Decomposing Centrally Symmetric Convex Polyhedral Surfaces into Parallelograms

Decomp osing Cen trally Symmetric Con v ex P olyhedral Surfaces in to P arallelograms Zili W ang ∗ Sc ho ol of Science Sun Y at-sen Universit y w angzli6@mail.sysu.edu.cn Cong W u Mathematics Departmen t Shenzhen Middle Sc ho ol cwu@shenzhong.net Marc h 31, 2026 Abstract Let M 2 N ( δ 1 , δ 2 , . . . , δ N ) be the mo duli space of centrally symmetric conv ex p olyhedral surfaces with 2 N lab eled v ertices and prescrib ed cone-deficits δ 1 , δ 2 , . . . , δ N . W e sho w that M 2 N ( δ 1 , δ 2 , . . . , δ N ) has the structure of a real h yp erbolic manifold of dimension 2 N − 3. When N = 4 and 5, w e show that every surface in M 2 N ( δ 1 , δ 2 , . . . , δ N ) can be decomp osed into at most 2  2 N − 2 2  parallelograms, and the decomp osition is inv ariant under the antipo dal map. Using the edge-lengths of these parallelograms as co ordinates, we show that the mo duli space of cen trally symmetric polyhedral surfaces with 8 unlabeled v ertices and cone-deficits π 2 is isometric to the quotient of a real h yp erbolic regular ideal 5-simplex by the dihedral group D 6 . 1 In tro ducton The surface of a p olyhedron is a Euclidean cone sphere. It has a metric that is locally Euclidean except at ev ery vertex v i , where it is lo cally isometric to a Euclidean cone with angle θ i . The v alue δ i = 2 π − θ i is called the cone-deficit of v i . It follows from the Gauss–Bonnet Theorem that the sum of all cone-deficits equals 4 π . A p olyhedron is conv ex if 0 < δ i < 2 π for all i . In [ 10 ], Thurston considered the mo duli space of conv ex p olyhedra with prescrib ed cone-deficits. By decomp osing the surfaces of p olyhedra in to triangles, he built co ordinates to describ e these p olyhedra, and show ed that their mo duli space admits a Hermitian metric, with resp ect to which it is a complex h yp erbolic manifold. In addition, under certain conditions, the metric completion of this mo duli space is an orbifold. This result provides a geometric wa y to in terpret the mono drom y groups of hypergeometric functions in vestigated by Picard [ 7 ][ 8 ], Mostow and Deligne [ 2 ][ 3 ][ 6 ]. In addition, this result has b een applied in counting and en umerating sp ecific structures on the sphere, such as counting con vex tilings [ 5 ] and enumerating fullerenes [ 4 ]. Th urston’s idea in building coordinates and the Hermitian form is also used to study the moduli space of con vex p olygons [ 1 ]. Inspired b y this work, we study the space of cen trally symmetric con v ex p olyhedra. A p olyhe- dron is centrally symmetric if it is isometric to a p olyhedron symmetric ab out the origin, whic h ∗ Zili W ang is supported b y the Natural Science F oundation of Guangdong Pro vince (Grant No. 2023A1515010658) 1 means that a p oin t x lies on its surface if and only if its antipo dal p oin t − x do es. The map that sends x to − x for all x on the surface is called the antipo dal map . Thus, a centrally symmetric p olyhedron has an ev en num b er 2 N of v ertices, and its cone-deficits are describ ed by N p ositiv e n umbers that sum up to 2 π . The main results of this pap er can b e summarized as follows. In Section 2, w e show that the mo duli space of centrally symmetric conv ex p olyhedra with prescrib ed cone-deficits has the structure of a real hyperb olic manifold. The w ay we compute the signature of the quadratic form is v ery similar to the wa y Thurston computed it for the Hermitian form in [ 10 ]. In addition, it is kno wn that every centrally symmetric conv ex p olygon can b e decomp osed in to finitely many parallelograms. This leads us to conjecture that a similar result holds for every cen trally symmetric con vex p olyhedral surface. In a previous work [ 12 ], we sho wed that the surface of every centrally symmetric o ctahedron has a parallelogram decomp osition that is inv ariant under the an tip o dal map. In Section 3, w e establish a metho d that relates a parallelogram decomposition of a polyhedral surface to a structure called “loop arrangement” on the sphere. Using this metho d, w e generalize this result to centrally symmetric conv ex p olyhedra with 8 and 10 vertices in Section 4 and 5. In these cases, it turns out that the edge-lengths of the parallelograms in the decomp osition can b e used as co ordinates to describ e p olyhedra. These co ordinates are differen t from Thurston’s co ordinates in [ 10 ], but may sometimes provide a more concrete geometric description of the mo duli spaces when the polyhedra are cen trally symmetric. F or example, in Section 4, we show that the mo duli space of centrally symmetric p olyhedra with 8 vertices whose cone-deficits are π 2 is isometric to the quotien t of a real h yp erbolic regular ideal 5-simplex by a group isomorphic to D 6 . 2 Space of Cen trally Symmetric Con v ex P olyhedra Let δ 1 , δ 2 , . . . , δ N b e N p ositiv e n umbers that sum up to 2 π . Let C 2 N ( δ 1 , δ 2 , . . . , δ N ) b e the space of centrally symmetric con vex p olyhedral surfaces with N pairs of antipo dal v ertices lab eled by i + and i − (where 1 ≤ i ≤ N ) with prescrib ed cone-deficit δ i . In this space, tw o surfaces are considered equiv alent if and only if they differ by a Euclidean isometry that preserves the lab els of all the v ertices. Theorem 2.1. The sp ac e C 2 N ( δ 1 , δ 2 , . . . , δ N ) has the structur e of a c omplex manifold of dimension N − 1 . Pr o of. Let S be a surface in C 2 N ( δ 1 , δ 2 , . . . , δ N ). First, w e will asso ciate S with N − 1 complex n umbers. Consider the set of directed cycles on S , each of which: 1) is inv ariant under the antipo dal map, 2) consists of directed edges on S (line segmen ts connecting the vertices of S ), and 3) visits all vertices of S except i + and i − for exactly one i ∈ { 1 , 2 , . . . , N } . Figure 1 shows tw o cycles satisfying the conditions ab o ve, where S is the surface of a cub e in C 8 ( π 2 , π 2 , π 2 , π 2 ). In the set ab o ve, let c b e a cycle of the shortest length. W e claim that the edges in c do not cross on S . Supp ose that u 1 v 1 and u 2 v 2 are t wo directed edges that cross on S , Without loss of generalit y , we assume that c visits the four v ertices in the order u 1 , v 1 , u 2 , v 2 . Then we can replace u 1 v 1 and u 2 v 2 with the directed edges u 1 u 2 and v 1 v 2 , rev erse the direction of the path from v 1 to u 2 , and p erform the same trick on their antipo dal images. This will make c a strictly shorter cycle satisfying the three conditions ab o ve, a con tradiction. F or example, in Figure 1 , if w e p erform this tric k on the cycle in the first picture, we obtain the cycle in the second picture, whic h is strictly shorter. 2 Figure 1: Two directed cycles on the surface of a cube. The second one is strictly shorter. Without loss of generalit y , supp ose that c visits the vertices of S in the order 1 + , 2 + , . . . , ( N − 1) + , 1 − , 2 − , . . . , ( N − 1) − , but not N + or N − . F or conv enience, we denote the directed edges in c by their starting points and endp oin ts, such as 1 + 2 + , 2 + 3 + , etc. W e cut S along c into t wo surfaces that are top ologically equiv alent to disks and are antipo dal to each other. Let S + b e the surface con taining N + . W e can cut S + op en b y cutting along an edge inside S + joining N + and a vertex of c . Without loss of generalit y , we assume that this line segmen t joins ( N − 1) − and N + , and we denote it by ( N − 1) − N + . In this wa y , we obtain a p olygon P S + that can b e unfolded on the complex plane C . In Figure 2 , S is the surface of a cub e with N = 4, S + is obtained b y cutting S along the cycle from the previous example, and P S + is obtained after cutting S + along 4 + 3 − and unfolding it on the plane. Figure 2: The surface S + is cut open and unfolded to a planar polygon P S + . Ev ery directed edge of P S + determines a v ector from its starting p oin t to its