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Ellipses in translation surfaces

We characterize subgroups of the mapping class group that stabilize a Teichmueller disk in terms of ellipses and strips that are immersed in the associated translation surface. In particular, we show that the space of immersed ellipses/strips that meet at least three cone points is naturally a (non-manifold) 2-dimensional cell complex. The topology of this complex and the geometry of its 0-cells determine the translation surface and its affine diffeomorphism group (up to the kernel of the differential).


Introduction
A translation structure µ on a (connected) topological surface X is an equivalence class of atlases whose transition functions are translations. Translation surfaces are fundamental objects in Teichmüller theory, the study of polygonal billiards, and the study of interval exchange maps.
Cylinders that are isometrically embedded in a translation surface play a central role in the theory. In Teichmüller theory, they appear as solutions to moduli problems [Strebel]. In rational billiards and interval exchange maps, cylinders correspond to periodic orbits [HMSZ06] [MsrTbc02] [Smillie00].
Each periodic geodesic γ on a translation surface belongs to a unique 'maximal' cylinder that is foliated by the geodesics that are both parallel and homotopic to γ. One method for producing such periodic geodesics implicitly uses ellipses. Indeed, if X admits an isometric immersion of an ellipse with area greater than that of X, then the image of the immersion contains a cylinder, and hence a periodic geodesic [MsrSml91] [Smillie00].
Ellipses interiors also serve to interpolate between maximal cylinders. The set, E(X, µ), of ellipse interiors isometrically immersed in X has a natural geometry coming from the space of quadratic forms on R 3 (see §6). The set of maximal cylinders is a discrete set lying in the frontier of the path connected space E(X, µ).
If the frontier of a translation surface X is finite, then each point in the frontier may be naturally regarded as a cone point with angle equal to an integral multiple of 2π. If the frontier of an immersed ellipse interior U contains a cone point c, then we will say that U meets x. If an ellipse interior meets a cone point, then the ellipse interior belongs to the frontier of E(X, µ). The remainder of the frontier consists of cylinders.
The first author thanks Indiana University for its hospitality. The second author thanks the Max-Planck Institut für Mathematik (Bonn), the Institut Fourier, and theÉcole Polytechnique Fédéral de Lausanne for their hospitality and support.
The number of cone points met by an ellipse interior determines a natural stratification of E(X, µ). 1 In §8 we demonstrate that E(X, µ) is homotopy equivalent to the stratum consisting of ellipse interiors that meet at least three cone points. In §9, we prove that the completion of this stratum is naturally a (non-manifold) 2-dimensional cell complex whose 2-cells are convex polygons.
We show that the topology of this polygonal complex and the geometry of the immersed ellipses and cylinders that serve as its vertices together encode the geometry of (X, µ) up to homothety.
Theorem 1.1. Suppose that there is a homeomorphism Φ that maps the polygonal complex associated to (X, µ) onto the polygonal complex associated to (X , µ ). If for each vertex U , the ellipses (or strips) U and Φ(U ) differ by a homothety, then (X, µ) and (X , µ ) are equivalent up to homothety.
Affine mappings naturally act on planar ellipses, and hence the group of (orientation preserving) affine homeomorphisms of (X, µ) acts on E(X, µ). Because µ is a translation structure, the differential of an affine homeomorphism is a well-defined 2 × 2 matrix of determinant 1 [Vch89]. The set of all differentials is a discrete subgroup of SL 2 (R) that is sometimes called the Veech group and is denoted Γ(X, µ). Using Theorem 1.1, one can characterize Γ(X, µ).
Theorem 1.2. The group Γ(X, µ) consists of the g ∈ SL 2 (R) for which there exists an orientation preserving self homeomorphism of the polygonal complex associated to (X, µ) such that for each vertex U there exist a homothety h U such that U differs from Φ(U ) by h U • g.
The group Γ(X, µ) is closely related to the subgroup of the mapping class group of X that stabilizes the Teichmüller disc associated to (X, µ) [Vch89]. To be precise, each mapping class in the stabilizer has a unique representative that is affine with respect to µ. The Veech group is the set of differentials of these affine maps, and is isomorphic to the stabilizer modulo automorphisms. In particular, if there are no nontrivial automorphisms in the stabilizer, then the quotient of the hyperbolic plane by a lattice Veech group is isometric to a Teichmüller curve.
There is a natural map that sends each ellipse interior U ⊂ R 2 to the coset of SO(2) \ SL 2 (R) consisting of g such that g(U ) is a disc. This map naturally determines a map from E(X, µ) onto the Poincaré disc. The image of the 1-skeleton of the polygonal cell complex determines a tessellation of the upper half-plane that coincides with a tessellation defined by William Veech [Vch99] [Vch04] and independently Joshua Bowman [Bwm08]. Indeed, our work began with a reading of a preprint of [Vch04]. In a companion paper [BrtJdg11], we will discuss these connections in more detail.
To develop the deformation theory of immersed ellipses, we use quadratic forms on R 3 . Each ellipse interior in an affine plane P ⊂ R 3 is a sublevel set, { x ∈ P | q( x) < 0}, of a quadratic form q. The space of quadratic forms on R 3 is a six dimensional real vector space, and two quadratic forms determine the same ellipse interior if and only if they differ by a positive scalar.
A cylinder is the image of an isometric immersion of a strip, the interior of the convex hull of two parallel lines in R 2 . A strip is also a sublevel set of a quadratic form q restricted to an affine plane P .
The language of quadratic forms unifies the treatment of immersed ellipses and cylinders. We define a planar subconic to be a sublevel set {(x, y) | q(x, y, 1) < 0} where q is a quadratic form on R 3 . The set of planar subconics includes ellipse interiors and strips, but also includes ellipse exteriors, parabola interiors, etc.
We will let S(X, µ) denote the space of immersions of subconics into a translation surface (X, µ). We let S n (X, µ) denote the stratum consisting of subconics that meet at least n cone points.
The translation structure µ determines a canonical Euclidean metric on the surface X. The completion of X with respect to this metric will be denoted by X. If A is a subset of X, then ∂A will denote the complement of the interior of A in its closure in X. In particular, ∂X we will denote the set of cone points.
Other notation can be found in the table that is located at the end of this introduction.
Outline of paper. In sections §2 through §5 we establish notation and introduce basic tools. In §2 we recall the basic theory of translation surfaces including the developing map, and in §3 we recall some basic facts about quadratic forms. In §4, we collect elementary facts about the subconics in the plane. In §5 we consider subconics in the plane determined by finite sets of points. In §6 we define the geometry of the space of immersed subconics in a translation surface. We show, for example, that S(X, µ) is naturally a 5-dimensional realprojective manifold.
Beginning in §7 we restrict attention to the translation surfaces that cover a precompact translation surface. For such surfaces, the only possible subconics are ellipse interiors and (immersed) strips. We show that S 5 (X, µ) is discrete and that only finitely many members of S 5 (X, µ) contain a given nonempty open subset of X.
In §8 we show that the subspace E 3 (X) ⊂ S 3 (X) consisting of ellipse interiors in homotopy equivalent to X. As a consequence, E 3 (X,μ) is the universal cover of E 3 (X, µ). This fact is used crucially in our proof of Theorem 1.1.
In §9 we define the cell structure on S 3 (X, µ). In particular, each 2-cell S Z corresponds to a triple Z ⊂ ∂X that defines a triangle in X. We show that S Z may be regarded as the convex hull of all subconics that lie in S Z ∩ S 5 (X, µ). In §10, we characterize those triples (resp. quadruples) in ∂X that determine a 2-cell (resp. 1-cell) in S 3 (X, µ).
The set S Z ∩ S 5 (X, µ) determines an orientation of each 2-cell. In §11, we relate this orientation to the ordering of ∂U ∩ ∂X where S Z ∩ S 5 (X, µ). In §12 we study the oriented link of each vertex in the 2-skeleton. For example, we show that the link of U is determined up to isomorphism by the cardinality of ∂U ∩ ∂X. In §13, we use the analysis of the oriented links to show that a homeomorphism Φ : S(X, µ) → S(X , µ ) determines a unique bijection β : ∂X → ∂X such that for each Z ⊂ ∂U ∩ ∂X, we have Φ(S Z ) = S β Z . Note that the results of these three sections are purely combinatorial.
In §14 we prove the main geometric lemma. Roughly speaking, the sides of a cyclic polygon are determined up to translation by the directions of the bisectors of successive diagonals in the polygon. In §15, we combine this geometric Lemma with the combinatorial results to prove Theorem 1.1. In §16, we prove Theorem 1.2.
Notation. For the convenience of the reader, we include the following list of notation.
§3 vector space of real quadratic forms on R n Q F §3 quadratic forms that vanish on F q §4 restriction of quadratic form q to first two coordinates ( §6 set of subconics in a translation surface S n (X, µ) §6 stratum of S(X, µ) E(X, µ) §8 set of ellipse interiors in a translation surface §14 orbit of U under the group generated by homotheties and translations.

