Unique ergodicity for non-uniquely ergodic horocycle flows

We consider the horocycle flow associated to a $\Z^d$-cover of a 
compact hyperbolic surface. Such flows have no finite invariant 
measures, and infinitely many infinite ergodic invariant Radon 
measures. We prove that, up to normalization, only one of these 
infinite measures admits a generalized law of large numbers, and 
we identify such laws.


Introduction
Motivation.The horocycle flow on the unit tangent bundle of a compact hyperbolic surface is uniquely ergodic: it admits exactly one invariant probability measure (Furstenberg [F]).In this paper we consider the horocycle flow on certain non-compact hyperbolic surfaces of infinite volume.The surfaces we consider (free Abelian covers of compact hyperbolic surfaces) are such that their associated horocycle flow does not preserve any invariant probability measure whatsoever; but it does preserve a family of infinite invariant Radon measures 1 [BL2], [S].
Our aim here is to prove that only one of these measures (up to normalization) is ergodic theoretically 'relevant' in a sense that is explained below.This can be viewed as a version of Furstenberg's unique ergodicity theorem in this context.
Generalized laws of large numbers.Consider a measure preserving flow ϕ t on a standard measure space (X, F, m).If m(X) < ∞, then the ergodic hypothesis lim m(X) holds for a.e.ω ∈ X.But if m(X) = ∞ then it fails in an essential way: For every A ∈ F there exists no normalization a(T ) s.t.lim ]dt exists a.e., other than those normalizations which make the limit zero or infinity [A2].
The failure of the ergodic hypothesis in the infinite measure setting suggests studying the following weaker -but possible -property, due to J. Aaronson [A1]: Definition 1.A measure preserving flow ϕ t on a standard measure space satisfies a generalized law of large numbers (GLLN), if there is a function L : {0, 1} R+ → R + , L = L[x(t)] s.t. for every A ∈ B, L[1 A (ϕ t ω)] = m(A) almost everywhere.
1 A Radon measure is a measure which is well-defined and finite on compact sets Example 1 (Finite measures).Any ergodic invariant probability measure admits the following generalized law of large numbers:

L[x(t)] = lim
T →∞ 1 T T 0 x(t)dt, when the expression makes sense, 0, when the expression does not make sense.
Proof.The example and its proof are due to Aaronson [A1].Suppose there were a generalized law of large numbers L. Fix A of positive finite measure and pick some x such that L[1 E (h s y)] = µ(E) for E = A, Q −1 A and y = x, Qx.Then in contradiction to c = 1.
Example 3 (Rational ergodicity [A3]).A flow ϕ t is called rationally ergodic if it is ergodic and there exists a measurable set A of positive finite measure such that A ( ).This condition implies the existence of a(T ) and T k → ∞ s.t.

L[x(t)] =
Cesaro-lim k→∞ 1 a(T k ) T k 0 x(t)dt, when the expression makes sense, 0, when the expression does not make sense, is a generalized law of large numbers.
The geodesic flow on the unit tangent bundle of a recurrent hyperbolic surface is rationally ergodic w.r.t. the volume measure (Aaronson & Sullivan [AS], see Roblin [Ro] for the variable curvature case).We prove below the rational ergodicity of the horocycle flow for a class of surfaces of negative curvature.
Horocycle flows on Z d -covers.Let M 0 be a compact connected orientable C ∞ Riemannian surface with negative curvature.Let T 1 (M 0 ) be the unit tangent bundle of M 0 , and g s : T 1 (M 0 ) → T 1 (M 0 ) the geodesic flow.Margulis [Mrg] and Marcus [Mrc] constructed a continuous flow h t : M 0 → M 0 for which the h-orbit of x is equal to W ss (x) := {y ∈ T 1 (M 0 ) : d(g s x, g s y) −−−→ s→∞ 0}, and for which ∃µ such that ∀s, t, g −s • h t • g s = h µ s t .This is the (stable) horocycle flow of M 0 .2Now let p : M → M 0 be a regular cover whose group of deck transformations (b) There exists µ such that ∀s, t, g −s • h t • g s = h µ s t ; (c) g s , h t commute with the deck transformations.
We call h t the horocycle flow of M .In the constant negative curvature case, µ = e, and we get the classical (stable) horocycle flow.
