aa r X i v : . [ m a t h . M G ] F e b TRIANGULATING METRIC SURFACES
PAUL CREUTZ AND MATTHEW ROMNEY
Abstract.
We prove that any length metric space homeomorphic to a sur-face may be decomposed into non-overlapping convex triangles of arbitrarilysmall diameter. This generalizes a previous result of Alexandrov–Zalgaller forsurfaces of bounded curvature. Introduction
Main results.
The theory of surfaces of bounded curvature was developedbeginning in the 1940s as a generalization of two-dimensional Riemannian geometry.One of the central results of this theory is that any surface of bounded curvature isthe limit of two-dimensional Riemannian manifolds of uniformly bounded integralcurvature. A key step in the proof of this result is to show that every surface ofbounded curvature admits a triangulation by convex geodesic triangles of arbitrarilysmall diameter. Although versions of the approximation and triangulation theoremsappeared earlier in works of Alexandrov [1] and Zalgaller [28], complete proofs haveonly been published in their monograph [2]. We also refer the reader to surveys byFillastre [16], Reshetnyak [25] and Troyanov [26] for an overview of the subject.While surfaces of bounded curvature remain an active research topic (see forinstance [3, 4, 6, 7, 9, 14, 15, 17]), various classes of metric surfaces that do notfall into this setting have also been widely studied in recent years. These includereversible Finsler surfaces [5, 11, 12, 23], minimal surfaces in spaces satisfying aquadratic isoperimetric inequality [13, 18, 19], metric minimizing disks [22], Ahlfors2-regular quasispheres [8], quasiconformal images of planar domains [24], and fractalspheres [10].In this paper, we generalize the theorem of Alexandrov–Zalgaller on the existenceof triangulations (see Theorem III.2 in [2]) to the case of arbitrary geodesic surfaces.In its simplest version, our result is the following.
Theorem 1.1.
Let X be a geodesic metric space homeomorphic to a closed surfaceand ε > . Then X may be decomposed into finitely many non-overlapping convextriangles, each of diameter at most ε . Here, a triangle is a subset of X homeomorphic to the closed disk whose boundaryis the union of three geodesics. We remark that, like the corresponding result forsurfaces of bounded curvature [2, Thm. III.2, p.59], our Theorem 1.1 does not givea triangulation of X in the classical sense. The difference is that we do not requireadjacent triangles to intersect along entire edges.One step in the original proof by Alexandrov–Zalgaller of the existence of tri-angulations is to show that any point in a surface of bounded curvature can beenclosed by an arbitrarily short piecewise geodesic curve. As noted in [2] and [25],this is the only step that relies on the assumption of bounded curvature. Simple Both authors were partially supported by DFG-grant SPP 2026, and the first-named authoralso by DFG-grant SFB/TRR 191 “Symplectic structures in Geometry, Algebra and Dynamics.”
Primary 53C45. Secondary 52A10, 53A05.
Keywords. triangulation, surfaces of bounded curvature, Alexandrov geometry. examples show that this property does not hold for general geodesic surfaces, andindeed it is difficult to prove even for surfaces of bounded curvature using the def-inition in [2]; see [2, Sec. III.5] and [25, p.81]. Instead, we give a relatively shortargument showing that every point has a neighborhood that may be covered byfinitely many polygons, each of arbitrarily small perimeter; see Lemma 5.2. Thusour approach also simplifies the original proof even in the bounded curvature case.In principle, except for this difference, we are able to follow the proof given in [2]for surfaces of bounded curvature. However, this proof contains several technicalerrors. These errors are related to the fact that geodesics at the present level ofgenerality can be highly non-unique and hence intersect in complicated ways. Tohandle this issue, Alexandrov–Zalgaller consistently use the notion of what theycall systems of geodesics without superfluous intersections . It turns out that theprinciple they use to pass to such systems is not valid in general, even in thebounded curvature setting; see the discussion in Section 4. Since this principle isapplied at numerous places, we choose to give a complete self-contained proof ofTheorem 1.1. In particular, Lemma III.6 in [2] is not correct, and this portion ofthe proof requires a more refined approach; see Sections 5.4 and 5.5.In the bounded curvature setting, the proof of approximation by Riemannian2-manifolds requires additional technical conclusions beyond those given in Theo-rem 1.1; compare [2, Thm. III.3, p.61]. Our proof equally allows for these conclu-sions, and thus we now give the following general version of our main result.
Theorem 1.2.
Let X be a length metric space homeomorphic to a surface suchthat every boundary component of X is a piecewise geodesic curve, and let ε > .Then X may be covered by a locally finite collection of non-overlapping triangles ( T i ) i ∈I such that the following hold for each i ∈ I .(i) The triangle T i is convex relative to its boundary.(ii) The diameter of T i is at most ε .(iii) The triangle T i is non-degenerate.(iv) ∂T i \ ∂X consists of transit points. Here, a triangle T is called non-degenerate if the corresponding Euclidean com-parison triangle is non-degenerate, or equivalently if all triangle inequalities for thesides of T are strict. See Sections 2 and 3 for definitions of the other terms usedin the statement of Theorem 1.2. Both in our paper and in [2], the conclusion (iii)that the triangles T i are non-degenerate can be achieved a posteriori by showingthat any degenerate triangle is decomposable into non-degenerate triangles; see [2,Lem. III.7, p.60] and Proposition 6.1 below. However, the proof of this fact in [2]relies heavily on the assumption of bounded curvature, and hence an original argu-ment is needed to obtain Proposition 6.1. This is the only step where the proof forsurfaces of bounded curvature turns out to be much simpler than the general case.Note that Theorem 1.2 also applies to non-compact surfaces, and that in this casea localized version of conclusion (ii) is possible. See Remark 5.5 below. The con-clusion (iv) about transit points is not included in the statement of [2, Thm III.2],although it is mentioned immediately after. This property is crucial for the proof ofthe approximation theorem; compare [2, Thm. II.11, p.47], [2, p.65] and [25, p.86].Finally, we remark that Theorem 1.2 can also be applied to prove approximationtheorems for metric surfaces that do not necessarily satisfy the bounded curvaturecondition. This will be discussed in a forthcoming article by the second author.1.2. Organization and outline of proof.
We first recall in Section 2 the basicnotions required from metric geometry. Convexity relative to the boundary and itsrole in the proof of our main result are discussed in Section 3. In Section 4, wediscuss the methods needed for handling superfluous intersections. We then prove
RIANGULATING METRIC SURFACES 3
Theorem 1.2 in Section 5, with the exception of the non-degeneracy conclusion (iii).Finally, in Section 6, we verify that one can further subdivide a triangulation sothat the non-degeneracy conclusion is satisfied.The proof of Theorem 1.2 consists of several steps. First, a relatively simpleargument, given in Section 5.1, shows that X is covered by polygons of smalldiameter. Next, in Section 5.2, we improve this to a cover by polygons havingboth small diameter and small perimeter. This is the main step where our proofdiffers from, and simplifies, the classical proof for surfaces of bounded curvature.In Section 5.3, we use an argument from [2] to find a cover by small polygons thatare also absolutely convex. The main remaining difficulty is to show that one canpass to a cover by polygons that are also non-overlapping. In [2], this is achievedby Lemma III.6. The proof of this lemma, unfortunately, is not correct. As areplacement, we use two intermediate steps. First, in Section 5.4, we show that wecan pass to a cover by boundary-convex polygons such that the boundary edgesform a locally finite graph. We then show in Section 5.5 that, from such a cover, wemay pass to one that consists of non-overlapping boundary-convex polygons. As afinal step, it suffices to show that every boundary-convex polygon may be cut intofinitely many non-overlapping boundary-convex triangles. This relatively simpleargument is provided in Section 5.6. Acknowledgments.
We thank Alexander Lytchak for encouraging us to work onthis topic and for his great support. We also thank Fran¸cois Fillastre, MikhailKatz, Christian Lange, Dimitrios Ntalampekos, Raanan Schul and Stephan Stadlerfor helpful comments and suggestions that improved the presentation of this article.2.
Preliminaries
Metric geometry.