endpoint. W e think of these v ectors as complex num b ers, and denote them by Z ( N − 1) − 1 + , Z 1 + 2 + , . . . , Z ( N − 2) + ( N − 1) + , Z ( N − 1) + 1 − , Z 1 − 2 − , . . . , Z ( N − 2) − ( N − 1) − . Among these complex num b ers, we take N − 1 of them ranging from Z 1 + 2 + to Z ( N − 1) + 1 − . By choosing an unfolding such that Z ( N − 1) − 1 + is a p ositive real n umber and P S + is on its right-hand side, these complex num b ers are uniquely determined. In this w ay , w e asso ciate S to a vector ( Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − ) ∈ C N − 1 . In Figure 2 , the num b ers Z 1 + 2 + , Z 2 + 3 + and Z 3 + 1 − are lab eled in the second picture. Con versely , the complex vector ( Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − ) also determines P S + and thus 3 S . First, the num b er Z 1 + 2 + determines the clo c kwise angle from Z ( N − 1) − 1 + to Z 1 + 2 + , which w e assume is π − θ (see Figure 2 for an example). Then the clo c kwise angle from Z ( N − 1) + 1 − to Z 1 − 2 − equals π − (2 π − δ 1 − θ ) = δ 1 + θ − π , so Z 1 − 2 − is uniquely determined by Z ( N − 1) + 1 − and δ 1 . Therefore, giv en Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − and the prescrib ed cone-deficits, we can similarly determine the remaining complex n umbers up to Z ( N − 2) − ( N − 1) − . The last tw o edges of P S + come from the cutting slit ( N − 1) − N + , so they hav e an equal modulus and mak e an angle of δ N . They are uniquely determined b ecause the p olygon P S + needs to close up. In particular, each of them is a complex linear combination of Z 1 + 2 + , Z 2 + 3 + , . . . , and Z ( N − 1) + 1 − . By v arying the complex num b ers Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − lo cally , w e can obtain polyhedral surfaces near S , so we ha ve constructed a lo cal co ordinate chart ϕ : U → C N − 1 , where U is a neigh b orho od of S in C 2 N ( δ 1 , δ 2 , . . . , δ N ). T o show that C 2 N ( δ 1 , δ 2 , . . . , δ N ) is a complex manifold, it remains to sho w that when tw o co ordinate charts ov erlap, the change-of-coordinate map is a complex linear transformation. It would b e helpful to think of ϕ as the restriction of a dev eloping map D : f S → C , where f S is the universal cov er of S 0 , the complement of the vertices of S . Let p : f S → S 0 b e the cov ering map. One can view f S as the space of homotopic paths in S 0 with a common basep oint. If e 1 and e 2 are t wo directed edges (not including the endp oin ts) of f S suc h that p ( e 1 ) = p ( e 1 ), then D ( e 1 ) and D ( e 2 ) can b e computed b y the analytic contin uation of U along paths with different homotop y classes in S 0 . It turns out that D ( e 1 ) and D ( e 2 ) differ by a rotation by a constant angle that is determined by the prescrib ed cone-deficits. One ma y refer to the discussion on the developing map and the holonom y in [ 11 ] (Section 3.5) or [ 10 ] (the paragraph after Prop osition 3.1) for more details. Let ϕ ′ : U ′ → C N − 1 b e another c hart such that S ∈ U ∩ U ′ . Supp ose that ϕ ′ is constructed b y cutting S along a different directed cycle c ′ and a new cutting slit. Up to a comp osition of translations and rotations, we may assume that ϕ ′ is also a restriction of D . Let e b e a directed edge of f S so that p ( e ) is in c ′ or is the new cutting slit. If p ( e ) is in c or is ( N − 1) − N + , we hav e seen that D ( e ) is a complex linear com bination of Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − . If p ( e ) is any other edge of S , then one can find edges e 1 , e 2 , . . . , e n in f S that form a cycle in f S with e , and p ( e i ) (or − p ( e i )) is ( N − 1) − N + or in c . Then D ( e ) is again a complex linear combination of Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − . Th us, the transformation from Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − to the coordinates giv en by ϕ ′ is complex linear, and is v alid in U ∩ U ′ . This sho ws that C 2 N ( δ 1 , δ 2 , . . . , δ N ) is a complex manifold of dimension N − 1. In the proof of Theorem 2.1 , we use N − 1 directed edges of S to construct a lo cal co ordinate c hart containing S . W e hav e seen that the co ordinates depend on which N − 1 directed edges of S w e c ho ose (also on ho w w e dev elop them on to the plane). The c hoice of directed edges do es not ha ve to b e restricted to the w ay in the pro of of Theorem 2.1 . In fact, if e 1 , e 2 , . . . , e N − 1 are directed edges of f S suc h that the complex linear transformation from ( Z 1 + 2 + , Z 2 + 3 + , . . . , Z ( N − 1) + 1 − ) to ( D ( e 1 ) , D ( e 2 ) , . . . , D ( e N − 1 )) is inv ertible, then w e can also use ( D ( e 1 ) , D ( e 2 ) , . . . , D ( e N − 1 )) as co ordinates to describ e surfaces near S . In this case, w e call ( D ( e 1 ) , D ( e 2 ) , . . . , D ( e N − 1 )) a local frame near S on C 2 N ( δ 1 , δ 2 , . . . , δ N ). F or example, in Figure 3 , we draw three different sets of directed edges on the surface of a cub e. By choosing a wa y to develop eac h set of directed edges on to the plane, we can get different lo cal frames near this surface on C 8 ( π 2 , π 2 , π 2 , π 2 ). W e will return to lo cal frames when we discuss parallelogram decomp ositions of p olyhedral surfaces in Section 4. Next, let M 2 N ( δ 1 , δ 2 , . . . , δ N ) b e the mo duli space of C 2 N ( δ 1 , δ 2 , . . . , δ N ), in whic h tw o p oly- hedral surfaces are equiv alent if and only if they differ by a similarity preserving the vertex- 4 Figure 3: Directed edges on the surface of a cub e that give rise to differen t local frames. lab els. Alternativ ely , we can regard M 2 N ( δ 1 , δ 2 , . . . , δ N ) as the space of p olyhedral surfaces in C 2 N ( δ 1 , δ 2 , . . . , δ N ) whose surface areas are equal to 1. In [ 10 ], it was shown that the mo duli space of conv ex p olyhedra with N prescrib ed cone-deficits has the structure of a complex hyperb olic manifold of dimension N − 3. The main idea of proof is to sho w that the surface area is a Hermitian form of signature (1 , n − 3) with r esp ect to the co ordinates arising from decomp osing the surfaces into triangles. The argumen t is based on induction on the n umber of vertices, where multiple v ertices of a surface may “collide” to pro duce a surface with few er v ertices. One ma y refer to [ 10 ] and [ 9 ] for more details of this op eration. Next, we are going to use collisions of vertices to prov e the follo wing result: Theorem 2.2. When N ≥ 3 , the sp ac e M 2 N ( δ 1 , δ 2 , . . . , δ N ) has the structur e of a r e al hyp erb olic manifold of dimension 2 N − 3 . Pr o of. The idea of pro of is to express the surface area as a quadratic form of signature (1 , 2 N − 3) on C 2 N ( δ 1 , δ 2 , . . . , δ N ). How ev er, the coordinates we use will b e based on induction and differen t from those in Theorem 2.1 . In the base case where N = 3, M 6 ( δ 1 , δ 2 , δ 3 ) is the mo duli space of centrally symmetric octa- hedra. W e sho wed in [ 12 ] that the surface area is a quadratic form of signature (1 , 3) with resp ect to the coordinates obtained by decomp osing the surfaces into parallelograms. When N > 3, let S b e a polyhedral surface in C 2 N ( δ 1 , δ 2 , . . . , δ N ). W e choose t wo non-an tip odal v ertices such that the sum of their cone-deficits is strictly less than 2 π . This is p ossible since N P i =1 δ i = 2 π . Without loss of generality , we assume that these tw o v ertices are 1 + and 2 + . Then w e choose an edge connecting 1 + and 2 + , cut S op en along this edge, and glue a cone to the slit. This cone can b e obtained b y gluing tw o congruent triangles with base angles δ 1 2 and δ 2 2 . In this wa y , w e kill the cone-deficits at 1 + and 2 + , and pro duce a new vertex V + with cone- deficit δ 1 + δ 2 2 . This op eration is demonstrated in Figure 4 . Simultaneously , W e p erform the same op eration on the vertices 1 − and 2 − . In this wa y , we construct a new p olyhedral surface S ′ in C 2 N − 2 ( δ 1 + δ 2 , δ 3 , . . . , δ N ). By induction hypothesis, there exists a local co ordinate c hart from an op en neigh b orhoo d of S ′ in C 2 N − 2 ( δ 1 + δ 2 , δ 3 , . . . , δ N ) to R 2 N − 4 with resp ect to whic h w e can write the surface area function in the form k 1 x 2 1 − k 2 x 2 2 − · · · − k 2 N − 4 x 2 2 N − 4 . The surface area of S equals that of S ′ min us the areas of the four congruen t triangles with base angles δ 1 2 and δ 2 2 . T o complete the induction, it suffices to show that the area of each triangle has the form k ( x 2 2 N − 3 + x 2 2 N − 2 ), where k > 0, and x 1 , x 2 , . . . , x 2 N − 4 , x 2 N − 3 , x 2 N − 2 form a set of real co ordinates for surfaces near S . 