The ergodic invariant Radon measures on Z d -covers of compact Riemannian surfaces of negative sectional curvature were identified in [BL2] and [S]: they form a family {cm ϕ : c > 0, ϕ : G → R d is a homomorphism}, where The parameter ϕ ≡ 0 corresponds to the Margulis measure, see [BL2], [BM].(In the constant negative curvature this is the volume measure.) The results.The measures m ϕ with ϕ ≡ 0 do not admit GLLN's, because -by (c) -they are all squashable.We show that the Margulis measure m = m 0 does admit such laws.We thus obtain: Theorem 1.Let M be a Z d -cover of a compact connected orientable C ∞ Riemannian surface of negative curvature.The horocycle flow on T 1 (M ) has, up to a factor, exactly one invariant Radon measure with a generalized law of large numbers: m 0 .
We proceed to describe this law of large numbers (in fact, we shall present two such laws).In what follows, M 0 ⊂ T 1 (M ) is some fixed precompact connected fundamental domain for the action of the group of deck transformations on T 1 (M ).
Theorem 2. The measure m 0 is rationally ergodic.Consequently, there exists a function a(T ) and a sequence T k 0 x(s)ds when the expression makes sense, and zero otherwise.The following theorem is a more explicit GLLN, in the spirit of the 'second order ergodic theorems' of [ADF]: Theorem 3.With the same a(T ) as in the previous theorem, for every Next, we describe a(T ).We first need to make some comments concerning the geodesic flow on a Z d -cover.
Let G := {D a : a ∈ Z d } an enumeration of the group of deck transformations of the cover p : It follows from the work of Ratner [R] and Katsuda & Sunada [KS2] that the (normalized) Margulis distribution of ξ ω (T )/ √ T as ω ranges over M 0 converges to the distribution of a multivariate Gaussian random variable N on R d , with a positive definite covariance matrix Cov (N ).Set Remark 1.The value of σ is known when M is the homology cover of a compact hyperbolic surface M 0 of genus g: σ = 1 2π (g − 1), see [KS1], [KS2] and [PS].Our method of proof also yields the following result, which together with the central limit theorem for ξ ω (T )/ √ T , explains the fluctuations of 1 f dm 0 = 1.For every ε and almost every ω, if T is large enough, then T 0 f • h s ds is sandwiched between two fluctuating quantities, which converge in distribution, but not pointwise.
Another consequence is the following equidistribution result for the geodesic flow, in the constant curvature case.Define for q ∈ M , T 1 q (M ) := {v ∈ T q (M ) : q = 1}, and S q (T ) := g T [T 1 q (M )].This is a circle with perimeter |S q (T )| = 2π sinh T , and the following result describes its distribution as T → ∞: Theorem 6. Suppose M 0 has constant curvature −1, and let f be a continuous function with compact support on T 1 (M ).Then, for all q ∈ M , Remark 2. In the variable curvature case, we can only prove the following: There is a function c(q) s.t. for all q ∈ M , lim , where m is the measure on T 1 M which is obtained by integrating the Lebesgue measure on the unstable manifolds with respect to the transverse invariant Margulis measure.
Other examples with unique non-squashable invariant Radon measure.
The phenomenon of having a unique non-squashable invariant Radon measure is not restricted to horocycle flows.We are aware of two other examples: . We have: (1) m = Haar×Counting Measure is a non-squashable ergodic invariant Radon measure [AK]; (2) All other ergodic invariant Radon measures are squashable, and there are infinitely many such measures [N], [ANSS].
In the particular case when α is a quadratic surd, the unique non-squashable measure is rationally ergodic, and therefore admits a GLLN [AK].
Multidimensional generalizations of these random walks with non-i.i.d.jumps are considered in [ANSS].They exhibit the same phenomenon.

The main estimate and its implications
The main estimate.Recall the definitions of ξ ω (T ), N , σ, and • H from introduction.We use the following shorthand notation: Our results are based on the following two lemmas Lemma 1.For every ε > 0, there exists E ⊆ M 0 Borel of finite positive m 0measure, s.t. for some Lemma 2. The following limits hold for any f : (1) lim We defer the proof of these lemmata to the next section, and explain first why they imply theorems 1,2,3,4 and 5. Henceforth we assume for simplicity that m 0 ( M 0 ) = 1 (this can always be arranged by normalization).
