CClassification of punctures on complete flatsurfaces ˙Ismail Sa˘glam ∗ Adana Alparslan T¨urke¸s Science and Technology University
Abstract
We investigate the behavior of a complete flat metric on a surfacenear a puncture. We call a puncture on a flat surface regular if it hasa neighborhood which is isometric to that of a point at infinity of acone. We prove that there are punctures which are not regular if andonly if the curvature at the puncture is 4 π . Keywords— flat surface, regular puncture, irregular puncture
Flat surfaces are obtained by gluing Euclidean triangles along their edges appro-priately. They appear in several areas of mathematics and physics. For example,they are studied in dynamics of billiard tables. It is known that each rational poly-gon can be covered by a flat surface with trivial holonomy group. Such surfacesare called translation surfaces and have been studied extensively [11]. Also, thesesurfaces are quite useful in Teichm¨uller theory. Together with quadratic differen-tials, they are used in the proofs of Teichm¨uller’s theorems [7]. They appear inquantum gravity and topological quantum field theory as well. See [1] and [3].These surfaces are interesting for their own sake. For example, Thurston ob-tained complex hyperbolic orbifolds from moduli spaces of certain flat spheres [13].Following Thurston, Bavard and Ghys obtained real hyperbolic orbifolds from the ∗ Electronic address: [email protected]
Mathematics Subject Classification : 51F99, 57M50 a r X i v : . [ m a t h . M G ] F e b oduli spaces of certain polygons in the plane [2]. Troyonov introduced certaingeometric structures on Teichm¨uller spaces by considering the moduli spaces offlat surfaces with prescribed curvature data [16].Compact flat surfaces are examples of length spaces. There is a length min-imizing geodesic between any two points of such a surface. [8], [4]. They canbe triangulated with finitely many triangles. In addition, Gauss-Bonnet formulaholds for these surfaces. See [14], [15].Flat surfaces with regular punctures have been studied in [10]. By a regularpuncture on a flat surface, we mean a puncture which has a neighborhood isometricto that of the point at infinity of a cone. Flat surfaces with possibly irregularpunctures have been studied in [12]. We now state the main results of [12]. Let ¯ S be a complete flat surface.1. ¯ S can be triangulated with finitely many types of triangles.2. Gauss-Bonnet formula holds for ¯ S .3. Each loop on ¯ S has a geodesic representative in its free homotopy class.Our objective is to understand the behavior of a complete flat metric neara puncture. More precisely, we classify complete flat metrics on a disk with apuncture up to the modification equivalence, where two flat metrics on a disk areequivalent if they are ”same” on a neighborhood of the puncture. Now we statethe main results of the present paper.Let ¯ S be a flat surface. • If the curvature at a punctured interior point equals to 4 π , then there areuncountably many modification-equivalence classes of complete flat metricsnear the puncture. • If the curvature at a punctured interior point is not equal to 4 π , then anycomplete flat metric is modification-equivalent to a cone near the puncture. In this paper, we use the notation in [12]. A doubly labeled surface is a compactsurface together with labeled points.
Definition 1.
