A genus 4 origami with minimal hitting time and an intersection property
aa r X i v : . [ m a t h . D S ] F e b A GENUS 4 ORIGAMI WITH MINIMAL HITTING TIME AND ANINTERSECTION PROPERTY
LUCA MARCHESE
Abstract.
In a minimal flow, the hitting time is the exponent of the power law, as r goes to zero, for the time needed by orbits to become r -dense. We show that on the so-called Ornithorynque origami the hitting time of the flow in an irrational slope equals thediophantine type of the slope. We give a general criterion for such equality. In general, forgenus at least two, hitting time is strictly bigger than diophantine type. Introduction An origami , also known as square-tiled surface , is a surface obtained glueing copies of thesquare [0 , along the boundaries. On a given origami, any α ∈ R defines a linear flow inslope α , whose dynamical properties are related to the diophantine properties of α . Thisreflects a more general principle in Teichm¨uller dynamics . [3] gives an introduction to thesubject and a selection of the many relevant references. In this paper we consider a specialgenus 4 origami called
Ornithorynque (see § hitting time in any slope α equals the diophantine type of α . This is the minimalpossible value for the hitting time (Lemma 1.2), and in many cases the equality does nothold, according to [6]. We prove Theorem 1.1 stating a general criterion based on a specificintersection property, namely Theorem 4.1, and showing that the Ornithorynque satisfiesthe intersection property (see § Eierlgende Wollmilchsau .1.1.
Origamis and linear flows.
Fix a finite set Q and a pair ( h, v ) of permutationsof Q generating a transitive subgroup h h, v i of the symmetric group. For any j ∈ Q let Q j := { j } × [0 , the j -th copy of the unit square. Denote by l j , r j , b j , t j the copies of thefour sides l := { } × [0 ,
1] , r := { } × [0 ,
1] , b := [0 , × { } , t := [0 , × { } . For any j ∈ Q paste the right side r j of Q j to the left side l h ( j ) of Q h ( j ) and the top side t j of Q j to the bottom side b v ( j ) of Q v ( j ) . An origami X is a surface arising in this way. It iscompact, connected, orientable and without boundary. We have a covering (1.1) ρ X : X → T over the standard torus T := R / Z , ramified only over [0] ∈ T , where [ x ] denotes the cosetof x ∈ R . The points p , . . . , p m in X where ρ X is ramified are in bijection with the cyclesof the commutator [ v, h ] := v − h − vh . Let k , . . . , k m in N be such that for any 1 ≤ j ≤ m the cycle of [ v, h ] corresponding to p j has length k j + 1. The surface X inherits a metric with Define it on
Q × [0 , as ρ X (cid:0) ( j, x ) (cid:1) := [ x ]. This gives a map on X because glued points have the sameimage. conical angle 2( k j + 1) π at any p j and which is flat outside these points. If g is the genusof X , then k + · · · + k m = 2 g −
2. Details can be found in [3], while § α ∈ R ∪ {∞} and set e α := (sin θ, cos θ ), where θ := arctan α ∈ ( − π/ , π/ e α with slope α . The linear flow φ α : R × X → X on X is the continuousflow determined for any p ∈ X and t ∈ R by(1.2) ρ X (cid:0) φ α ( t, p ) (cid:1) = ρ X ( x ) + te α mod Z . Equation (1.2) determines k j + 1 trajectories starting at any conical point p j , which may ormay not be defined for any t ≥
0, where the trajectory stops at t = t if φ α ( t , p j ) is alsoa conical point. Similarly we have k j + 1 trajectories ending in p j . We call singular leaves such trajectories. The flow φ α is a regular R -action outside singular leaves. If α ∈ R \ Q ,then φ α is uniquely ergodic , that is the Lebesgue measure of X is the only invariant measure.This implies that any positive-infinite orbit is dense. On the other hand, if α ∈ Q then anyinfinite orbit is periodic, moreover periods take finitely many values.1.2. The
Ornithorynque origami.
Consider the set Q := Z / Z × Z / Z × Z / Z and let X O be the origami defined by the pair ( h, v ) of permutations of Q given by h ( i, , i, , i, , i, , := ( i + 1 , , i − , , i, , i, , and v ( i, , i, , i, , i, , := ( i − , , i, , i + 1 , , i, , . Figure 1 represents the origami X O . Half of the 24 pairs of identified sides are representedby dotted lines. The other 12 pairs are named by letters A i , B i , C i , D i with i ∈ Z / Z . Thereare three conical points p , p , p with orders k = k = k = 2, that is a conical angle 6 π at each conical point. Figure 1 shows 3 big squares with size 2 ×
2. The 12 vertices ofthese big squares are identified to p , the 6 middle points of the horizontal sides correspondto p and the 6 middle points of the vertical sides correspond to p . From the relation2 g − k + k + k we get that X O has genus g = 4. A i +1 B i A i − B i − D i C i +1 D i − C i − A i − B i +1 A i B i D i +1 C i − D i C i ( i, , i, ,
0) ( i, , i, , A i B i − A i +1 B i +1 D i − C i D i +1 C i +1 Figure 1.