We first review the relevant definitions from metric geom-etry. Let (
X, d ) be a metric space. For each pair of subsets
A, B ⊂ X and x ∈ X ,let d ( A, B ) = inf a ∈ A,b ∈ B d ( a, b ) and d ( x, A ) = d ( { x } , A ). The diameter of A isdefined by diam( A ) = sup a,a ′ ∈ A d ( a, a ′ ). A family ( A i ) i ∈I of subsets of X is locallyfinite if every compact set K ⊂ X intersects at most finitely many of the sets A i .A curve is a continuous map γ : I → X , where I ⊂ R is an interval. We denotethe image of the curve γ by | γ | . A curve γ : I → X is compact if I is compact.A compact curve γ is closed if both its endpoints coincide and simple if it doesnot have self-intersections, except possibly coinciding endpoints. A simple closedcurve is called a Jordan curve . The length of a curve γ is denoted by ℓ ( γ ). Theconcatenation of two curves γ , γ is denoted by γ ∗ γ . The reverse of the curve γ is denoted by ¯ γ .The metric space X is a length space if d ( x, y ) = inf γ ℓ ( γ ) for all x, y ∈ X , wherethe infimum is taken over all compact curves γ joining x to y . The space X is a geodesic space if additionally this infimum is attained for all pairs of points x, y .A compact curve is a geodesic if its length equals the distance between its end-points. A compact curve is piecewise geodesic if it is the concatenation of finitelymany geodesics. A non-compact curve is piecewise geodesic if its restriction to eachcompact interval is piecewise geodesic.A surface is a topological 2-manifold with boundary. A surface is closed if it iscompact and its boundary is empty. We recall that the boundary of a 2-manifoldis a possibly disconnected 1-manifold, hence the countable union of disjoint curveseach homeomorphic to either the circle or the real line.A point p ∈ X is called a transit point if there is a geodesic passing through p within X . The following simple observation shows that such points are abundantin the setting of Theorem 1.2. PAUL CREUTZ AND MATTHEW ROMNEY
Lemma 2.1 (cf. [25], p.80) . Let X be as in Theorem 1.2. Then transit pointsare dense in X . More generally, transit points are dense within any simple curve γ ⊂ X . Note that, for the latter conclusion to hold, it is important that X be a surfacewith ∂X composed of piecewise geodesic curves. We remark that transit points donot play any role in the proof of Theorem 1.1. Thus a reader who is not interestedin conclusion (iv) of Theorem 1.2 may ignore all statements about transit pointsthroughout the paper.2.2. Disks and polygons.
Throughout this section, let X be as in Theorem 1.2.That is, X is a length space homeomorphic to a surface such that every componentof ∂X is piecewise geodesic. A set U ⊂ X is a neighbourhood of x ∈ X if x liesin the topological interior of U within X , and a disk if U is homeomorphic to aclosed ball in R . If U is a disk, then we denote by ∂U its boundary as a manifold,rather than its topological boundary within X , and by U ◦ its interior as a manifold.In particular, if x ∈ ∂X , then for every disk neighborhood U of x we have that x ∈ ∂U . A family ( U i ) i ∈I of disks is non-overlapping if U ◦ i ∩ U ◦ j = ∅ for all distinct i, j ∈ I .If U is a disk then, after fixing an orientation on U , to every p ∈ U and everyclosed curve c in U \ { p } we can associate its winding number w ( c, p ) ∈ Z . Thewinding number is characterized, up to sign, by the following properties: • If c is simple and non-contractible in U \ { p } , then | w ( c, p ) | = 1. • The winding numbers of two curves agree precisely when they are homo-topic within U \ { p } . • The winding number is additive with respect to composition of closedcurves.We further observe that the winding number w ( c, p ) is continuous as a functionof c with respect to uniform convergence. We say that c winds around p if c isnon-contractible in U \ { p } or, equivalently, if w ( c, p ) = 0. We also say that c windsaround the set A ⊂ U \ | c | if c winds around every x ∈ A .A polygon is a disk P ⊂ X with piecewise geodesic boundary ∂P , together witha representation of ∂P as a piecewise geodesic curve e ∗ · · · ∗ e n . Each geodesic e i is called an edge of P , and each initial point of an edge is called a vertex . If e i ∗ e i +1 is a geodesic for some i ∈ { , . . . , n } (taking e n +1 = e ), then e i and e i +1 can be consolidated into a single edge e e i , thus forming a new polygon with n − reduced if no such consolidation is possible. For anypolygon P , repeating this process of consolidation gives a reduced polygon. Apolygon is a triangle if it has at most 3 vertices and a bigon if it has exactly 2vertices. A triangle is called degenerate if it can be reduced to a bigon. Note thatwhether a triangle T is degenerate or not depends not only on T as a set but also onthe choice of boundary geodesics. For example, let T be the square [0 , equippedwith the ℓ -metric. Taking { (0 , , (1 , } as the vertex set gives a representationof T as a bigon, while { (1 / , , (0 , / , (1 , } gives a representation of T as anon-degenerate triangle. 3. Boundary convexity
Throughout this section, let X be as in Theorem 1.2. That is, X is a lengthspace homeomorphic to a surface such that every component of ∂X is piecewisegeodesic. A set K ⊂ X is convex if for every x, y ∈ K some geodesic from x to y iscontained in K , and completely convex if for every x, y ∈ K every geodesic from x to y is contained in K . The following convexity property plays a fundamental rolein the proof of Theorem 1.1. RIANGULATING METRIC SURFACES 5
Definition 3.1.
A disk K ⊂ X is convex relative to its boundary or boundaryconvex if there is a disk U containing K such that the following hold:(1) d ( K, ∂U \ ∂X ) > · ℓ ( ∂K ),(2) diam( U ) ≤ diam( X ) /
3, and(3) for every subarc γ of ∂K and every curve η in U \ K ◦ that is path homotopicto γ within U \ K ◦ , one has ℓ ( γ ) ≤ ℓ ( η ).In the situation of Definition 3.1, we also say that K is boundary convex withrespect to U . It is easy to see that boundary convexity implies convexity. Finally,the disk K is absolutely convex if it is boundary convex and completely convex.Note that boundary convexity is called “bounded convexity” in [25]. Our defi-nition is slightly more restrictive than the ones given on p. 48 of [2] and on p. 80of [25]. The main difference is that we have added condition (2). This ensures thatif U and U are both ambient disks satisfying conditions (1) and (2) for a givendisk K , then condition (3) holds for U if and only if it holds for U . Condition (1)alone does not suffice to guarantee this independence, because without (2) a curve η ⊂ U ∩ U may homotope to different subarcs of ∂K in the respective ambientdisks U and U . This unpleasant behavior can also be avoided by assuming that X is not homeomorphic to S .The proof of Theorem 1.1 depends in an essential way on boundary convexity, asopposed to convexity or complete convexity. The reason is that boundary convexityis preserved by certain operations of both intersection and subdivision. This is thecontent of the next two lemmas. Lemma 3.2.
Let K and K be boundary-convex disks. If W is the closure of aconnected component of the interior of K ∩ K , then W is a boundary-convex disk. Note that Lemma 3.2 fails when boundary convexity is replaced by mere con-vexity. This intersection property does not appear explicitly in [2], but we need itto work around Lemma III.6 in [2]. See Section 5.5 below.
Proof.