Let S be a connected compact topological surface perhaps withboundary B . Let l , p , l (cid:48) , p (cid:48) be finite disjoint subsets of S so that • l and p are subsets of the interior of S , p (cid:48) , l (cid:48) are subsets of B .An element in l will be called a labeled interior point. An element in p willbe called a punctured interior point. Other points in the interior of S are calledordinary interior points. An element in B will be called a boundary point. Anelement in l (cid:48) will be called a labeled boundary point. An element in p (cid:48) will be calledpunctured boundary point. Other points in the boundary will be called ordinaryboundary points. A doubly labeled surface, shortly DL surface, is the tuple ( S, B, l , p , l (cid:48) , p (cid:48) )Also we will use the following notation:1. S B = S − B .2. S l = S − l S B, l = S − ( B ∪ l )4. . . .We will denote a doubly labeled surface ( S, B, l , p , l (cid:48) , p (cid:48) ) as S L . Underlyingcompact surface of S L will simply be denoted by S . Note that DL surfaces canbe considered as punctured surfaces with puncture set p ∪ p (cid:48) . Indeed, S p , p (cid:48) is thepunctured surface that we consider. We point out that the punctured and labeledpoints may lie in the boundary.A cone having angle θ >
0, or equivalently curvature κ = 2 π − θ , is the set { ( r, ψ ) : r ∈ R ≥ , ψ ∈ R /θ Z } (1)with the metric µ = dr + r dψ . (2)We can consider a cone as a DL sphere with one punctured and one labeledinterior point. The point (0 ,
0) is called vertex or the origin of the cone. We willdenote it by v . Let θ ( v ) = θ and κ ( v ) = 2 π − θ . Note that we may talk aboutthe point at infinity or the punctured point. We shall denote this point by v ∞ . Definition 2.
Consider a cone with angle θ > .1. κ ( v ∞ ) = 2 π + θ is called the curvature at v ∞ .2. θ ( v ∞ ) = − θ is called the angle at v ∞ . e will denote a cone with angle θ by C θ . Definition 3.
A cylinder of width r , C r , is a metric space obtained by identifyingedges of an infinite strip in the Euclidean plane having width r through oppositepoints. Observe that a cylinder can be considered as DL sphere with two punctured points.By convention, the angles at these punctures are 0. We also call a cylinder as acone of angle 0. Also, again by convention, the curvature at each of the puncturedpoints, is 2 π . Definition 4.
A (flat) cone metric on a DL surface S L is a metric on S p , p (cid:48) sothat each point x in S p , p (cid:48) has a neighborhood isometric to a neighborhood of theapex of the cone C θ = C θ x or a section of a cone V θ = V θ x , and1. l = { y ∈ S p ,B : θ y (cid:54) = 2 π } ,2. l (cid:48) = { y ∈ B − p (cid:48) : θ y (cid:54) = π } .Angle at x , θ ( x ) is defined to be θ x . If x ∈ S p ,B , then the curvature at x , κ ( x ) , isdefined as π − θ ( x ) . If x ∈ B − p (cid:48) , then the curvature is κ ( x ) = π − θ ( x ) . x iscalled singular if κ ( x ) (cid:54) = 0 . Otherwise it is called non-singular. Note that the conditions and assure that the set of singular points and l ∪ l (cid:48) are same. A flat surface with a flat cone metric is called a flat DL surface. Definition 5.
A punctured interior point on a flat DL surface is called regular ifit has a neighborhood isometric to a neighborhood of the point at infinity of a cone.Otherwise, it is called irregular.
An example of a flat DL surface with an irregular punctured interior point isgiven in [12]. Also, the curvature at a punctured interior (or boundary) point of aflat DL surface was defined in [12]. For a DL surface, the following theorem holds.See [12] for a proof.