The Ornithorynque origami X O . he surface X O was discovered by Forni and Matheus in the preprint [2], and then includedin a larger family of surfaces in [4]. After Delecroix and Weiss, the origami X O was named Ornithorynque (french for Platypus), as a rare example of surface with totally degenerate
Lyapunov spectrum . Previously, in [1], Forni discovered the only other known example withsuch property, which is a genus g = 3 surface X E called in german Eierlegende Wollmilchsau .The surface X E was first introduced in [5] and its name was given by Herrlich, M¨oller andSchmith¨usen, referring to its peculiar algebro-geometrical properties, which make X E a sourceof counterexamples in Teichm¨uller theory.1.3. Main statement.
Recall that the diophantine type of α ∈ R is w ( α ) := sup (cid:26) w > (cid:12)(cid:12) α − p/q (cid:12)(cid:12) < q w +1 for infinitely many p/q ∈ Q (cid:27) , where as usual any fraction p/q is written with co-prime p and q . We always have w ( α ) ≥ w ( α ) = 1 for almost any α . Fix an origami X and α ∈ R .For any p ∈ X and r >
0, the time needed by the positive φ α -orbit of p to become r -dense is T ( X, α, p, r ) := sup (cid:26)e p ∈ X : inf (cid:8) t > r : Dist (cid:0) φ α ( t, p ) , e p (cid:1) < r (cid:9)(cid:27) , where Dist( · , · ) is the distance on X , which equals the euclidean distance on small enoughdiscs in X \ { p , . . . , p m } . Minimality implies that T ( X, α, p, r ) is defined for any p outsidesingular leaves. In general, the scaling law of T ( X, α, p, r ) as r → r − H , where the bestexponent H = H ( X, α, p ), called hitting time , is defined by H ( X, α, p ) := lim sup r → + log T ( X, α, p, r ) − log r . Theorem 1.1.
Let X O be the Ornithorynque origami. Then for any α irrational and any p outside of singular leaves we have H ( X O , α, p ) = w ( α ) . Theorem 4.1 below proves the identity H ( X, α, p ) = w ( α ) in a more general setting.The non-trivial inequality is H ( X, α, p ) ≤ w ( α ), which holds for any origami X satisfying aspecific intersection property. Proposition 3.1 and Corollary 3.2 below show that X O satisfiessuch property. The same is true for the Eierlegende Wollmilchsau X E ( § Cycliccovers in [9] are a natural candidate for testing the assumption of Theorem 4.1 and thusproving the identity between diophantine type and hitting time. The easier inequality inTheorem 4.1 and Theorem 1.1 is implicit in [6]. We state it as follows (a proof is in § A).
Lemma 1.2.
Let X be any origami and α be an irrational slope. For any p outside singularleaves we have H ( X, α, p ) ≥ w ( α ) . For any origami X and any α irrational, the function p H ( X, α, p ) is invariant under φ α (Lemma 4.2 in [6]). Thus H ( X, α, · ) is constant almost everywhere. Theorem 1.1 wasproved on the standard torus X = T in [8]. Proposition 2.5 in [6] extends the same result tothe Eierlegende Wollmilchsau X E . On the other hand, for any origami X with genus g = 2 nd an unique conical point of order k = 2, Theorem 2.2 in [6] proves that for any λ ∈ [1 , α with H ( X, α, p ) = w ( α ) λ for almost any p ∈ X. For X with the same topological data, we have H ( X, α, p ) ≤ w ( α ) for any α and any p outside singular leaves (Theorem 2.1 in [6]). Proposition 4.6 in [6] proves that for any origami X and α irrational we havelim inf r → log (cid:0) inf (cid:8) t > r : Dist (cid:0) φ α ( t, p ) , e p (cid:1) < r (cid:9)(cid:1) − log r = 1 for almost any p, e p ∈ X. Combining the last result and Theorem 1.1, and recalling that generically w ( α ) = 1, we getthat for almost any α and almost any p, e p in X O there exists the limitlim r → log (cid:0) inf (cid:8) t > r : Dist (cid:0) φ α ( t, p ) , e p (cid:1) < r (cid:9)(cid:1) − log r = 1 . The limit above was established for generic interval exchange transformations in [7]. Mostresults quoted from [6] are proved in the general setting of translation surfaces . Contents of this paper. In § , Z ) over the set of origamis,which fixes X O . In § X O satisfies the intersection property in Theorem 4.1. In § , Z ) and use them as a renormalization tool to prove Theorem 4.1.The proof of Theorem 1.1 is resumed in § § A we prove Lemma 1.2.
Acknowledgements.
The author is grateful to D. H. Kim, S. Marmi and C. Matheus.2.