Let W be the closure of a connected component of the interior of K ∩ K ,and let U , U denote the ambient disk neighborhoods of K , K , respectively, as inDefinition 3.1. We assume without loss of generality that ℓ ( ∂K ) ≥ ℓ ( ∂K ). Then ∂K ⊂ U ∩ U . Note that, by condition (2), whenever c ⊂ U ∩ U is a Jordancurve, then the disk bounded by c within U is the same as the disk boundedby c within U . Compare also the discussion after Definition 3.1. Applying thisobservation to ∂K , we conclude that K ⊂ U .Note that W is also the closure of some complementary component of ∂K ∪ ∂K in U . Since ∂K and ∂K have more than one point in common, Ker´ekj´art´o’stheorem (see e.g. [21, p.168]) implies that W is a disk.It remains to show that W is boundary convex with respect to U . Certainlycondition (2) is satisfied. Assume that condition (3) fails. Then there are a subarc e γ of ∂W and a curve e η ⊂ U \ W ◦ that are path homotopic within U \ W ◦ and suchthat ℓ ( e η ) < ℓ ( e γ ). By the Arzel`a–Ascoli theorem, lower semicontinuity of lengthand continuity of the winding number, we may assume that e η is shortest among allcurves that are path homotopic to e γ within U \ W ◦ . By the boundary convexityof K with respect to U , we may assume that | e η | ⊂ K . There must be subcurve η of e η such that η intersects ∂W only in its endpoints and ℓ ( η ) < ℓ ( γ ), where γ is thesubarc of ∂W that is path homotopic to η within U \ W ◦ . Otherwise, e η would bepath homotopic within U \ W ◦ to a curve e ν that is contained in ∂W and satisfies ℓ ( e ν ) ≤ ℓ ( e η ). This would be a contradiction since then | e γ | ⊂ | e ν | and hence ℓ ( e γ ) = H ( | e γ | ) ≤ H ( | e ν | ) ≤ ℓ ( e ν ) ≤ ℓ ( e η ) . PAUL CREUTZ AND MATTHEW ROMNEY
Note that γ must be contained in ∂K . Thus, if η intersected the entire polygon K only at its endpoints, then we could apply the boundary convexity of K toderive that γ is a geodesic and hence obtain a contradiction. However, since thismight a priori not be true, we must work harder.First, we claim that η is simple. To verify this, note that both endpoints of η lieon a simple arc A ⊂ K ∩ ∂K ∩ ∂W that contains | γ | and separates | η | \ A from W ◦ within K . Thus, if η had a self-intersection, then we could shorten η by deletingsome subcurve of it. Note that A would still separate the resulting curve from W ◦ within K and thus this curve must also be path homotopic to γ within U \ W ◦ .This would contradict the length minimality of η . Let c = η ∗ ¯ γ and denote by O the complementary component of | η | ∪ ∂K ∪ ∂K that is adjacent to γ and differsfrom W . Note that | c | ⊂ U \ O and c winds around O within U . Denote by( η i ) i ∈I the closures of the connected components of | η | \ K , and for each i denoteby γ i the respective subarc of ∂K which is path homotopic to η i within U \ K ◦ .Note that, again by condition (2), γ i is also path homotopic to η i within U \ K ◦ ,and hence we conclude by the boundary convexity of K that ℓ ( γ i ) ≤ ℓ ( η i ). Thisin turn implies that each γ i is path homotopic to η i within U \ O , for otherwise | γ | ⊂ | γ i | and hence ℓ ( γ ) ≤ ℓ ( γ i ) ≤ ℓ ( η i ) ≤ ℓ ( η ) . Thus the curve b c obtained by replacing in c each η i with γ i is path homotopic to c within U \ O . However, | b c | is contained in K , and hence we conclude that c iscontractible in U \ O . This gives a contradiction, since we had initially observedthat c winds around O within U .Finally, we verify condition (1). It suffices to show that ℓ ( ∂W ) ≤ ℓ ( ∂K ), sincethen d ( W, ∂U \ ∂X ) ≥ d ( K , ∂U \ ∂X ) > · ℓ ( ∂K ) ≥ · ℓ ( ∂K ) ≥ ℓ ( ∂W ) . To this end, let ( α i ) i ∈I be the countable family of closures of the connected com-ponents of ∂W \ ∂K . Then each α i is a subarc of ∂K and contained in K andintersects ∂K precisely at its endpoints. Let β i be the subarc of ∂K which ispath homotopic to α i within U \ W ◦ . By condition (3), we have ℓ ( α i ) ≤ ℓ ( β i ).Note that distinct β i and β j can intersect at most in their endpoints and that β i intersects ∂W only in its endpoints. We conclude that ℓ ( ∂W ) ≤ ℓ ( ∂K ). (cid:3) Lemma 3.3.
Let P ⊂ X be a boundary-convex polygon and γ a geodesic in X withendpoints in X \ P ◦ . If Q is the closure of some connected component of P \ | γ | ,then Q is a boundary-convex polygon. Furthermore, ∂Q \ ∂P consists of transitpoints. See [2, Lem. III.2, p.49] for a slightly weaker result. Note that Lemma 3.3holds when boundary convexity is replaced by convexity, but not when replaced bycomplete convexity.
Proof.
Let U be an ambient disk such that P is boundary convex with respect to U .Since otherwise the claim is obvious, we may assume that Q = P and, by possiblydeleting initial and terminal subcurves, that γ has endpoints in ∂P . Notice that Q still appears as the closure of some complementary component when replacingall portions of γ that lie outside P by the respective homotopic subcurves of ∂P .Hence, by the boundary convexity of P , we may assume that | γ | ⊂ P . Then Q isa polygon with boundary comprised of a subarc η of ∂P and a subcurve α of γ .We show that Q is boundary convex with respect to U . Clearly Condition (2) issatisfied. That α is a geodesic implies Condition (1), since d ( Q, ∂U \ ∂X ) ≥ d ( P, ∂U \ ∂X ) > · ℓ ( ∂P ) ≥ · ( ℓ ( α ) + ℓ ( η )) = 4 · ℓ ( ∂Q ) . RIANGULATING METRIC SURFACES 7 x yγ γ γ γ Figure 1
To check condition (3), let c be a simple subcurve of ∂Q and ν be a simple curvein U \ Q ◦ that is path homotopic to c within U \ Q ◦ . By the boundary convexityof P , we may find a curve ν ⊂ P \ Q ◦ that is path homotopic to ν within U \ Q ◦ and satisfies ℓ ( ν ) ≤ ℓ ( ν ). The closure of each connected component of ν \ Q ismoreover path homotopic within U \ Q ◦ to some subarc of the geodesic α . Thus, wemay find a curve ν in ∂Q that is path homotopic to ν within U \ Q ◦ and satisfies ℓ ( ν ) ≤ ℓ ( ν ). Since c and ν are homotopic within U \ Q ◦ , both contained in ∂Q and c is simple, we must have that | c | ⊂ | ν | . Then ℓ ( c ) = H ( | c | ) ≤ H ( | ν | ) ≤ ℓ ( ν ) ≤ ℓ ( ν ) ≤ ℓ ( ν ) . We conclude that Q is boundary convex with respect to U . (cid:3) Superfluous intersections of geodesics
Let X be as in the statement of Theorem 1.2. That is, X is a length spacehomeomorphic to a surface such that every component of ∂X is piecewise geodesic.A family ( γ i ) i ∈I of geodesics in X does not have superfluous intersections if forevery i, j ∈ I the intersection | γ i | ∩ | γ j | is connected. Similarly, we say that anadditional geodesic γ does not have superfluous intersections with ( γ i ) i ∈I if forevery i ∈ I the intersection | γ | ∩ | γ i | is connected. It is claimed on p. 51 of [2] andp. 79 of [25] that, given points x, y ∈ X and a finite system of geodesics ( γ i ) ki =1 thatdoes not have superfluous intersections, then one can always find a geodesic joining x to y that does not have superfluous intersections with ( γ i ) ki =1 . This claimedobservation is frequently used in [2]. However, it turns out to be false in general.As a counterexample, consider a surface X containing geodesics γ , . . . , γ thatintersect as pictured in Figure 1, with the property that any geodesic connectingthe pictured points x and y is contained in | γ | ∪ · · · ∪ | γ | . Then one can checkthat Γ = ( γ ) i =1 does not have superfluous intersections, but any geodesic from x to y must have superfluous intersections with Γ.To overcome this complication, we prove two weaker results. The first is thefollowing. Lemma 4.1.
Let P ⊂ X be a polygon with edges ( e , . . . , e n ) , where n ≥ , and let p , p , p ∈ P . Assume further that P is contained in a disk U with d ( P, ∂U \ ∂X ) > diam( P ) . Then there is a geodesic γ from p to p that does not have superfluousintersections with ( e , . . . , e n ) . In addition, for any such γ , there is a geodesic γ from p to p that does not have superfluous intersections with ( e , . . . , e n , γ ) . Note that the respective result for three geodesics emanating from p fails by acounterexample similar to the construction in Figure 1. Proof.
Since U is compact and d ( P, ∂U \ ∂X ) > diam( P ), the points p and p are joined by some geodesic η . Now we inductively construct geodesics η i from p to p so that η i does not have superfluous intersections with ( e , . . . , e i ). The firstpart of the claim then follows by setting γ = η n . Assume that such η i has beenconstructed for some 0 ≤ i ≤ n −
1. If η i does not intersect e i +1 , we set η i +1 = η i .Otherwise, let l be the first point of intersection of η i and e i +1 , and r be the last PAUL CREUTZ AND MATTHEW ROMNEY one. Now η i +1 is obtained from η i by replacing the portion of η i between l and r bythe respective one of e i +1 . Clearly | η i +1 | ∩ | e i +1 | is connected. Furthermore, since n ≥
3, we have for j ≤ i that | e j | ∩ | η i +1 | is either empty or equal to | e j | ∩ | η i | , andhence connected.Now fix a geodesic γ of the described type. The previous argument also pro-vides us with a geodesic e η from p to p not having superfluous intersections with( e , . . . , e n ). Let q be the last point of intersection of e η and γ . If q = p , we maychoose γ as a subcurve of γ , and, if q = p , we may set γ = e η . Otherwise, let η bethe geodesic which is obtained from e η upon replacing the portion of e η between p and q with the respective portion of γ . By reparametrizing, we may assume that η is parametrized on the interval [0 ,
2] so that η (0) = p , η (1) = q and η (2) = p .If η does not have superfluous intersections with ( e , . . . , e n , γ ) then we may set γ = η . Otherwise, since η does not have superfluous intersections with γ and therestrictions of η to [0 ,
1] and to [1 ,
2] respectively do not have superfluous intersec-tions with ( e , . . . , e n ), there must exist s < t > η ( s ) and η ( t ) lieon a common edge e i with q / ∈ | e i | . We may choose s as the minimal one for whichsuch t and edge e i exists and choose t maximal for the given value s . The curve γ is obtained from η by replacing η | [ s,t ] with the respective portion of e i . It remainsto show that γ does not have superfluous intersections with ( e , . . . , e n , γ ).First of all, since | γ ∩ η | = η ([0 , γ ([0 , s ]) ⊂ | γ | ∩ | γ | ⊂ γ ([0 , t ]) . However, γ ([ s, t ]) is a subgeodesic of e i and hence, since γ does not have super-fluous intersections with e i , it follows that γ ([ s, t ]) ∩ | γ | is connected and hence,by (1), that so is | γ | ∩ | γ | . Next, we have by the minimal and maximal choice of s and t that | γ | ∩ | e i | = γ ([ s, t ]) and hence that | γ | ∩ | e i | is connected. Finally,let j = i . Since the edges of P only intersect in their endpoints we must have | e j | ∩ | γ | ⊂ | γ | \ γ (( s, t )). However, by the maximal and minimal choice of s and t , we must indeed have that | e j | ∩ | γ | is either contained in γ ([0 , s ]) or in γ ([ t, γ to these subintervals do not have superfluousintersections with e j , it follows also that | e j | ∩ | γ | is connected. (cid:3) Remark 4.2.