Preliminaries on translation surfaces
Recall that a translation atlas on a surface X is an atlas {µ α : U α → R 2 } such that each transition map µ β • µ −1 α is a translation. A translation structure µ on X is an equivalence class of translation atlases where two atlases are equivalent if the they agree on their common refinement up to translations.
Let (X, µ) and (Y, ν) be translation surfaces. A mapping f : X → Y is called a translation map with respect to µ and ν-or simply a translation-iff for each chart µ α in µ and ν β in ν, the map ν β • f • µ −1 α is a translation. One may define a metric structure on (X, µ) in several equivalent ways. One approach is to simply pull back the Riemannian metric tensor dx 2 + dy 2 on the Euclidean plane via a translation atlas. Since translations preserve dx 2 + dy 2 , the resulting metric tensor on X is well-defined. Thus, µ determines a (Riemannian) distance function d µ : X × X → R.
Let X denote the metric completion of X with respect to d µ .
Definition 2.1. The frontier of X with respect to µ is the set ∂X := X \ X.
A translation structure pulls-back under a covering map. In particular, one associates to each translation structure µ on X a translation structureμ on the universal coverX.
The following assumption will be implicit throughout this paper.
Assumption 2.2. The frontier ∂X is a discrete subset of X. Moreover, the frontier ∂X of the universal coverX is a discrete subset of the completion ofX.
Assumption 2.2 implies that the frontier of a precompact 2 translation surface is finite. In addition, a precompact translation surface has finite genus.
A translation structure is an example of a geometric structure in the sense of [Thurston]. If X is simply connected, for each translation structure µ on X, there exists a map dev µ : X → R 2 such that the set of restrictions to convex sets is an atlas that represents µ. This developing map is unique up to post-composition by a translation. Note that dev µ extends uniquely to a continuous function on X. We will abusively use dev µ to denote this extension.
In general, the developing map is not injective. But the restriction of dev µ to a 'star convex' subset is injective. To be precise, a set Y ⊂ X will be said to be star convex with respect to x if and only if for each x ∈ Y , there exists a geodesic segment in Y joining x to x. A subset Y ⊂ X will be called convex if and only if it is star convex with respect to each of its points.
It will be convenient to extend the notions of convexity and star convexity to subsets of the completion X. We will say that Y ⊂ X is star convex with respect to y if and only if for each y ∈ Y , there exists a convex subset C ⊂ X such that y, y ∈ C. A set Y ⊂ X will be called convex if and only if it is star convex with respect to each y ∈ Y .
Remark 2.3. Our notions of convexity for subset of X are defined in a way that assures that the restriction of the developing map is injective. Note that the definition differs significantly from the one that is natural if X is regarded as a length space. If ∂X is discrete, then the geodesics in the length space X are finite concatenations of completions of geodesic segments in X. From the perspective of length spaces, each geodesic is (star) convex, but from our viewpoint many geodesics are not (star) convex.
Let (X, µ) be a simply connected translation surface. Given A ⊂ X, define the inradius of A with respect to µ to be the supremum of the radii of disks that can be embedded into A.
Proposition 2.4. If (X, µ) is the universal cover of a precompact translation surface with nonempty discrete frontier, then the inradius of X is finite.
Proof. By assumption we have a precompact translation surface (Y, ν) and a cov- denote the respective distance functions on the respective completions of X and Y . Note that p is a local isometry with respect to these distance functions. The For each x ∈ X, we have d X (x, ∂X) = d Y (p(x), ∂Y ). Indeed, since ∂Y is finite, there exists a geodesic segment γ joining p(x) to ∂Y whose length equals d Y (p(x), ∂Y ). This geodesic segment lifts under p to a unique pathγ that joins x to ∂X. Since one endpoint ofγ lies in ∂X and (X, µ) is a translation surface, the restriction of p toγ is an isometry onto γ. In particular, the length ofγ equals the length of γ.
Therefore, for each x ∈ X, we have d X (x, ∂X) < K Y . It follows that if D is a Euclidean disk embedded in X, then the radius of D is less than K Y .

Quadratic forms on R 3
We recall some basic facts about quadratic forms defined over the real numbers. See chapter 14 in [Berger].
Let Q(V ) denote the vector space of all real-valued quadratic forms on a real vector space V . Given q ∈ Q(V ) we let q(·, ·) denote its polarization. There is a basis for V such that the matrix associated to the polarization is diagonal with entries belonging to {+1, −1, 0} (Sylvester's law). Let n + (q), n − (q), and n 0 (q) denote, respectively, the number of diagonal entries that are +1, −1, and 0. We say that q has signature (n + (q), n − (q)). If n 0 (q) = 0, then we say that q is degenerate.
The radical, rad(q), of (the polarization of) the quadratic form q is the set of v such that q(v, w) = 0 for all w ∈ R 3 . The form q is nondegenerate iff only rad(q) = {0}. If rad(q) = {0}, then 0 is an eigenvalue and rad(q) is the associated eigenspace.
We will be interested in quadratic forms that vanish on a prescribed subset F ⊂ R 3 . Define Note that Q F is a vector subspace.
To each point v ∈ R 3 \ {0} we may associate the unique line (v) containing v and the origin. We say that a set of points F ⊂ R 3 is in general position if no three of the lines in (F ) are coplanar.
To construct quadratic forms that vanish on subsets of F ⊂ R 3 , one can use the collection, (R 3 ) * of linear forms on R 3 . Given linearly independent vectors v, w ∈ R 3 , let L vw ⊂ (R 3 ) * be the collection of linear forms that vanish on the plane spanned by v and w. Note that a product η · η of linear forms η ∈ L xy and η ∈ L zw is a quadratic form in Q {x,y,z,w} .
Let Card(F ) denote the cardinality of a set F .
Proof. Let F = {x 1 , . . . , x n } where n = Card(F ). Note that for each i, the map q → q(v i ) is a linear functional on Q(R 3 ). The set Q F is the kernel of the homomorphism φ : Q(R 3 ) → R n defined by φ(q) = (q(v 1 ), . . . , q(v n )) .
Since dim(Q(R 3 )) = 6, it suffices to show that φ has full rank. If n = 1, then this is true. If 1 < n ≤ 5, then suppose that there exist (a 1 , . . . , a n ) ∈ R n such that for each q ∈ Q(R 3 ) we have i a i · q(v i ) = 0. Since F is in general position, for each j there exists q j ∈ Q(R 3 ) such that q j (v j ) = 0 and q j (v i ) = 0 if i = j. For example, if n = 5 and j = j 1 , then choose η ∈ L vj 2 ,vj 3 , η ∈ L vj 4 vj 5 , and set q = η · η .
Hence a j · q j (v j ) = 0 and therefore a j = 0 for each j. Thus, φ has full rank.
We also have the following variant of the preceding proposition that will be used in §10. Let dq x : R 3 → R denote the differential of q at x.
Proposition 3.2. Let n = 3 or 4, and let {v 1 , . . . , v n } ⊂ R 3 \ {0} be in general position. Let {w 1 . . . w 5−n } be a subset of R 3 \ {0} such that for each i, k, the vector w k does not belong to the plane spanned by v k and v i . Then the set of quadratic forms q with q(v i ) = 0 and dq vi (w i ) = 0 is a 1-dimensional vector space.
Proof. The set in question is the kernel of the homomorphism φ : It suffices to show that φ has full rank.
Suppose that there exist (a 1 , . . . , a n ) ∈ R n and (b 1 , . . . , b 5−n ) such that for each Since F is in general position and each w k does not belong to the plane spanned by v k and v i , for each i = 1, . . . , n, there exists q i ∈ Q(R 3 ) such that q i (v i ) = 0 but q i (v j ) = 0 for each j = i and d(q i ) v k (w k ) = 0 for each k. Similarly, for each k = 1, . . . , 5 − n, there exists q k so that d(q k ) v k (w k ) = 0 but q i (v j ) = 0 for each j and dq vj (w j ) = 0 for each j = k.
It follows that a i = 0 for each i and b k = 0 for each k. Thus, φ has full rank.
Let F ⊂ R 3 be a triple. For each x ∈ F , define D x ⊂ Q F to be the set of quadratic forms d such that d(x, v) = 0 for all v ∈ R 3 . Note that the set D x is a 1-dimensional subspace. If d ∈ D x and d = 0, then where x, y is the plane spanned by x and y.
We will call {d 1 , d 2 , d 3 } a natural basis for Q F .
Proof. Suppose that there exist a i ∈ R such that a i · d i (v) = 0 for all v ∈ R 3 . Let {i, j, k} be distinct indices. Note that both d i and d j vanish identically on the span of {v i , v j }, but since (F ) is not coplanar, the form d k does not. It follows that a k = 0. Hence the forms {d 1 , d 2 , d 3 } are independent, and the claim follows from Proposition 3.1.
Corollary 3.4. If the triple F is in general position and q ∈ Q F \{0} is degenerate, then q is of type (1, 1).
If F is a triple in general position, then the vector space Q F has a canonical orientation. Indeed, note that since F is in general position, then F has a canonical cyclic ordering: If F = {x, y, z}, then we say that y follows x if and only if (y−x, z − x, x) is an oriented basis for R 3 . We say that a natural basis {q x ∈ Q x | x ∈ F } is negative if and only if for each p in the interior of the convex hull of F and each x ∈ F we have q x (p) < 0. We say that an ordered negative natural basis (d 1 , d 2 , d 3 ) is oriented if and only if the corresponding ordered triple (x 1 , x 2 , x 3 ) ∈ F 3 is cyclically ordered. In general, a basis of Q F is oriented if and only if it differs from an oriented negative basis by a linear transformation having positive determinant.
We now turn to quadruples F ⊂ R 3 .
Lemma 3.5. If the quadruple F is in general position, then the set of degenerate quadratic forms in Q F is a union of three distinct lines.
, define Q ij,k to be the set of pairwise products of linear forms from L vivj and L vj v k . Each Q ij,k is a line in Q F , and there are three distinct such lines. Let q ∈ Q F be degenerate. It follows from Corollary 3.4 that q is of type (1, 1). Thus, there exist linear forms η, η ∈ (R 3 ) * such that q = η · η . Since q ∈ Q F , for each i = 1, 2, 3, either η(v i ) = 0 or η (v i ) = 0. Since F is in general position, η and η each vanish on exactly two points in F and since q ∈ Q F these pairs are distinct. Thus, q belongs to one of the distinct lines.

Subconics in the plane
Recall that a conic section is the intersection of a hyperplane in R 3 and the zero level set of a quadratic form q ∈ Q(R 3 ). In the context of translation surfaces, we are mainly interested in the intersection of the sublevel set q −1 {(−∞, 0)} and a hyperplane. For simplicity, we choose this hyperplane to be {(x 1 , x 2 , 1) : (x 1 , x 2 ) ∈ R 2 }. The following notation will be convenient.
Definition 4.2. The subconic associated to q is the set The subconic U q said to be nontrivial if and only if U q is a proper subset of R 2 . The boundary of U q is the set Given a quadratic form q ∈ Q(R 3 ), let q denote the restriction of q to the the plane {(x, y, 0) | (x, y) ∈ R 2 }. Subconics can be classified according to the signatures of q and q. Definition 4.3. A subconic U q is called an ellipse interior if and only if q has signature (2, 1) and q has signature (2, 0). Definition 4.4. A subconic U q is called a strip if and only if q has signature (1, 1) and q has signature (0, 0).
Definition 4.5. A subconic U q is called a half-plane if and only if q has signature (1, 1) and q has signature (0, 0). Definition 4.6. A subconic U q is called a parabola interior if and only if q has signature (2, 1) and q has signature (1, 0).
Remark 4.7. In this paper, we will be almost exclusively concerned with ellipse interiors and strips. See Proposition 7.2.
The following propositions are elementary.
Proposition 4.8. A non-trivial subconic is convex if and only if it is an ellipse interior, a parabola interior, a strip, or a half-plane.  Proposition 4.11. The set of quadratic forms asociated to ellipse interiors is convex and is preserved by the R + action.

