Horizon saddle connections and Morse-Smale dynamics of dilation surfaces

Without loss of generality, we can assume that $d$ is vertical. Since separatrices travel between singularities and boundary components, they have to be compact segments. In a given direction, there are finitely many such separatrices (each conical singularity has a finite angle). Cutting along them, we decompose $X$ into connected components $X_{1},\dots,X_{k}$.

We then consider one of these components $X_{i}$. It is a dilation surface with boundary $\partial X_{i}$. Let $x$ be a singularity of $X_{i}$. There are two cases:

• (i)

  $x$ is the intersection of a vertical separatrix of $X$ with a regular point of $\partial X$;

• (ii)

  $x$ is a conical singularity of $X$.

In case (i), the conical angle of $X_{i}$ at $x$ is strictly smaller than $\pi$.
In case (ii), if the conical angle of $X_{i}$ at $x$ is strictly larger than $\pi$, then the interior of $X_{i}$ contains a vertical separatrix (this contradicts the definition of $X_{i}$).
Consequently, at each singularity of $X_{i}$ the magnitude of the angular sector is at most $\pi$. In particular, every singularity of $X_{i}$ belongs to $\partial X_{i}$ (conical angles of interior singularities are integer multiples of $2\pi$).
Up to adding an arbitrary number of marked points, we can decompose $X_{i}$ into flat triangles and apply the Gauss-Bonnet formula. The total angle defect of $X_{i}$ is nonnegative so that $X_{i}$ is either a polygon (contractible domain with a boundary formed by saddle connections) or a topological cylinder such that the angle at each singularity of $\partial X_{i}$ is equal to $\pi$.
In the latter case, this implies that singularities of $\partial X_{i}$ are singularities of $X_{i}$ and therefore $\partial X_{i}$ is formed by saddle connections of $X$. These saddle connections are vertical (otherwise the interior of $X_{i}$ would contain some vertical separatrix). The hypothesis that every vertical separatrix crosses the boundary in a regular point implies that there is no vertical saddle connection. We get a contradiction and thus $X_{i}$ is a polygon.
Since polygons are simply connected, their dilation structure reduces to a translation structure. In a translation surface of finite area, a trajectory can hit a singularity, cross the boundary, be minimal in some domain or be periodic (see Proposition 5.5 in [3] for the classification of invariant components). Minimal trajectories only appear in surfaces of genus at least one and every loop in a polygon is contractible. Therefore, in each polygon, every vertical trajectory crosses $\partial X_{i}$. Boundary segments of $\partial X_{i}$ are either subsets of saddle connections of $X$ or vertical separatrices of $X$. However, a vertical trajectory cannot intersect a vertical separatrix. Consequently, every vertical trajectory in $X_{i}$ crosses a boundary saddle connection of $X$.
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We simply have to note that for any trajectory $T$ having $k$ intersection points with $L$, slight enough deviations of trajectory $T$ provide trajectories with at least $k$ intersection points with $L$. ∎

Let $k$ be the largest integer such that $U\cap S(k)$ is not empty. Integer $k$ is the maximal number of intersection points a trajectory may have with saddle connection $L$ when the direction of the trajectory belongs to $U$.
Let trajectory $T$ belonging to direction $d\in\mathbb{S}^{1}$ be a trajectory with $k$ intersection points with $L$. Let $x_{1},\dots,x_{k}\in L$ be the intersection points (with the order induced by direction $d$). There is an open neighborhood $\Omega$ of $d$ such that every trajectory starting from $x_{k}$ in a direction chosen in $-\Omega$ has at least $k$ intersection points with $L$. This implies that trajectories in $P(x_{k},\Omega)$ cannot intersect $L$. Otherwise the bound property that defines $k$ would be violated.
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Let $P(x_{0},U)$ be an exceptional pencil in $X$. We assume that there is a hyperbolic cylinder $C$ and an open interval $V\subset U$ such that the directions of closed geodesics of $C$ coincide with $V$ or $-V$ ($C$ may be only a portion of a bigger cylinder covering a bigger interval of directions). Since $P(x_{0},U)$ is an exceptional pencil, $P(x_{0},V)$ is also an exceptional pencil.
If $x_{0}\in C$, then $x_{0}$ belongs to some hyperbolic closed geodesic $\alpha$ of $C$. The direction of $\alpha$ belongs to $V$ or $-V$ so $\alpha$ coincides with some trajectory in pencil $P(x_{0},V)$. This implies that $P(x_{0},V)$ is not an exceptional pencil.
If $x_{0}\notin C$ and a trajectory $T$ of $P(x_{0},V)$ enters $C$, then Lemma 3.1 implies that $T$ accumulates on some hyperbolic closed geodesic of $C$. This would also contradict the assumption that $P(x_{0},V)$ is an exceptional pencil. Consequently, in any case, $P(x_{0},V)$ is disjoint from cylinder $C$.
We can remove $C$ from $X$ and obtain a new dilation surface that still contains an exceptional pencil. If this leads to the existence of boundary components without any singular points, we can simply mark points on the boundary. Since there are at most finitely many hyperbolic closed geodesics in any given direction (see Lemma 3.2), after finitely many modifications, we obtain a dilation surface $X_{1}$ with nonempty boundary and an exceptional pencil $P(x_{0},W)$ such that no hyperbolic closed geodesic of $X_{1}$ belongs to a direction of $W$ or $-W$.
Dilation surface $X_{1}$ could contain a hyperbolic cylinder of angle greater than or equal to $\pi$ because such a cylinder would contain a closed geodesic in a direction of $W$ or $-W$. Applying Veech’s criterion (Theorem 3.4) then implies that $X_{1}$ decomposes into finitely many flat triangles whose sides are saddle connections. Lastly, if $x_{0}$ is not already a singularity of $X_{1}$, we can still make it a marked point. Therefore $X_{1}$ belongs to class $(\Delta)$.
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We consider a regular point $b$ in boundary saddle connection $B$ such that $P(b,\epsilon U)$ is nonempty for some $\epsilon=\pm 1$. We start by proving that no trajectory in pencil $P(b,\epsilon U)$ can hit a singularity.
Otherwise such a trajectory $\gamma$ hits a singularity $A$, and Lemma 3.5 proves the existence of a flat triangle $AMN$ (where $M,N$ are singularities of $X$) such that:

• •

  denoting by $]\theta_{0},\theta_{1}[$ the open angular sector of vertex $A$ inside $AMN$, trajectory $\gamma$ (with the opposite orientation) belongs to pencil $P(A,]\theta_{0},\theta_{1}[)$;

• •

  every trajectory in pencil $P(A,]\theta_{0},\theta_{1}[)$ crosses boundary saddle connection $B$ at some regular point.

Now we consider an open subset $V\subset U$ such that $-\epsilon V\subset]\theta_{0},\theta_{1}[$. Since $P(x_{0},U)$ is an exceptional pencil, $P(x_{0},V)$ is also an exceptional pencil. We are going to prove that $P(x_{0},V)$ is disjoint from triangle $AMN$.
We start by considering the case where $\epsilon=-1$. Every trajectory traveling in a direction of $]\theta_{0},\theta_{1}[$ (including $V$ in this case) that crosses triangle $AMN$ eventually crosses boundary saddle connection $B$ (see Figure 4). Consequently, if $\epsilon=-1$, pencil $P(x_{0},V)$ is disjoint from triangle $AMN$.
If $\epsilon=1$, assuming that some trajectory in pencil $P(x_{0},V)$ crosses triangle $AMN$, its starting point $x_{0}$ belongs to pencil $P(A,]\theta_{0},\theta_{1}[)$ (see Figure 4). However, we know by hypothesis that no trajectory in pencil $P(A,]\theta_{0},\theta_{1}[)$ can hit a singularity. Thus $P(x_{0},V)$ is also disjoint from triangle $AMN$ in that case.
Now since $P(x_{0},V)$ is disjoint from triangle $AMN$, we can remove $AMN$ from surface $X$ and obtain a surface $X^{\prime}$ in class $(\Delta)$ that contradicts the fact that $X$ belongs to $(\Delta_{min})$. Indeed, pencil $P(x_{0},V)$ is still an exceptional pencil and $X^{\prime}$ does not contain any hyperbolic closed geodesic in any direction in $V\cap-V$. Thus, no trajectory in pencil $P(b,\epsilon U)$ can hit a singularity.

Unless some trajectory in pencil $P(b,\epsilon U)$ crosses a boundary saddle connection, Lemma 3.3 proves the existence of a hyperbolic closed geodesic in some direction of $U\cup-U$. Starting from a trajectory of $P(b,\epsilon.U)$ crossing some boundary saddle connection $B^{\prime}$, a continuity argument involving Theorem 3.4 and similar to the proof of Lemma 3.5 then proves that every trajectory of the pencil crosses $B^{\prime}$ (otherwise one trajectory would hit a singularity). A similar statement holds for any starting point in $B$. Since these pencils intersect each other, it is straightforward that the trajectories of these pencils eventually cross the same boundary saddle connection $B^{\prime}$. ∎

If there exists a dilation surface with nonempty boundary and an exceptional pencil, then Proposition 4.4 proves that class $(\Delta)$ is nonempty. We will consider a surface of subclass $(\Delta_{min})$ and obtain a contradiction.
Set $X$ a dilation surface in $(\Delta_{min})$ with an exceptional pencil $P(x_{0},U)$ and no hyperbolic closed geodesic that has its direction in $U\cup-U$. Up to taking an open subset of $U$, we can assume that no direction of any boundary saddle connection in $X$ belongs to $U\cup-U$. Then Lemma 4.5 proves that for every regular point $b$ of a boundary saddle connection $B$, trajectories of the nonempty pencils $P(b,\epsilon.U)$ (for $\epsilon=\pm 1$) cross the same boundary saddle connection $B^{\prime}$ in a regular point.
Looking at these trajectories with the opposite orientation, we use Lemma 4.5 once again to prove that for every regular point $b^{\prime}$ chosen from boundary saddle connection $B^{\prime}$, trajectories of the nonempty pencils $P(b^{\prime},-\epsilon.U)$ cross saddle connection $B$. This automatically implies that one trajectory of the pencils hits one endpoint of the opposite boundary saddle connection.
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Such a pencil $P(x_{0},U)$ is not exceptional because this would contradict Proposition 4.2. Thus, at least one of its trajectories $\gamma$ accumulates on some hyperbolic closed geodesic. These geodesics belong to a hyperbolic cylinder. Even if the hyperbolic closed geodesic is singular, we can perturb it to one side or the other so that the perturbed trajectory $\gamma^{\prime}$ accumulates on a closed geodesic in the same cylinder. ∎

Lemma 2.6 implies that in any open set of directions, we can find a pencil of trajectories in $X\setminus L$ such that none of these trajectories cross the boundary. Corollary 4.6 then implies that some trajectories of this pencil accumulate on hyperbolic closed geodesics. Therefore, there are directions of hyperbolic closed geodesics in any open subset of $\mathbb{S}^{1}$.
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