5 Figure 4: V ertices 1 + and 2 + “collide” to v 12 , resulting in a surface with one few er vertex. Fix a ray r emanating from v 12 on S ′ as a reference. The surface S is uniquely determined b y b oth the distance d b et ween v 12 and 1 + , and the clo c kwise angle θ from the ray r to the ray from v 12 to 1 + . Then the area of each triangle is a multiple of d 2 b y a constant determined by δ 1 2 and δ 2 2 . Thus, w e can take x 2 N − 3 = d cos θ and x 2 N − 2 = d sin θ . The co ordinates x 1 , x 2 , . . . , x 2 N − 4 , x 2 N − 3 , x 2 N − 2 remain v alid for surfaces near S in C 2 N ( δ 1 , δ 2 , . . . , δ N ). This completes the induction. Therefore, surfaces near S in C 2 N ( δ 1 , δ 2 , . . . , δ N ) with area 1 corresp ond to vectors in R 2 N − 2 of length 1 with resp ect to a quadratic form of signature (1 , 2 N − 3). This establishes lo cal co ordinate charts from M 2 N ( δ 1 , δ 2 , . . . , δ N ) to the real hyperb olic space H 2 N − 3 . Finally , an y c hange-of-co ordinate map on C 2 N ( δ 1 , δ 2 , . . . , δ N ) keeps the surface area in v arian t. This sho ws that M 2 N ( δ 1 , δ 2 , . . . , δ N ) is a real hyperb olic manifold of dimension 2 N − 3. Let M 2 N ( δ 1 , δ 2 , . . . , δ N ) b e the metric completion of M 2 N ( δ 1 , δ 2 , . . . , δ N ) with resp ect to the real hyperb olic metric ab o v e. F or example, the surface in the third picture of Figure 4 is not in M 2 N ( δ 1 , δ 2 , . . . , δ N ) but in M 2 N ( δ 1 , δ 2 , . . . , δ N ) if it has area 1. This is because one can construct a Cauch y sequence of surfaces in M 2 N ( δ 1 , δ 2 , . . . , δ N ) conv erging to it by colliding tw o vertices and normalizing the surface area at the same time. W e will return to this space in Section 4. 3 Lo op Arrangemen ts In this section, we show that there is a natural corresp ondence b et ween a parallelogram decompo- sition of a centrally symmetric polyhedral surface and a “lo op arrangemen t” on the sphere S 2 . Supp ose that we are giv en 2 N distinct vertices on S 2 lab eled by 1 + , 1 − , 2 + , 2 − , . . . , N + and N − , where i + and i − are an tip o dal for 1 ≤ i ≤ N . By a lo op w e mean a great circle on S 2 that do es not pass through any lab eled vertices. A set of lo ops form a lo op arrangemen t if any tw o of them are non-homotopic with resp ect to the labeled vertices. Since loops are great circles, every tw o of them intersect exactly t wice at tw o antipo dal p oin ts. W e assume without loss of generalit y that no three lo ops are concurren t, otherwise we can p erturb an y one of the them. As an example, in Figure 5 , we sketc h the “front view” of a lo op arrangement with 6 lo ops on a sphere with 8 lab eled v ertices. Each red curv e represents a loop. W e use black dots to represent the v ertices 1 + , 2 + , 3 + , and 4 + . In addition, we use gray dots to represent the v ertices 2 − , 3 − , and 4 − (ev en though they should b e invisible from the actual front view) b ecause it will b e helpful for our demonstrations later. 6 Figure 5: A loop arrangemen t with 6 lo ops (red) on a sphere with 8 lab eled vertices. Let A be a lo op arrangement on a sphere with 2 N lab eled vertices. Regarding it as a graph on the sphere, we can take its dual graph D ∗ . Since no three lo ops are concurrent, the faces of D ∗ are all quadrilaterals. As we sho w next, D ∗ is closely related to the decomp osition of p olyhedral surfaces in C 2 N ( δ 1 , δ 2 , . . . , δ N ) into parallelograms. More precisely , we will first turn each quadrilateral face in to a parallelogram by sp ecifying its side lengths and angles, and then ch eck that we do obtain a p olyhedral surface in C 2 N ( δ 1 , δ 2 , . . . , δ N ). W e first lab el all lo ops in A . F or example, if A is the loop arrangement in Figure 5 , we can denote its 8 loops by l a , l b , l c , l d , l e , and l f . F or conv enience, these loops are marked by a , b , . . . , f in the first picture of Figure 6 . In the second picture, w e dra w the dual graph D ∗ of A in blue. Figure 6: The loop arrangemen t A with its dual graph D ∗ (blue). W e say that an edge in D ∗ is transverse to the lo op l i if it is dual to an arc of this lo op. F or eac h lo op l i , W e assign a p ositiv e num ber i to all edges in D ∗ transv erse to l i . These num b ers will b e the edge-lengths of the quadrilaterals. F or example, in Figure 6 , the thick ened edges are all transv erse to l a and assigned length a . Next, we assign angles to each quadrilateral. Let Q b e a quadrilateral in D ∗ . Consider tw o edges of Q that meet at a vertex v in Q . W e are going to assign a v alue θ as the angle made b y these tw o edges in Q . T o compute θ , let l i and l j b e the tw o lo ops to which these t wo edges are transv erse. Then l i and l j divide the sphere in to 4 lunes, one of which contains v in its in terior. W e denote this lune by L . F or example, in the first picture of Figure 7 , w e sketc h a quadrilateral Q (blue), the angle θ we wan t to compute in Q , and the lune L (red) w e consider to compute θ . 7 Figure 7: Assigning angles to quadrilaterals in D ∗ . Consider all the lab eled vertices enclosed b y L . W e compute θ by first taking the sum of the cone-deficits associated to the enclosed lab eled vertices, then dividing this angle sum by half, and finally taking its supplementary angle. That is, θ = π − 1 2 X n + or n − ∈L δ n (1) F or example, consider the t wo quadrilaterals in the second picture of Figure 7 . F or the top quadrilateral, we hav e θ 1 = π − 1 2 δ 1 , since L encloses 1 + only . F or the b ottom quadrilateral, we ha ve θ 2 = π − 1 2 ( δ 1 + δ 4 ) = 1 2 δ 2 + 1 2 δ 3 . By symmetry , the v alue of the opp osite angle in Q is also θ . As the sum of all δ i ’s equals 2 π , Equation 1 implies that θ > 0. Let θ ′ b e the v alue assigned to any one of the remaining tw o angles in Q , and let L ′ b e the lune we consider to compute θ ′ . Note that L ∪ L ′ is a hemisphere. Therefore, the sum of the cone-deficits of the lab eled vertices enclosed in L and those enclosed in L ′ equals 2 π . Applying Equation 1 , we get θ + θ ′ = π . Since θ , θ ′ > 0, with the edge-lengths we assigned previously , Q is no w a parallelogram. Therefore, we regard all quadrilaterals in D ∗ as solid parallelograms. In this wa y , we obtain a p olyhedral surface (Euclidean cone sphere) that is almost everywhere flat except p ossibly at the v ertices of the parallelograms. These parallelograms form a parallelogram decomposition of the surface. By v arying the edge-lengths of the parallelograms, we obtain different p olyhedral surfaces. Our next goal is to sho w that these surfaces all b elong to C 2 N ( δ 1 , δ 2 , . . . , δ N ). That is: Theorem 3.1. On a spher e S 2 with 2 N lab ele d vertic es, let A b e a lo op arr angement with 2 N − 2 lo ops ( N ≥ 3 ). L et D ∗ b e the dual gr aph of A , and S b e any surfac e arising fr om D ∗ by assigning e dge-lengths and angles to the quadrilater als in D ∗ in the way discusse d ab ove. Then S ∈ C 2 N ( δ 1 , δ 2 , . . . , δ N ) . Pr o of. The key step of pro of is to establish Equation 2 b elo w and generalize it to Equation 4 b y induction. Then w e will use Equation 4 to c heck that every v ertex in S has the correct cone-deficit. Since 2 N − 2 ≥ 4, we can choose three distinct lo ops in A , denoted by l i , l j and l k . By assumption, they are not concurrent, so they divide the sphere in to 8 geo desic triangles. Denote one of these geo desic triangles by T . Let u i , u j , and u k b e the v ertices of T suc h that u i is opp osite to the edge of T contained in l i , and similarly for u j and u k . The first picture of Figure 8 is a sk etch of T with its vertices. 