Subconics determined by a finite set of points
Given a finite set Z ⊂ R 2 , let S Z denote the set of nontrivial subconics whose boundary contains Z, and let Q Z denote the set of quadratic forms q such that q(Ẑ) = {0}.
Note that S Z ⊂ {U q | q ∈ Q Z }. The opposite inclusion does not hold in general. For example, if Z lies on a line, then let η be a nontrivial linear form that vanishes on a plane containingẐ. The quadratic form η 2 belongs to Q Z but U η 2 is the empty set.
Lemma 5.1. Suppose that Z ⊂ R 2 is not collinear. If U q is empty for some q ∈ Q Z , then q ≡ 0. In particular, Proof. Suppose that U q = ∅ for some q ∈ Q Z . Given a line ⊂ R 2 , the restriction of q toˆ is a nonnegative quadratic function. Thus, if x and y are distinct points in such that q(x) = 0 = q(ŷ), then q vanishes identically onˆ .
Let z 1 , z 2 , and z 3 be three noncolinear points in Z. Since q ∈ Q Z , the form q vanishes on the lines determined by the pairs z i , z j . It follows that q vanishes on the plane {(x 1 , x 2 , 1) | x 1 , x 2 ∈ R}. Hence q ≡ 0.
Let Z ⊂ R 2 be a noncolinear triple. ThenẐ is in general position. Let d = (d 1 , d 2 , d 3 ) be an oriented negative natural basis for Let r d : T d → S Z denote the restriction of the map q → U q to T d .
Proof. Since d = {d 1 , d 2 , d 3 } is a basis for Q Z , Lemma 5.1 implies that map q → U q is onto S Z . Note that U q = U q if and only if there exists λ ∈ R + such that λ·q = q . It follows that r q is injective. The continuity of q → U q and the inverse of its restriction follow from linearity and the definitions of the various topologies.

Define a linear map
Let E Z denote the collection of ellipse interiors whose boundaries contain Z.
Proposition 5.4. Let Z ⊂ R 2 be a noncollinear triple, let d be a natural basis for Q Z , and let T = T d . The set {q ∈ T | U q ∈ E Z } is a bounded, convex subset of the plane T . In particular, E Z is compact, and q ∈ T | U q ∈ E Z is a compact convex subset of T .
Proof. The set Q Z is a vector subspace, and hence it follows from Proposition 4.11 that the set of all q such that U q ∈ E Z is a convex subset of Q(R 3 ). Thus, Let (ẑ 1 ,ẑ 2 ,ẑ 3 ) be the cyclic ordering of the elements ofẐ with respect to the orientation of R 3 . Let d = (d 1 , d 2 , d 3 ) be a corresponding oriented negative natural basis for Q Z . To show that q ∈ T | U q ∈ E Z is bounded, it suffices to show that if U q ∈ E Z and q ∈ T d , then each coordinate of q with respect to d is positive.
For {i, j, k} = {1, 2, 3}, let σ i denote the segment joining z j and z k . Since U q ∈ E Z is strictly convex and Z ⊂ ∂U q , we have σ i ⊂ U q and σ i ∩ ∂U q = {x j , x k }. In particular, for each y ∈ σ i \Ẑ we have q(ŷ) < 0.
Let (t 1 , t 2 , t 3 ) be the coordinates of q with respect to d. Since d is a negative natural base, for each y ∈ σ i \ Z we have q(ŷ) < 0. It follows that t i > 0 for each i = 1, 2, 3.

Subconics in a translation surface
In this section, we make precise the notion of subconic in a translations surface (X, µ), we define the space of subconics in (X, µ) and its stratification, and we derive some basic properties.
Let Gal(X/X) denote the group of covering transformations associated to the universal covering p :X → X. Each covering transformation is a translation mapping of (X,μ) and hence Gal(X/X) acts on the set of convex subsetsŨ ⊂X such that dev µ (Ũ ) is a subconic in R 2 . Definition 6.1. A subconic in X with respect to µ is an orbit of the action of Gal(X/X) on the collection of convex subsetsŨ ofX such that dev µ (Ũ ) is a subconic in R 2 .
There is a natural one-to-one correspondence between the subconics U in X and equivalence 3 classes of isometric immersions from a subconic U ⊂ R 2 into X. Indeed, each isometric immersion f : U → X lifts to an immersionf : U →X. The lift is unique up to the action of Gal(X/X). On the other hand, given a subconicŨ ⊂X, the map p • dev −1 µ restricted to dev µ (Ũ ) gives an immersion. Each representativeŨ of a subconic U will be called a lift of U . In most cases, a subconic in X is determined by the image a lift under p. For example, if X is simply connected, then each subconic has a unique lift. In this case, we will identify the singleton set with the element that it contains.
On the other hand, there are situations in which one must keep track of the orbit in the universal cover (or, equivalently, the immersion). For example, if X is the once-punctured torus, (R 2 \ Z 2 )/Z 2 , then the set X is itself the image of two ellipses that do not differ by an element of Gal(X/X). 4 In practice, we will be working with a particular liftŨ in the universal cover, and we will regard the image p(Ũ ) as a subconic in X.
We will use the terminology used in the classification of subconics in the plane to describe subconics in X with respect to µ.
Remark 6.2. An isometric embedding of a cylinder, [a, b] × (R/c · Z), corresponds to a strip in a translation surface (X, µ).
If X is simply connected, the developing map furnishes a natural topology on S(X, µ). More generally, we endow S(X, µ) = S(X,μ)/Gal(X/X) with the quotient topology. The action of the deck group Gal(X/X) on S(X,μ) is discontinuous, and hence the quotient map p * : S(X,μ) → S(X, µ) is a covering map with deck group Gal(X/X).
Roughly speaking, the space S(X, µ) has a natural stratification determined by the number of points in the boundary, ∂X, that are met by a liftŨ . To be more precise we make the following definition.
Definition 6.4. For n ∈ Z + , define the n-stratum, S n (x, µ), to be the collection of all subconics U ⊂ X such that Card(∂Ũ ∩ ∂X) ≥ n and dev µ (∂Ũ ∩ ∂X) has maximal span in R 2 .
If one component of the boundary of a strip U contains at least three points in ∂X while the other other component contains none, then the strip does not belong to S n (X, µ) for any n. Indeed, in this case, the span of dev µ (∂Ũ ∩ ∂X) is 1-dimensional.
Proof. Since dev µ is a local embedding, it follows that the map U → dev µ is a local The action of the translations on X is dev µ -equivariant action of a subgroup of translations on R 2 . In turn this action is f -equivariant to an action of a subgroup of P GL(Q(R 3 )).