Background
Let SL(2 , Z ) be the group of 2 × A with coefficients in Z and determinantdet( A ) = 1. In particular we consider the following elements(2.1) T := (cid:18) (cid:19) ; V := (cid:18) (cid:19) ; R := (cid:18) −
11 0 (cid:19) . Any A ∈ SL(2 , Z ) acts projectively on points α ∈ R ∪ {∞} by A · α := aα + bcα + d where A = (cid:18) a bc d (cid:19) . Action of
SL(2 , Z ) . Fix an origami X , defined by permutations ( h, v ) of a finite set Q .Fix A ∈ SL(2 , Z ) and consider the parallelogram P := A ([0 , ). For j ∈ Q the j -th copy P j := { j } × P has sides e l j := { j } × A ( l ) ; e r j := { j } × A ( r ) ; e b j := { j } × A ( b ) ; e t j := { j } × A ( t ) , where the sides l, r, b, t of [0 , are defined in § j ∈ Q , paste the side e r j of P j to the side e l h ( j ) of P h ( j ) and the side e t j of P j to the side e b v ( j ) of P v ( j ) . Let A · X bethe corresponding surface, which is compact, connected, orientable and without boundary.Moreover A · X is an origami, corresponding to a pair ( h ( A ) , v ( A ) ) of permutations of Q . It ispossible to see from the commutator [ h ( A ) , v ( A ) ] that A · X has the same number of conical oints as X , with same orders k , . . . , k m , and thus also the same genus (see [3] for details).For the matrix T in Equation (2.1) we have h ( T ) = h and v ( T ) = v ◦ h − , while for the matrix V in Equation (2.1) we have h ( V ) = h ◦ v − and v ( V ) = v. Since
T, V generate SL(2 , Z ), we can compute ( h ( A ) , v ( A ) ) from ( h, v ) for any A ∈ SL(2 , Z ). A i − B i A i − B i − D i A i − B i +1 A i B i D i C i A i B i +2 A i +1 B i +1 C i B i A i − B i − A i − e D i B i +1 A i − B i − A i e D i e C i B i +2 A i − B i A i +1 e C i Figure 2.
Cut the dotted triangles in the above line and paste them alongthe sides C i , D i for i = 0 , ,
2, as in the line below. It follows T · X O = X O . Proposition 2.1.
We have A · X O = X O for any A ∈ SL(2 , Z ) .Proof. Recall that
T, R in Equation (2.1) generate SL(2 , Z ). Figure 2 shows that T · X O = X O ,while it is clear from Figure 1 that R · X O = X O . See [3] for more details. (cid:3) Affine homeomorphisms.
Fix an origami X and A ∈ SL(2 , Z ). For j ∈ Q , the affinemaps ( j, x ) (cid:0) j, A ( x ) (cid:1) of Q j onto P j agree on glued sides, where we use the same notationof § ψ A : X → A · X sending { p , . . . , p m } bijectively onto the set of conical points of A · X . Local inverses ϕ : U → X \{ p , . . . , p m } of the covering ρ X in Equation (1.1), defined over simply connectedopen sets U ⊂ T , give smooth charts for X \ { p , . . . , p m } . Change of charts are indeedtranslations . Similar translation charts exist on A · X (minus its conical points). In thesetranslation charts ψ A is a diffeomorphism, which is locally affine. The linear part Dψ A is the linear part of ψ A computed in translation charts. We have of course Dψ A = A .The automorphisms group Aut( X ) is the set of orientation preserving homeomorphisms ψ : X → X which preserve { p , . . . , p m } and are affine in translation charts, with Dφ = Id.In general Aut( X ) is non trivial, thus for a given A ∈ SL(2 , Z ) there exist more than one ψ A Thus they are holomorphic, and one can extend them to an holomorphic atlas over the entire X , see [3]. s in Equation (2.2). We have Aut( X O ) ≃ Z / Z , which acts by translation on the big 2 × § The intersection property of X O Let X be any origami and p , . . . , p m be its conical points. Let ρ X : X → T be thecovering in Equation (1.1). A straight segment in X , or simply segment , is a smooth path S : ( a, b ) → X \ { p , . . . , p m } such that there exists a vector v ∈ R with ddt ρ X (cid:0) S ( t ) (cid:1) = v for any t ∈ ( a, b ) . If v = ( x, y ) ∈ R , then the slope Slope( S ) ∈ R ∪ {±∞} of such S isSlope( S ) := xy . The length | S | of such segment is | S | := | b − a | · k v k , where k · k is the euclidean norm in R .Observe that segments do not contain conical points in their interior. Endpoints of straightsegments can be conical points. A saddle connection of the surface X is a straight segmentconnecting conical points. Proposition 3.1 is the main result in this section. Its proof isresumed in § § § Proposition 3.1.
Let X O be the Ornithorynque origami. Fix segments H, V in X O with < Slope( V ) < and Slope( H ) < − . If both segments have length | H | , | V | ≥ √ then H ∩ V = ∅ . Let S : R → R acting by S ( x, y ) := ( − x, y ). The same construction as in § S · X O , obtained glueing copies { j } × S (cid:0) [0 , (cid:1) of the reflected square S (cid:0) [0 , (cid:1) ,where j ∈ Q and where identifications in S · X O are induced by identifications in X O . It iseasy to see that indeed we have S · X O = X O . As in § f S : X O → X O with linear part Df S = S . If H, V are segments in X O with − < Slope( H ) < V ) >
1, then 0 < Slope (cid:0) f S ( H ) (cid:1) < (cid:0) f S ( V ) (cid:1) < − Corollary 3.2.