Theorem 1 (Gauss-Bonnet formula) . Let S L be a complete flat DL surface. Thefollowing formula holds: (cid:88) x ∈ S κ ( x ) = 2 πχ ( S ) , (3) where χ ( S ) is the Euler characteristics of S . .2 Modification If we have a complete flat metric on a DL surface with boundary, then we cancut a triangle having one edge incident to the boundary to get another completeflat metric. Note that the behavior of the metric near the punctures remainsunchanged after this operation. By a modification of flat DL surface, we meanthe surface obtained by removing finitely many triangles which are incident to theboundary.We denote a closed disk with one punctured interior point by D L . We assumethat D L has no labeled interior points and punctured boundary points. As usual,we denote the underlying closed disk by D . Let ¯ D L be a closed disk with onepunctured boundary point. We assume that ¯ D L has no labeled interior points orpunctured interior points. We denote the underlying closed disk by ¯ D . Proposition 1 ([12]) . Each complete cone metric on D L can be modified so thatresulting disk does not have any points with positive curvature on its boundary. Recall that a modification of a flat metric on D L is a flat metric obtained bysuccessively cutting Euclidean triangles which are incident to its boundary. In thissection, we will classify complete flat metrics on D L up to modification equivalence.Two flat complete metrics µ and η are called the modification equivalent if theycan be modified so that there is an orientation preserving isometry between theresulting complete flat disks. Lemma 1 implies that any flat complete metric on D L contains a metric with non-positive curvature data in its equivalence class.Thus, from now on, by a complete flat metric on D L , we mean a metric withnon-positive curvature data. We will also study the case of the disc with onepunctured boundary point. Note that one can define modification equivalence forflat complete metrics on ¯ D L in a similar way. Let µ be a complete cone metric on D L and K be the total curvature on itsboundary. If K = 0 then D L is modification isometric to a half-cylinder. Thereforewe need to consider the case K <
0. We start with a simple fact. i -th vertex b i becomes K n . Lemma 1.
Let a , . . . , a k be real numbers so that not all of them are equal. Let a = (cid:80) ki =1 a i k (4) be their avarage. There exists i so that • a i ≥ a and a i +1 < a , or • a i < a and a i +1 ≥ a ,where a k +1 = a . Lemma 2.
Assume that D L has k labeled boundary points. D L can be modifiedso that resulting metric has also k singular points and these points have samecurvature.Proof. First label singular points as b , . . . , b k so that they are in a cyclic order onthe boundary. Assume that not all of the curvatures are equal. By above lemma,there exists i so that either κ i < K k and κ i +1 ≥ K k . r, κ i ≥ K k and κ i +1 < K k , Let us consider the first case. Remove the triangle with vertices b i , b i +1 , b (cid:48) i +1 havingangles | κ i | − | K | k , | κ i +1 | , π + | K | k − | κ i | − | κ i +1 | from D L , respectively. See Figure 1. • if κ i +1 > K k , then resulting metric has less singular points having curvaturenot equal to K n , • if κ i +1 = K k , then i -th vertex has curvature κ i +1 = K k and i + 1-th vertex hascurvature κ i in the resulting disc.Observe that if κ i ≥ K k and κ i +1 < K k , there is a similar cutting operation withthe following properties: • if κ i > K k then resulting metric resulting metric has less singular pointshaving curvature not equal to K k . • if κ i = K k , then i + 1-th vertex has curvature K k and i -th vertex has curvature κ i +1 in the resulting disc.Apply the following algorithm to D L repeatedly.1. If there is an index i so that either κ i > K k and κ i +1 < K k , or, κ i < K k and κ i +1 > K k . apply cutting operation described above. Observe that the number of sin-gular points having curvature which is not equal to K k decreases after thisoperation. Do this repeadetly so that there are no index i satifying any ofthe properties above.2. After the first step, if all curvatures of the singular points of the resultingmetric are equal, then we are done. If this is not the case, we can permutecurvatures to get a modification for which there exists i so that κ i > K k and κ i +1 < K k , or, κ i < K k and κ i +1 > K k . L , L and the boundary of D L toobtain a cone metric with one singular point having curvature K .
3. Apply the first step to the resulting metric.Since the number of singular points having curvature not equal to K k decreases ateach run of the algorithm, we get a cone metric of desired type in finitely manysteps. Lemma 3.
Assume that ≤ | K | < π . Let x be a boundary point on D L and L be a half-line originating from x and directing toward its interior. This disc canbe modified so that resulting metric has at most one singular point and no pointsexcept x on L is in the triangles removed during the modification.Proof. The case K = 0 is trivial since there are no singular points for this case.Desired modification for the cone metrics having singular points is described inFigure 2. Draw two half lines L , L originating from x making an angle of π with each other. Gauss-Bonnet formula implies that L and L intersect onlyat 2 points, one of which is x . These lines, together with the boundary of S L ,bound a compact region with polygonal geodesic boundary. We can obtain desiredmodification by removing this region. Proposition 2.