Subconics in coverings of precompact surfaces with finite frontier
In this section and the ones that follow, we make the following assumption.
Assumption 7.1. We consider only translation surfaces (X, µ) that cover a precompact translation surface with nonempty frontier.
This assumption limits the types of subconics that can be immersed in X. Recall that an embedded cylinder is a strip. See Remark 6.2.
Proposition 7.2. Each subconic U ⊂ X is either a strip or an ellipse interior.
Proof. LetŨ be a lift of U to the universal coverX. It follows from Proposition 2.4 and Assumption 7.1 that the inradius ofX is finite. The only nontrivial subconics that have finite inradius are strips and ellipse interiors.
In order to deal with a strip U ⊂ X, it will prove convenient to construct a canonical 'maximal' neighborhood M of U such that the restriction dev µ | M to M is an injective map.
Definition 7.3. Let M be the collection of all x ∈ X such that there exists y ∈ U and a geodesic segment α joining x to y such that the vector dev µ (x) − dev µ (y) is orthogonal to the radical of q where U q = dev µ (U ). We will call M the canonical neighborhood of U .
The following is elementary.
Proposition 7.4. If U ⊂ X is a strip, then there exists a unique strip that contains every strip that contains U . If τ : X → X is a nontrivial translation and τ (U ) = U , then for each connected component C of ∂U , the set C ∩ ∂X is infinite and τ (C ∩ ∂X) = C ∩ ∂X.
We will call the strip described in Proposition 7.4 the maximal strip containing U .
Proposition 7.5. If (X, µ) is simply connected and U ⊂ X is a maximal strip, then there exists a nontrivial translation τ : X → X such that τ (U ) = U . Proof. Let p : X → Y be as in Assumption 7.1. Since X is simply connected, p is the universal covering. Since ∂Y is finite, Y is precompact, and hence p(U ) is precompact. Since U is a strip, it is not precompact (Proposition 4.9). Therefore, there exists x, x ∈ U and a nontrivial deck transformation τ : X → X such that τ (x) = x . Because U is maximal and τ (∂X) = ∂X, we find that τ (U ) is also maximal.
The intersection τ (U ) ∩ U is a strip. Since U ∩ τ (U ) belongs to both U and τ (U ), if follows from the uniqueness of maximal strips-Proposition 7.4-that U = τ (U ).
By combining Proposition 7.4 and 7.5 we obtain the following.
Proof. By assumption there exists a precompact translation surface (X, ν) with finite frontier and a covering p : X → Y such that µ = p • ν. It suffices to assume that p is a universal covering. Note that p extends uniquely to a continuous map from X to Y that we will abusively call p.
Let U be a subconic such that ∂U ∩ ∂X has at least five points. Given δ, define the δ-neighborhood of U by Let q ∈ Q(R 3 ) be such that dev µ (U ) = U q . Let M ⊂ R 2 denote the image of N under the developing map. Since dev µ is an open mapping, the set M is open. By Proposition 7.2, U is either an ellipse interior or a strip. In each of these cases, we will show that U is isolated in S 5 (X, µ).
Suppose that U , and hence U q , is an ellipse interior. Let x 0 ∈ U q . Let A be the set of all r ∈ Q(R 3 ) such that either r( x 0 ) ≥ 0 or there exists x ∈ ∂M such that r( x) ≤ 0. Since ∂M is compact, the set A is closed. Since q(x 0 ) < 0 and U q ⊂ M , the form q does not belong to A.
Let U be a subconic in X such that ∂U ∩ ∂X has at least five elements. Let q ∈ Q(R 3 ) so that U q = dev µ (U ). We claim that if U = U , then q ∈ A. Note that if q (x 0 ) ≥ 0 held, then we would have q ∈ A. Thus, we assume, without loss of generality, that x 0 ∈ U .
It follows from (1) that each element of dev µ (∂U ∩ ∂X) either belongs to dev µ (∂U ∩ ∂X) or to the complement of M . If dev µ (∂U ∩ ∂X) were a subset of dev µ (∂U ∩ ∂X), then since dev µ (∂U ∩ ∂X) contains at least five points, Proposition 3.1 would imply that U = U . In other words, if U = U , then there exists x ∈ ∂U ∩ ∂X that does not belong to M . In particular, x ∈ U q . Since x and x 0 both belong to the convex set U q , there exists a path α : If U ∈ S 5 (X, µ) is a strip, then it follows from Propositions 7.4 and 7.5 that there exists a nontrivial planar translation τ such that τ Let w denote the distance between the boundary components of the strip U q . Let be the 'center-line' of U q , namely : Let A ⊂ Q(R 3 ) be the set consisting of all r such that for some x ∈ σ, we q(x) ≥ 0. Note that A is a closed subset of Q(R 3 ). Indeed, given a sequence {r n } ⊂ A, let x n ∈ σ such that r n (x n ) ≥ 0. Since σ is compact, a subsequence of x n converges to some x ∈ σ. Thus, if r n converges to r, then r(x) ≥ 0.
For each x ∈ σ, we have q(x) < 0, and hence q / ∈ A. Let U = U be a subconic such that ∂U ∩ ∂X contains at least five elements. Let q ∈ Q(R 3 ) be such that U q = dev µ (U ). It suffices to show that q ∈ A. Indeed, A is closed and q / ∈ A. An argument similar to the one given in the case of ellipse interiors shows that since q = q there exists x such that dist(x , U ) = and such that q(x) = 0. Since U q is convex, the set σ = ∩ U q is a compact connected set. If σ = ∅, then q (x) = 0 for all x ∈ , and so q ∈ A. Therefore, it suffices to assume that σ = ∅.
Let T be the convex hull of x and the endpoints of σ . Observe that T is a (perhaps degenerate) triangle. Since U q is convex, the triangle T is a subset of U q . Since σ is parallel to ∂U , we may use a similar triangles argument to show that the segment α = T ∩ ∂U has length |α| = |σ | · + w .
Proposition 7.7 leads us to make the following definition.
Definition 7.8. The elements of S 5 (X, µ) will be called rigid subconics.
The following proposition ensures that each polygonal 2-cell in the cellulation of S 3 (X, µ) has finitely many vertices. (See §9).
Proposition 7.9. Let V ⊂ X be open and nonempty. The set of all rigid subconics that contain V is finite.
Proof. By Proposition 7.7, it suffices to show that the set of all rigid subconics that contain V is compact. Without loss of generality, we may assume that (X, µ) is a universal covering of a precompact translation surface that has finite frontier.
Let U n be a sequence of rigid subconics containing V . Since V is nonempty and each U n is convex, the union W = n U n is star convex. Thus, the restriction of dev µ to W is injective.
Let · be a norm on the finite dimensional vector space Q(R 3 ). For each n, let q n be such that U qn = dev µ (U n ) and q = 1. Since the unit sphere is compact, the sequence q n has a limit point q. Without loss of generality, the sequence q n converges to q for otherwise take a subsequence. We have U q ⊂ dev µ (W ), and hence dev µ | −1 W (U q ) is a subconic in X. It suffices to show that U is rigid. We first claim that U q is either an ellipse interior or a strip. To prove this, we will eliminate the other possibilities: U q is empty, is the complement of a line, is a parabola, or is the plane. For each n, we have dev µ (V ) ⊂ U qn . Thus, since V is open and q n → q, we have V ⊂ U q and hence U q is nonempty. By Proposition 2.4, X has finite inradius R. Thus for each n, the inradius of U qn is at most R, and hence the inradius of U q is at most R. This eliminates the remaining cases. Define In particular, if n > N , then the set Z n = dev µ (∂U n ∩ ∂X) is a subset of M .
Suppose that U q is an ellipse interior. Since q n → q, Proposition 4.10 implies that-by passing to a tail if necessary-we may assume that for each n, the subconic U qn is an ellipse that lies in M 1 . In particular, dev µ (W ) ⊂ M 1 . Since M 1 is compact, ∂X is discrete, and dev µ | W is a homeomorphism onto its image, the set For each n, let x 1 n ∈ Z n . Since Z n ⊂ Z, there exists an infinite set A 1 ⊂ N and If U is a strip, then by Proposition 7.4, U belongs to a maximal strip U . For each > 0, we have Z n ⊂ M . Thus for each > 0, the set dev µ (U ) ⊂ M . Hence dev µ (U ) ⊂ dev µ (U ) and thus U = U . Thus, by Corollary 7.6, U is a rigid conic.

The homotopy type of the space of ellipse interiors
Let E(X, µ) denote the space of ellipse interiors in a translation surface (X, µ).
Proposition 8.1. The space E(X, µ) is homotopy equivalent to X.
Proof. Given an ellipse interior U in X, letc(U ) be the center of a lift of U toX. Define a map c : E(X, µ) → X by setting c(U ) = p(c(u)). Given a point x ∈ X, letD(x) be largest disk contained inX with center at somex ∈ p −1 (x). (Since X is an open manifold, a disk exists, and by Assumption 7.1 and Proposition 2.4 there is a largest such disk). Let D(x) be the orbit ofD(x) under Gal(X/X). It is straightforward to show that the maps D : X → E(X, µ) and c determine a homotopy equivalence.
Proof. We first define the retraction under the asumption that X is simply connected.
Let X be simply connected. Let E n (X, µ) denote the collection of ellipses U ⊂ X such that ∂X ∩ ∂U contains at least n points. The desired retraction is a concatenation of three retractions: from E(X, µ) to E 1 (X, µ), from E 1 (X, µ) to E 2 (X, µ), and from E 2 (X, µ) to E 3 (X, µ).
The first retraction consists of dilating the ellipse interior until the boundary meets ∂X. To be precise, let τ be the planar translation that sends the center of dev(U ) to the origin. Let q be the quadratic form such that U q = τ (dev(U )). Since the center of U q is the origin, we have . Note that the area of U qt increases to infinity at t tends to 1. By Assumption 7.1 and Proposition 2.4, the supremum, t 0 , of the set of t such that τ −1 U (U qt ) is the dev-image of a lift of a subconic is less than one. By setting we obtain a retraction f 1 : [0, 1] × E(X, µ) → E 1 (X, µ). The deformation retract from E 1 (X, µ) onto the E 2 (X, µ) is defined by dilating and translating the center while maintaining contact with ∂X. To be precise, given an ellipse U and x ∈ ∂U ∩ ∂X, let τ and q be as before. Let η be a linear form on R 3 defined as the differential of q at dev µ (τ (x)) In particular, η dev µ (τ (x)) = 0. Define It follows from Assumption 7.1 and Proposition 2.4 that the supremum, t 0 , of the set of t such that τ −1 U (U qt ) is the dev-image of a lift of a subconic is finite. By defining f 3 as in (2), we obtain a retraction f 2 : [0, 1] × E 1 (X, µ) → E 2 (X, µ).
The deformation retract from E 2 (X, µ) onto the E 3 (X, µ) is defined by dilating and translating the center while maintaining contact with two points in ∂X. To be precise, given an ellipse U and x 1 , x 2 ∈ ∂U ∩ ∂X, let τ and q be as before. Let η be a linear form on R 3 whose kernel is spanned by dev µ (τ (x 1 )) and dev µ (τ (x 2 )) and so that the value at the midpoint of these two vectors equals 1. Define q t as in (3). It follows from Assumption 7.1 and Proposition 2.4 that the supremum, t 0 , of the set of t such that τ −1 U (U qt ) is the dev-image of a lift of a subconic is finite. By defining f 3 as in (2), we obtain a retraction f 3 : [0, 1] × E 1 (X, µ) → E 2 (X, µ).
By concatenating f 1 , f 2 and f 3 , we obtain a retract f : E(X, µ) → E 3 (X, µ). Suppose thatX is the universal cover of a translation surface (X, µ). Let σ : X →X be a translation mapping. Since the translation τ associated to U differs from the translation associated to σ(U ) by σ, we find that f i (s, σ(U )) = f i (s, U ) for each U ∈ E i−1 . Since each covering transformation is a translation, the retraction f descends to a homotopy retraction from E(X, µ) onto E 3 (X, µ).
Corollary 8.4. E(X, µ) is simply connected if and only if X is simply connected.
The deck group Gal(X/X) acts onX as translations. In particular, Gal(X/X) acts on each stratum S n (X,μ). The action is free and discontinuous, and hence the natural projection map p * : S n (X,μ) → S n (X, µ) is a covering.
From henceforth we assume that X is connected.