Proof of Theorems 1 and 2. All measures m ϕ with ϕ ≡ 0 are squashable, and therefore have no GLLN's (Example 2).Rationally ergodic measures admit GLLN's (Example 3), so it is enough to show that m 0 is rationally ergodic.Choose E to be the set given by lemma 1 for some ε > 0 (it doesn't matter which), and set T satisfy the large deviations property [K].Using the fact that all norms on R d are equivalent, we deduce the existence of α > 0 such that m On the other hand, lemmas 1 and 2 give the following asymptotic lower bound for 1 E\Ω log µ T I T 1 : Comparing this to the upper bound of 1 E∩Ω log µ T I T 1 , we see that In the same way one shows that ), proving rational ergodicity.
Proof of theorem 4. The return sequence of m 0 is defined by [A1]).The proof of theorems 1 and 2 shows that the sets E given by lemma 1 are such sets.That proof also shows that for T large enough The normal distribution is such that , and fix f ∈ L 1 (m 0 ).For every ε > 0, construct a set E as in lemma 1.The ergodic theorem for the geodesic flow on T 1 (M 0 ) implies that ξω(T ) for almost all ω, the conditions of lemma 1 are satisfied for T large enough, whence 3 The case d = 1 is trivial; the general case can be obtained from it by orthogonal diagonalization of Cov (N ), and a suitable change of coordinates.
The ratio ergodic theorem says that f dm 0 almost everywhere.
Consequently, for almost every ω and T large Therefore, since ε > 0 is arbitrary, it is enough to check that for a.e.ω, Proof of theorem 5. Fix ε > 0 and let E be as in lemma 1.The case f = 1 m0(E) 1 E is that lemma.The general case follows from the ratio ergodic theorem.
Proof of theorem 6.Given a continuous function with compact support f and ε > 0, we first choose simple 4 functions f The first claim is that there is δ 0 such that, if T is large enough, ω T is a point of S q (T ) and C ω,T the arc of S q (T ) of length 2δ 0 sinh T centered at ω T , then where h s un denotes the unstable horocycle flow (in the previous theorems, we used stable horocycle flows).To check (1), let r θ denote the rotation by θ radians around q, and let ω ∈ T 1 q (M ) be the point such that ω T = g T (ω).We have Representing h s un , r θ , g T as matrix actions on PSL(2, R), 5 one calculates and sees that d In particular, any ω ∈ T 1 q (M ) has some δ 0 = δ 0 (ω) such that this upper bound is less than δ for all |θ| < δ 0 .Since ω ranges over the compact set T 1 q (M ), we can make this δ 0 uniform in ω. (1) now follows from the definition of f 1 , f 2 and the fact that sinh T ∼ 1 2 e T as T → ∞ and cos 2 θ ∼ 1 as θ → 0. Theorem 5 applied to the unstable horocycle flow and the simple functions ´, `eT/2 0 0 e −T /2 ´, and ´.
(The minus in front of ω T is because we are working with the unstable, not the stable horocycle flow.)It follows that the average over as T → ∞.If we choose δ 0 so small that tan δ0 δ0 = e ±ε , then for T large, We now average this estimate over ω ∈ S q (T ).The average of the LHS is just The average of the RHS can be determined from the following lemma Lemma 3. The distribution of ξω(T ) √ T , when ω runs over T 1 q (M ), convreges as T → ∞ to the distribution of N .
Proof.Let W s δ (ω) be the neighbourhood of size δ of the point ω on the weak stable leaf of ω.
The set E has positive measure (it contains an open neighborhood of (x 0 , y 0 )), and the measure on E projects on T 1 q (M ) to the Lebesgue measure.It follows that the asymptotic distribution of ξω(T ) √ T , when ω runs over T 1 q (M ) and T goes to infinity, is the same as the asymptotic distribution of ξ ω (T ) √ T , when ω runs over E. This latter distribution converges to the distribution of N by lemma 2, part 1.

Proof of lemmas 1 and 2
Symbolic dynamics.The proof of lemma 1 will make use of the symbolic coding of the geodesic flow.Before describing this coding, we recall some of the basic notions of symbolic dynamics that will be needed below.
A subshift of finite type with set of states S and transition matrix . There is a one-sided version σ : Σ + → Σ + obtained by replacing Z by N ∪ {0}.A subshift of finite type is topologically mixing iff ∃m s.t.all the entries of A m are positive.