8 Figure 8: 1) The geo desic triangle T ; 2) The vertices v ∗ i , v ∗ j , and v ∗ k ; 3) The region L i ∪ L j ∪ L k . Let p ∗ i b e the parallelogram dual to u i in D ∗ , and v ∗ i b e its unique v ertex in the interior of T . Let θ i b e the angle of p ∗ i at v ∗ i . Similarly , we define p ∗ j , p ∗ k , v ∗ j , v ∗ k , θ j and θ k . The second picture of Figure 8 sho ws an example where v ∗ i , v ∗ j , and v ∗ k are mutually distinct, but they may also rep eat in some cases. No w w e wan t to establish the follo wing Equation 2 : θ i + θ j + θ k = 2 π − X n + or n − ∈T δ n (2) Let L i b e the lune b ounded by l j and l k and con taining T , and similarly define L j and L k . Consider the region L i ∪ L j ∪ L k . Note that its complemen t in S 2 is just the interior of its image under the antipo dal map. In the third picture of Figure 8 , the region L i ∪ L j ∪ L k is painted in red and its complement in S 2 is white. In addition, L i ∩ L j ∩ L k = T . Thus, by symmetry and the Gauss-Bonnet Theorem, w e hav e X n + or n − ∈L i δ n + X n + or n − ∈L j δ n + X n + or n − ∈L k δ n − 2 X n + or n − ∈T δ n = 2 π (3) Therefore, applying Equation 1 and Equation 3 , w e hav e θ i + θ j + θ k = ( π − 1 2 X n + or n − ∈L i δ n ) + ( π − 1 2 X n + or n − ∈L j δ n ) + ( π − 1 2 X n + or n − ∈L k δ n ) = 3 π − 1 2 (2 π + 2 X n + or n − ∈T δ n ) = 2 π − X n + or n − ∈T δ n W e hav e thus deriv ed Equation 2 . No w w e focus on all the loops in A . These loops divide S 2 in to con vex geo desic p olygons. Let P m b e one of these p olygons. Without loss of generality , we assume that it is b ounded by the lo ops l 1 , l 2 , . . . , and l m in A . Figure 9 shows an example when m = 6. Consider the m parallelograms in D ∗ dual to the m vertices of P m . Let v ∗ 1 , v ∗ 2 , . . . , v ∗ m b e the v ertices of these parallelograms inside 9 P m , whic h ma y be distinct or rep eat. Let Θ m b e the sum of the m angles of these parallelograms at the v ∗ i ’s. In the example in Figure 9 , the 6 parallelograms are drawn in blue, and the sum of all the angles mark ed in red equals Θ 6 . Figure 9: An example of P 6 . The sum of all the angles marked in red equals Θ 6 . Next, we wan t to establish the following equation: Θ m = 2 π − X n + or n − ∈P m δ n (4) Since the base case is Equation 2 , it remains to complete the inductive step. Consider the con vex geodesic ( m − 1)-gon made b y l 1 , l 2 , . . . , l m − 1 that con tains P m . W e denote this polygon by P m − 1 , and let T be the triangle P m − 1 \ P m . In the previous example of P 6 , T is the green triangle in Figure 10 . Let Θ 3 b e the sum of the 3 angles for T as w e defined previously , and let Θ m − 1 b e the sum of the m − 1 angles for P m − 1 . In the first picture of Figure 10 , Θ 3 is the sum of the three angles marked in green. In the second picture, Θ m − 1 is the sum of the five angles mark ed in blue. Figure 10: Computing Θ m using Θ 3 (sum of the green angles) and Θ m − 1 (sum of the blue angles). By induction h yp othesis, we hav e Θ m − 1 = 2 π − X n + or n − ∈P m − 1 δ n 10 No w we add up the m angles for P m (whose sum equals Θ m ) and the 3 angles for T (whose sum is 2 π − P n + or n − ∈T δ n b y Equation 2 ). Among these angles, there are tw o pairs of complementary angles, so their sum equals 2 π . These four angles app ear in the tw o parallelograms dual to the tw o v ertices that b elong to b oth P m and T . In Figure 10 , w e draw these t wo parallelograms and mark the four angles in the third picture. Apart from these four angles, the sum of the remaining angles equals Θ m − 1 . Therefore, we obtain Θ m + Θ 3 = Θ m − 1 + 2 π . Com bined with the induction h yp othesis, we ha ve Θ m + 2 π − X n + or n − ∈T δ n = Θ m − 1 + 2 π = 2 π − X n + or n − ∈P m − 1 δ n + 2 π After rearrangement, it b ecomes Θ m = 2 π − ( X n + or n − ∈P m − 1 δ n − X n + or n − ∈T δ n ) = 2 π − X n + or n − ∈P m δ n where the second equality follows from the fact that P m − 1 = P m ∪ T . This completes the induction and the pro of of Equation 4 . Finally , we chec k that every vertex in S has the right cone-deficit. Let v ∗ b e any vertex in D ∗ . W e need to show that the cone-angle at v ∗ in S is 2 π − δ n if v ∗ is the lab eled v ertex n + or n − , and 2 π otherwise. Let P m b e the conv ex geo desic p olygon cut by the lo ops in A that is dual to v ∗ . Then the parallelograms dual to the vertices of P m meet at v ∗ , so Θ m is just the cone-angle at v ∗ . In addition, v ∗ is the only vertex in D ∗ that is inside P m . Th us, by Equation 4 , if v ∗ is the lab eled v ertex n + or n − , then Θ m = 2 π − δ n . If it is not a lab eled v ertex, then the sum of cone-deficits inside P m is 0, so Θ m = 2 π . This completes the pro of of the theorem. In preparation for the next section, we introduce some terminologies. Recall that all quadrilaterals in D ∗ are parallelograms whose angles are determined by the prescrib ed cone-deficits. W e can assign arbitrary p ositiv e num b ers to the edge-lengths of those parallelograms, provided that all the edges transverse to the same lo op hav e the same length. In this wa y , we obtain a p olyhedral surface. Let O b e the collection of all p olyhedral surfaces that can b e obtained in this wa y . By Theorem 3.1 , O is a subset of C 2 N ( δ 1 , δ 2 , . . . , δ N ). If we allow some edges in D ∗ to hav e 0 length, what we can obtain is a larger set denoted by O . Note that O ⊈ C 2 N ( δ 1 , δ 2 , . . . , δ N ), since w e can assign length 0 to all edges in D ∗ . F rom no w on, w e will call O the space arising from A , and O the closure of O . By a face σ i of O we mean the collection of elemen ts in O where the lengths of all edges in D ∗ transv erse to the loop l i are 0. No w supp ose that A ′ is also a lo op arrangement on a sphere with 2 N lab eled vertices. W e sa y that A and A ′ are equiv alen t if there exists a homeomorphism b et ween the tw o spheres that maps vertices to vertices and lo ops to lo ops preserving their lab els. If A and A ′ differ by only one lo op (that is, they b ecome equiv alent after we replace one lo op in A or A ′ b y another), we say that A and A ′ are adjacent . F or example, consider the lo op arrangements A and A ′ in Figure 11 . Note that A ′ can b e obtained from A by moving the vertices 2 + , 3 + , 2 − and 3 − across the loop l a sim ultaneously . In fact, A and A ′ are adjacen t. One can see this by dragging the vertices 2 + , 3 + , 2 − and 3 − in A ′ to their positions in A , and distorting the lo op l a in A ′ (without changing its homotopy class) to the green curv e in the third picture. This sho ws that A and A ′ differ only b y l a . 11 Figure 11: The loop arrangemen ts A and A ′ are adjacent and differ b y l a . Let O ′ b e the space arising from A ′ and O ′ b e its closure. W e say that O and O ′ are adjacent if A and A ′ are adjacent lo op arrangements. Supp ose that A and A ′ differ only by the loop l i . Then σ i is a common face of O and O ′ . In this case, O and O ′ can b e on the same side or on differen t sides of σ i . W e say that O and O ′ are on the same side of σ i if O and O ′ o verlap in p olyhedral surfaces sufficiently close to σ i . Otherwise, w e say that they are on differen t sides of σ i , or O ′ is on the other side of σ i with resp ect to O . Finally , each lo op in A can b e equipp ed with an orientation . Since a loop divides the sphere in to tw o parts, it can hav e tw o p ossible orientations dep ending on which part is considered its inside or outside. W e can choose one orientation from the tw o possibilities, and represent it b y an arro w on that lo op p oin ting outside. The orientation of a lo op induces a direction on each edge transv erse to that lo op in D ∗ . F or example, consider the lo op arrangement in Figure 12 . The first picture sho ws the orientation we choose for the lo op l a , and the second picture shows the induced directions of some edges transverse to l a in D ∗ . By c ho osing an orientation for eac h lo op in A , w e can turn D ∗ in to a directed graph. Figure 12: 1) The orientation of l a is marked b y a blue arrow; 2) The induced directed edges in D ∗ . 