The conclusion of Lemma 4.1 is false whenever n = 2 and p , p arethe two vertices of P . Nevertheless, the proof also shows that the result remainstrue for bigons P if we additionally require that none of the points p , p , p is avertex point of P . In many situations, it is not necessary to achieve that | γ i |∩| γ j | is connected, but itsuffices that the intersection | γ i |∩| γ j | has finitely many connected components. Wesay that a system Γ = ( γ i ) ki =1 of geodesics is a finite graph if | Γ | := | γ | ∪ · · · ∪ | γ k | isa finite topological graph when endowed with the subspace topology. Equivalently,( γ i ) ki =1 is a finite graph if | γ i | ∩ | γ j | has only finitely many connected componentsfor every 1 ≤ i, j ≤ k . Lemma 4.3.
Let
Γ = ( η i ) ki =1 be a finite graph and γ ∗ an additional geodesic in X .Then there is a geodesic γ satisfying the following:(i) γ has the same endpoints as γ ∗ ,(ii) γ ∪ Γ is a finite graph, and(iii) every connected component of | γ | \ | γ ∗ | is contained in the image of somegeodesic η i ∈ Γ .Proof. We prove the claim by induction on k . For the base case k = 0, we simplyset γ = γ ∗ . So assume the claim of the lemma holds for all finite graphs comprisingat most k − η i ) ki =1 be a finite graph and set b Γ = ( η i ) k − i =1 .By the induction assumption, we can find a geodesic b γ with the same endpoints RIANGULATING METRIC SURFACES 9 as γ ∗ such that b γ ∪ b Γ is a finite graph and each connected component of | b γ | \ | γ ∗ | is contained in the image a single geodesic η i ∈ b Γ.If | b γ |∩| η k | has only finitely many connected components, then we may simply set γ = b γ . So assume otherwise. Then, since η k ∪ b Γ and b γ ∪ b Γ are finite graphs, all exceptfinitely many of these connected components are contained in | b γ |\| b Γ | ⊂ | γ ∗ | . Let l bethe first point in | b γ |∩| γ ∗ |∩| η k | and r be the last one. Let γ be the geodesic obtainedfrom b γ upon replacing the portion between l and r by the respective portion of η k .Then γ has the same endpoints as b γ , and hence also as γ ∗ . The intersection of γ with each η i ∈ Γ has only finitely many connected components, since γ is thecomposition of three curves, each having this property. In particular γ ∪ Γ is a finitegraph. Finally, each connected component of | γ | \ | γ ∗ | is either contained in | η k | ,or is equal to a connected component of | b γ | \ | γ ∗ | and thus contained in the imageof a single geodesic η i ∈ b Γ ⊂ Γ. (cid:3) Proof of the main theorem
We now proceed with the proof of Theorem 1.2, with the exception of the non-degeneracy conclusion (iii), which is postponed until Section 6. Throughout thissection, let X be as in Theorem 1.2. That is, X is a length surface with ∂X composed of piecewise geodesic curves.5.1. Covering by polygons of small diameter.
We begin by showing that X may be covered by polygons having arbitrarily small diameter. Lemma 5.1 (cf. [2], Lem. III.3, p.51) . Let x ∈ X and ε > . Then there is apolygonal neighbourhood P of x such that diam( P ) ≤ ε and ∂P \ { x } consists oftransit points. The idea of the proof is the following. We take a small disk neighbourhood V of x and then choose a sufficiently fine finite set of points in ∂V . Connecting eachconsecutive pair of these points by a geodesic gives a piecewise geodesic curve γ .If γ is a Jordan curve, then P is the disk bounded by γ . In general, however, wemust carefully delete some portions of γ to turn it into a Jordan curve. Proof.
Let U be a disk neighbourhood of x such that diam( U ) ≤ ε and U ∩ ∂X contains no vertices of ∂X except for possibly x itself. Let V ⊂ U be a diskneighbourhood of x such that d ( V, ∂U \ ∂X ) > δ satisfying 0 < δ < d ( V, ∂U \ ∂X ), to be determined later. Nowtake a finite collection of distinct points { y j } mj =0 ⊂ ∂V , labelled in cyclic order, suchthat m ≥ | η j | ) < δ for all j , where η j denotes the arc of ∂V between y j and y j +1 . Furthermore, if x ∈ ∂X , we assume that y = x . Applying Lemma 2.1,by perturbing the points if needed, we can choose each y j , except for possibly x , tobe a transit point. Connect each y j to y j +1 by a geodesic e j . Such e j exists and iscontained in U since d ( y j , y j +1 ) < d ( V, ∂U \ ∂X ). Whenever | η j | ⊂ ∂X , then η j isa geodesic and we choose e j = η j . Let γ = e ∗ · · · ∗ e m . Then γ defines a closedpiecewise geodesic curve that is contained in U and such that | γ | \ { x } consistsof transit points. By Lemma 4.3, we may furthermore assume that | γ | is a finitetopological graph.We now consider two cases. First, suppose that x ∈ X \ ∂X . In this case, set µ = 1 / · d ( x, ∂V ) >
0. As noted in the proof of [22, Lemma 9.4], we may choose0 < δ < µ such that a subset of U of diameter at most δ cannot separate a subsetof U of diameter at least µ from the boundary curve ∂U . Otherwise, we wouldbe able to find a sequence of such subsets converging to a point that separates anon-empty subset of U from ∂U . This contradicts the fact that U is a disk. For each j ∈ { , . . . , m } , set c j = e j ∗ η j . Then diam( | c j | ) ≤ δ and hence, since d ( x, | c j | ) ≥ d ( x, ∂V ) − diam( | η j | ) ≥ µ − δ > µ, the curve | c j | cannot separate x from ∂U . Thus η j is path homotopic to e j within U \ { x } , and hence ∂V is path homotopic to γ within U \ { x } . In particular, | γ | must separate x from ∂U . Since | γ | is a finite topological graph, there is a Jordancurve b γ with | b γ | ⊂ | γ | also separating x from ∂U ; see e.g. [27, Theorem IV.6.7].The curve b γ must be piecewise geodesic curve as well. The claim follows by taking P ⊂ U to be the disk bounded by b γ .Next, suppose that x ∈ ∂X . Denote by ν the connected component of ∂V ∩ ∂X that contains x . In this case, let δ < d ( x, ∂V \ ν ). Then each curve e j is either equalto η j , or cannot pass through x . Hence the curve γ passes through x exactly once.Furthermore, note that e = η and e m = η m . Again, by [27, Theorem IV.6.7] wemay find a piecewise geodesic Jordan curve b γ such that | b γ | ⊂ | γ | and | b γ | containsa neighbourhood of x within ∂X . The claim follows by taking P ⊂ U to be thepolygon corresponding to b γ . (cid:3) Covering by polygons of small perimeter.
The next step is to cover X bypolygons that have not only small diameter but also small perimeter. In the originalproof of Alexandrov–Zalgaller for surfaces of bounded curvature, this is deducedfrom the fact that every point in such a surface has a polygonal neighbourhood ofsmall perimeter; see [2, Lem. III.5, p.53] or [25, Lem. 6.3.3]. However, as notedin [2] and [25], this fact does not generalize to arbitrary metric surfaces. Onepossible counterexample is to take X to be the quotient metric space obtainedfrom the cylinder S × [0 ,
1] by collapsing one of its boundary circles to a singlepoint. Instead, we show the following lemma, which is the main novel ingredient ofour proof.
Lemma 5.2.
Let x ∈ X and ε > . Then there is a neighbourhood U of x such that diam( U ) ≤ ε and U = T ∪ · · · ∪ T n where each T i is a triangle such that ∂T i \ ∂X consists of transit points. Note that each triangle T i has perimeter at most 3 ε and that we do not requirethe triangles to be non-overlapping. Proof.