Assume that n > and K ≤ so that ( n − π ≤ | K | < nπ. lso assume that the metric has n + m, m ≥ singular points. Let x be one ofthem and L be a half-line originating from x and pointing interior of the disc. Themetric can be modified so that the resulting metric has n singular points of negativecurvature and the removed triangles contain no points on L other than x .Proof. We will prove the statement by induction on the number n . The base case n = 1 is done by the Lemma 3. Assume that n ≥ nπ ≤ | K | < ( n + 1) π. Take a cone metric on the disc having n + 1 + m singular points. Take aboundary segment with vertices x, y and half-lines L , L making angles π withthe segment. See Figure 3. Cut the half plane and glue the half-lines L and L .By this way we get a new cone metric on the punctured disc so that the totalcurvature at its boundary is K (cid:48) = K + π , thus satisfies the inequality below:( n − π ≤ | K (cid:48) | < nπ. Let L the half line formed by gluing L with L . We show the vertex obtainedby x and y by xy . Observe that by induction hypothesis, we can modify this newmetric so that the resulting metric has n singular points and removed triangles donot contain any point of L except x . Now cut this new punctured disc togetherwith the induced metric through L , and glue the half-plane that we removed asin the figure. Resulting cone metric has n + 1 singular points and is a modificationof the metric we started with. Also observe that compact part removed during themodification does not intersect with half-line L except at the vertex x . Theorem 2.
Every complete cone metric on D L with total boundary curvature ( n − π ≤ | K | < nπ can be modified so that resulting metric has n singular points, and curvature ofeach of these singular points is K n .Proof. First modify the metric as in Proposition 2 to get a metric with n singularpoints of negative curvature. Modify this new metric as in Lemma 2, to get ametric of desired type. yL L L L xy
Figure 3: Remove half plane determined by L , L and boundary segment[ x, y ] and glue L and L to get a cone metric of total curvature curvature K (cid:48) ,( n − π ≤ | K (cid:48) | < nπ . Modify it, then cut it thorough L and add the halfplane removed to get a desired modification of the cone metric we started. ≤ | K | < π Lemma 4. If ≤ | K | < π , then the puncture on any complete flat metric on D L is regular.Proof. We know that if K = 0 then D L is isometric to a half-cylinder. Thus thepuncture is regular. First assume that 0 < | K | < π . By Theorem 2, this flat metriccan be modified so that there is only one singular point on its boundary. Let l be the length of its boundary. Take a isosceles triangle with angles − K , π + κ , π + κ with the length of the edge opposite to the vertex with angle − K is equal to l . Ifyou glue the equal edges of this triangle and glue the resulting flat disk with D L ,you get the cone C − κ . Hence the puncture is regular. See Figure 4.Assume that π ≤ | K | < π . Then we can modify D L so that it has twosingular points of curvature K . Call the singular points b and b . Observe thatthe boundary has two components and these components connect b and b . Let l and l be length of these components. Then we can find two isosceles triangleswith the following properties: • First triangle has edges of length l , a, a and the angles at its vertices are α, α, γ . Also the vertex with angle γ is opposite the edge having length l . D L to get a cone. • Second triangle has edges of length l , a, a and the angles at its vertices are β, β, γ . Also the vertex with angle γ (cid:48) is opposite the edge having length l . • γ + γ (cid:48) = − K .Now glue the triangles along the edges having length a to get a flat disk onesingular interior point and two singular boundary points. Note that the angle atthe singular interior point is − κ . If you appropriately glue this disk with D L alongtheir boundaries, you get the cone C − κ . See Figure 5. Hence the puncture isregular. Regarding Theorem 2 , for each K < n − π ≤ | K | < nπ , we will studymodification-equivalence on the set C ( K , n ) = { Flat metrics on D L with n singular boundary points of curvature K n } / isometryNote that two elements µ and η are equivalent, µ ∼ η , if there is an orientationpreserving isometry between them which respects the labeling of the vertices. Each a toget a flat disk with one singular interior point and two singular boundarypoints. Then glue this disk with D L to get a cone. element in C ( K , n ) is uniquely determined by the lengths of the boundary segmentsof the punctured disc. See [12][Theorem 1]. Let Ld ( µ ) = ( l [ b , b ] , . . . , l [ b n , b n +1 ]), where l [ b i , b i +1 ] is the length of the boundarysegment joining b i and b i +1 . We call Ld ( µ ) as length data of µ . Therefore belowmap is a bijection: Ld : C ( K , n ) → R n + µ → Ld ( µ ) , where R + is the set of positive real numbers.We will denote the set of equivalence classes of cone metrics on the disc by M ( K ) = M ( K , n ) := C ( K , n ) / Modification . Now we define principal operations which can be thought as maps C ( K , n ) → C ( K , n ).Let n ≥
3. Take an element in µ ∈ C ( K , n ). Fix an index j ∈ { , . . . n } anda non-negative real number r . From the punctured disc (together with the metric µ ), subtract a quadrangle having angles | K n | , π − | K n | , π − | K n | , | K n | ) , and edge lengths ( r, l [ b j , b j +1 ] + 2 r cos ( π − K n ) , r, l [ b j , b j +1 ]) , which has the segment [ b j , b j +1 ] as an edge. Since | K | n + | K | n = 2 | K | n > π , for each r > C ( K , n ), denote this map byΘ j,r : C ( K , n ) → C ( K , n ) . From the Figure 6, description of this map in terms of length data is clear:Θ j,r : R n + → R n + Θ j,r ([ l , . . . l n ]) = [ l , . . . , l j − , l j − + r, l j + 2 r cos( π − K n ) , l j +1 + r, l j +2 , . . . ] (5)= [ l , . . . , l n ] + [0 , . . . , , j − ↓ r , r cos( π − K n ) , r, , . . . ,
0] (6)Observe that above formulas imply the following:1. Θ j, is identity map on C ( K , n ) (or on R n ),2. Θ j,r ◦ Θ j,r (cid:48) = Θ j,r + r (cid:48) ,3. Θ j,r ◦ Θ j (cid:48) ,r (cid:48) = Θ j (cid:48) ,r (cid:48) ◦ Θ j,r . Definition 6.
We call semi-group generated by Θ j,r ’s, either as maps on C ( K , n ) or R n + , the principal semigroup , and denote it as T = T ( K ) . Remark 1.
Let C be a compact set of D L and µ ∈ C ( K , n ) . There exists anelement T in T ( K ) so that C is a subset of the removed part of the once puncturedsphere after the modification with respect to T . l j = l [ b j , b j +1 ] We recall the basic properties of the circulant matrices. A circulant matrix C isa m × m matrix obtained from one column vector c = [ c , . . . , c m − ] T so thatcolumns of C are determined by cyclic permutations of c as below: C = c c m − . . . c c c c c m − c ... c c . . . ... c m − . . . . . . c m − c m − c m − . . . c c . See [5] for the basic properties of circulant matrices. f C ( x ) = c + c x + c x + · · · + c n − x n − is called associated polynomial of C .Let ω j = exp (cid:16) πıjn (cid:17) for each j = 0 . . . , n −