12 4 P olyhedra with 8 V ertices In this section, we fo cus on C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ), the space of centrally symmetric conv ex p olyhedral surface with prescrib ed cone-deficits δ 1 , δ 2 , δ 3 , and δ 4 . The goal is to show that every polyhedral surface in this space has a parallelogram decomp osition that is in v ariant under the antipo dal map. Restricted to this section, a lo op arrangement alwa ys contains 6 lo ops labeled by l a , l b , l c , l d , l e , and l f on a sphere with 8 v ertices labeled b y ± 1, ± 2, ± 3 and ± 4. Let A 1 b e the lo op arrangement in the first picture of Figure 6 , and D ∗ 1 b e its dual graph. Let O 1 the space arising from A 1 , and O 1 b e its closure. Let a , b , c , d , e and f b e p ositiv e v ariables. Recall from Section 3 that all quadrilaterals in D ∗ 1 are considered parallelograms whose angles are determined b y the prescribed cone-deficits. Th us, we hav e a map F 1 that sends ev ery vector ( a, b, c, d, e, f ) to a polyhedral surface S in O 1 b y assigning length i to all edges in D ∗ 1 transv erse to l i , where i ∈ { a, b, c, d, e, f } . W e denote b y (0 , ∞ ) 6 the p ositiv e orthant of R 6 , which is the space of vectors in R 6 where all comp onen ts are p ositiv e. Similarly , denote b y [0 , ∞ ) 6 the non-negative orthant of R 6 . By definition, F 1 maps (0 , ∞ ) 6 and [0 , ∞ ) 6 on to O 1 and O 1 , resp ectiv ely . Lemma 4.1. The set O 1 is op en in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) . Pr o of. It is clear that F 1 is contin uous on (0 , ∞ ) 6 . W e wan t to sho w that it is also injective on (0 , ∞ ) 6 . Let S b e a polyhedral surface in O 1 whose parallelogram decomp osition constructed from D ∗ 1 is sketc hed in the first picture of Figure 13 . Suppose that S = F 1 ( a, b, c, d, e, f ), where a , b , . . . , f are the lengths of the edges transverse to l a , l b , . . . , l f , resp ectiv ely . Analogous to the pro of of Theorem 2.1 , w e cut S op en along the green edges (three of which are directed) in the first picture of Figure 13 and unfold it to obtain a planar p olygon P S + . This time we identify the plane with R 2 instead of C . W e denote the three green directed edges by 2 + 3 + , 3 + 4 + and 4 + 2 − . Their images in P S + determine three vectors in R 2 , denoted by Z 2 + 3 + , Z 3 + 4 + and Z 4 + 2 − . F rom the first picture, w e see that there exist unit v ectors  a x a y  ,  b x b y  ,  c x c y  ,  d x d y  ,  e x e y  and  f x f y  in R 2 suc h that Z 2 + 3 + = a  a x a y  + b  b x b y  , Z 3 + 4 + = c  c x c y  + d  d x d y  and Z 4 + 2 − = e  e x e y  + f  f x f y  (5) No w we v ary S in O 1 . The green edges in the first picture of Figure 13 v ary contin uously with S without c hanging their homotop y classes. This allows us to define a consistent wa y of unfolding all surfaces in O 1 . Since such a wa y of unfolding exists throughout O 1 , w e can demonstrate it on a picture of A 1 rather than on a particular surface only . In the second picture in Figure 13 , we use green edges that join the labeled vertices of the sphere to represen t our consistent choices of the edges on the surfaces in O 1 to cut along. In this wa y , w e can define P S + consisten tly (up to Euclidean isometries) for all surfaces in O 1 . Eac h P S + giv es rise to a vector ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) ∈ R 6 . Up to rotations, we may assume that Z 2 + 3 + , Z 3 + 4 + , and Z 4 + 2 − can b e expressed in terms of a , b , . . . , f in Equation 5 , where the unit v ectors remain constant vectors. Let ψ 1 b e the map on (0 , ∞ ) 6 that sends ev ery vector ( a, b, c, d, e, f ) to ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) via unfolding surfaces in O 1 . Then we can write do wn the form ula of ψ 1 as follows: 13 Figure 13: W e can demonstrate a consistent w ay of unfolding p olyhedral surfaces in O by dra wing segmen ts on a sphere with A 1 . ψ 1 ( a, b, c, d, e, f ) =         a x b x 0 0 0 0 a y b y 0 0 0 0 0 0 c x d x 0 0 0 0 c y d y 0 0 0 0 0 0 e x f x 0 0 0 0 e y f y                 a b c d e f         Denote the matrix ab o ve by M 1 . Then det( M 1 ) =     a x b x a y b y     ·     c x d x c y d y     ·     e x f x e y f y     = sin( δ 1 2 ) · sin( δ 1 2 ) · sin( δ 1 2 ) > 0 where the determinant of each 2 × 2 matrix is computed based on the angles in three parallelograms, whic h can b e calculated by looking at the vertices enclosed in some lune b ounded b y t wo lo ops. One may refer to Equation 1 in Section 3 for more details. Therefore, ψ 1 is contin uous and inv ertible. Since ψ 1 is a comp osition of F 1 b y another map, F 1 is injective on (0 , ∞ ) 6 . By Theorem 2.1 , C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) has real dimension 6. By the inv ariance of domain, O 1 = F 1 ((0 , ∞ ) 6 ) is open in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). F rom the proof of Lemma 4.1 , we see that ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) is a local frame defined on O 1 , so w e can also visualize this local frame with A 1 . In the third picture of 13 , w e visualize its three comp onen ts Z 2 + 3 + , Z 3 + 4 + , and Z 4 + 2 − as three green directed edges joining their corresp onding lab eled v ertices on the sphere. This visualization is sometimes helpful to see the existence of the constan t unit v ectors  a x a y  ,  b x b y  , . . . ,  f x f y  in Equation 5 . F or example, let us orient the lo ops in A 1 according to the blue arro ws in the third picture, and consider the comp onen t Z 2 + 3 + . The orien tations of l a and l b induce directions on the edges transverse to them in D ∗ 1 , including the tw o edges lab eled a and b in the first picture of Figure 13 . W e can then consider the triangle formed b y these t wo edges and the edge 2 + 3 + , and see the existence of  a x a y  and  b x b y  to make Equation 5 hold. 14 W e say that the map ψ 1 in the pro of of Lemma 4.1 is represented b y the matrix M 1 . It can b e defined and is inv ertible on [0 , ∞ ) 6 . In addition, we know that F 1 maps [0 , ∞ ) 6 on to O 1 . This implies that F 1 is a homeomorphism from [0 , ∞ ) 6 to O 1 . Therefore, every elemen t in O 1 corresp onds to a unique v ector in [0 , ∞ ) 6 through F 1 . This vector will b e called its co ordinates with respect to A 1 , or simply co ordinates when the underlying lo op arrangemen t is clear. Similarly , we can also consider its a -co ordinate, b -co ordinate, etc. The collection of elements in O 1 where at least tw o coordinates v anish is called its co dimension- t wo b oundary . F or example, if an element in O 1 has zero a -co ordinate and b -co ordinate, while the remaining co ordinates are p ositiv e, then it b elongs to the codimension-tw o boundary . In this case, it is a surface in C 6 ( δ 1 , δ 2 + δ 3 , δ 4 ). In fact, outside the co dimension-t w o b oundary , elemen ts in O 1 b elongs to C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ), so the lo cal frame defined ab o ve is still v alid in the interior of ev ery face of O 1 . Later, we will show that outside the co dimension-t w o b oundary , every element in O 1 has an open neigh b orhoo d in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). According to the definition of O 1 , ev ery surface in O 1 has a parallelogram decomp osition that is inv ariant under the antipo dal map. Our goal of