A two dimensional cell complex
In this section we will assume that (X, µ) is the universal cover of a precompact translation surface with finite frontier. Here we show that the set, S 3 (X, µ) is naturally a 2-dimensional polyhedral complex. Moreover, S 4 (X, µ) is the 1-skeleton and S 5 (X, µ) is the 0-skeleton. To simplify the exposition, we will sometimes abbreviate S n (X, µ) by S n .
Given Z ⊂ ∂X, let S Z (X, µ) denote the set of all subconics U such that Z = ∂U ∩ ∂X and such that Z intersects each component of ∂U . Note that S Z (X, µ) is the set of all subconics such that Z ⊂ ∂U ∩ ∂X. We will abbreviate S Z (X, µ) by S Z if the context makes the choice of (X, µ) clear.
Abusing notation slightly, we will let dev µ denote the map from S(X, µ) to S(R 2 ) defined by U → dev µ (U ).
Lemma 9.1. If Card(Z) ≥ 2, then the restriction of dev µ to S Z is a homeomorphism onto its image in S(R 2 ).
Proof. Suppose that S Z = ∅ and let {z, z } ⊂ Z. Each U ∈ S Z contains the segment σ joining z and z . It follows that the union, W , of all U ∈ S Z is star convex with respect to, for example, the midpoint of σ. In particular, the restriction of the developing map to W is injective. Thus, dev µ determines an injection from S Z into E(R 2 ). Hence since dev µ defines the topology of S(X, µ), the restriction is a homeomorphism onto its image.
Let Z ⊂ ∂X. If there exists a convex subset P ⊂ X with nonempty interior such that dev µ (P ) equals the convex hull of dev µ (Z), then we will say that Z defines the polygon P . 5 We will say that a set of points A ⊂ X is noncollinear if and only if dev µ (A) is not a subset of some line.
Proposition 9.2. If Z ⊂ ∂X contains three noncollinear points and S Z (X, µ) = ∅, then Z defines a polygon.
Proof. Let U be a subconic such that Z = ∂U ∩ ∂X. By Proposition 7.2, the set U is convex. Thus, dev µ | U is injective. Since Z contains three noncollinear points, the convex hull H of dev µ (Z) has nonempty interior. The set P = dev µ | −1 U (H) is the desired set.
Remark 9.3. Note that S Z = ∅ does not imply that Z defines a polygon. For example, let Z be the intersection of ∂X and a single boundary component of a strip. On the other hand, the assumption that Z defines a polygon does not imply that S Z = ∅. For example, suppose that U is a subconic such that dev µ (Z) is the unit disc and dev µ (∂U ∩ ∂X) is the set of roots of z 6 − 1. (Here we have made the usual identification of R 2 with the complex plane). If Z is the set of roots of z 3 − 1, then S Z = ∅.
Proposition 9.4. If Z defines a polygon, then the closure S Z (X, µ) is compact.
Proof. By Proposition 7.2, we have By Proposition 5.4, the set E devµ(Z) (R 2 ) is compact. Thus, by Proposition 9.1 it . Each U n contains the interior of the polygon P defined by Z. In particular, the intersection n U n is nonempty, and therefore the union W = n U n is star convex. Therefore, the restriction dev µ | W is an injection and the image is the union dev µ (W ) = n dev µ (U n ).
Let q ∈ Q(R 3 ) such that U q = V . Since dev µ (U n ) converges to V , there exists sequence q n converging to q such that U qn = dev µ (U n ). If x ∈ R 2 and q(x) < 0, then there exists N such that for all n > N , we have q n (x) < 0. Thus, V ⊂ dev µ (W ) and U = dev µ | −1 (V ) is a subconic belonging to S Z (X, µ).
If Z defines a triangle, then W = dev µ (Z) is a noncollinear triple. Thus we can apply the results of §4. In particular, let (d 1 , d 2 , d 3 ) be an oriented negative natural basis for Q devµ(Z) and let T be the plane consisting of linear combinations i t i · d i with t i = 1. By Lemma 5.2 the restriction of q → U q to T is a homeomorphism onto S devµ(Z) . Let h : S devµ(Z) → T denote the inverse of this homeomorphism. Define Since dev µ and h are homeomorphisms onto their respective images, the map f is a homeomorphism onto its image. In general, if Z defines a polygon, then there exists a triple Z ⊂ Z that defines a triangle. Note that S Z ⊂ S Z and hence one may restrict the map f associated to Z to the set S Z .
Proposition 9.5. If Z defines a triangle, then f (S Z (X, µ)) is open in T .
Proof. Let x belong to the interior of the triangle P defined by Z. Let M be the maximal star convex neighborhood of x, and let D ⊂ R 2 be the set dev µ (M x ∩∂X).
Since ∂X is discrete, the set D is discrete. It follows that the set is open in Q devµ(Z) . By Proposition 4.10 and Lemma 5.1, Suppose that U ∈ S Z (X, µ). Since U ∩ ∂X = Z, the set dev µ (U ) does not intersect D\dev µ (Z). It follows that f (U ) ∈ W . Since Z contains only three points, it follows from Proposition 7.6 that f (U ) ∈ W . In sum, f (S Z (X, µ)) ⊂ W ∩W ∩T .
Let q ∈ W ∩ W ∩ T . Since q ∈ W , the set U q is convex, and therefore contains P . It follows that U q contains dev µ (x). In other words, q dev µ (x) < 0. Since Since the signature of q is (2, 0), the function γ z has a unique global minimum t 0 and is strictly increasing for t > t 0 . Thus, since q dev µ (x) < 0 and q(ẑ) ≥ 0, we Proposition 9.6. If Z ⊂ ∂X defines a polygon, then f (S Z (X, µ)) is convex.
Proof. Let U, U ∈ S Z (X, µ). Since U (resp. U ) is convex, U (resp. U ) contains the interior of the polygon defined by Z. In particular, U ∩ U = ∅. Therefore, the union U ∪ U is star convex, and hence the restriction of dev µ to U ∪ U is injective.
Proposition 9.7. Let Z define a polygon. The point f (U ) is an extreme point of f S Z (X, µ) if and only if U is a rigid subconic that belongs to S Z (X, µ).
Proof. Let U be a rigid subconic in S Z . Suppose to the contrary that f (U ) is not an extreme point. That is, assume that there exists V and V ∈ S Z such that f (U ) = 1 2 (f (V ) + f (V )). Since S Z is convex and the set, S 5 , of rigid subconics is discrete, we may assume that neither f (V ) nor f (V ) is a rigid subconic. It follows that there exists x ∈ ∂U ∩ ∂X such that x / ∈ V ∪ V . Thus, f (V )(x) > 0 and f (V )(x) > 0, and therefore f (U )(x) > 0. This contradicts the fact that x ∈ ∂U ∩ ∂X.
Suppose that U ∈ S Z (X, µ) is not a rigid subconic. In particular, letting Z = ∂U ∩ ∂X, we have Card(Z ) = 3 or 4. By Proposition 9.2, the setẐ is in general position. Thus, by Proposition 3.1, the vector space Q devµ(Z) has dimension 2 or 3. Thus, by Lemma 5.1, there exists a nontrivial linear family t → q t of quadratic forms in T such that U q0 = dev µ (U ) and dev µ (Z) ⊂ ∂U qt . By Lemma 9.5, there exists δ > 0 such that if |t| < δ, then U qt ∈ dev µ (S Z (X, µ)). Therefore, f (U ) is not an extreme point.
The following theorem summarizes the preceding material and provides the basis for the cell complex. In particular, if Z defines a triangle and S Z = ∅, then S Z is homeomorphic to a cell.
Theorem 9.8. Let Z ⊂ ∂X define a triangle. The map f is a homeomorphism from S Z (X, µ) onto a compact convex planar polygon with finitely many sides. Each side of f (S Z (X, µ)) equals f (S Z (X, µ)) where Z ⊂ ∂X defines a quadrilateral and Proof. By Proposition 9.6, the image f (S Z ) is convex. By Proposition 9.4 the set f (S Z ) is compact and hence closed. Therefore, by the Krein-Milman theorem and Proposition 9.7, the set f (S Z ) is the convex hull of the f -images of the rigid subconics that belong to S Z . Each such rigid subconic contains the nonempty interior of the polygon P Z . Therefore, by Corollary 7.9, the set of such subconics is finite. In sum, f (S Z ) is the convex hull of finitely many points in the 2-dimensional plane T .
Since f is a homeomorphism onto its image, Proposition 9.5 implies that f (S Z ) is the interior of the polygon f (S Z ). In particular, the boundary ∂S Z consist of those subconics U with Card(∂U ∩ ∂X) ≥ 4. Since the rigid subconics correspond to extreme points, each side corresponds to a subset Z ⊂ X with Card(∂U ∩∂X) = 4.
Remark 9.10. The cell complex S 3 (X, µ) is not a CW -complex. Indeed, the topology on S 3 (X, µ) does not give the weak topology with respect to the cells. If U is a rigid strip, then there exists a sequence of 1-cells S Zn and points U n ∈ S Zn such that U n converges U . The set {U n } is closed in the weak topology but not in the natural topology chosen here. On the other hand, Theorem 7.9 implies that the cell complex S 3 (X, µ) has the closure finiteness property.