Let X O be the Ornithorynque origami. Fix segments H, V in X O with − < Slope( H ) < and Slope( V ) > . If both segments have length | H | , | V | ≥ √ then H ∩ V = ∅ . Preliminary Lemmas.
The proof of Lemma 3.3 below is left to the reader.
Lemma 3.3.
Let Q := [0 , and Q := [1 , × [0 , . Let ℓ H and ℓ V be two lines with < Slope( ℓ V ) < and Slope( ℓ H ) < − and set P := ℓ H ∩ ℓ V . If both ℓ V and ℓ H intersect { } × [0 , , then either P ∈ Q or P ∈ Q . Lemma 3.4.
Let X be any origami labelled by a finite set Q . Fix a square Q j with j ∈ Q and let H and V be segments in X with < Slope( V ) < and Slope( H ) < − , such thatboth H, V have endpoints in S l = j ∂Q l . If both H ∩ Q k = ∅ and V ∩ Q k = ∅ then H ∩ V = ∅ .Proof. Let ℓ H , ℓ V be lines as in Lemma 3.3 and R be the matrix in Equation (2.1). Thelines R ( ℓ H ) , R ( ℓ V ) satisfy the same assumption of Lemma 3.3, with inverted roles. ThusLemma 3.3 holds replacing Q by any of the four unitary squares sharing a side with Q . Such xtended version of Lemma 3.3 implies the statement, observing that if H, V are segmentsas in the statement, then there must be a side of Q j intersecting both H and V . (cid:3) Cutting sequences.
Recall Figure 1 and consider the twelve letters alphabet A := { A i , B i , C i , D i : i = 0 , , } . Geometrically, any element γ ∈ A is a saddle connection of X O . Symbolically, elements γ ∈ A are letters which compose words ( γ , . . . , γ n ). Such words arise as cutting sequences of straight segments S in X . Fix a straight segment S ⊂ X , and abusing the notationdenote S : (0 , → X its parametrization with constant speed. Define recursively integers k = 1 , . . . , n and instants 0 ≤ t < · · · < t n ≤ t := min { t ≥ ∃ γ ∈ A : S ( t ) ∈ γ } t k := min { t > t k − : ∃ γ ∈ A : S ( t ) ∈ γ } for k ≥ , where t n = max { ≤ t ≤ ∃ γ ∈ A : S ( t ) ∈ γ } . Then define the cutting sequence[ S ] := ( γ , . . . , γ n )of S as the word in the letters of A such that S ( t k ) ∈ γ k for k = 1 , . . . , n. In the notation of § i = 0 , , tile T i ⊂ X by T i := Q ( i, , ∪ Q ( i, , ∪ Q ( i, , ∪ Q ( i, , . Lemma 3.5.
Let V be a segment with < Slope( V ) < and assume that its cutting sequence [ V ] = ( γ , . . . , γ n ) contains n ≥ letters. Then V ∩ T i = ∅ for i = 0 , , .Proof. Assume without loss of generality that V does not cross the tile T . Then we have γ k = A , B , C , D , C , D , A , B for k = 1 , . . . , n − . Observing that γ k = C ⇒ γ k +1 = A we get γ k = C for k = 1 , . . . , n − . Since γ k = B ⇒ γ k +1 ∈ { D , C } it follows γ k = B for k = 1 , . . . , n − . Moreover we have γ k = A ⇒ γ k +1 ∈ { A , B , C } , therefore γ k = A for k = 1 , . . . , n − . Finally γ k = D ⇒ γ k +1 ∈ { A , B } , which implies γ k = D for k = 1 , . . . , n − . Since n ≥
6, the conditions above imply that there is no value left for γ , which is absurd. (cid:3) Lemma 3.6.
Let H be a segment with Slope( H ) < − and assume that its cutting sequence [ H ] = ( γ , . . . , γ n ) contains n ≥ letters. Then H ∩ T i = ∅ for i = 0 , , Proof.
The Lemma follows by an argument similar to Lemma 3.5. Alternatively consider R in Equation (2.1), observe that V := R ( H ) satisfies the assumption of Lemma 3.5, and recall R · X O = X O . (cid:3) emma 3.7. Fix segments
H, V with
Slope( H ) < − and < Slope( V ) < and cuttingsequences [ H ] = ( γ , . . . , γ n ) and [ V ] = ( ν , . . . , ν m ) with n ≥ and m ≥ . Fix i = 0 , , and assume that there exists ≤ k ≤ n − with γ k ∈ { C i +2 , A i } and γ k +1 ∈ { B i +1 , D i } . Then H ∩ V = ∅ .Proof. Let i ∈ { , , } be as in the statement. Let e V be the minimal subsegment of V with cutting sequence [ e V ] = ( ν , . . . , ν m − ). Lemma 3.5 implies e V ∩ T i = ∅ . Let e H bethe minimal subsegment of H with cutting sequence [ e H ] = ( γ , . . . , γ n − ). The assumptionon [ H ] implies that e H intersects at least 3 of the 4 squares Q ( i, , , Q ( i, , , Q ( i, , , Q ( i, , composing the tile T i , where we recall the squares in an origami are closed and they overlapalong the boundaries. Moreover the square missed by e H can only be either Q ( i, , or Q ( i, , ,and it is crucial to observe that none of these two squares can contain e V ∩ T i . It followsthat e V ∩ Q = ∅ and e H ∩ Q = ∅ , where Q is one of the four squares composing the tile T i .Lemma 3.4 gives H ∩ V = ∅ . (cid:3) Lemma 3.8.