Let G be a group, assumed for simplicity to be Abelian.The skew-product over T : X → X with the cocycle f : The suspension semi-flow over T : X → X and height function r * : X → R + is the semi-flow ϕ s : X r * → X r * , where If T is invertible, then this semiflow has a unique extension to a flow.The suspension (semi)-flow can be identified with the (semi)-flow (x, t) → (x, t + s) on (X × R)/ ∼ (respectively X × R + / ∼) where ∼ is the orbit relation of the skew-product T −r * .Both descriptions shall be used below.Now let p : M → M 0 be a Z d -cover of a compact connected orientable C ∞ Riemannian surface M 0 , and let G ∼ = Z d be the group of deck transformations of the cover.Enumerate We describe the geodesic flow on g s : T 1 (M ) → T 1 (M ) as a suspension flow, whose base is a skew-product, whose base is a subshift of finite type.This description is well-known [BL2], [Po] and can be obtained by a straightforward lifting argument from Bowen's symbolic dynamics of g s : There exist a topologically mixing two-sided subshift of finite type (Σ, T ), Hölder continuous r : Σ → R which depends only on the non-negative coordinates, f : Σ → Z d s.t.f (x) = f (x 0 , x 1 ), h : Σ → R Hölder continuous, and a Hölder continuous surjection π : Σ × Z d × R → T 1 (M ) with the following properties: (1) ) is a bounded-to-one surjection, where Remark 3.This construction can be made so that ε * := sup r * is arbitrarily small.
Remark 4. [Sh, C] In the context of geodesic flows of Z d -covers of compact connected surfaces of negative curvature, (−r n (x), f n (x)) : Next we describe the Margulis measure in these coordinates.There exists a unique P such that P top (−P r) = 0 (as it turns out P = ln µ).Let L −P r : C(Σ + ) → C(Σ + ) be the operator (L −P r F )(x) = σy=x e −P r(y) F (y) (Ruelle's operator).By Ruelle's Perron-Frobenius theorem, there exists ψ : Σ + → R + Hölder continous, and a Borel probability measure ν such that The measure ψdν is a shift invariant probability measure which can be extended to the two-sided shift Σ.Denote this extension by ν.
Lemma 5. [BM] The Margulis measure on T 1 (M ), subject to the normalization The symbolic local stable manifold of ω = π(x, ξ, s + h(x)) is defined by 5) implies that this is a subset of W ss (ω).
(Replace the alphabet S by S n for n large enough.) Proof of lemma 1.We normalize m 0 , for simplicity, to satisfy m 0 ( M 0 ) = 1.Fix some 0 < ε 0 < min r * .Using remark 5, we make sure that log µ sup x0=y0 ψ(x) ψ(y) < ε 0 .Our set is going to be With this set in mind, define Step 1. Fix T 0 large, and T * := log µ T − log µ T 0 .For all ω and T > T 0 , there are T 0 can be chosen s.t.
because of the definition of ω .By the definition of T * and the commutation relation between the horocycle flow and the geodesic flow, g T * [A T ] is also a horocyclic arc, and its length is T 0 .
Let N − be the number of different symbolic local stable manifolds which are contained in g T * [A T ], and let N + be the number of symbolic local stable manifolds which intersect it with positive measure.These numbers are finite, because the measures of symbolic local stable manifolds is bounded below (Remark 5), and because for almost every ω, any two symbolic local stable manifolds are either equal or disjoint modulo ω .(The disjointness is because ω are the conditional measures of the m on the leaves of the strong stable foliation, and |π −1 (ω)| = 1 for m-a.e.ω.) Choose We have: Assume by way of contradiction that δ is infinite.Since d ω • dD = d dD(ω) for all D ∈ G, it is is possible to work on (the compact) T 1 (M 0 ) and find ω n , ω n ∈ T 1 (M 0 ) on the same symbolic local stable manifold such that Writing ω n , ω n in symbolic coordinates, we see that there must exist a subsequence n k such that ω n k → ω, ω n k → ω and ω, ω are on the same symbolic local stable manifold.In particular, ω = h t0 (ω ) for some t 0 (which we assume w.l.o.g. to be non-negative).For every t > 0, h t (ω n k ) → h t (ω ).Since d ωn k (ω n k , ω n k ) > t for all k large, h t (ω n k ) must be between 7 ω n k and ω n k for all k large enough.It follows that h t (ω ) is between ω and ω = h t0 (ω), whence t 0 > t.But this cannot be the case for all t > 0, a contradiction.