Let V be a disk neighbourhood of x such that diam( V ) ≤ ε . By Lemma 5.1,there exists a polygonal neighbourhood P of x such that diam( P ) < d ( P, ∂V \ ∂X )and ∂P \ { x } consists of transit points. Note that x ∈ ∂P only when x ∈ ∂X . Let ∂P = e ∗ · · · ∗ e n +1 be a representation of ∂P as a piecewise geodesic curve, and let v , . . . , v n +1 be the corresponding vertices of P , where v i is the initial point of e i .By iterated application of Lemma 4.1, we may choose for each i = 1 , . . . , n + 1 ageodesic γ i from v to v i such that ( e , . . . , e n +1 , γ i , γ i +1 ) does not have superfluousintersections whenever i ≤ n . Note that necessarily γ = e and γ n +1 = e n +1 .For each i = 1 , . . . , n , set c i = γ i ∗ e i ∗ ¯ γ i +1 . Thus the image of c i comprisesthree geodesics whose complement in V consists of an outer component containing ∂V \ ∂X and at most one inner component. All possible topological types of theclosed curve c i are shown in Figure 2; see also Figure 17 on page 51 of [2]. If | c i | bounds an inner component, let T i ⊂ V be the triangle bounded by the Jordancurve obtained by deleting the inward- and outward-pointing ends of c i when suchexist. Otherwise, we set T i = ∅ . We now set U = T ∪ · · · ∪ T n . Note thatdiam( U ) ≤ ε , since U ⊂ V .It remains to check that U is a neighbourhood of x . To do so, it suffices to showthat P ⊂ U . Let O = P ◦ \ S n +1 i =1 | γ i | . Then O is a dense subset of P . Since thetriangles are compact, the claim reduces to proving that O ⊂ T ∪ · · · ∪ T n . To RIANGULATING METRIC SURFACES 11
Figure 2 this end, consider a point y ∈ O . First, observe that ∂P winds around y within U .However, ∂P = e ∗ · · · ∗ e n +1 is path homotopic to e ∗ e ∗ ¯ γ ∗ γ ∗ e ∗ ¯ γ ∗ γ ∗ e ∗ · · · ∗ e n − ∗ ¯ γ n ∗ γ n ∗ e n ∗ e n +1 within V \ { y } , which is in turn equal to c ∗ · · · ∗ c n . Thus, by additivity of thewinding number, there must be some i such that c i winds around y within U . Thisimplies that y ∈ T i . (cid:3) Note that, even for surfaces of bounded curvature, the proof of Lemma 5.2 isshorter and conceptually simpler than that of [2, Lem. III.5, p.53].5.3.
Covering by absolutely convex polygons.
At this point, Lemma 5.2 givesa cover of X by polygons of small perimeter. In this section, we improve this to acover by absolutely convex polygons. Proposition 5.3.
Let ε > . Then X may be covered by a locally finite collectionof absolutely convex polygons ( P i ) i ∈I such that diam( P i ) ≤ ε and ∂P i \ ∂X consistsof transit points for each i ∈ I . The proof relies on Lemma 5.2 and the following lemma.
Lemma 5.4 (cf. [2], Lemma III.4, p.51) . Let x ∈ X and ε > . Then there is δ > such that any polygon P with x ∈ P and ℓ ( ∂P ) ≤ δ is contained in an absolutelyconvex polygon Q such that diam( Q ) ≤ ε and ∂Q \ ∂P consists of transit points. The idea of the proof is to take Q to be a polygon of least perimeter among allpolygons containing P and contained in some fixed ambient disk U . This guaranteesthe boundary convexity of Q . Complete convexity is achieved by taking, among allpolygons of least perimeter, the maximal one with respect to set inclusion. Proof of Lemma 5.4.
We assume without loss of generality that ε ≤ diam( X ) / U be a disk neighbourhood of x such that diam( U ) ≤ ε and set ζ = d ( x, ∂U \ ∂X ) / > . Let 0 < δ < ζ be such that any Jordan curve c in U of length at most δ whicheither winds around or passes through x is contained in B ( x, ζ ). Such a value δ mustexist, since otherwise there would be a point separating B ( x, ζ/
2) from ∂U \ ∂X .Compare also the respective step in the proof of Lemma 5.1.Now let P be a polygon with x ∈ P and ℓ ( ∂P ) ≤ δ . By the Arzel`a–Ascolitheorem, lower semicontinuity of length and continuity of the winding number,there exists a closed curve c that is shortest among all curves that wind around P ◦ within U . Denote by C the collection of all such shortest curves c . By the additivityof the winding number, every c ∈ C must be a Jordan curve, and we denote itsenclosed disk within U by Q c . Then Q c ⊂ U is a disk which contains P . Next, we show that Q c is boundary convex with respect to U . Certainly U is ofdiameter at most diam( X ) /
3. Furthermore, since ℓ ( ∂Q c ) ≤ δ , we have that ∂Q c iscontained in B ( x, ζ ) and hence that(2) d ( Q c , ∂U \ ∂X ) ≥ d ( x, ∂U \ ∂X ) − ζ = 4 ζ > δ ≥ · ℓ ( ∂P ) ≥ · ℓ ( ∂Q c ) . Now let γ be a proper subcurve of c and η ⊂ U \ Q ◦ c be a simple curve which is pathhomotopic to γ within U \ Q ◦ c . Then we must have that ℓ ( γ ) ≤ ℓ ( η ). Otherwise,the curve obtained from c when replacing γ with η would be shorter than c andnon-contractible within U \ P ◦ . This would contradict to the assumption that c ∈ C .We conclude that Q c is boundary convex with respect to U .We claim that every c ∈ C is piecewise geodesic and hence that the correspondingdisk Q c is a polygon. Since c is a compact curve, it suffices to show that c islocally piecewise geodesic. By symmetry, we may furthermore assume that c isparametrized by arc length on the interval [ − l, l ] and only show that c is locallypiecewise geodesic at 0. First, assume that c (0) ∈ U \ ∂P . In this case, therestriction of c to [ − µ, µ ] is a geodesic, where µ = 18 · min { d ( c (0) , ∂P ) , d ( c (0) , ∂U \ ∂X ) , l } > . Otherwise since every geodesic from c ( − µ ) to c ( µ ) must be contained in U \ P ◦ , wecould shorten c within the admissible class by replacing one of the arcs of c by thegeodesic. Next, if c (0) ∈ ∂P , then let ν be the union of the (at most two) edges of ∂P which contain c (0). Then the restrictions of c to [ − µ,
0] and [0 , µ ] are geodesics,where µ = 12 · min { d ( c (0) , ∂P \ ν ) , d ( c (0) , ∂U \ ∂X ) , l } > . Namely, by our choice of µ , every geodesic γ from c (0) to c ( − µ ) (respectively,to c ( µ )) is contained in U and can intersect ∂P only in ν . Thus, by replacing asubcurve of γ with a subarc of P , we can also find a geodesic η from c (0) to c ( − µ )(respectively, to c ( µ )) which is contained in U \ P ◦ . Now, if the restriction of c to [ − µ,
0] (respectively, to [0 , µ ]) were not a geodesic, then we could, as before,shorten c within the admissible class by replacing one of its arcs with η . Note thatour argument also shows that ∂Q c \ ∂P consists of transit points.It remains to show that we can find c ∈ C such that Q c is completely convex. Wefind such a curve c as an application of Zorn’s lemma. To do this, we introduce thepartial ordering (cid:22) on C defined by c (cid:22) c whenever Q ◦ c ⊂ Q ◦ c . To apply Zorn’slemma, we must show that every chain C ⊂ C has an upper bound. Since X is asecond countable space, we may assume that C = ( c i ) ∞ i =1 with c (cid:22) c (cid:22) c (cid:22) · · · ;see e.g. [20, Thm. 30.3]. Then, by the Arzel`a–Ascoli theorem, lower semicontinuityof length and continuity of the winding number, the sequence ( c i ) ∞ i =1 subconvergesto some c ∈ C , which must certainly be an upper bound for C with respect to (cid:22) .We conclude by Zorn’s lemma that there is a maximal element c m ∈ C with respectto (cid:22) . We claim that Q = Q c m is completely convex. Otherwise, there would bea geodesic γ with endpoints in Q which is not entirely contained in Q . By (2),we would have | γ | ⊂ U and, by passing to a subgeodesic, we could assume that γ intersects Q only in its endpoints. However, then we could enlarge Q by replacingone of the arcs of ∂Q with γ . This would contradict the maximality of c m . (cid:3) Proof of Proposition 5.3.