Fix segments
H, V with
Slope( H ) < − and < Slope( V ) < and cuttingsequences [ H ] = ( γ , . . . , γ n ) and [ V ] = ( ν , . . . , ν m ) with n ≥ and m ≥ . Fix i = 0 , , and assume that (3.1) ( γ k , γ k +1 , γ k +2 ) = ( C i +2 , C i , C i +1 ) or ( γ k , γ k +1 , γ k +2 ) = ( D i , D i +2 , D i +1 ) . Then H ∩ V = ∅ .Proof. Let e V be the minimal subsegment of V with cutting sequence [ e V ] = ( ν , . . . , ν m − ).Assume first ( γ k , γ k +1 , γ k +2 ) = ( C i +2 , C i , C i +1 ). Let e H a , e H b be respectively the minimalsubsegments of H with [ e H a ] = ( C i +2 , C i ) and [ e H b ] = ( C i , C i +1 ), so that in particular e H a ⊂ T i and e H b ⊂ T i +1 . We have e V ∩ T i +1 = ∅ by Lemma 3.5. If e V ∩ e H b = ∅ then there exists k with 1 ≤ k ≤ m − ν k , ν k +1 ) = ( B i +2 , D i +1 ). Therefore the subsegment of e V encodedby ( ν k +1 , ν k +2 ) satisfies ν k +1 = D i +1 and ν k +2 ∈ { A i , B i } , and this last property impliesthat such subsegment intersects e H a . Thus H ∩ V = ∅ . Now assume ( γ k , γ k +1 , γ k +2 ) =( D i , D i +2 , D i +1 ). Let e H c , e H d be respectively the minimal subsegments of H with [ e H c ] =( D i , D i +2 ) and [ e H d ] = ( D i +2 , D i +1 ), so that in particular e H c ⊂ T i +2 and e H d ⊂ T i +1 . Wehave e V ∩ T i +1 = ∅ by Lemma 3.5. If e V ∩ e H d = ∅ then there exists k with 1 ≤ k ≤ m − ν k , ν k +1 ) = ( C i , A i +1 ). Therefore the subsegment of e V encoded by ( ν k +1 , ν k +2 ) satisfies ν k +1 = A i +1 and ν k +2 ∈ { A i +2 , B i +2 } , and this last property implies that such subsegmentintersects e H c . Again it follows H ∩ V = ∅ . The Lemma is proved. (cid:3) Proof of Proposition 3.1.
Let [ H ] = ( γ , . . . , γ n ) and [ V ] = ( ν , . . . , ν m ) be thecutting sequences of H, V respectively. Since | H | , | V | ≥ √ n ≥ m ≥
12. Assume that the cutting sequence [ H ] of H does not satisfy Equation (3.1) forany i = 0 , ,
2. Then we must have − < Slope( H ) < −
1. Since n ≥
12, then H satisfiesthe assumption of Lemma 3.7. Proposition 3.1 follows. . The general criterion
Theorem 4.1.
Let X be an origami and assume that there exists a constant K > such thatfor any origami Y ∈ SL(2 , Z ) · X and any pair of segments H, V ⊂ Y we have H ∩ V = ∅ whenever they have length | H | , | V | ≥ K and satisfy • either Slope( H ) < − and < Slope( V ) < • or − < Slope( H ) < and Slope( V ) > .Then H ( X, α, p ) = w ( α ) for any α irrational and any p outside ( X, α ) -singular leaves. Proof of Main Theorem 1.1.
Recall that SL(2 , Z ) · X O = X O by Proposition 2.1.Theorem 1.1 follows combining Theorem 4.1 with Proposition 3.1 and Corollary 3.24.2. Continued fractions.
Let
T, V be as in Equation (2.1). Consider positive integers a , . . . , a n and define g ( a , . . . , a n ) ∈ SL(2 , Z ) by(4.1) g ( a , . . . , a n ) := (cid:26) V a ◦ · · · ◦ V a n − ◦ T a n for even n ; V a ◦ · · · ◦ T a n − ◦ V a n for odd n. Let [ α ] := max { k ∈ Z , k ≤ α } be the integer part and { α } := α − [ α ] be the fractionalpart of α ∈ R , where 0 ≤ { α } <
1. The
Gauss map G : [0 , → [0 ,
1) is defined by G ( α ) := { α − } for α ∈ [0 , . Any irrational α ∈ (0 ,
1) admits an unique continued fraction expansion (4.2) α = [ a , a , . . . ] := 1 a + 1 a + . . ., where we set α := α and α n := G ( α n − ) for n ≥
1, so that the n -th partial quotient of α isgiven by a n := (cid:20) α n − (cid:21) that is 1 α n − = a n + α n . The n -th convergent p n /q n := [ a , . . . , a n ] of α is obtained truncating Equation (4.2) to its n -th partial quotient a n . We get(4.3) g ( a , . . . , a n − ) = (cid:18) p n − p n − q n − q n − (cid:19) and g ( a , . . . , a n ) = (cid:18) p n − p n q n − q n (cid:19) from the recursive relations q n = a n q n − + q n − and p n = a n p n − + p n − . Therefore(4.4) p n /q n = (cid:26) g ( a , . . . , a n ) · ng ( a , . . . , a n ) · ∞ for odd n. We have α − n = a n +1 + α n +1 ⇔ α n = V a n +1 · α − n +1 ⇔ α − n = T a n +1 · α n +1 . Hence(4.5) α = g ( a , . . . , a k ) · α k = g ( a , . . . , a k , a k +1 ) · α k +1 for any k ∈ N . .3. Proof of Theorem 4.1.