this section is to show that this result is true for all surfaces in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). Since there exist polyhedral surfaces in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) but outside O 1 , we need to consider spaces arising from different lo op arrangemen ts. Let A 2 b e the lo op arrangemen t A ′ in the second picture of Figure 11 . According to the paragraph b efore Figure 11 , A 1 and A 2 are adjacent lo op arrangements since they differ by l a only . In Figure 14 , we sketc h A 2 in the first picture. In fact, A 2 is combinatorically equiv alen t to A 1 , which means that A 2 b ecomes equiv alent to A 1 after we relabel its lo ops and vertices. One can see this by rotating the sphere so that 4 + is at the center of our view, as the second picture shows. Then one can chec k that each lo op in the first picture can b e mov ed to the lo op with the same lab el in the second picture without changing its homotopy class. It may b e helpful to consider which v ertices are on the same side with resp ect to eac h lo op. Then it is clear that A 2 is combinatorically equiv alent to A 1 b y comparing the second and the third picture. Figure 14: The loop arrangemen ts A 1 and A 2 are combinatorically equiv alent. Let O 2 b e the space arising from A 2 and O 2 b e its closure. Since A 2 is combinatorically equiv alent to A 1 , O 2 is also op en in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) b y Lemma 4.1 and symmetry . Since A 1 and A 2 differ by l a only , O 1 and O 2 are adjacent and coincide on the face σ a . Then w e hav e the following result: Lemma 4.2. The sp ac es O 1 and O 2 ar e on differ ent sides of σ a . 15 Pr o of. Let ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) b e the lo cal frame defined on O 1 in the pro of of Lemma 4.1 . It is also defined in the interior of σ a , since every element there still represents a surface in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). In Lemma 4.1 , this lo cal frame is obtained via unfolding every surface to a p olygon P S + . When w e think of this unfolding as the restriction of a dev eloping map, ev ery v ertex of P S + comes from a path on S we dev elop along from a base-p oint of S to a vertex of S . When S mo ves across the in terior of σ a in to O 2 , we contin uously v ary this path with S without changing its homotopy class. This allo ws us to extend the definition of ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) to surfaces in O 2 , even though the actual cutting and unfolding ma y not alw ays exist there. In Figure 15 , we visualize this lo cal frame with A 1 and A 2 . In A 1 , the orientations of all the lo ops are still the same as in Lemma 4.1 . In A 2 , the orien tations from l b to l f are the same as in A 1 . Finally , we orien t l a in A 2 according to the blue arrow marked on it. Figure 15: Visualizing the lo cal frame ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) with A 1 and A 2 . Let  b x b y  ,  c x c y  ,  d x d y  ,  e x e y  , and  f x f y  b e the unit vectors in Lemma 4.1 . Then observe from Figure 15 that there exists a constant unit vector a ′ x a ′ y ! suc h that we can write Z 2 + 3 + = − a  a ′ x a y  + b  b x b y  , Z 3 + 4 + = a  a ′ x a y  + c  c x c y  + d  d x d y  and Z 4 + 2 − = − a  a ′ x a y  + e  e x e y  + f  f x f y  in terms of a , b , . . . , f . Here a ′ x a ′ y ! is different from  a x a y  since the loop l a has b een changed. Th us, let ψ 2 b e the map defined on (0 , ∞ ) 6 that sends every v ector ( a, b, c, d, e, f ) to ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) via surfaces in O 2 . Then we hav e ψ 2 ( a, b, c, d, e, f ) =          − a ′ x b x 0 0 0 0 − a ′ y b y 0 0 0 0 a ′ x 0 c x d x 0 0 a ′ y 0 c y d y 0 0 − a ′ x 0 0 0 e x f x − a ′ y 0 0 0 e y f y                  a b c d e f         16 Denote the matrix in the abov e equation b y M 2 . Then det( M 2 ) =      − a ′ x b x − a ′ y b y      ·     c x d x c y d y     ·     e x f x e y f y     = − sin( δ 4 2 ) sin 2 ( δ 1 2 ) < 0 where the determinan ts of the three 2 × 2 matrices can b e calculated using Equation 1 . Let ψ 1 and M 1 b e defined as in Lemma 4.1 . Supp ose that there exists a surface S near σ a in O 1 ∩ O 2 . This surface has (probably different) co ordinates in (0 , ∞ ) 6 with resp ect to both A 1 and A 2 , which are mapp ed to the same image ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) b y ψ 1 and ψ 2 , resp ectiv ely . W e can then use Cramer’s Rule to compute and compare its a -co ordinates with resp ect to A 1 and A 2 . Since M 1 and M 2 differ only by their first column, we obtain the same matrix by replacing their first columns with ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ). How ev er, the determinants of M 1 and M 2 ha ve opp osite signs. By Cramer’s Rule, it is imp ossible that S has positive a -co ordinates with resp ect to b oth A 1 and A 2 , which contradicts the fact that S ∈ O 1 ∩ O 2 . Therefore, O 1 and O 2 are on differen t sides of σ a . Remark. It turns out that the pr o of of L emma 4.1 and L emma 4.2 works r e gar d less of the lo c al fr ame we cho ose. Supp ose that we cho ose a differ ent lo c al fr ame. Then by definition of lo c al fr ames, the change-of-c o or dinate map fr om ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 2 − ) to the new lo c al fr ame is given by an invertible matrix. Ther efor e, we just multiply this matrix to the left of M 1 and M 2 . Then we stil l obtain two invertible matric es whose determinants have opp osite signs. Finally , w e turn to the pro of that ev ery polyhedral surface in C 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) can b e decomp osed in to parallelograms. It suffices for us to pro ve this result for M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ), the metric completion of its moduli space. Giv en a space O arising from a lo op arrangement A , w e can consider the space of p olyhedral surfaces in O with surface area 1. W e call this space the mo duli space arising from A , and denote it by ∆. Let ∆ ⊆ M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) be the metric completion of ∆ with respect to the real h yp erb olic metric in Theorem 2.2 . W e denote a face of ∆ by σ i (1) (where 1 refers to the surface area) if it is a subset of the face σ i in O . If A ′ is another loop arrangement that differs from A b y l i only , we say that ∆ and ∆ ′ are on different sides of σ i (1) if and only if O and O ′ are on differen t sides of σ i . Let ∆ 1 b e the metric completion of the moduli space arising from A 1 . Lemma 4.3. The sp ac e ∆ 1 has the structur e of a r e al hyp erb olic r e gular ide al 5 -simplex. Pr o of. Combinatorically , ∆ 1 is a polytop e with six v ertices and six facets. Each vertex of ∆ 1 has fiv e v anishing co ordinates among six with resp ect to A 1 . This corresp onds to the case where the surface degenerates to a line. Note that each vertex of ∆ 1 b elongs to five facets of ∆ 1 , and each facet contains five vertices. Therefore, ∆ 1 is a 5-simplex by Prop osition 2.16 in [ 13 ]. Geometrically , we sho w that ∆ 1 is b ounded b y totally geodesic h yp erplanes. Consider the set F − 1 1 (∆ 1 ) in (0 , ∞ ) 6 . According to Theorem 2.2 , b y some linear transformation, F − 1 1 (∆ 1 ) is mapp ed to a subset in the parab oloid x 2 1 − x 2 2 − · · · − x 2 6 = 1. This subset is b ounded b y the intersections of the parab oloid with six planes through the origin, whic h are the images of the co ordinate planes under this linear transformation. Therefore, this subset and hence ∆ 1 is b ounded by totally geo desic h yp erplanes. Finally , ∆ 1 is ideal b ecause w e ha ve seen that its vertices do not b elong to the real hyperb olic space H 5 . It is regular due to symmetry . This completes the pro of of the result. 