We first assume that X is compact. For each x ∈ X ,choose δ x > x and ε . By Lemma 5.2, there is aneighbourhood V x of x such that diam( V x ) ≤ δ x / V x = T x ∪ · · · ∪ T n x x , wherethe T ix are triangles such that ∂T ix \ ∂X consists of transit points. By deleting someof the triangles if necessary, we may assume that each T ix contains x . Then, byLemma 5.4, each triangle T ix is contained in an absolutely convex polygon P ix such RIANGULATING METRIC SURFACES 13 P P (a) P P (b) Figure 3 that diam( P ix ) ≤ ε and ∂P ix \ ∂X consists of transit points. Since X is compactand the interiors of the neighbourhoods V x cover X , by choosing a subcollection ofthe P ix , we find our desired finite cover ( P i ) ki =1 .If X is non-compact, then we find a sequence (Ω j ) ∞ j =1 of relatively compact opensets such that X = S ∞ j =1 Ω j and Ω j ⊂ Ω j +1 for each j . For each x ∈ X , choose0 < ε x ≤ ε so that B ( x, ε x ) ⊂ Ω j +1 \ Ω j − whenever x ∈ Ω j \ Ω j − . Now, we choose δ x as in the compact case but according to ε x instead of ε . As before, we choosethen the sets V x , T ix and P ix . Now for each j , by choosing a subcollection of the P ix ,we can find finitely many absolutely convex polygon P j , . . . , P jk j such thatΩ j \ Ω j − ⊂ P j ∪ · · · ∪ P jk j ⊂⊂ Ω j +1 \ Ω j − and each P ji is of diameter at most ε and such that ∂P ji \ ∂X consists of transitpoints. Then ( P ji ) j ∈ N , ≤ i ≤ k j is locally finite and a cover of the desired type. (cid:3) Remark 5.5.
When applying Theorem 1.2 to non-compact surfaces, it might behelpful to have further control on the diameters of the triangles. Indeed, withoutserious additional difficulties, one can replace (ii) by the following stronger conclu-sion: Let (Ω j ) ∞ j =1 be an exhaustion of X by relatively compact open sets as in thepreceeding proof and ( ε j ) ∞ j =1 be a sequence of positive reals. Then every triangle T i that intersects X \ Ω j has diameter at most ε j . This is possible since the respectivediameter bound can be achieved in Proposition 5.3, and the latter steps will proceedby subdividing this given cover. Handling superfluous intersections.
By Proposition 5.3, we are now ableto cover X by small absolutely convex polygons. Our next objective is to find acover by small non-overlapping boundary-convex polygons. The argument givenin [2] relies on a lemma stating that if P , . . . , P n are non-overlapping boundary-convex polygons and P n +1 is absolutely convex, then one can subdivide S n +1 j =1 P j into finitely many non-overlapping boundary-convex polygons. This is found asLemma III.6 in [2]. If the initial polygons P , . . . , P n +1 do not have superfluousintersections, then this claim is correct and the proof given in [2] applies. However,if these polygons have superfluous intersections, the procedure given in the proof ofLemma III.6 in [2] may not work, and it is not clear whether there exists a generalprocedure that remedies this.Figure 3a gives an example of a configuration consisting of a boundary-convexpolygon P and absolutely convex polygon P with edges intersecting in a Cantor setfor which the procedure in [2] does not apply. In Figure 3a, P cannot be enlargedwithout potentially interfering with its boundary convexity. Note, however, thatin our situation we have the stronger property that each polygon P i is absolutelyconvex, not just boundary convex. This allows us to avoid configurations such as the one in Figure 3a. On the other hand, a configuration such as that in Figure 3bis possible for absolutely convex polygons and must be accounted for in the proof.These kinds of overlaps especially complicate the situation when attempting todecompose the union of more than two absolutely convex polygons.Our argument proceeds in two steps. The first is the following lemma. It allowsus to turn a cover by absolutely convex polygons into one by boundary-convexpolygons whose boundary edges form a locally finite topological graph. Lemma 5.6.
Let P be a locally finite family of absolutely convex polygons in X .Then there is a locally finite family of boundary-convex polygons Q such that(1) S Q = S P ,(2) S Q ∈Q ∂Q is a locally finite topological graph,(3) each Q ∈ Q is contained in some P ∈ P , and(4) S Q ∈Q ∂Q \ S P ∈P ∂P consists of transit points. The idea of the proof is to enumerate the edges of the polygons in P and in-ductively replace them by means of Lemma 4.3 to obtain a locally finite graph.Lemma 3.3 then guarantees that the arising polygons are boundary convex. Proof.
We first consider the case of a finite collection P = { P , . . . , P k } and provethe statement by induction on k . The base case k = 1 is trivial. For the inductionstep, assume that the statement of the lemma holds for all collections of at most k − P = { P , . . . , P k } of k polygons as in the statementof the lemma. By the induction assumption, we may find a finite collection b Q of boundary-convex polygons satisfying properties (1)-(4) for b P = { P , . . . , P k − } .Denote by b E the finite graph formed by the the edges of the polygons Q ∈ b Q andfix a disk U such that P k is absolutely convex with respect to U .Denote by ( e i ) mi =1 the edges of P k . Applying Lemma 4.3 for each i = 1 , . . . , m we can find a new geodesic e ∗ i such that(i) e ∗ i has the same endpoints as e i ,(ii) | b E| ∪ | e ∗ i | is a finite topological graph, and(iii) every connected component of | e ∗ i | \ | e i | is contained in the image of someedge e ∈ b E .By the aboslute convexity of P k , we must have | e ∗ i | ⊂ P k . Furthermore, by condi-tion (iii), the interiors of the edges e ∗ i and e ∗ j for i = j can only intersect within | b E| .Thus, it follows from condition (ii) that | b E|∪| e ∗ |∪· · ·∪| e ∗ m | is also a finite topologicalgraph.Applying Lemma 3.3 inductively, it follows that the geodesics e ∗ , . . . , e ∗ m subdi-vide P k into a family of non-overlapping boundary-convex polygons e Q k such that S Q ∈ e Q k ∂Q \ ∂P k consists of transit points. Let Q k be the subcollection of those Q ∈ e Q k which do not intersect ∂P k \ ( | e ∗ | ∪ · · · ∪ | e ∗ m | ). Note that Q k is finite since | e ∗ | ∪ · · · ∪ | e ∗ m | is a finite topological graph. Set Q = b Q ∪ Q k . It follows readilythat the properties (2)-(4) are satisfied for Q . Since S b Q = P ∪ · · · ∪ P k − and S e Q k = P k , to verify (1) it suffices to show that every Q ∈ e Q k \ Q k is alreadycontained in some P j with j < k .To this end, let Q ∈ e Q k \ Q k . Then Q intersects | e i |\ | e ∗ i | for some 1 ≤ i ≤ m . ByLemma 3.3, e ∗ i subdivides P k into a family of nonverlapping polygons R . Certainlythere must be some R ∈ R with Q ⊂ R . The boundary of R is composed of asubcurve of e i together with a connected component of | e ∗ i | \ | e i | , denoted by e ′ i . By(iii), e ′ i is a subcurve of some edge of a polygon b Q ∈ b Q , and hence, by property (3)for b Q , e ′ i is contained in some P j with j < k . Since P k and P j are completelyconvex, we have ∂R ⊂ P k ∩ P j . Since ∂R is a Jordan curve, it must bound some RIANGULATING METRIC SURFACES 15 P P Figure 4 disk within P j . If this disk were not equal to R , then we would have X = R ∪ P j and hence X = P k ∪ P j . This would be a contradiction since X is connected and P j and P k are both of diameter at most diam( X ) /
3. Thus we conclude that R ⊂ P j and hence also that Q ⊂ P j .In the case of an infinite collection P = { P , P , P , . . . } , inductively apply theabove construction setting Q = Q ∪ Q ∪ · · · . Note here that, throughout theprocess, we only add polygons, and the previously constructed polygons remainunchanged. Thus, certainly Q satisfies (1) as well as (3) and (4). Within anycompact set, by the local finiteness of P and since S Q k ⊂ P k , only finitely manysteps are of interest. Hence the local finiteness of Q and of the graph formed by itsedges also follow. (cid:3) Covering by non-overlapping polygons.
By the previous results, we areable to cover X by small boundary-convex polygons such that the boundary edgesof the polygons form a locally finite graph. Using the following lemma, we canimprove this to a cover of X by small non-overlapping boundary-convex polygon. Lemma 5.7.