Let X be an origami as in Theorem 4.1. Write real numbersas α = a + e α , where a := [ α ] and e α := { α } are the integer and fractional part respectively.Set Y := T − a · X and let ψ : X → Y be an affine homeomorphism with Dψ = T − a as in § κ > φ e α (cid:0) t, ψ ( p ) (cid:1) = ψ (cid:0) φ α (cid:0) κt, p ) (cid:1) for any t ∈ R and p ∈ X. Thus H (cid:0) Y, e α, ψ ( p ) (cid:1) = H ( X, α, p ) and obviously any Y ∈ SL(2 , Z ) · X satisfies the sameassumption as X . On the other hand w ( e α ) = w ( α ). Hence it is no loss of generality toconsider α ∈ (0 , § Proposition 4.2.
Let X and K > be an origami and a constant as in Theorem 4.1. Fixa slope α = [ a , a , . . . ] ∈ (0 , . For any p ∈ X outside singular leaves and n ∈ N we have T ( X O , α, p, r n ) ≤ K · q n where r n := 2( K + 1) q n Set w := w ( α ), so that q n ≤ K · q wn − for some K and all n . Fix p ∈ X outside singularleaves. For any r > n with r n − ≤ r < r n . Proposition 4.2 giveslog T ( X O , α, p, r ) | log r | ≤ log T ( X O , α, p, r n ) | log r n − | ≤ log 4 K + log q n log q n − − log 2( K + 1) ≤ log 4 K + log K + w · log q n − log q n − − log 2( K + 1) → w for n → + ∞ . Hence H ( X, α, p ) ≤ w . Lemma 1.2 gives the other inequality. Theorem 1.1 is proved. (cid:3) Cylinder decompositions.
Let X be any origami. A closed geodesic is a straightsegment σ : [ a, b ] → X with σ ( a ) = σ ( b ), where such point is not conical. If ρ X is thecovering in Equation (1.1), then ρ X ◦ σ is a closed geodesic in T and must have rationalslope. Thus Slope( σ ) ∈ Q ∪ {∞} . Given any p/q rational, a cylinder in slope p/q is amaximal open and connected subset C ⊂ X foliated by closed geodesics σ with same lengthand Slope( σ ) = p/q . Set Slope( C ) := p/q and | C | := | σ | , where σ is any closed geodesic asabove. The boundary ∂C is union of saddle connections with slope p/q .Referring to Figure 1, the vertical path σ : [0 , → X O such that σ (2 i ) is the middlepoint of A i for i = 0 , , X O . We have | σ | = 6 andSlope( σ ) = 0. The two vertical cylinders of X O are C (+)0 := [ i =0 , , Q ( i, , ∪ Q ( i, , and C ( − )0 := [ i =0 , , Q ( i, , ∪ Q ( i, , We have a decomposition X O = C (+)0 ∪ C ( − )0 , where the boundaries of the two cylinders aremade by vertical saddle connections.Referring to [3], recall that any origami X admits a cylinder decomposition in the verticalslope p/q = 0, with a number l ≥ C (1)0 , . . . , C ( l )0 . For i = 1 , . . . , l any cylinderhas Slope( C ( i )0 ) = 0, integer length L i := | C ( i )0 | and integer width W i , where W i is definedas the length of an horizontal segment in C ( i )0 with endpoints in ∂C ( i )0 . Fix p/q ∈ Q ∪ {∞} ,take A ∈ SL(2 , Z ) with A · ∞ = p/q and an origami Y with A · Y = X . Let ψ : Y → X e an affine homeomorphism with Dψ = A , as in § Y = C (1)0 ∪ · · · ∪ C ( l )0 induces the cylinder decomposition of X in slope p/q , that is(4.6) X = C (1) p/q ∪ · · · ∪ C ( l ) p/q where C ( i ) p/q := ψ ( C ( i )0 ) for i = 1 , . . . , l. Lemma 4.3.
Consider an origami X , a slope p/q ∈ Q ∪ {∞} and the decomposition inEquation (4.6) . Let H be a segment in X crossing the cylinders C j p/q , . . . , C j n p/q . We have | H | ≤ W j + · · · + W j n p q + p cos (cid:12)(cid:12) arctan (cid:0) Slope( H ) (cid:1) − arctan( − q/p ) (cid:12)(cid:12) . Proof.