17 By Lemma 4.1 and Lemma 4.2 , we also obtain the following result: Corollary 4.4. Every element in ∆ 1 or in the interior of a fac e of ∆ 1 has an op en neighb orho o d in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) . Pr o of. Let x b e an elemen t in ∆ 1 . If x ∈ ∆ 1 , the result follows from Lemma 4.1 . If x is in the in terior of a face of ∆ 1 , without loss of generality , w e assume that this face is σ a (1). Let ∆ 2 b e the mo duli space arising from A 2 and ∆ 2 b e its metric completion. W e can choose a small ϵ -neighborho o d of x in b oth ∆ 1 and ∆ 2 . When ϵ is sufficiently small, the tw o neigh b orho ods are b oth op en half-balls with radius ϵ , and they hav e no in tersection outside σ a (1) by Lemma 4.2 . Therefore, their union is an open ball in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) containing x . Similar to A 2 , by p erm uting the the lab els of vertices and lo ops in A 1 , we can obtain more lo op arrangements that are different from but com binatorically equiv alen t to A 1 . It is clear that there are finitely man y such lo op arrangemen ts. W e consider the metric completions of the mo duli spaces arising from these lo op arrangemen ts. F or conv enience, w e will call them “5-simplices” due to Lemma 4.3 . Later, we are going to construct infinitely many copies of such 5-simplices and regard them as distinct spaces, though some of them arise from the same lo op arrangemen t. Our goal is to glue these 5-simplices to form a branched cov er of M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). Let X b e a space constructed in the following wa y: • Initially , X is just a copy of the 5-simplex ∆ 1 . • Supp ose that X has a b oundary face that is the face σ i (1) of some 5-simplex ∆. By Lemma 4.2 and symmetry , we can construct a copy of some 5-simplex ∆ ′ so that ∆ and ∆ ′ are on differen t sides of σ i (1). Then we glue this cop y of ∆ ′ to X along σ i (1). • W e rep eat the pro cess abov e so that every face in X is incident to exactly t wo 5-simplices, one on eac h side. Let X 2 b e the union of the co dimension-t wo b oundaries of the 5-simplices in X . Since every face in X is inciden t to exactly tw o 5-simplices, X \ X 2 is a manifold. There is a contin uous map ι : X → M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) such that, restricted to every 5-simplex in X , ι is the inclusion map by viewing the 5-simplex as a subset of M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). Although X contains infinitely man y 5-simplices, there are finitely many loop arrangements in volv ed, so ι ( X ) is a finite union of closed sets. Hence, ι ( X ) is closed in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ), which implies that ι ( X ) \ ι ( X 2 ) is closed in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) \ ι ( X 2 ). By Corollary 4.4 , ev ery surface in ι ( X ) \ ι ( X 2 ) has a neigh borho o d con tained in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ). Since ι ( X 2 ) has co dimension tw o, the neighborho o d is aw ay from ι ( X 2 ) if it is sufficiently small. Therefore, ι ( X ) \ ι ( X 2 ) is also op en in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) \ ι ( X 2 ). Finally , M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) \ ι ( X 2 ) is non-empt y due to their differen t dimensions, so we hav e sho wn that ι ( X ) \ ι ( X 2 ) = M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) \ ι ( X 2 ). Th us, every surface in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) \ ι ( X 2 ) has a parallelogram decomposition. By definition of X 2 , every surface in ι ( X 2 ) has a parallelogram decomp osition as well. The decomp osition is in v ariant under the antipo dal map by our construction. In addition, there are 6 lo ops in each lo op arrangement, ev ery t wo of which meet at t w o an tip odal p oints since they are great circles. In conclusion, we hav e pro ved the following result: Theorem 4.5. Every surfac e in M 8 ( δ 1 , δ 2 , δ 3 , δ 4 ) c an b e de c omp ose d into at most 2  6 2  = 30 p ar- al lelo gr ams, and the de c omp osition is invariant under the antip o dal map. 18 In summary , w e hav e established tw o metho ds to build co ordinate charts on the space of poly- hedral surfaces. One is through unfolding the surfaces, and the other is through decomp osing them in to parallelograms. Although the first metho d applies to more general conv ex p olyhedral surfaces, the second metho d is also in teresting b ecause it sometimes provides a more concrete geometric description of moduli spaces. F or instance, consider the space of centrally symmetric p olyhedral surfaces with 8 unlab eled v ertices, whose cone-deficits all equal to π 2 . In this unlab eled space, tw o surfaces are equiv alent if they differ by a Euclidean isometry . Denote its unlabeled mo duli space by M 8 ( π 2 ), and its metric completion by M 8 ( π 2 ). The following result is a direct consequence of Lemma 4.3 : Corollary 4.6. The sp ac e M 8 ( π 2 ) has the structur e of the quotient sp ac e ∆ 1 / Γ , wher e ∆ 1 is define d as b efor e, and Γ is isomorphic to the dihe dr al gr oup D 6 with or der 12 . Pr o of. W e can interpret M 8 ( π 2 ) as the space of equiv alence classes in ∆ 1 . The main task is to figure out the equiv alence relation. Figure 16 shows the front view of a parallelogram decomp osition of a surface in ∆ 1 , whose co ordinates are ( a, b, c, d, e, f ) with resp ect to A 1 . Some edge-lengths of the parallelograms are lab eled. One can think of the co ordinates as reading the edge-lengths consecutively in the red p olygonal path in Figure 16 whose direction is indicated b y the red arrow. Figure 16: Parallelogram decomp osition of a surface in ∆ 1 with co ordinates ( a, b, c, d, e, f ). In the unlab eled mo duli space M 8 ( π 2 ), we don’t distinguish vertices of the surfaces. Thus, if w e rotate the parallelogram decomp osition so that the red p olygonal path starts at a differen t v ertex, we get an equiv alent surface in M 8 ( π 2 ) despite that their co ordinates with resp ect to A 1 are differen t. F or example, ( a, b, c, d, e, f ) and ( c, d, e, f , b, a ) corresp ond to an equiv alent surface in M 8 ( π 2 ). In addition, we can reflect the parallelogram decomp osition so that the co ordinates are read in reverse order. F or example, ( a, b, c, d, e, f ) and ( f , e, d, c, b, a ) corresp ond to an equiv alent surface in M 8 ( π 2 ). Thus, the group Γ is isomorphic to the symmetry group of a regular hexagon, whic h is the dihedral group D 6 . 5 P olyhedra with 10 V ertices In this section, w e study the space of cen trally symmetric conv ex p olyhedral surfaces with 10 vertices with pres cribed cone-deficits δ 1 , δ 2 , δ 3 , δ 4 and δ 5 . Analogous to the last section, the goal of this 19 section is to show that every surface in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) can b e decomp osed into parallelograms. Restricted to this section, a loop arrangement alwa ys has 8 loops lab eled by l a , l b , l c , . . . , l g , l h on a sphere with 8 vertices lab eled by ± 1, ± 2, . . . , ± 5. The main idea is still to construct spaces arising from different lo op arrangemen ts. This time, w e consider the four lo op arrangements A 1 , A 2 , A 3 , and A 4 in Figure 17 . Let O 1 , O 2 , O 3 , and O 4 b e the spaces arising from A 1 , A 2 , A 3 , and A 4 , respectively . In addition, we say that a space O has Type 1 if it arises from a lo op arrangement combinatorically equiv alent to A 1 up to relab eling v ertices or lo ops. Similarly , we define spaces of T yp e 2 , Type 3 , and T yp e 4 . Theorem 3.1 implies that these spaces are subsets of C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ). Figure 17: Loop arrangements A 1 , A 2 , A 3 , and A 4 . The num b ers in b o xes refer to the types of spaces arising from these lo op arrangements. Lemma 5.1. Every T yp e 1 sp ac e is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . Pr o of. It suffices to pro ve the result for O 1 . The pro of is similar to that of Lemma 4.1 . In the first picture of Figure 18 , we demonstrate a consistent