Realizability and adjacency
In this section (X, µ) is the universal cover of a precompact translation surface with finite and nonempty frontier.
The cells of S 3 (X, µ) are indexed by the subsets Z ⊂ ∂X such that S Z (X, µ) = ∅ and Card(Z) ≥ 3. We will call such a set Z a realizable set. By Theorem 9.8, the inclusion of closed cells corresponds to the inclusion of realizable subsets of ∂X.
In this section, we define the notion of 'adjacency' and use it to characterize the subsets of a realizable set that are realizable.
Definition 10.1. Let U be a subconic. We say that x, y ∈ ∂U ∩ ∂X are adjacent in ∂U ∩ ∂X if and only if there exists a connected component of ∂U \ {x, y} that does not intersect ∂X.
Proposition 10.2. Let U be a subconic with Card(∂U ∩ ∂X) ≥ 3. Each x ∈ ∂U ∩ ∂X is adjacent to exactly two points in ∂U ∩ ∂X.
Suppose that Z is realizable with Card(Z) ≥ 5. If Z ⊂ Z is also realizable, then it follows from Proposition 3.1 that Card(Z ) = 3 or 4.
Proposition 10.3. Let Z ⊂ ∂X be realizable and let Z ⊂ Z with Card(Z ) = 4. The set Z is realizable if and only if there exists U ∈ S Z such that Z intersects each component of ∂U and there exists a partition of Z into pairs {x − , y − }, {x + , y + } such that each pair {x ± , y ± } is adjacent in ∂U ∩ ∂X.
Proof. We first note that the claim is true if Z = Z . Indeed, if Z = Z , then Card(Z) = 4, and hence each subconic U ∈ S Z is an ellipse interior. Thus, ∂U is homeomorphic to the unit circle, and each point in Z = ∂X ∩ ∂U is adjacent to exactly two other points in Z . In particular, there exists a partition of Z into adjacent pairs. Conversely, if Z = Z , then since Z is realizable, Z is realizable.
Thus, we may assume that Z = Z.
(⇒) If Z is realizable, then there exists U ∈ S Z . Since Z = Z , we have U = U . Let q ∈ Q(R 3 ) (resp. q ∈ Q(R 3 )) such that U q = dev µ (U ) and U q = dev µ (U ). Since U = U , the forms q and q do not belong to the same line. Let F = dev µ (Z). Consider the components of the complement ∂U q \ F . Since ∂U is a 1-manifold, the intersection C ∩ C of the closures of two components, C, C , is either empty or is a singleton. In the latter case, we will say that the components are 'adjacent'.
By Proposition 3.1, the zero locus of the restriction q | ∂Uq equals F . (Indeed, if it were not the case, then q and q would both vanish at 5 common points.) Thus, since q | ∂Uq is continuous, the restriction to each component q | C is either positive or negative. We claim that if C and C are adjacent, then the restrictions q | C and q | C have opposite signs.
Since Z is realizable, Z intersects each component of ∂U . We now identify the desired points {x ± , y ± }.
If U is a strip, then let + , − denote the two components. Since U ∈ S Z and Card(Z) = 4, we have Card(Z ∩ ± ) = 2. Indeed, otherwise U would contain three collinear points, and U would be a strip. But if U were a strip, then by Proposition 7.6, the set Z = ∂U ∩ ∂X would be infinite. Since ± is homeomorphic to a line, the complement ± / \ Z has exactly two unbounded components and one bounded component. The frontier of each unbounded component C is a singleton and hence U ∩ C = ∅. Thus, q is positive on the corresponding ray. Therefore, the form q is negative on the segment that corresponds to the bounded component, B ± ⊂ ± . Thus U ⊃ B ± , and we define {x ± , y ± } to be the frontier points of B ± .
If U is an ellipse interior, then ∂U is homeomorphic to the unit circle. Thus, ∂U \ Z has four components that we index with Z/4Z so that C i and C i+1 are adjacent for each i ∈ Z/4Z. For some i ∈ Z/4Z, we have that the restriction of q to arc corresponding to C i is negative, and hence the restriction of q to the arc corresponding to C i+2 is negative. In particular, U ⊃ C i ∪ C i+2 . Set {x − , y − } to be the frontier points of C i , and set {x + , y + } equal to the frontier points of C i+2 .
(⇐) Suppose that there exists U ∈ S Z such that Z admits a partition into two pairs of points adjacent in Z and such that Z intersects each component of ∂U . Let F = dev µ (Z) and F = dev µ (Z ). Let q ∈ Q F so that dev µ (U ) = U q .
Since Z intersects each component of ∂U , the set Z defines a polygon P . Since F = dev µ (Z ) is the set of extreme points of K = dev µ (P ), the set K is the intersection of four closed half planes. Thus R 2 \ K is the union of four open half planes. Since F = F and F ⊂ R 2 \ K, one of these open half planes, H 1 , intersects F . Let σ 1 be the side of K that is contained in the closure of H 1 . Let σ 3 be the side of P opposite to σ, and let H 3 be the open half plane whose closure H contains σ 3 . Let H 2 and H 4 be the other two half-planes.
Let p − and p + be the endpoints of σ 1 . Then p − , p + ∈ F . We claim that the corresponding points x − = dev µ | −1 U (p − ) and x + = dev µ | −1 U (p + ) are not adjacent in Z = ∂U ∩ ∂X.
If U q is a strip and x − and x + were adjacent, then p − and p + would belong to the same component of ∂U q . The component would coincide with the boundary of H 1 , and ∂U q would then intersect both components of R 2 \ . Since U q is a strip, could not be the boundary of U q . Thus, in this case, x − and x + are not adjacent. If U q is an ellipse interior, then the intersection ∂H i ∩ ∂U q consists of two points. Since H 1 ∩ ∂U q = ∅, the component C of ∂U q \ F that joins p − , p + ∈ ∂H 1 ∩ ∂U q belongs to H 1 . But C ∩ F = ∅ and so x − and x + are not adjacent in this case.
Since Z has a partition into two adjacent pairs, x + (resp. x − ) is adjacent to an endpoint y + (resp. y − ) of σ 3 . Let C + and C − denote the arcs with endpoints {x ± , y ± } such that ∂X ∩ C ± = ∅.
Let η i be a linear 1-form whose kernel contains the span of ∂H i and such that η i is negative on K • where K • is the interior of K. Then the quadratic form q = −η 1 ·η 3 belongs to Q F . Note that H 1 ∪ H 3 = (q ) −1 ((0, ∞)), and q is negative on C ± .
For t ∈ R, define If t ∈ [0, 1], then U qt ⊂ U q ∩ U q . Indeed, if q t (x) < 0, then either q(x) < 0 or q (x) < 0. Let x ∈ K • , and let M ⊂ X be the maximal star convex neighborhood of x. Since U is convex and U q = dev µ (U ), we have U q ⊂ dev µ (M ).
Therefore since t → q t is continuous and D is discrete, there exists ∈ (0, 1) such that if |t| < , then q −1 t ((−∞, 0]) ∩ D = F . We claim that there exists > 0 such that if t ∈ [0, ), then U qt is an ellipse interior. If U q is an ellipse, then this follows from Proposition 4.11 and the continuity of t → q t .
Suppose that U q is a strip. Then q| Uq is bounded from below. Indeed, q = η + · η − where η − are linear 1-forms. Since kernel, ± , of η ± is parallel to the kernel, ∓ , of ± , it follows that the form η ± | Uq is bounded. Hence q| Uq > −N for some N > 0. Let v belong to the kernel of η ± . For each p ∈ R 2 define f p (s) = q (p + s · v). Note that f p is a quadratic polynomial and we claim that f p is nontrivial with leading coefficient positive. Since Card(U q ∩ U q ) = 4, each line in q −1 {0} intersects (q ) −1 {0}. It follows that η i (v) = 0 for i = 1 or 3, and hence f p (s) = (η 1 (p) + s · η 1 (v)) · (η 3 (p) + s · η 3 (v)) is nonconstant. Since q | σi ≡ 0 and q | K 0 ≤ 0, the leading coefficient of the quadratic polynomial f p is positive for each p ∈ σ 1 . Since σ 1 is compact for each t, there exists N t such that if |s| > N t and p ∈ σ 1 , then q t (p + s · v) > (N + 1)/t. It follows that U qt ∩ U q is bounded for each t > 0. Since U q ∩ U q is bounded, there exists ∈ (0, 1) so that if |t| < , then U qt ∩ U q is bounded. Thus, for t ∈ (0, ), we have that U qt is bounded, and hence by Proposition 4.9, the subconic U q is an ellipse interior.
Fix t ∈ (0, min{ , }). We have U qt \ U q ⊂ U q ⊂ dev µ (M ). Thus, it suffices to show that U qt ∩ U q ⊂ dev µ (M ). For then dev| −1 M (U qt 0 ) ∈ S Z . If y ∈ U qt ∩ U q , then y / ∈ D. Since U qt is convex, the line segment σ joining y and dev µ (x) belongs to U qt . Therefore, D∩σ = ∅, and it follows that y ∈ dev µ (M ).
Proposition 10.4. Let Z ⊂ ∂X be realizable and Z ⊂ Z with Card(Z ) = 3. The triple Z is realizable if and only if there exists U ∈ S Z and a pair {x, x } ⊂ Z that is adjacent in ∂U ∩ ∂X.
Proof. The proof is similar to the proof of Proposition 10.3. We leave it to the reader.

Orientation and succession
In this section, we first observe that each 2-cell of S(X, µ) has a canonical orientation. We then reinterpret this orientation in terms of succession, the natural refinement of adjacency. We will assume that (X, µ) is the universal covering of a precompact translation surface with nonempty and finite frontier.
Let Z ⊂ X be a realizable triple, and thus, in particular, dev µ (Z) is noncollinear. Let d is an oriented negative natural basis for Q devµ(Z) , and let r d : T d → S devµ(Z) be the homeomorphism defined in §5. Recall that the plane T d has a canonical (outward normal) orientation, and hence the homeomorphism r d induces an orientation on S Z . By Proposition 5.3, this orientation does not depend on the choice of oriented negative natural basis. As a result, the cell S Z has a canonical orientation.
The orientation of S Z induces an orientation of its boundary ∂S Z . In particular, it induces a cyclic ordering of the 1-cells lying in the boundary of S Z . These 1-cells are in one-to-one correspondence with the set, ∂Z, of realizable quadruples Z that contain Z.
Definition 11.1. Let Z , Z ∈ ∂Z. We will say Z follows Z if and only if S Z immediately follows S Z in the canonical ordering of ∂Z.
The cyclic ordering has an alternate description that uses the oriented refinement of adjacency. In particular, if Z follows Z , then Z and Z share a vertex and this vertex corresponds to the unique rigid conic U such that Z ∪ Z ⊂ ∂U ∩ ∂X. In Proposition 11.6 below, we reinterpret 'following' in terms of an ordering of ∂U ∩ ∂X.
Definition 11.2. Suppose that Card(∂U ∩ ∂X) ≥ 3. Given adjacent points x, y ∈ ∂U ∩ ∂X, let C be the unique component of ∂U \ ∂X such that {x, y} ⊂ C. If there exists an oriented path α : [−1, 1] → C such that α(−1) = x and α(+1) = y, then we say that y is the successor of x in ∂U ∩ ∂X, and we write s U (x) = y.
Proposition 11.3. The map s U is a permutation of ∂U ∩ ∂X.

Proof. Straightforward.
In the remainder of this section, we will assume that U is a rigid subconic. That is, we assume that Card(∂U ∩ ∂X) ≥ 5.
Notation 11.4. Given x, x ∈ ∂U ∩ ∂X, set If x = x and {x, x } are nonadjacent, then Z(x, x ) is a quadruple, and hence by Proposition 10.3, this quadruple is realizable. Conversely, if Z is realizable and U ∈ S Z , then Z can be partitioned into adjacent pairs {x − , y − } and {x + , y + }. For each pair either s U (x ± ) = y ± or s U (y ± ) = x ± , but not both. Thus, (x, y) → Z U (x, y) is a bijection from the set of nonadjacent pairs in ∂U ∩ ∂X onto the realizable quadruples Z ⊂ ∂U ∩ ∂X.
Definition 11.5. We will say that Z ⊂ ∂U ∩ ∂X is consecutive in ∂U ∩ ∂X if and only if there exists x ∈ Z and k ∈ Z + such that Z = {x, s U (x), . . . , s k U (x)}. Let x ∈ Z. If for all y ∈ Z, we have s U (y) = x and s U (x) = y, then we will say that x is Z-isolated in ∂U ∩ ∂X.
Let Z ⊂ ∂U ∩ ∂X be a realizable triple. Then either Z is consecutive, namely, Z = {x, s 2 U (x), s 3 (x)} for a unique x ∈ Z, or there exists y ∈ Z that is isolated, that is Z = {x, s U (x), y} for a unique x ∈ Z.
Proposition 11.6. Let U be a rigid subconic and let Z = ∂U ∩ ∂X be a realizable triple.
We first note that the ordered triple (v 0 , v 1 , v 2 ) is cyclically ordered with respect to the standard orientation of R 2 . Indeed, Z belongs to a single component of ∂U , and thus, since Z is realizable, ∂U has only one component. In particular, U is an ellipse interior. It is then straightforward to construct a homotopy of piecewise smooth embeddings f t : R/Z → R 2 such that f 0 is an oriented parametrization of dev µ (∂U ), f 1 is a parametrization of the boundary of the convex hull of {v 0 , v 1 , v 2 }, and f t ([i/3]) = x i for all t ∈ [0, 1] and i = 0, 1, 2.
We claim that a i = q for i = 1, 2. Since the set {x, s(x), s 2 (x)} is consecutive, this set is a subset of one boundary component of ∂U . Thus, since {x, s(x), s 2 (x)} is realizable, U is an ellipse interior and q is nondegenerate. Therefore since a i is degenerate, a i = q.
Define q t = (1−t)·q+t·d 1 . We claim that for all t ≥ 0, the quadratic form q t does not lie in . Recall that d 1 = −η 0,1 · η 1,2 . The pointsv 2 ,v 3 , andv −1 lie in the same component of R 2 \ v 0 ,v 1 , and hence η 0,1 (v 3 ) > 0 and η 0, By the choice of orientation of T , we have that (d 2 − d 1 , d 0 − d 1 ) is an oriented basis for R 2 . Since t 0 , t 1 ∈ (0, 1), the ordered pair (a 1 − d 1 , a 0 − d 1 ) is an oriented basis for R 2 . If t ≥ 0, the quadratic form q t does not lie in the line containing a 0 and a 1 . Therefore, for t ≥ 0, the set {a 1 − q t , a 0 − q t } is a basis and by continuity, (a 1 − q, a 0 − q) is an oriented basis. Thus, f (S Z(x,s 2 (x)) ) follows f (S Z(s −1 (x),s(x)) ).
The proof of the other claim is similar.