Let P be a locally finite family of boundary-convex polygons, and let Γ ⊂ X be a locally finite topological graph with S P ∈P ∂P ⊂ Γ . Then there is alocally finite family of non-overlapping boundary-convex polygons Q such that(1) S Q = S P ,(2) Γ ∪ S Q ∈Q ∂Q is a locally finite topological graph,(3) each Q ∈ Q is contained in some P ∈ P , and(4) S Q ∈Q ∂Q \ Γ consists of transit points. We follow the idea suggested by the proof of Lemma III.6 in [2], which worksprovided that one assumes that the union of the boundary edges of the polygonsform a finite graph. By Lemma 5.6, one can make this additional assumption.However, when doing so, one can no longer guarantee that each of the polygons isabsolutely convex, and hence the intersection of two polygons P i , P j ∈ P may failto be convex. Instead, to adapt the argument in [2], we employ Lemma 3.2 above.The decomposition procedure used to prove Lemma 5.7 is illustrated in Figure 4. Proof.
We first consider the case of a finite collection P = { P , . . . , P n } . We mayassume further assume that the polygons P , . . . , P n − are non-overlapping. Thelatter is possible since the case of a general finite collection follows from this specialcase by induction.By Lemma 3.2, and since S P ∈P ∂P is a finite topological graph, for each i =1 , . . . , n the closure of the interior of P i ∩ P n is a union of finitely many non-overlapping boundary-convex polygons W i , . . . , W ik i . For each j = 1 , . . . , k i , welet v ij , . . . , v m ij ij be a cyclic enumeration of the topological vertices of ∂P i ∪ ∂P n thatlie in ∂W ij . Note that W ij is a geodesic surface with piecewise geodesic boundary and Γ ∩ W ij is a finite graph. Thus, by iterated application of Lemma 4.3, we mayfind for each 1 ≤ i ≤ n , 1 ≤ j ≤ k i and 0 ≤ l ≤ m ij a geodesic γ lij ⊂ W ij from v lij to v l +1 ij such that Γ ∪ S ni =1 S k i j =1 S m ij l =0 | γ lij | is a locally finite topological graph. Let G be the finite topological graph that one obtains by deleting from ∂P ∪ · · · ∪ ∂P n alltopological edges which are contained in the interior of some P i and adding in allthe geodesics γ lij . Now we set Q , . . . , Q r to be the closures of the complementarycomponents of G in S ni =1 P i and take Q := { Q , . . . , Q r } .Then certainly the Q i are non-overlapping and form a decomposition of S ni =1 P i .Furthermore, since each component of ∂P i \ S m ij l =0 | γ lij | in W ij is separated from ∂P n by S m ij l =0 | γ lij | , and similarly for ∂P n , it follows that the boundary of eachcomplementary component of ∂P i ∪ ∂P n ∪ S m ij l =0 | γ lij | in W ij is contained either in ∂P i ∪ S m ij l =0 | γ lij | or contained in ∂P n ∪ S m ij l =0 | γ lij | . Thus we deduce that (3) holds.After noticing the latter, iterated application of Lemma 3.3 shows that each Q i isa boundary-convex polygon. Finally, (2) and (4) hold, since n [ i =1 ∂P i ∪ r [ i =1 ∂Q i = n [ i =1 ∂P i ∪ n [ i =1 k i [ j =1 m ij [ l =0 | γ lij | and the curves γ lij are geodesics with endpoints in S ni =1 ∂P i ⊂ Γ.Now consider the case of a locally finite collection P = { P , P , . . . } . We claimthat for each j ∈ N there is a finite collection Q j that satisfies the conclusions ofthe lemma for P j := { P , . . . , P j } , with the additional property that { Q ∈ Q j : Q ⊂ P i } = { Q ∈ Q j − : Q ⊂ P i } whenever i < j and the polygons P i and P j are non-overlapping. For the case j = 1,we simply set Q := { Q } . Next, assume the family Q j has been constructed. De-note by e Q j the subfamily of those Q ∈ Q j that overlap with P j +1 . The family Q j +1 is obtained from Q j by replacing e Q j with the family of those polygons obtainedwhen applying the finite case of the result to e Q j ∪ { P j +1 } . It is not hard to checkthat the desired properties carry over.Set Q := T ∞ j =1 S ∞ i = j Q i . Certainly, Q is a family of non-overlapping boundaryconvex polygons which satisfies (3) and (4). Now let K ⊂ X be compact. By thelocal finiteness of P , there is a maximal l ∈ N such that P l intersects K . Also bythe local finiteness of P , there is a maximal s ∈ N such that P s intersects S P l .Then the collection of those Q ∈ Q that intersect K is the same as the collectionof those Q ∈ Q s that intersect K , and hence finite. We conclude that Q is locallyfinite. Similar arguments show (1) and (2). (cid:3) Decomposing polygons into triangles.
After covering X by a locally finitecollection of small non-overlapping boundary-convex polygons, the final step is tocut these polygons into triangles. This is achieved by the following lemma. Comparealso [2, p.60]. Lemma 5.8.
Let P ⊂ X be a boundary-convex polygon. Then one can decompose P into non-overlapping boundary-convex triangles T , . . . , T n such that ∂T i \ ∂P con-sists of transit points for each i .Proof. We prove this by induction on the number of vertices k of P . The basecases k = 2 and k = 3 are trivial. Now assume the claim holds for polygonswith at most k − k ≥
4. Let P be a boundary-convex polygonwith k ≥ v and w of ∂P . Since P isconvex and has piecewise geodesic boundary, we may apply Lemma 4.1 with X replaced by P . Thus we can find a geodesic γ from v to w within P which does not RIANGULATING METRIC SURFACES 17 have superfluous intersections with ∂P . By Lemma 3.3 the curve γ subdivides P into finitely many non-overlapping boundary-convex polygons P , . . . , P m . Theboundary of each P i is composed of a subgeodesic γ i of γ and a subarc η i of ∂P .Since v and w are nonconsecutive we can arrange that on the interior of η i therelie at most k − ∂P . Thus we may represent each P i as a polygon withat most k − ∂P i \ ∂P consists of transit points.Thus we can derive the claim by applying the inductive assumption to each of thepolygons P i . (cid:3) Proposition 5.3, Lemma 5.6, Lemma 5.7 and Lemma 5.8 together complete theproof of Theorem 1.2, except for the non-degeneracy conclusion (iii).6.
Decomposing bigons into non-degenerate triangles
Let X be a length surface as in Theorem 1.2. In Section 5, we found a cover( T i ) i ∈I of X by triangles satisfying all the conclusions of Theorem 1.2 with thepossible exception of property (iii), namely that each triangle is non-degenerate.To complete the proof of Theorem 1.2, we apply the following proposition to eachdegenerate triangle. Proposition 6.1.
Let B ⊂ X be a boundary-convex triangle. Then B may be de-composed into finitely many non-overlapping non-degenerate boundary-convex tri-angles T , . . . , T n such that ∂T i \ ∂B consists of transit points for each i . Proposition 6.1 is proved for surfaces of bounded curvature X in [2, Lemma III.7,p.60], and the proof of this special case is easier. The remainder of this section isdedicated to the proof of Proposition 6.1.6.1. Preliminary remarks.
The conclusion is trivial if B is a non-degenerate tri-angle. Thus, to prove Proposition 6.1, we may assume that B is a bigon. We calla boundary-convex bigon B ⊂ X indecomposable if no subdivision as in Propo-sition 6.1 is possible. The strategy is to analyze the structure of a hypotheticalindecomposable bigon until we reach a contradiction. Before giving the proof, wemake several preliminary observations.Consider a bigon B with bottom vertex b , top vertex t , left side L and right side R as shown in Figure 5a. Assume that the curves L and R are parametrized by arclength on the interval [0 , a ], beginning at b and ending at t . We refer to ( b, t, L, R, a )as the data associated with B . If ∂B is locally a geodesic at the vertex b , then wecan turn B into a non-degenerate triangle by what we call the vertex perturbationtrick ; see the proof of Lemma III.7 in [2, p.60-61]. Namely, choose a small value δ > b with the points b l = L ( δ ) and b r = R ( δ ). One checks that thethree triangle inequalities for the vertices t, b l , b r are strict, and hence we obtain arepresentation of the set B as a non-degenerate triangle. In particular, if a bigon B is indecomposable, then ∂B is not locally a geodesic at each of its vertices.By Remark 4.2, any pair of points L ( l ) , R ( r ) with l, r ∈ (0 , a ) is joined bya geodesic that does not have superfluous intersections with L and R . Such ageodesic γ will be called horizontal , and we denote l γ = l and r γ = r . We saythat a horizontal geodesic is transverse if it intersects ∂B only in its endpoints.From the previous paragraph, we see that any indecomposable bigon B contains anabundance of transverse geodesics. Namely, every horizontal geodesic γ contains aunique transverse subcurve, denoted by b γ , which we refer to as the transverse part of γ . We say that γ points upward if l b γ < r b γ and points downward if r b γ < l b γ .By Lemma 3.3, a transverse geodesic γ splits B into two boundary-convex trian-gles B γb and B γt such that b ∈ B γb , t ∈ B γt and ( ∂B γb ∪ ∂B γt ) \ ∂B consists of transitpoints. If B is an indecomposable bigon, then at least one of B γt and B γb must be btL R (a) b = b ′ tL ( s ) R ( s ) t ′ L ( m ′ ) γ B ′ α (b) btL ( m ) R ( r n ) α n γ (c) Figure 5 an indecomposable bigon. Assume without loss of generality that this is the casefor B γb . Then B γb can be represented as a triangle with vertices { b, L ( l γ ) , R ( r γ ) } .Since B γb is indecomposable, this representation must reduce to a bigon. However,we cannot delete b , since ∂B is not locally a geodesic at b . Hence we conclude that ℓ ( γ ) = | r γ − l γ | , and in particular that also B γt is a bigon.If γ , γ are two horizontal geodesics, we say that γ is below (respectively, above ) γ if | b γ | is contained in B b γ b (respectively, in B b γ t ). Given l, r ∈ (0 , a )we can find a bottommost and a topmost geodesic among all horizontal geodesics γ with l γ = l and r γ = r . To construct these curves, one proceeds as in theconstruction of the outermost curve in the proof of Lemma 5.4.6.2. Proof of Proposition 6.1.
