e−2S.