Backward Rauzy-Veech algorithm and horizontal saddle connections

We study the combinatorial and dynamical properties of translations surfaces with horizontal saddle connections from the point of view of backward Rauzy-Veech induction. Namely, we prove that although the horizontal saddle connections do not rule out existence of the infinite orbit under backward Rauzy-Veech algorithm, they disallow the ∞completeness of such orbit. Furthermore, we prove that if an orbit under backward Rauzy-Veech algorithm is infinite, then the minimality of the horizontal translation flow is equivalent to the eventual appearance of all horizontal saddle connections as sides of the polygonal represenation of a surface. The main goal of this note is to study the relations between horizontal saddle connections and the combinatorics of the inverse RauzyVeech algorithm for translation surfaces as well as dynamics of the horizontal translation flows. In [2] (Proposition 4.3) Marmi, Ulcigrai and Yoccoz prove that if a translation surface does not have horizontal saddle connections, then its backward Rauzy-Veech induction orbit is indefinitely well-defined and ∞-complete, that is every symbol is a backward winner infinitely many times. In the same article the authors pose a question, whether these two conditions are equivalent. We answer affirmatively to this question in Theorem 11. The proof utilizes only combinatorics and geometry of translation surfaces. However, before proving Theorem 11, we prove Proposition 7 which states that, typically, possessing horizontal saddle connections does not rule out that the backward orbit with respect to the inverse RauzyVeech algorithm is well defined. Moreover, in Theorem 12 we prove that appearance of horizontal connections as sides of polygonal representations of translations surfaces is closely tied to the minimality of the horizontal translation flow. More precisely, we show that the horizontal translation flow is minimal if and only if all (if any) horizontal saddle connections appear as sides of a polygonal representation of a surface after applying a finite number of backward Rauzy-Veech induction steps. MSC CLASSIFICATION: 37E05, 37E35 1 ar X iv :2 10 9. 13 69 1v 2 [ m at h. D S] 2 3 Fe b 20 22 P. BERK HORIZONTAL SADDLE CONNECTIONS Acknowledgments: The author would like to thank Corinna Ulcigrai for pointing out the problem and her continuous support and Frank Trujillo for many useful remarks. The research that lead to this result was supported by Swiss National Science Foundation Grant 200021 188617/1 and Narodowe Centrum Nauki Grant OPUS 142017/27/B/ST1/00078. 1. Interval exchange transformations and translations surfaces We recall first basic notions and properties related to IETs and translation surfaces. Let A be an alphabet of #A ≥ 2 elements. For more information and basic properties, including the ergodic properties of interval exchange transformations, translation surfaces and Rauzy-Veech algorithm we refer the reader e.g. to [4] and [5]. Let SA 0 :={π = (π0, π1) : A → {1, . . . ,#A} × {1, . . . ,#A}; π1 ◦ π−1 0 {1, . . . , k} = {1, . . . , k} ⇒ k = #A} be the set of irreducible permutations, where π0 and π1 are bijections. Let us also denote by R>0 the set of all d-dimensional positive real vectors and for every λ ∈ R>0 let |λ| := ∑ α∈A λα. An interval exchange transformation on [0, |λ|) (IET) T = (π, λ) ∈ SA 0 × R>0 is a bijective piecewise translation, where the intervals Iα :=  ∑ β∈A;π0(β)<π0(α) λβ, ∑ β∈A;π0(β)≤π0(α) λβ  for α ∈ A are rearranged inside [0, |λ|) with respect to the permutation π. More precisely, for every α ∈ A, we have T (x) = x+ δα if x ∈ Iα, where δα = ∑ β∈A; π1(β)<π1(α) λα − ∑ β∈A; π0(β)<π0(α) λα. Note that T preserves Lebesgue measure. We denote by Ω = [ωαβ]α,β the associated translation matrix, with coefficients given by ωαβ :=  +1 if π0(α) < π0(β) and π1(α) > π1(β); −1 if π0(α) > π0(β) and π1(α) < π1(β); 0 otherwise. Then, if δ := [δα]α∈A, we get δ = Ωπ · λ. 2 P. BERK HORIZONTAL SADDLE CONNECTIONS On the space SA 0 × R>0 we consider an operator R called RauzyVeech induction, defined as R(π, λ) = (π, λ), where (π, λ) is the first return map of (π, λ) to the interval [0, |λ| −min{λπ−1 0 (d), λπ−1 1 (d)}). If λπ−1 0 (d) > λπ −1 1 (d) we say that R is of ”top” type and we say that it is of ”bottom” type if λπ−1 0 (d) < λπ −1 1 (d) . We denote the symbol corresponding to the longer interval as w (the winner) and to the shorter one as l (the loser). The map R(π, λ) is properly defined as an interval exchange transformation of d intervals if and only if λπ−1 0 (d) 6= λπ−1 1 (d). Keane [1] gave an equivalent condition on (π, λ), for the iterations of Rauzy-Veech induction to be defined indefinitely. More precisely, we say that IET T satisfies Keane’s condition if for every two discontinuities a and b of T equality T (a) = b for some n ∈ N implies n = 1, a = T−1(0) and b = 0. In particular, if the vector λ is rationally independent, that is for every choice of cα ∈ Z, α ∈ A we have ∑ α∈A cαλα = 0 ⇒ cα = 0 for every α ∈ A, then (π, λ) satisfies Keane’s condition. When it is well defined, we denote R(π, λ) = (π, λ) for every n ∈ N. We say that the orbit of (π, λ) via Rauzy-Veech induction is ∞-complete if every symbol in A appears infinitely many times in the sequence of winners {w}. Note that λ = A(π, λ)λ, where a matrix A(π, λ) is defined in the following way Aαβ =  1 if α = β; −1 if α = w and β = l; 0 otherwise. Inductively, for every n ∈ N we define A(π, λ) = A1(πn−1, λn−1)An−1(π, λ). Then λ = A(π, λ)λ. We will refer to A(π, λ) as Rauzy-Veech matrices. Note that for every n ∈ N, the matrix (A(π, λ))−1 is non-negative. For every π ∈ SA 0 let ΘA = ΘA(π) = { τ ∈ RA; ∑ α∈A;π0(α)≤k τα > 0 and ∑ α∈A;π1(α)≤k τα < 0 for every k ∈ {1, . . . , d− 1} } . Then every (π, λ, τ) ∈ SA 0 × ΛA × ΘA1 may be see as a translation surface as follows. More precisely, first we consider two broken line 1Note that this space is not really a product space since ΘA depends on π and thus SA 0 ×ΛA×ΘA = ⋃ π∈SA 0 {π}×ΛA×ΘA(π). However, we shall use this notation for simplicity. 3 P. BERK HORIZONTAL SADDLE CONNECTIONS Figure 1. A translation surface and one step of backward Rauzy-Veech induction. The parallel segments are identified via translation. The winning segment is the one which is first crossed by a rightward separatrix starting from (0, 0).