Finiteness conditions on translation surfaces

# Finiteness conditions on translation surfaces

Joshua P. Bowman IMS, Department of Mathematics
Stony Brook University
Stony Brook, NY 11794
###### 2010 Mathematics Subject Classification:
Primary 30F30. Secondary 57M50, 32G15, 14K20

## Introduction

Throughout this note, let denote a translation surface, i.e., a (connected) topological surface with a translation atlas. Then is automatically endowed with a conformal structure and a flat metric, and so it is both a Riemann surface and a Riemannian manifold [HM]. An orientation-preserving homeomorphism is called affine if it is affine in local charts. We use to denote the group of affine maps of . Any element of has a well-defined global derivative . The image of the homomorphism is called the Veech group of [Ve, Vo, EG, GJ].

The existence of affine self-maps of a translation surface has applications in the study of mapping class groups, Teichmüller theory, algebraic geometry, and dynamical systems (for a small sampling of such applications, see, e.g., [Th, HS, Mc, , LR, De]). They measure a kind of “symmetry” more general than that of isometries, which nonetheless has consequences for such systems as geodesic flow on the surface and geodesics in Teichmüller space. Veech first observed the importance of the group of derivatives of affine maps [Ve].

Let denote the metric completion of . The classical study of translation surfaces assumes that is itself a compact surface and is finite. If these conditions are satisfied, we will say that has finite affine type. Here we wish to consider four other “finiteness” conditions that may be placed on :

1. has finite analytic type as a Riemann surface, meaning that it is obtained from a compact Riemann surface by making finitely many punctures.

2. has finite area as a Riemannian manifold, meaning that the integral of the induced area form over all of is finite.

3. is bounded as a metric space, meaning that there exists a constant such that for every pair of points and in .

4. is totally bounded as a metric space, meaning that for any fixed , can be covered by finitely many balls of radius (equivalently, is compact).

We will prove two main results about these conditions, one negative and one positive.

###### Theorem 1.

Except for the trivial implication “totally bounded bounded”, none of the conditions (1)–(4) on implies any of the others. However, if has finite analytic type, then the other three conditions are equivalent and imply that has finite affine type.

###### Theorem 2.

Suppose the ideal boundary of is empty. If has at least one periodic trajectory and is totally bounded or has finite area, then its Veech group is a discrete subgroup of . However, there exist bounded surfaces and surfaces of finite analytic type with non-discrete Veech groups.

###### Remark.

It is likely that the condition of having a periodic trajectory follows from the assumptions of having empty ideal boundary and being totally bounded or of finite area, in which case it can be dropped in Theorem 2.

Translation surfaces of infinite analytic type appear, for example, in [EG, CGL, HLT, HHW, Va, Bo], and it is such examples that motivated the study presented here. We will prove Theorem 1 in §1 and Theorem 2 in §2.

## 1. Inequivalence of finiteness conditions

We begin with the trivial, and only, implication among the finiteness conditions (1)–(4).

###### Proposition 1.1.

is totally bounded” is bounded”.

###### Proof.

This is a generality about metric spaces. Pick , and cover with balls of radius . Then the distance between any two points is at most . ∎

The rest of the first part of Theorem 1 is proved through a series of examples. One general construction will be quite useful and flexible, so we describe it first and establish some notation.

###### Example 1 (An infinite “stack of boxes”).

Let be a sequence of positive numbers, and let be a strictly decreasing sequence of positive numbers tending to zero. Then we construct a translation surface as follows (see Figure 1):

• For each , let be a rectangle with horizontal side and vertical side .

• Place the sequence of rectangles in the plane , starting with having its lower left corner at the origin, and with immediately above so that its left edge is along the -axis.

• Identify the right and left sides of each with each other via horizontal translation, and identify the portion of the top of not covered by (of length ) with the portion of the bottom edge of directly below via vertical translation. (We omit the vertices.)

The genus of is infinite, as can be seen by considering the (pairwise non-homotopic) horizontal core curves of the . The area of is

 Area(XH,W)=∞∑n=1Area(Rn)=∞∑n=1hnwn.

In particular, the area of is finite if and are sequences in , but this is not necessary. Let denote the metric completion of this surface.

###### Lemma 1.1.

has only one point.

###### Proof.

The translation structure has been defined by taking a quotient of the union of the rectangles except for their vertices. The vertices are all collapsed to a single point, as is evident in Figure 1: using the notation of that figure, observe that . ∎

###### Lemma 1.2.

is bounded if and only if is bounded.

###### Proof.

Suppose is bounded by and by . Then every point of every rectangle is within of a corner. Since the vertices are identified to a single point in , is an upper bound for the distance between any two points of .

Now suppose is not bounded. Then the centers of the rectangles become arbitrarily far from the vertices, and so is not bounded. ∎

###### Lemma 1.3.

is totally bounded if and only if tends to zero.

###### Proof.

Let denote the unique point in .

Suppose tends to zero. Then, because also tends to zero, for every there exists such that and for all . This implies that the -neighborhood of covers all for . The complement of in is compact, being a finite union of compact pieces, and can therefore be covered by finitely many -balls.

Suppose does not tend to zero. Then there exists such that for infinitely many . Fix such an and cover by the following open sets:

• the -neighborhood of ;

• for each not fully covered by (of which there are infinitely many), the interior of this cylinder;

• for each edge between and , a neighborhood of radius .

Any finite subcover of this open cover would fail to cover infinitely many interior points of the s, and so cannot be compact. ∎

With these observations about in mind, we proceed to our counterexamples.

Take with .

Take with and .

Take with and .

###### Example 5 (bounded ⇏ finite analytic type).

Take Example 2 or 4.

Take with and .

###### Example 7 (bounded ⇏ totally bounded).

Take Example 4 or 6.

Take Example 2.

Take with .

###### Example 10 (finite analytic type ⇏ finite area or bounded).

The Riemann surface has finite analytic type, since it is obtained from the Riemann sphere by removing two points. However, the translation structure given by the differential makes isometric to an infinite cylinder, in which case it does not have finite area, and it is not bounded.

Example 10 shows that the essential way a surface of finite analytic type can fail to have finite affine type is that one could take a meromorphic differential on a compact Riemann surface and remove the zeroes and poles to obtain a translation surface of finite analytic type. However, if we pair the “finite analytic type” condition with any of the others, then the rest follow. This fact is likely to be well-known, but we prove it here for completeness and to show how the analytic structure of the translation surface plays a role.

###### Proposition 1.2.

If has finite analytic type and finite area, then it is totally bounded.

###### Proof.

The translation structure is given by an abelian differential on . Let denote the compact surface from which is obtained as a Riemann surface. Because has finite area, the differential can be extended to ; each point of is either a regular point or a zero of the differential. Thus is canonically homeomorphic to , so is totally bounded. ∎

###### Proposition 1.3.

If has finite analytic type and it is bounded, then it has finite area.

###### Proof.

The translation structure is given by an abelian differential on that is meromorphic on the compact Riemann surface from which it is obtained by punctures. Because is bounded, none of the punctures is at an infinite distance from any other point of . Therefore the differential has no poles on , and so it has finite area. ∎

###### Proof of second part of Theorem 1.

Immediate from Propositions 1.1, 1.2, and 1.3. ∎

To conclude this section, we observe that other collections of conditions do not imply any of the remaining ones, except as trivially follows from what has been established.

Take Example 2.

Take Example 4.

## 2. Discreteness of Veech groups

Recently, it has become apparent that translation surfaces of infinite analytic type allow for Veech groups of much greater complexity than occurs in the case of finite type. Specifically, it is well-known that the Veech group of a translation surface of finite affine type is always a Fuchsian (i.e., discrete) subgroup of and is never co-compact. In contrast, it has been shown by direct construction that any countable subgroup of (in fact, of ) that avoids the set of matrices with operator norm less than can occur as the Veech group of a translation surface whose topological type is that of a “Loch Ness Monster”, meaning it has infinite genus and one topological end [PSV]. Other “naturally occurring” examples (e.g., the surface obtained by “unfolding” an irrational polygon [Va]) also demonstrate that one cannot in general expect the Veech group of a translation surface of infinite type to be discrete. In this section, we show that this phenomenon of non-discreteness relies essentially on the failure of a surface to be totally bounded or to have finite area; i.e., it is not enough that the surface simply have infinite analytic type.