We can now choose T 0 .We have already explained that for almost every ω, any two symbolic local stable manifolds are either equal or disjoint modulo ω .Now, T 0 = ω [g T * A T ], and so | 1 uniformly on a set of full measure. Step and H(0) = ln µ, ∇H(0) = 0, and H (0) = −Cov(N ) −1 .
µ rn(y)−T # i = µ ±ε0 ψ(y 0 , y 1 , . ..).Consequently, where the y's in this sum take values in the one-sided shift Σ + , and δ ij is Kronecker's delta.The asymptotic behaviour of this sum is described by theorem 4 of [BL2] (this estimation is reproduced in the appendix), and the step follows.
Step 3. Completion of the proof. Proof.
Recall that ∇H(0) = 0 and that H (0) = Cov(N ) −1 is negative definite.The function w H := −w t H (0)w is a norm on R d , and H ) as w H → 0 (all norms on R d are equivalent).Fixing an ε > 0, we see that there exists δ 0 such that −( Recall that for every i, uniformly as T → ∞.Therefore, we can make δ 0 so small that ξω(T ) Next recall that by the definition of ψ (Lemma 6) and the choice of T 0 (step 1), Using the identities e H(0) = µ, µ T * = T /T 0 , m 0 (E) = ε0 R rdν , we obtain These estimate are valid and uniform for ω in [ ξω(T * ) Recalling that ε * , ε 0 can be made arbitrarily small, and that m 0 was assumed to be normalized so that m 0 ( M 0 ) = 1, we see that lemma 1 is proved.
Since this integral is uniformly bounded (by m 0 (E)), and since f is absolutely integrable, we can replace the order of the limit and the integral, and obtain proving the first part of the lemma.
The second step is done in a similar way, but using the Birkhoff ergodic theorem.We begin by considering the case f (x) = e iϕη(x) for some fixed η ∈ R d .For every ω which satisfies (4) w.r.t.this η: Applying the Birkhoff ergodic theorem for dW η , we see that: N ) ] for a.e.ω .
By Fubini's theorem, the following set has full measure in M 0 ×R d with respect to m×Lebesgue: (N ) ] .

It follows that
For almost every ω lim N ) ] for a.e.η .
Integrating this limit against f (η) (assumed to be in L 1 (R d )), we have by the dominated convergence theorem and the inversion formula for the Fourier transform that for almost every ω ∈ M 0 .

Appendix
Let Σ + be a one-sided subshift of finite type, and let r, ψ : Σ + → R, f : Σ + → Z d two Hölder continuous functions, with r cohomologous (as a function on Σ) to a positive function r * : Σ → R. We assume using remark 4 that and explain how to obtain the asymptotic expansion of that was crucial to the proof of lemma 1.What follows is an expanded form of the the proof of theorem 4 in [BL2], and is not new.We decided to include it for completeness, and because we needed to emphasize that the following estimates are uniform.
Step 1. Rewriting A(x, ξ, T ) in terms of complex Ruelle operators.
Fix δ 0 > 0, and introduce two parameters P ∈ R and u ∈ R d (which shall later be calibrated to obtain optimal asymptotics).Construct two even functions in such a way that their Fourier transforms γ 1 , γ 2 have compact support, belong to C N (R) for N > 2d + 10, and satisfy e −δ0 ≤ γ 1 (0)/ γ 2 (0) < e δ0 .We have: Set T d e i w,ξ−fn(y) dw.

Fourier's inversion formula gives
e (−P +iα)rn(y)+ u−iw,fn(y) ψ(y) dαdw We simplify this expression using Ruelle operators [Bo2].Define for s := P − iα, z := u − iw the following operator on C(Σ + ): One easily checks that (L n s,z ϕ)(x) = σ n y=x e −srn(y)+ z,fn(y) ϕ(y).Consequently, with s := P − iα, z := u − iw.It is much easier to study the quantities with s := P − iα, z := u − iw, because of the possibility of bringing in operator theoretic methods to study 8 Here and throughout the Fourier transform is b We shall do so below, and analyze A i (x, ξ, T ); The information that we will gather in the process will eventually allow us justify the summation under in the integral and show that A i (x, ξ, T ) = A i (x, ξ, T ) (step 7).