Any cylinder in Equation (4.6) has length | C ( j ) p/q | = L j p q + p and euclidean area L j W j . Let e H j ⊂ C ( j ) p/q be a segment with endpoints in ∂C ( j ) p/q . If Slope( e H j ) = − q/p , whichis orthogonal to p/q , then | e H j | = W j ( q + p ) − / . If e H j has a different slope, then itslength increases by the inverse of the cosinus of the angle between Slope( e H j ) and − q/p . Thesegment H is union of n segments e H j , . . . , e H j n as above. The Lemma follows. (cid:3) Proof of Proposition 4.2.
Let X be an origami as in Theorem 4.1 and α = [ a , a , . . . ]irrational. Fix any two points p, e p in X , with p outside ( X, α )-singular leaves.Consider first the case n = 2 k . Set A := g ( a , . . . , a k ) and let X k ∈ SL(2 , Z ) · X be thesurface with A · X k = X . Let ψ : X k → X be an affine homeomorphism with Dψ = A , asin § α k := A − · α and p k /q k := A · i = 1 , . . . , l let C ( i )0 be the cylinder in the decomposition of X k in verticalslope p/q = 0. Let W i be the width of C ( i )0 . The cylinder decomposition of X in slope p k /q k is X = C (1) p/q ∪ · · · ∪ C (1) p/q , where C ( i ) p/q := ψ ( C ( i )0 ). Consider β irrational such that (cid:26) A − · β < − (cid:12)(cid:12) arctan( β ) − arctan( − q k /p k ) (cid:12)(cid:12) > / . The slope e β = − q k /p k is orthogonal to p k /q k and satisfies the first condition above, indeedrecalling Equation (4.3) we have A − · − q k p k = (cid:18) q k − p k − q k − p k − (cid:19) · − q k p k = − ( q k + p k ) q k q k − + p k p k − < − a k < − . The same condition is satisfied by some irrational slope β close to e β , by continuity of theprojective action of A . The second condition on β is easily satisfied.Let e H ⊂ X be a straight segment passing through e p with Slope( e H ) = β . Consider thesegment H := ψ − ( e H ) ⊂ X k . We have Slope( H ) = A − · β , which is irrational since β isirrational. Since H has irrational slope, it is not a subsegment of a saddle connection of X k . Therefore H can be extended along the slope A − · β and we can assume that it haslength | H | = K . If H crosses the vertical cylinders C ( j )0 , . . . , C ( j n )0 of X k , then we have W j + · · · + W j n ≤ K + 1. The second condition on β and Lemma 4.3 imply | e H | ≤ K + 1) p q k + p k ≤ K + 1) q k = r k . Since p does not belong to any ( X, α )-singular leaf, then ψ − ( p ) does not belong to any( X k , α k )-singular leaf and it has infinite positive orbit. Let V be a segment in X k which has n endpoint in ψ − ( p ), with Slope( V ) = α k , length | V | = K . By assumption V intersects H . In other words, we have t > φ α k (cid:0) t, ψ − ( p ) (cid:1) ∈ H and 0 ≤ t ≤ | V | , Consider
T > ψ ◦ φ α k ( t, · ) = φ α ( T, · ) ◦ ψ , so that we have φ α ( T, p ) = φ α (cid:0) T, ψ (cid:0) ψ − ( p ) (cid:1)(cid:1) = ψ (cid:0) φ α k (cid:0) t, ψ − ( p ) (cid:1)(cid:1) ∈ ψ ( H ) = e H. Both e p and φ α ( T, p ) belong to e H . Hence (cid:12)(cid:12) φ α ( T, p ) − e p (cid:12)(cid:12) ≤ | e H | ≤ r k . We have T ≤ | ψ ( V ) | ≤ k A k · | V | ≤ ( p k + q k + p k − + q k − ) · | V | ≤ K · q k . Since e p is arbitrary, we get T ( X, α, p, r k ) ≤ K · q k . Now consider the case n = 2 k −
1. Set A := g ( a , . . . , a k − , a k − ) and let X k ∈ SL(2 , Z ) · X be the surface with A · X k = X . Let ψ : X k → X be an affine homeomorphism with Dψ = A .Let α k − be the slope related to α by Equation (4.5), that is α = A · α − k − . We have A · ∞ = p k − /q k − by Equation (4.4). Moreover − < A − · ( − q k − /p k − ) <
0, indeedEquation (4.3) gives A − · − q k − p k − = (cid:18) q k − − p k − − q k − p k − (cid:19) · − q k − p k − = − q k − q k − − p k − p k − q k − + p k − . Therefore we can chose an irrational slope β such that (cid:26) − < A − · β < (cid:12)(cid:12) arctan( β ) − arctan( − q k − /p k − ) (cid:12)(cid:12) > / . Let e H ⊂ X be a segment passing through e p with Slope( e H ) = β such that H := ψ − ( e H ) is asegment in X k with length | H | = K . Let V ⊂ X k be a segment having an endpoint in ψ − ( p ),with Slope( V ) = 1 /α k − and length | V | = K . By assumption we have H ∩ V = ∅ . Theremaining part of the argument is as in case n = 2 k and is left to the reader. Proposition 4.2is proved. (cid:3) Appendix A. Proof of Lemma 1.2