The link of a vertex
Recall that if v is a vertex in a 2-dimensional cell complex, then the link of v, Lk(v), is the abstract graph defined as follows: The vertices are the 1-cells that contain v. Two 1-cells C, C ⊂ ∂U ∩ ∂X are joined by an edge iff there exists a 2-cell D such that v ∈ ∂C ∩ ∂C and C ∪ C ⊂ ∂D.
In this section, we study the link Lk(U ) of a rigid conic U in S 3 (X, µ). The vertices of Lk(U ) may be regarded as realizable quadruples Z ⊂ ∂U ∩ ∂X. Two quadruples Z and Z are joined by an edge if and only Z ∩ Z is a realizable triple. The orientation of the 2-cell S Z∩Z determines a direction of this edge. Hence, Lk(U ) is naturally a directed graph.
We will assume throughout this section that (X, µ) is the universal covering of a precompact translation surface with nonempty and finite frontier.
By Proposition 10.3, the set of 1-cells that contain U is in bijection with the set of unordered pairs {x, y} of non-adjacent points in ∂U ∩ ∂X. If U is an ellipse, then the convex hull of ∂U ∩ ∂X is an n-gon with n = Card(∂U ∩ ∂X) < ∞. The non-adjacent vertices define 'diagonals' in the n-gon. The classic count of diagonals gives the following.
To simplify the notation in what follows, we will write x + k in place of s k U (x). Proposition 12.2. Each vertex of Lk(U ) has degree 4. If the quadruple Z(x, y) is consecutive, then the neighbours of Z(x, y) are If the quadruple Z(x, y) is not consecutive, then the neighbours of Z(x, y) are Proof. A quadruple Z is a neighbour of Z U (x, y) if and only if the intersection Z ∩ Z U (x, y) consists of three points. If |x − y| = 2, then either Z U (x, y) = {x, x + 1, x + 2, x + 3} or Z U (x, y) = {y, y + 1, y + 2, y + 3}. Without loss of generality, the former holds. In this case, the possibile intersections are {x, x + 1, x + 2}, {x + 1, x + 2, x + 3}, {x, x + 2, x + 3}, and {x, x + 1, x + 3}. By Proposition 11.6, the only other quadruple that contain these triples are listed in (5).
We next consider the 3-cycles in the undirected graph Lk(U ).
Suppose that Z U (x, y) is not consecutive with respect to U . If a 3-cycle contains Z U (x, y), then either x = y + 3 and the cycle is {Z(x, y), Z(x, y + 1), Z(x − 1, y)} or y = x + 3 and the cycle is Then by Proposition 12.2, the neighbours of Z are Z(x − 1, x + 2), Z(x − 1, x + 1), Z(x, x + 3), and Z(x + 1, x + 3). Inspection shows that the first two intersect in 3 points and the last two intersect in three points. No other pairs intersect in three points. Thus, we have exactly two 3-cycles containing A. Suppose that Z = Z(x, y) is not consecutive. Then by Proposition 12.2 the neighbours of Z(x, y) are Z(x + 1, y), Z(x, y + 1), Z(x − 1, y) and Z(x, y − 1). Since Z(x, y) is not consecutive, we have x = y − 2, y − 1, y, y + 1, y + 2. Since Card(∂U ∩ ∂X) ≥ 5 we have x = x − 2, x − 1, x, x + 1, x + 2. Inspection shows that In sum, if y + 2 = x − 1 and y + 2 = x − 1, then Z(x, y) does not belong to a 3-cycle. If y + 2 = x − 1 or y + 2 = x − 1 but not both, then Z(x, y) belongs to exactly one 3-cycle. Finally, y + 2 = x − 1 and y + 2 = x − 1 if and only if Z(x, y) belongs to exactly two 3-cycles.
In particular, if Z(x, y) is nonconsecutive and belongs to two 3-cycles, then x = x + 6 and hence Card(∂U ∩ ∂X) = 6.
Corollary 12.4. Each 3-cycle contains at most one non-consecutive vertex.
Corollary 12.5. If U is a maximal strip, then Lk(U ) has no 3-cycles.
Proof. Indeed, if a quadruple Z is consecutive, then Z is a subset of one component of the boundary of the strip. Hence Z is collinear and is not realizable.
Proposition 12.6. Let {Z, Z , Z } be a 3-cycle. Then Z is consecutive if and only if either both Z and Z follow Z or Z follows both Z and Z .
Next we turn to the analysis of Lk(U ) in the case that U is a maximal strip. First recall that the n-dimensional infinite grid, Grid n , is the graph whose vertex set is Z n and two vertices m and n are joined by an edge iff m and n differ in exactly one coordinate and this difference has absolute value 1. This graph can be geometrically realized as the set of points in R n all but one of whose coordinates equals an integer. Any graph that is isomorphic to the n-dimensional infinite grid will be called an n-dimensional grid graph.
Proposition 12.7. If U ⊂ X is a maximal strip, then Lk(U ) is a 2-dimensional grid graph. In particular, if x and y are points in the frontier of X that belong to distinct components of ∂U , then f (m, n) = Z(x + m, y + n) defines an isomorphism f : Grid 2 → Lk(U ).
Proof. Let x (resp. y ) denote the component of ∂U containing x (resp. y.) For each x ∈ x (resp. y ∈ y ) there exists a unique m (resp. n) such that x = x + m (resp. y = y + n). It follows that f is a bijection. Since U is a strip, each 1-cell is non-consecutive. Therefore (6) implies that f is a graph isomorphism.
We will say that a subset L of a grid graph G is a line if and only if L is isomorphic to a 1-dimensional grid and no two consecutive edges of L belong to a cycle of order 4. Let L(G) denote the set of lines in G. The lines in the infinite grid correspond exactly to the lines in R n that are contained within its geometric realization.

The map of frontiers
Let (X, µ) and (X , µ ) be universal covers of precompact translation surfaces with ∂X finite and nonempty. Let Φ : S 3 (X, µ) → S 3 (X , µ ) be an isomorphism of 2-complexes. In this section, we will assume that Φ preserves the natural orientation of each 2-cell. Such an isomorphism will be called orientation preserving.
The goal of this section is to prove the existence of a bijection β : ∂X → ∂X so that for each cell in the 2-complex S 3 (X, µ) we have We will reduce the construction of β to a 'local' problem associated to a rigid subconic U . In particular, we say that a bijection β U : ∂U ∩ ∂X → ∂Φ(U ) ∩ ∂X is adapted to Φ if and only if for each realizable Z ⊂ ∂U ∩ ∂X we have The following proposition tells us that the 'global' construction of β is equivalent to the 'local' problem of constructing β U adapted to Φ for each rigid conic U .
Proof. By Proposition 9.11, it suffices to suppose that U and U are endpoints of a 1-cell in S 3 (X, µ). In particular, Z = ∂U ∩ ∂U is a quadruple {x 1 , x 2 , x 3 , x 4 }. Let Z i be the triple Z \ {z i }. The triple Z i is realizable, and hence Notation 13.2. Since each cell is determined by a subset of the frontier, the map Φ may be regarded as a map from certain subsets of ∂X to subsets of ∂X . Abusing notation slightly, we will sometimes use Φ to denote this mapping of subsets. For example, if Z ⊂ ∂X is a realizable triple and S Z is the associated 2-cell, then we will let Φ(Z) denote the triple that determines the image 2-cell Φ(S Z ).
We first define β U in the case that U is a maximal strip. Let x ∈ ∂U ∩ ∂X. The isomorphism Φ determines a graph isomorphism of Lk(U ) onto Lk(Φ(U )). Since Lk(U ) is infinite, Lk(Φ(U )) is infinite, and hence Φ(U ) is a maximal strip. Since U and Φ(U ) are maximal strips, each quadruple Z ⊂ ∂U ∩ ∂X with S Z = ∅ is of the form Z U (x, y) where x and y belong to different components of ∂U . Let x denote the component containing x and let ⊥ x denote the component that does not contain x.
Given x ∈ ∂U ∩ ∂X, let L x be the line in Lk(U ) consisting of 1-cells Z U (x, y) with y ∈ ⊥ x . The image of L x is a line in Lk(Φ(U )) and hence there exists unique Proposition 13.3. Let U be a maximal strip. The map β U defined in (9) is adapted to Φ. Moreover, Proof. In this proof, we will suppress the subscripts 'U ' and 'Φ(U )' since they will be clear from the context. Let y ∈ ⊥ x . Then Z(x, y) is a 1-cell that belongs to L x , and hence Φ(Z(x, y)) belongs to the line L β(x) . In particular, there exists y ∈ ⊥ β(x) such that Φ(Z(x, y)) = Z(β(x), y ). Note that Z(x, y) = Z(y, x) and hence Φ(Z(y, x)) = Z(β(y), x) for some x ∈ ⊥ β(y) . Hence since Z(β(x), y ) = Z(y , β(x)), we have (11) Z(β(y), x ) = Φ(Z(y, x)) = Z(y , β(x)).
Lemma 13.4. A quadruple Z is consecutive with respect to the rigid conic U if and only if Φ(Z) is consecutive with respect to Φ(U ).
Proof. If Z is consecutive, then exactly two 3-cycles contain Z. Let {Z, Z , Z } be such a 3-cycle. By Proposition 12.6, either both Z and Z follow Z or Z follows Z and Z . Since Φ is orientation preserving, we have that either both Φ(Z ) and Φ(Z ) follow Φ(Z) or Φ(Z) follows Φ(Z ) and Φ(Z ). Hence by applying Proposition 12.6 to {Φ(Z), Φ(Z ), Φ(Z )} we find that Φ(Z) is consecutive.
The converse follows by considering Φ −1 .
We summarize this section with the following.
Theorem 13.6. If Φ : S 3 (X, µ) → S 3 (X , µ ) is an orientation preserving cell complex isomorphism, then the exists a unique bijection β : ∂X → ∂X such that for each realizable Z ⊂ ∂X we have Moreover, for each rigid conic U ⊂ X, we have