Step 2. Summation of ∞ n=0 (L n s,z ψ)(x), leading to the first constraint on P and u.Let κ be a (common) exponent of Hölder continuity of r, ψ and f .It is well-known that L s,z acts continuously on The following is known [PP]: (a) L has a positive eigenvalue λ.This eigenvalue is equal to exp P top (−P r + u, f ), where and the spectral radius of N is strictly less than λ; (c) The spectral radius of L −P +iα,u−iw is smaller than λ when (α, w) = (0, 0).The first and second statements are a re-formulation of Ruelle's Perron-Frobenius theorem.The third statement follows from (5) and theorem 4.5 in [PP].
The map (s, z) and N s,z has spectral radius strictly less than |λ s,z |.Calculating, we see that . Now, the spectral radius of N s,z is strictly less than λ = λ(P, u).If this number were equal to one, then N n s,z would converge in norm.This is our first constraint on P and u: P and u should satisfy P top (−P r + u, f ) = 0. We will discuss the possibility of choosing such P and u later.For the moment, we note that if this condition is satisfied, then we can sum over n and obtain where A 1 (s, z) := P s,z ψ and B 1 (s, z) := (I − N s,z ) −1 ψ.Note that these functions are analytic in an ε 1 -neighbourhood of s = P, z = u.We can also arrange for A 1 (s, z) not vanish in this neighbourhood.9We have: We now discuss the constraint P top (−P r+ u, f ) = 0.The following are standard properties of the topological pressure functional (see e.g.[Bo2]): (a) Fix u and consider p u (t) := P top (−tr + u, f ).The first property says that p u (t) is convex, whence continuous.The third says that we can replace r by the positive r * .
The second now implies that p u (t) −−−−→ t→±∞ ∓∞.It follows that ∃t such that p u (t) = 0. Another look at the second property shows that p u (t) is strictly decreasing, and therefore this t is unique.Consequently, for every u ∈ R d , there exists a unique P = P (u) for which P top (−P r + u, f ) = 0.This function can be easily seen to be continuous in u.
One conclusion is that the set Λ := {(P, u) ∈ R × R d : P top (−P r + u, f ) = 0} is pathwise connected.In particular, Λ is connected.It follows that the analytic functions A 1 (s, z), B 1 (s, z), λ(s, z) which were defined locally in the neighbourhoods of (P, u) ∈ Λ patch up to well-defined H κ -valued analytic functions in an open complex neighbourhood of Λ, and (9) holds in this neighbourhood.
In particular, P (0) is positive definite.
Summarizing, we see that there is an open set U ⊃ Λ and holomorphic functions A(s, z), B(s, z) ∈ H κ , R(z) on U such that for every u ∈ R d and α, w such that (s, z) : where R(u − iw) = 1 2 w t P (u)w + o( w 2 ) as w → 0, and A(s, z) := − A1(s,z) A2(s,z) = 0. We also note that, by equations ( 10) and (11), We now explain how to modify (12) so that it holds for all (α, w) ∈ R × T d .We have already mentioned that thanks to (5), the spectral radius of L s,z is strictly less than one for (α, w) = (0, 0).Consequently, the left hand side of ( 12) is holomorphic outside U , and is equal to the right hand side inside U .It is now a standard procedure using C ∞ -bump functions to redefine A, B and R in such a way that (12) holds on R×T d .The only sacrifice we need to make is that A, B, R are become C ∞ , but not necessarily holomorphic.
Next, note that Re R(u − iw) is positive for w small because P (u) is positive definite.Using the method of C ∞ -functions we can modify A(s, z), B(s, z) and R(z) so that Re R(u − iw) is positive for all w ∈ T d and all u in any given compact neighbourhood.(The neighbourhood we need is (∇P ) −1 (K 0 ) where K 0 is described by the next step.) Step 4. A useful change of variables, leading to the second constraint on P and u, the definition of H(•), and the choice of K 0 .
The expansion we got for (L n s,z ψ)(x) gives for (s, z) = (P (u) − iα, u − iw): The change of variables α → α + w, ξ T would make the R dα integral the Fourier transform at T of two functions, which give rise to integrals whose asymptotic behavior can then be determined.This change of variables would be particularly convenient, if we require the following second constraint on P and u: P and u should satisfy: ∇P (u) = ξ/T , because together with the condition P = P (u), it would the lead to R×T d e −iT α b(α, w)(x)dαdw, where (14) • H :=minus the Legendre transform of P , so H (0) = −P (0 . We now discuss the possibility of choosing P and u such that ∇P (u) = ξ T and P = P (u) (⇔ P top (−P r + u, f ) = 0).As shown in [BL1] Once we have this u, choose P = P (u).We see that as long as ξ T ∈ K 0 , our expansion is valid.We also note that a(α, w)(x), b(α, w)(x), R(u−iw) all depend in a C N way on ξ T (via their dependence on u and P ).