The usual proof of discreteness in the case of finite affine type is carried out by showing that the Veech group acts on the set of holonomy vectors of saddle connections, which is a discrete subset of (see, e.g., [Vo]). For surfaces not of finite affine type, this last clause no longer holds: in many examples, the holonomy vectors of saddle connections do not have their lengths bounded away from zero. We find another subset of on which the Veech group acts and which, under the conditions of Theorem 2, is also discrete. Our proof holds also for surfaces of finite affine type, and bypasses considerations of whether the holonomy vectors of saddle connections form a discrete set or not.

Observe, first of all, that if has finite area, then any element of must preserve this area, and so the condition that follows automatically. Similarly, we have the following.

###### Lemma 2.1.

If is totally bounded, then .

###### Proof.

We use the compactness of to establish a kind of Poincaré recurrence, which will permit us to define a first return map. Let . For any open subset of with piecewise smooth boundary, we observe that the images cannot all be disjoint: for otherwise, we could take them together with one more open subset, formed by the union of their complement and regular neighborhoods of their boundaries, and we would have an open cover of with no finite subcover. Therefore, by a standard argument, for some . Proceeding inductively, we obtain a first return map into , defined on an open subset of whose complement has measure zero. Choose so that it has finite area. The area of the image is

 Area(Rϕ)=∫U(detDRϕ)dArea≤Area(U).

If had determinant greater than , then the Jacobian determinant in the above integral would be greater than on the entire domain, and the given inequality would not hold. We conclude that any element of must have a derivative in . ∎

Now we proceed to the main ideas in the proof of Theorem 2. Throughout this section, we take cylinders in to be open subsets of ; that is, they do not include their boundaries.

###### Lemma 2.2.

Let and be two maximal cylinders in a translation surface whose respective circumferences are and and whose respective heights are and , and suppose that they intersect but do not coincide. Then the angle between the core curves of and satisfies

 |tanθ|>min{h1w1,h2w2}.
###### Proof.

Let and be the core curves of and , respectively. If and meet at right angles, then we are done. So suppose they do not meet at right angles. We note that each time crosses one boundary component of , it must cross the other boundary component before returning to the first, and likewise for crossing the boundary of . Therefore the connected components of are Euclidean parallelograms whose sides are arcs of the boundaries of and . Let be one such parallelogram (see Figure 2). The angles of are and , so it suffices to consider the smaller of these angles. Note that and are also the two heights of . Let and be the distances from the vertex at to the orthogonal projections of the adjacent vertices onto the adjacent sides of in the directions of and , respectively. At least one of the following inequalities holds: or . But , from which the desired result follows. ∎

###### Lemma 2.3.

If satisfy , then the angle between and satisfies

 |tanθ|<ε√|v0|2−ε2.
###### Proof.

Under the given conditions, the largest angle a vector can make with is when is tangent to the circle with radius centered at ; the assumptions imply that this angle is strictly smaller than in absolute value. The result now follows by direct calculation of the tangent of the angle in this extreme case and monotonicity of the tangent function on . ∎

The values and in Lemma 2.2 are, of course, the moduli of the cylinders. The basic idea behind the next lemma is that if two cylinders have the same area and almost the same circumference, then their moduli are not very different; we can thus play the two inequalities of Lemmata 2.2 and 2.3 against each other.

###### Notation.

Given a translation surface and , we denote by the set of maximal cylinders on with area , and by the set of holonomy vectors of core curves of elements of .

###### Lemma 2.4.

Let be a translation surface that either is totally bounded or has finite area, and let . Then is either empty or a discrete subset of .