Let B ⊂ X be an indecomposable bigon with data( b, t, L, R, a ). As the first step, we show that we can replace B by another inde-composable bigon in a way that yields additional information on certain geodesics.See (i) and (ii) below.Fix s ∈ (0 , a ). Let γ be a horizontal geodesic with l γ = r γ = s . Then we musteither have that l b γ ≤ s and r b γ ≤ s, or that l b γ ≥ s and r b γ ≥ s. Otherwise, b γ would split B in such a way that both B b γt and B b γb are decomposableusing the vertex perturbation trick. By interchanging b and t and reorienting L and R if needed, we may assume that l b γ ≤ s and r b γ ≤ s .Now choose γ to be bottommost among all horizontal geodesics with l γ = s and r γ = s . After possibly interchanging L and R , we may also assume that b γ pointsupward. Then B b γt can be decomposed by means of the vertex perturbation trick,and hence B b γb is indecomposable. Thus B ′ = B b γb is a new indecomposable bigonwith data ( b ′ , t ′ , L ′ , R ′ , a ′ ), where b ′ = b , t ′ = R ( r b γ ), L ′ is the composition of therestriction of L to [0 , l b γ ] and b γ , R ′ is the restriction of R to [0 , r b γ ], and a ′ = r b γ .Additionally, for m ′ = l b γ ∈ (0 , a ′ ) and the indecomposable bigon B ′ , we have thefollowing properties:(i) The restriction of L ′ to [ m ′ , a ′ ] is the unique geodesic within B ′ from L ( m ′ )to t ′ , and(ii) every horizontal geodesic α ⊂ B ′ with l α = m ′ and r b α ∈ ( m ′ , a ) mustsatisfy l b α < m ′ . RIANGULATING METRIC SURFACES 19
Property (i) follows immediately from our assumption that γ is downmost betweenits endpoints. For the second property, assume that α ⊂ B ′ is a horizontal geodesicwith m ′ = l α ≤ l b α and r b α ∈ ( m ′ , a ′ ). We may assume without loss of generality that r α = r b α . Then α is a transverse geodesic when considered not with respect to B ′ butwith respect to B . See Figure 5b. In particular, we must have that ℓ ( α ) = r α − m ′ .Thus we obtain a geodesic within B ′ from L ( m ′ ) to t ′ by composing α with thesubarc of R ′ from R ′ ( r α ) to t . This contradicts (i).From now on, by replacing B with B ′ , we may assume that our indecomposablebigon has the properties (i) and (ii) for some m ∈ (0 , a ). Lemma 6.2.
Let B be an indecomposable bigon with data ( b, t, L, R, a ) that satisfiesproperties (i) and (ii) for some m ∈ (0 , a ) . Then any geodesic γ ⊂ B that intersects L ([ m, a ]) must have at least one endpoint in L ([ m, a ]) .Proof. Assume that γ ⊂ B is a geodesic that intersects L ([ m, a ) and has bothendpoints in B \ L ([ m, a ]).Let ( r n ) be a sequence satisfying r n ր a . For each n , let α n ⊂ B be a horizontalgeodesic with l α n = m and r α n = r n . Then, by the Arzel`a–Ascoli theorem and (i),the sequence ( α n ) must converge uniformly to the restriction of L to [ m, a ]. So wemust eventually have that r b α n > m and hence by (ii) that l b α n < m . Thus, if n issufficiently large, then α n must intersect γ before and after γ intersects L ([ m, a ]).See Figure 5c. Thus, choosing such n and replacing some portion of α n by therespective one of γ , we obtain a geodesic β from L ( m ) to R ( r n ) which initiallymoves downward along L but then on its way intersects L (( m, a ]). However, such β cannot exist by (i). (cid:3) The remaining goal is to construct a curve γ as in the statement of Lemma 6.2and hence reach a contradiction. Before doing so, we need the following additionalobservation. Lemma 6.3.
Let B be an indecomposable bigon with data ( b, t, L, R, a ) that satisfiesproperties (i) and (ii) for some m ∈ (0 , a ) . Then there is a horizontal geodesic γ ⊂ B with l b γ < m < l γ .Proof. Reasoning as in the proof of Lemma 6.2, we find a horizontal geodesic α such that l b α < m = l α and r := r α = r b α ∈ ( m, a ). For each s ∈ ( m, r ), let γ s be ahorizontal geodesic with l γ s = s and r γ s = r . By replacing a subcurve of γ s with asubcurve of b α if needed, we may assume that r b γ s ≥ r > m . If l b γ s < m for some s ,then we may set γ = γ s .So suppose that l b γ s ≥ m for all s ∈ ( m, r ). Then b γ s must point downward foreach s . Otherwise, we would obtain a contradiction to (i) by considering the curvewhich goes along L from L ( m ) to L ( l b γ s ), then along b γ s and then along R from R ( r b γ s )to t . In particular, we must have l b γ s > r and hence that L ([ s, r ]) ⊂ | γ s | . Thus,by the Arzel`a–Ascoli theorem, we can find a geodesic η from L ( m ) to R ( r ) with L ([ m, r ]) ⊂ | η | . Again, we could assume that η is horizontal and that r b η ≥ r > m .However, then η would contradict (ii). (cid:3) We are finally prepared to prove Proposition 6.1.
Proof of Proposition 6.1.
Suppose there exists an indecomposable bigon B ⊂ X .As discussed at the beginning of this subsection, we may assume that B satisfiesproperties (i) and (ii).First, we claim that there is a horizontal geodesic α such that m < l α < l b α . Tothis end, choose l ∈ ( m, a ) sufficiently large so that, for any l > l , no geodesicfrom L ( l ) to R ( l ) can intersect L ([0 , m ]). As observed in the proof of Lemma 6.2,there must exist a transverse geodesic β with l β < m and l < r β . Let α ⊂ B be btL ( l α ) L ( l b α ) L ( m ) α (a) bt = b ′ L ′ ( m ′ ) t ′ L ′ B ′ R ′ (b) bt = b ′ L ′ ( m ′ ) t ′ γ L ( m ) (c) Figure 6 a horizontal geodesic from L ( r β ) to R ( r β ). After replacing a subcurve of α witha subcurve of β if needed, we can assume that r b α ≥ r β . Since α cannot intersect L ([0 , m ]), we deduce as in the proof of Lemma 6.3 that b α must point downward.In particular, it follows that l b α > r b α ≥ r β = l α > m. This verifies the claim.So choose such a geodesic α . We may furthermore assume that α is topmostamong all curves with the same endpoints. Then B αb is decomposable by the vertexperturbation trick. See Figure 6a. Thus B ′ = B αt is indecomposable. Furthermore,as we did in the beginning of this subsection, we may deduce that B ′ itself satisfiesthe properties (i) and (ii) when we choose L ′ , R ′ , b ′ , t ′ , a ′ and m ′ appropriately; seeFigure 6b. Note in particular that the restriction of L ′ to [ m ′ , a ′ ] equals b α . Thus,by applying Lemma 6.3 to B ′ , we may find a geodesic γ ⊂ B ′ that starts in theinterior of b α , ends in R ′ ((0 , a ′ )) ⊂ R ((0 , a )) and intersects L ′ ((0 , m ′ )) ⊂ L (( m, a )) inbetween. However such γ cannot exist by Lemma 6.2. Compare also Figure 6c. (cid:3) References [1] A. D. Aleksandrov. Foundations of the inner geometry of surfaces.
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Department of Mathematics and Computer Science, University of Cologne, Weyer-tal 86-90, 50931 K¨oln, Germany.
Email address : [email protected] Mathematics Department, Stony Brook University, Stony Brook NY, 11794, USA.
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