Finally we make K 0 small enough that if ξ T ∈ K 0 then u is so close to zero that A(P (u), u) = e ±δ0 A(P (0), 0).
Step 5. Asymptotic analysis of the integrals in (14).Throughout this step ξ/T ∈ K 0 , P , and u are fixed so that ∇P (u) = ξ T and P top (−P r + u, f ) = 0.
The integral R×T d e −iT α b(α, w)(x)dαdw is, up to a constant, the Fourier transform of . Consequently, it has 2d + 10 many derivatives (all of which are elements in H κ ), and uniformly in x, ξ, and T such that ξ T ∈ K 0 .
The estimation is done precisely as in the proof of lemma 2.3 in [BL1].Using the change of coordinates 1 z = − ∞ 0 e zT dT valid for every z ∈ C such that Re z < 0, and the expansion R(u − iw) = 1 2 w t P (u)w + o( w 2 ) we see that , where the big Oh is uniform in x and ξ T (we have used Re R(u − iw) > 0 for u ∈ (∇P ) −1 (K 0 ), w ∈ T d ).Since c depends continuously on ξ T , we get that uniformly in uniformly in x ∈ Σ + and ξ/T ∈ K 0 .
Now that we have obtained the asymptotic behaviour of A i (x, ξ, T ), we show that A i (x, ξ, T ) = A i (x, ξ, T ).Since A(x, ξ, T ) is sandwiched between A i (x, ξ, T ), i = 1, 2, this will give us our final goal: an (approximate) asymptotic expansion for A(x, ξ, T ).
Starting from the definition of A i (x, ξ, T ) in ( 8), we see that it is enough to show that γ i (α) n≥N (L n s,z ψ)(x) are dominated by a (single) absolutely integrable function of (u, w).
We shall show that this holds (uniformly) for x ∈ Σ + , ξ/T ∈ K 0 .Recalling the definition of • κ , the proof of step 2, and the convention P top (−P r + u, f ) = 0, we see that We make c so small that |x + iy| ≥ c(|x| + |y|) for all x, y ∈ R. Using the identity (1 + v 3 )dv < ∞.
Absolute integrability, and with it step 7, is proved.
an isometry, and p • D = p} is isomorphic to Z d .Such covers are called Z d -covers.The geodesic flow and the horocycle flow of M 0 lift to continuous flows g s , h t on T 1 (M ) for which: (a) The h-orbit of x is equal to W ss (x) := {y ∈ T 1 (M ) : d(g s x, g s y) −−−→ s→∞ 0}; and let ω and d ω (•, •) be the measure and metric on W ss (ω) given by ω {h s (ω) : a < s < b} = b − a and d ω h s (ω), h t (ω) = |s − t|.

.
The constant is uniform in N , because |λ s,z | ≤ |λ P,u | = 1 due to the first constraint on P and u, and because Q s,z , N s,z are bounded on compacts (e.g. the support of γ i times T d ).It follows that it is enough to show the absolute integrability of∞ n=0 (L n s,z ψ)(x) on compact subsets of R × T d .Equation (12) shows that it is enough to prove to absolute integrability ofA(s, z)(•) κ i(α − ∇P (u), w ) − R(u − iw)on a compacts.After the change of coordinates α → α + w, ξ T suggested in step 4, this becomes the absolute integrability ofa(α, w)(•) κ iα − R(u − iw) = O 1 iα − R(u − iw)on compacts.(Note that this change of coordinates preserves the property of being absolutely integrable on compacts, because of the assumption that ξ/T ∈ K 0 where K 0 is compact.)Recalling the choice of K 0 at the end of step 3, we see that iα − R(u − iw) vanishes only at the origin.Therefore, it is enough to check the convergence of a −a dα [−b,b] d |1/[iα − R(u − iw)]|dw for some a, b > 0. We choose b > 0 so small that for some c > 0 Re R(u − iw) ≥ c w 2 Im R(u − iw) ≤ c w 3 (w ∈ [−b, b] d ).