###### Proof.

Let . If and is any vector such that , then the modulus of any corresponding cylinder is bounded below by

 f1(ε)=A(|v0|+ε)2.

On the other hand, Lemma 2.3 implies that if and , then the absolute value of the tangent of the angle between and is bounded above by

 f2(ε)=2ε√|v0|2−ε2|v0|2−2ε2.

Note that, as , tends to , while tends to . We can therefore choose small enough that . Then Lemma 2.2 implies that, for any pair of distinct elements such that and , the corresponding cylinders in must be disjoint.

If has finite area, there can only be finitely many such cylinders, and so there can only be finitely many elements of within of .

If is totally bounded, there again can be only finitely many such cylinders; otherwise we could find infinitely many disjoint balls of some fixed positive radius on , which is impossible in a totally bounded space.

In either case, is an isolated point in ; since was arbitrary, is discrete. ∎

###### Lemma 2.5.

Let be a translation surface that either is totally bounded or has finite area, and let . Then preserves and (X) preserves .

###### Proof.

The affine image of a cylinder is a cylinder, and the maximality of a cylinder is preserved because saddle connections are sent to saddle connections by elements of . We have already observed that an affine self-homeomorphism of must preserve area, and so the first claim is proved. The second follows immediately. ∎

###### Lemma 2.6.

Let be a translation surface without ideal boundary, and let be a maximal cylinder of finite area. Suppose that is not a torus. Then the stabilizer of in is a cyclic group, hence discrete.

###### Proof.

Because has finite area and is not a torus, has an ideal boundary. Because the ideal boundary of is empty, does not consist only of . Therefore each boundary component of contains a saddle connection; call these and . Any element of must also preserve the lengths of and . Because the boundary of has finite length, we may assume, up to taking a finite index subgroup, that every element in fixes and . But this implies that every element of fixes the entire boundary of and thus is a power of a full Dehn twist in . Ergo is isomorphic to a subgroup of , from which the result follows. ∎

###### Proof of first part of Theorem 2.

The result is already known if is a torus, because then the Veech group is (conjugate to) , so assume this is not the case.

Let be a periodic trajectory on . Then the image of is contained in some maximal cylinder. This cylinder must have some finite area : this is immediate if the area of is finite, and if is totally bounded, it follows because the points of the cylinder must be a bounded distance apart. Thus the set is non-empty. By Lemma 2.4, it is therefore discrete. By Lemma 2.5, acts on . So it suffices to show that the stabilizer inside of a point is discrete in . To see this, we observe that, up to taking a finite index subgroup, the stabilizer of in may be identified with the stabilizer inside of some cylinder in . Lemma 2.6 now implies the desired result. ∎

To prove the second part of Theorem 2, we again turn to examples.

###### Example 13.

Let be an “irrational” rhombus, meaning its angles are not rational multiples of . The surface obtained by unfolding is such that consists of four points, arising from the vertices of . is therefore bounded. Its Veech group, however, is an indiscrete subgroup of , generated by rotations through the angles of .

###### Example 14.

The infinite cylinder of Example 10 has finite analytic type. This surface has one homotopy class of periodic trajectories; these are the images of vertical lines in the plane under the universal covering map from to , which is made into a translation covering by taking the differential on the domain. The parabolic map of is affine with respect to for any , and it descends to as an affine map with respect to , acting as a Dehn twist on each annulus , . The Veech group of therefore contains a copy of , and so it is not discrete.

###### Remark.

It is still not known whether a surface of infinite genus that has finite area or is totally bounded can have a lattice Veech group—in particular, whether the Veech group of such a surface can be co-compact.

### Acknowledgements

The author wishes to thank the Hausdorff Research Institute for Mathematics in Bonn as well as the organizers of the trimester program “Geometry and Dynamics of Teichmüller Spaces”, where much of this work was carried out. Thanks also to Pascal Hubert, Gabriela Schmithüsen, and Ferrán Valdez for helpful conversations and feedback, and to the referee for useful suggestions.

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