1, where ı = √− • The set of eigenvalues of C is { λ j = c + c n − ω j + c n − ω j + . . . + c ω n − j : j = 0 , , . . . , n − } . (7) Determinant of C , det( C ) , is n − (cid:89) j =0 ( c + c ω j + c ω j + · · · + c n − ω n − j ) . (8) • Rank of C is the n − d where d is the degree of greatest common divisor ofthe polynomials f C ( x ) and x n − • Eigenvector with eigenvalue λ j is v j = [1 , ω j , ω j , . . . , ω n − j ] T , j = 0 , , . . . , n − . Observe that a circulant matrix is diagonalizable and the eigenvectors ofsuch a matrix do not depend on the coefficients c , c , . . . , c n − . Each Θ j,r can be thought as a translation map R n → R n . The vector space oftranslation maps can be identified with R n and its canonical basis can be identifiedwith the basis of R n that consists of the vectors e = (1 , , . . . , e = (0 , , . . . , . . .e n = (0 , , . . . , j, has coordinates[0 , . . . , , j − ↓ , π − K n ) , , , . . . , T Therefore the matrix that the coordinates of Θ , , Θ , , . . . , Θ n, form is a circulantmatrix with c = 1 , c = 2 cos( π − K n ) , c = 1 and c j = 0 if j (cid:54) = 0 , ,
2. Call thismatrix C . Lemma 5. If K (cid:54) = − π , then π − K n ) ω j + ω j (cid:54) = 0 .Proof. Let ω j = exp (cid:16) πıjn (cid:17) . Assume that ( n − π ≤ | K | < nπ and n >
3. Then( n − n + 1) π ≤ π − K n < π os πn ≤ cos ( π − K n ) < . In particular, < cos ( π − K n ). Now assume that 1 + 2 cos ( π − K n ) ω j + ω j =0. It follows that | ω j | = 2 cos ( π − K n ) >
2, which is impossible. Hence1 + 2 cos ( π − K n ) ω j + ω j (cid:54) = 0 for all j . A similar argument shows that 1 +2 cos ( π − K n ) ω j + ω j (cid:54) = 0 when n = 3 and κ (cid:54) = − π . Proposition 3. If κ < − π , then M ( κ ) consists of one single point, that is, thereare no irregular punctures on D L when the total curvature on the boundary is notequal to − π .Proof. We will show that C ( K , n ) / T ( K ) consists of a single point. Clearly this im-plies that M ( κ ) consists of one point. Lemma 5 implies that det( C ) (cid:54) = 0. Thereforethe group generated by Θ , , . . . , Θ n, is the full group of translations of R n . Sothis group has one orbit. It follows that C ( K , n ) / T ( K ) = R n + / T ( K )has only one point.Now we consider the case K = − π . Proposition 4.
There is a bijection between M ( − π ) and R .Proof. Consider the map R → M ( − π ) sending ( α, β ) to the modification-equivalence class of the metric in C ( − π,
3) having boundary segments of length1 , α, β . Let us denote the modification-equivalence class of the metric on D L withboundary segment of length a, b, c by [ a, b, c ]. Since cos( π − π ) = , it follows thatΘ ,r sends [ a, b, c ] to [ a + r, b + r, c + r ]. Therefore it is easy to see that this mapis surjective. Now assume that [1 , b, c ] and [1 , b (cid:48) , c (cid:48) ] are modification equivalent. Itfollows that there is an orientation preserving isometry sending the flat disk D L whose boundary segments have length 1 + r, b + r, c + r to the flat disk D L whoseboundary segments have length 1 + r (cid:48) , b (cid:48) + r (cid:48) , c (cid:48) + r (cid:48) , where r, r (cid:48) ≥
0. It followsthat r = r (cid:48) , b = b (cid:48) and c = c (cid:48) . So the map is injective.Now we collect the results in Lemma 4, Proposition 3 and Proposition 4 in asingle theorem. Theorem 3.
1. If K (cid:54) = − π , then M ( K ) consists of one singular point. Inother words there are no complete flat metrics on D L so that the punctureis irregular when curvature at the puncture is not equal to π .2. M ( − π ) ≡ R . cknowledgements I am really grateful to Ahmet Refah Torun for his remarks and corrections. Thiswork is supported by Research Fund of Adana Alparslan T¨urke¸s Science and Tech-nology University. Project Number: 18119001.
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