Configurations, oriented bisectors, and eigenlines
In this section we will prove a geometric lemma that will be used in the following section to prove Theorem 1.1. We will assume throughout that (X, µ) is the universal covering of a precompact translation surface with nonempty and finite frontier.
Definition 14.1. We will say that two subconics in the plane are equivalent up to homothety if and only if they differ by a homothety and/or a translation. Let [U ] denote the equivalence class of the subconic [U ].
Two (non-circular) ellipse interiors are equivalent if and only if their major axes are parallel and they have equal eccentricities. 6 A configuration of subconics about an ellipse consists of an ellipse interior U , a finite subset Z ⊂ ∂U with Card(Z) ≥ 5, and for each nonconsecutive pair x, y ∈ Z, a subconic U x,y ∈ E(R 2 ) such that Z ∩ ∂U x,y = Z(x, y).
Lemma 14.2 (Geometric Lemma). Let (U, Z, U x,y ) and (U , Z , U x,y ) be two configurations of subconics about an ellipse. If [U ] = [U ] and there exists a bijection β : Z → Z such that β • s U = s U • β, and for each nonconsecutive pair x, y ∈ Z we have [U x,y ] = U β(x),β(y) , then for each x ∈ Z, the line (x, s(x)) is parallel to the line (β(x), β(s(x))). 6 The referee has informed us that some refer to this data as the 'complex dilatation'.

U
x 2 x 3 x 4 x 5 The remainder of this section is dedicated to proving Lemma 14.2. We first discuss the geometry of eigenvectors for degenerate quadratic forms in R 2 . Then at the end of the section, we use this discussion to prove Lemma 14.2.
Let q ∈ Q(R 2 ) be a quadratic form of signature (2, 0). Let SO(q) denote the group of orientation preserving linear transformations that preserve q. This group acts transitively and freely on the 'ellipse' q −1 (1), and thus SO(q) is homeomorphic to a circle. There exists a unique orientation preserving universal covering map e : R → SO(q) such that e is a group homomorphism and 2π = inf{θ > 0 | e(θ) = Id}. Of course, one should be thinking of the example q(x 1 , x 2 ) = x 2 1 + x 2 2 in which case e(θ) = cos(θ) sin(θ) − sin(θ) cos(θ) .
Let (x, y) be an ordered pair of distinct vectors belonging to q −1 (r 2 ), the q-circle of radius r. Let θ = inf {θ > 0 | e(θ ) · x = y} , Note that v xy = −v yx and hence α vw = −α wv . We will refer to v xy as the oriented q-bisector of (x, y).
Definition 14.4. Let F = (x 0 , . . . , x n ) ⊂ q −1 (r 2 ) be an ordered n-tuple of distinct points. We will say that the (ordering of ) F is compatible with e if and only if there exist 0 = θ 0 < · · · < θ n ≤ 2π such that For example, let x, y be distinct points in q −1 (r 2 ). The ordered triple (x, v xy , y) is compatible with e, but (x, v yx , y) is not.
Let q(·, ·) denote the polarization of the quadratic form q. For each ordered pair (x, y) of distinct points in q −1 (r 2 ), define the linear form α xy that is q-dual to v xy by (16) α xy (w) = q(w, v xy ).
Proposition 14.5. The vector v x2x3 − v x0x1 belongs to L + Q .
Proof. The polarization of q Q is given by Thus, using (16) and the fact that q(v x0x1 ) = r 2 = q(v x2x3 ), we find that for each Thus, by definition, v x2x3 − v x0x1 is an eigenvector with eigenvalue r 2 − q(v x2x3 − v x0x1 ). Since, by assumption, v x2x3 = v x0x1 , this eigenvalue is positive.
x 1 x 2 x 3 x 4 x 5 x Figure 4. The geometric content of Proposition 14.7: The directions of the bisectors (↔) of successive diagonals determine the directions of the dotted lines. The direction of the bisector of x i x i+2 and x i+1 x i+3 is determined by the average of the four angles, (θ i + θ i+1 + θ i+2 + θ i+3 )/4. The direction of the dotted line x i x i+1 is determined by (θ i + θ i+1 )/2. We now apply the preceding discussion to obtain an intermediate form of the geometric lemma. Let Z be a finite subset of q −1 (r 2 ) that contains at least five points. Let s : Z → Z be the successor function associated to the counter-clockwise orientation. In particular, s(x) = y if and only if there exists θ > 0 such that e(θ · x) = y and e((0, θ ) · x) ∩ Z = ∅.
Recall that, given two nonsuccessive points in Z, the symbol Z(x, y) denotes the ordered quadruple (x, s(x), y, s(y)).
Proposition 14.7. Let Z and Z be finite subsets of the unit circle ∂U q0 each with cardinality at least five. If there exists a bijection β : Z → Z such that β • s = s • β and for each pair of nonsuccessive vertices x, y ∈ Z, we have Since Card(Z) ≥ 5, we have Similarly, we obtain so that e(θ i ) = s i (β(x)).
It follows from the hypothesis that By taking the alternating sum of the left hand sides of these equations and the alternating sum of the right hand sides, we find that The claim follows.
To prove the main geometric lemma, we shift the context to quadratic forms on R 3 . In particular, q will denote an element of Q(R 3 ) and q will denote the 'restriction' to the first two coordinates q(x 1 , x 2 ) = q(x 1 , x 2 , 0). Proposition 14.8. Let q ∈ Q(R 3 ). If q has signature (2, 1) and q has signature (2, 0), then there exists a (unique) homothety h and a (unique) translation τ so that h • τ (∂U q ) = q −1 (1).
where a = (a 13 , a 23 ) and A = a 11 a 12 a 12 a 22 .
Proof of Lemma 14.2. Note that the set of weights on a configuration is invariant under homotheties and translations. 7 Also note that parallelism is invariant under homotheties and translations. In particular, without loss of generality, we have U = U . Since U is an ellipse interior, there exists a quadratic form q ∈ Q(R 3 ) of signature (2, 1) such that U q = U and such that q has signature (2, 0). By Proposition 14.8, there exists a homothety h and a translation τ such that h • τ (U ) = q −1 (1). It follows that without loss of generality, U = q −1 (1). Let x, y ∈ Z be distinct, nonsuccessive points. We will consider three quadratic forms that belong to the plane, Q Z(x,y) , consisting of quadratic forms that vanish on the quadruple Z(x, y).
The first form is q.
The second form is a quadratic form r xy such that U rxy = U xy . Note that since ∂U xy ∩ Z = Z(x, y), we have r xy ∈ Q Z(x,y) . Since Card(Z) ≥ 5 and U xy = U , the restriction of r xy is negative on the arc in ∂U that joinsx to s(x).
The third quadratic form is a degenerate form defined as follows. Let η xy ∈ (R 3 ) * be a linear form such that η xy (x) = 0 = η xy ( s(x)) and η xy is negative on the interior of the convex hull of Z(x, y). Let η xy ∈ (R 3 ) * be such that η xy (ŷ) = 0 = η xy ( s(y)) and η xy is negative on the interior of the convex hull of Z(x, y). The third quadratic form is defined by p xy = −η xy · η xy . Note that p x,y ∈ Q Z(x,y) and p xy is positive on the arc in ∂U that joinsx to s(x).
Note that all three forms are nonpositive on the convex hull of Z(X, Y ). Thus, since q vanishes on the arc in ∂U that joinsx to s(x) and dim(Q Z(x,y) ) = 2, there exist positive a, b ∈ R + such that (20) q = a · r x,y + b · p x,y .
Since r xy is not a multiple of q (resp. p x,y is not a multiple of q), the form r x,y (resp. p x,y ) has two distinct eigenvalues µ − < µ + (resp. ν + < ν−) with respect to q. Let M ± x,y (resp. N ± x,y ) denote the eigenspace of r x,y (resp. p x,y ) associated to µ ± (resp. ν ∓ ). It follows from (20) that M ± x,y = N ∓ x,y and b · ν ± = 1 − a · µ ∓ .
The discussion above applies equally well to a pair of nonadjacent distinct points x , y ∈ Z . We define forms, r x ,y , p x ,y , and eigenspaces, M ± x ,y = N ∓ x ,y , in an analogous fashion.

Isomorphisms and affine homeomorphisms
In this section, we prove Theorem 1.1. We begin with a lemma that reduces the construction of a homeomorphism to a local problem.
Proof. Let Z, Z ⊂ ∂X be realizable quadruples such that Card(Z ∩ Z ) = 3. For each x ∈ dev µ (Z ∩ Z ), we have ψ Z (x) = ψ Z (x). Since each element of Aff + (R 2 ) is determined by its values at three points, we have ψ Z = ψ Z . By Proposition 9.11, the graph S 4 (X, µ) is connected, and hence the function Z → ψ Z is constant. Let ψ ∈ Aff + (R 2 ) denote this constant value.
As a first application, we have the following.
Proof. Let β : ∂X → ∂X be the bijection that is compatible with Φ.
Each realizable quadruple Z ⊂ ∂X is the intersection of the boundaries of two strips. In particular, dev µ (Z) is the vertex set of a parallelogram. Since Φ is an isomorphism, each rigid subconic U ∈ S 5 (X, µ) is also a strip. Thus, it follows that dev µ (β(Z)) is also the vertex set of a parallelogram. Since β is orientation preserving, there exists a ψ Z ∈ Aff + (R 2 ) such that ψ Z (dev µ (z)) = dev µ (β(z)). The claim now follows from Lemma 15.1.
Recall that the developing map dev µ is determined up to post-composition by translations. In particular, each subset A ⊂ X determines a unique class [A] := [dev µ (A)].
We will say that a pair {x, y} ∈ ∂X is realizable if and only if there exists a subconic U such that {x, y} = ∂U ∩∂X. Note that since ∂X is discrete, realizability of a pair is equivalent to the existence of a geodesic segment joining x and y. Indeed, given a geodesic segment σ joining x and y and > 0, there exists an ellipse interior U such that {x, y} ⊂ ∂U and the Hausdorff distance between σ and U is less than .
If U is an ellipse, then Φ(U ) is an ellipse, and the claim then follows from Lemma 14.2.