Let
X, α, p be as in the statement. For 0 ≤ w < r > r -neighbourhood of the orbit segment { φ α ( t, p ) : 0 ≤ t ≤ r − w } has area bounded by 2 · r − w .Thus H ( X, α, p ) ≥
1. Lemma 1.2 is proved for those slopes with w ( α ) = 1. Now assume w ( α ) > w with 1 < w < w ( α ). We can assume 0 < α <
1, as in § α = [ a , a , . . . ]. There exist infinitely many n with(A.1) a n +1 ≥ q w − n . It is not a loss of generality to assume that all n as above are even, that is n = 2 k (otherwiserepeat the proof replacing the vertical slope p/q = 0 by the horizontal p/q = ∞ ). Modulosubsequences, assume that there exists X in the orbit SL(2 , Z ) · X such that g ( a , . . . , a k ) · X = X for any k. Recall Equation (4.6) and let X = C (1)0 ∪ · · · ∪ C ( l )0 be the cylinder decomposition of X invertical slope p/q = 0, where any C ( i )0 has width W i and length L i . Let e p ∈ X be a point in he boundary of some vertical cylinder and not on any ( X , α k )-singular leaf. According toEquation (6.9) in [6], if α k < min ≤ i ≤ l L − i then there exists i with(A.2) φ α k ( t, e p ) ∈ C ( i )0 for 0 < t < W i · p α k α k . Since α k = ( a k +1 + α k ) − <<
1, then Equation (A.2) holds. Equation (A.1) gives(A.3) W i · p α k α k ≥ W i α k ≥ a k +1 W i ≥ a k +1 ≥ q w − k . Set r := 1 /
4. Equation (A.2) and Equation (A.3) imply that for any e p ∈ X there exists acylinder C ( i )0 and vertical closed geodesic σ ⊂ C ( i )0 such that(A.4) φ α k ( t, e p ) N ( σ, r ) for 0 ≤ t ≤ q w − k / , where N ( σ, r ) is the r -neighbourhood of σ . Set A := g ( a , . . . , a k ) and let ψ : X → X be an affine homomorphism with Dψ = A . Recall that p k /q k = A · α = A · α k .Moreover ψ ◦ φ α k ( t, · ) = φ α ( κt, · ) ◦ ψ , where the stretching factor of A on vectors with slope α k satisfies κ > q k / √ p ∈ X there exists a cylinder C ⊂ X with Slope( C ) = p k /q k and a closed geodesic e σ ⊂ C with φ α ( t, p ) N (cid:0)e σ, r · ( q k + p k ) − / (cid:1) for 0 ≤ t ≤ ( q w − k / · ( q k / √ , where the size of the neighbourhood of e σ is derived from Lemma 4.3. Since α < p k < q k , setting r k := ( q k √ − we obtain T ( X, α, p, r k ) ≥ q w k / √ H ( X, α, p ) ≥ lim sup k →∞ log T ( X, α, p, r k ) | log r k | ≥ lim sup k →∞ w log q k − log √ q k + log √
32 = w. Therefore H ( X O , α, p ) ≥ w ( α ) since w < w ( α ) is arbitrary. Lemma 1.2 is proved. (cid:3) References [1] G. Forni:
On the Lyapunov exponents of the Kontsevich-Zorich cocycle.
Handbook of DynamicalSystems (eds. B. Hasselblatt and A. Katok), Vol. 1B, Elsevier (2006), 549-580.[2] G. Forni, C. Matheus:
An example of a Teichm¨uller disk in genus 4 with degenerate Kontsevich-Zorichspectrum . Preprint, arXiv:0810.0023.[3] G. Forni, C. Matheus:
Introduction to Teichm¨uller theory and its applications to dynamics of intervalexchange transformations, flows on surfaces and billiards , Journal of Modern Dynamics, Vol. 8 (2014)271-436.[4] G. Forni, C. Matheus, A. Zorich:
Square-tiled cyclic covers . Journal of Modern Dynamics, Vol. 5(2011), 285-318.[5] F. Herrlich and G. Schmith¨usen:
An extraordinary origami curve , Math. Nachr. 281 (2008), 219-237.[6] D. H. Kim, L. Marchese, S. Marmi:
Long hitting time for translation flows and L-shaped billiards .Journal of Modern Dynamics, Vol. 14, 2019, 291-353.[7] D. H. Kim, S. Marmi:
The recurrence time for interval exchange maps . Nonlinearity, 21 (2008),2201-2210.[8] D.H. Kim, B.K. Seo:
The waiting time for irrational rotations . Nonlinearity, 16 (2003), 1861-1868.[9] C. Matheus, J.-C. Yoccoz:
The action on the affine diffeomorphisms on the relative homology groupof certain exceptionnally symmetric origamis . Journal of Modern Dynamics, Vol. 4, 2010, 453-486.
Dipartimento di Matematica, Universit`a di Bologna, Piazza di Porta San Donato 5, 40126,Bologna, Italia
Email address : [email protected]@unibo.it