w ay of unfolding surfaces in O 1 b y cutting along the edges of these surfaces represen ted b y the green segments joining the lab eled v ertices on the sphere. This gives rise to a lo cal frame ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 5 + , Z 5 + 2 − ) on O 1 , which is visualized in the second picture. W e orien t the loops in A 1 according to the blue arrows. Then w e can observe from Figure 18 that there exist constant unit vectors  a x a y  ,  b x b y  , . . . ,  h x h y  in R 2 , using which we can express Z 2 + 3 + , Z 3 + 4 + , Z 4 + 5 + and Z 5 + 2 − in terms of a , b , . . . , h as we did in Lemma 4.1 . Similar to the constructions in Section 4, let F 1 b e the map that sends ev ery v ector ( a, b, . . . , g , h ) ∈ (0 , ∞ ) 8 to a surface in O 1 . Let ψ 1 b e the map that sends every v ector ( a, b, . . . , g , h ) ∈ (0 , ∞ ) 8 to ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 5 + , Z 5 + 2 − ) via unfolding surfaces in O 1 . Then we hav e ψ 1 ( a, b, c, d, e, f , g , h ) =             a x b x 0 0 0 0 0 0 a y b y 0 0 0 0 0 0 0 0 c x d x 0 0 0 0 0 0 c y d y 0 0 0 0 0 0 0 0 e x f x 0 0 0 0 0 0 e y f y 0 0 0 0 0 0 0 0 g x h x 0 0 0 0 0 0 g y h y                         a b c d e f g h             20 Figure 18: Visualization of the unfolding and the local frame it gives rise to with A 1 . Denote the matrix in the abov e equation b y M 1 . Then det( M 1 ) =     a x b x a y b y     ·     c x d x c y d y     ·     e x f x e y f y     ·     g x h x g y h y     = sin 4 ( δ 1 2 ) > 0 Th us, ψ 1 is contin uous and in vertible. Since ψ 1 is the composition of F 1 with another map, F 1 is contin uous and injective on (0 , ∞ ) 8 . As O 1 = F 1 ((0 , ∞ ) 8 ) and C 10 ( δ 1 , δ 2 .δ 3 , δ 4 , δ 5 ) has dimension 8, the result follows from the inv ariance of domain. Let O 1 b e the closure of O 1 . Analogous to Section 4, we define the co ordinates with resp ect to A 1 and then the co dimension-t wo b oundary of O 1 . Awa y from the co dimension-t wo b oundary of O 1 , every element in O 1 represen ts a surface in C 10 ( δ 1 , δ 2 .δ 3 , δ 4 , δ 5 ), and the lo cal frame ( Z 2 + 3 + , Z 3 + 4 + , Z 4 + 5 + , Z 5 + 2 − ) in Lemma 5.1 is defined. Next, we will establish t wo technical results that generalize Lemma 4.2 . Let A and A ′ b e tw o adjacen t lo op arrangements that differ b y the loop l i . Let O and O ′ b e the spaces arising from A and A ′ , resp ectiv ely . Let O and O ′ b e their closures. Supp ose that there is a lo cal frame defined on O . Since O and O ′ are adjacent and meet on the face σ i , the lo cal frame on O can b e extended contin uously to O ′ through the interior of σ i in a wa y we describ ed in Lemma 4.2 . Let ψ and ψ ′ b e the maps from (0 , ∞ ) 8 to this local frame via surfaces in O and O ′ , resp ectiv ely . Let M and M ′ b e the matrices represen ting ψ and ψ ′ , resp ectiv ely . Let Z T U and Z V W b e tw o comp onents in the lo cal frame ab ov e, where T , U , V and W are four lab eled v ertices. The first technical result in volv es Z T U only . Lemma 5.2. Supp ose that ther e exist c onstant unit ve ctors  i x i y  , i ′ x i ′ y ! and  j x j y  such that Z T U = i  i x i y  + j  j x j y  in the formula of ψ , and Z T U = i i ′ x i ′ y ! + j  j x j y  in the formula of ψ ′ . Then O and O ′ ar e on differ ent sides of σ i if and only if the determinants     i x j x i y j y     and      i ′ x j x i ′ y j y      have opp osite signs. 21 Pr o of. The main idea of pro of is to show that M and M ′ ha ve opp osite signs in their determinants. Without loss of generality , w e may assume that Z T U is the first comp onen t of the lo cal frame, and i , j are the first tw o comp onen ts of the vectors in (0 , ∞ ) 8 . Then the first tw o ro ws in M are  i x j x 0 . . . 0 i y j y 0 . . . 0  . Similarly , the first t wo rows in M ′ are i ′ x j x 0 . . . 0 i ′ y j y 0 . . . 0 ! . Note that M and M ′ differ only b y the first column. Therefore, by deleting the first tw o ro ws and the first tw o columns from M and M ′ , respectively , we get the same sub-matrix. By multiplying the determinan t of this sub-matrix to     i x j x i y j y     and      i ′ x j x i ′ y j y      , resp ectiv ely , w e get the determinan ts of M and M ′ . Thus, M and M ′ ha ve opp osite signs if and only if     i x j x i y j y     and      i ′ x j x i ′ y j y      ha ve opp osite signs. The rest of the pro of is same as that of Lemma 4.2 in principle. T o apply lemma 5.2 in practice, we can visualize the comp onent Z T U of the lo cal frame with b oth A and A ′ . There is a natural w ay to orien t the lo ops l i and l j in b oth A and A ′ according to the formula of Z T U in Lemma 5.2 . It will b e helpful to lo ok at the arro ws on these lo ops that indicate their orien tations. F or example, consider l i and l j in A and A in the first picture of Figure 19 . Based on the form ula of Z T U in Lemma 5.2 , the lo ops l i and l j should b e oriented by the blue arrows. In A , the arro w on l j o verlaps with the arro w on l i after a counterclockwise rotation by an angle < π . This means     i x j x i y j y     > 0. In A ′ , the arro w on l j o verlaps with the arrow on l i after a clo ckwise rotation b y an angle < π , which means     i x j x i y j y     < 0. By Lemma 5.2 , O and O ′ are on different sides of σ i . Similarly , the reader ma y chec k that for A and A ′ in the second picture, O and O ′ are on the same side of σ i . Figure 19: Applying Lemma 5.2 by chec king the orien tations of l i and l j in A and A ′ The second tec hnical result in volv es b oth Z T U and Z V W . Lemma 5.3. Supp ose ther e exist c onstant unit ve ctors  i x i y  , i ′ x i ′ y ! ,  j x j y  ,  k x k y  and  m x m y  such that Z T U = i  i x i y  + j  j x j y  + m  m x m y  and Z V W = k  k x k y  − m  m x m y  in the formula of ψ . In 22 addition, Z T U = j  j x j y  + m  m x m y  and Z V W = i i ′ x i ′ y ! + k  k x k y  − m  m x m y  in the formula of ψ ′ . Then O and O ′ ar e on differ ent sides of the fac e σ i if and only if the pr o ducts of the determinants     i x j x i y j y     ·     k x m x k y m y     and      i ′ x k x i ′ y k y      ·     j x m x j y m y     have opp osite signs. Pr o of. Again, we need to sho w that the determinants of M and M ′ ha ve opp osite signs. Without loss of generalit y , w e assume that Z T U and Z V W are the first tw o comp onen ts of the lo cal frame, and i , j , k , m are the first four comp onents of the v ectors in [0 , ∞ ) 8 . Then the first four rows of M and M ′ are     i x j x 0 m x 0 . . . 0 i y j y 0 m y 0 . . . 0 0 0 k x − m x 0 . . . 0 0 0 k y − m y 0 . . . 0     and      0 j x 0 m x 0 . . . 0 0 j y 0 m y 0 . . . 0 i ′ x 0 k x − m x 0 . . . 0 i ′ y 0 k y − m y 0 . . . 0      , resp ectiv ely . Since M and M ′ differ only by the first column, their sub-matrices after deleting the first four ro ws and columns are equiv alen t. Thus, to compare the signs of their determinan ts, it suffices to compare the signs of their 4 × 4-minor on the upper-left corner. F or M , this minor is         i x j x 0 m x i y j y 0 m x 0 0 k x − m x 0 0 k y − m y         = −     i x j x i y j y     ·     k x m x k y m y     . F or M ′ , it is          0 j x 0 m x 0 j y 0 m y i ′ x 0 k x − m x i ′ y 0 k y − m y          = −      i ′ x k x i ′ y k y      ·     j x m x j y m y     . Thus, the determinan ts of M and M ′ ha ve opp osite signs if and only if these t wo pro ducts hav e opp osite signs. The rest of the pro of is similar to that of Lemma 4.2 . Again, to apply Lemma 5.3 in practice, it will b e helpful to visualize Z T U and Z V W with A and A ′ . There is a natural wa y to orien t the lo ops l i , l j , l k and l m in A and A ′ according to the form ulas of Z T U and Z V W in Lemma 5.3 . The sign of the pro ducts in Lemma 5.3 can b e computed more quickly by lo oking at the arrows that indicate their orien tations. Figure 20: Applying Lemma 5.3 by chec king the orien tations of l i , l j , l k and l m F or example, Figure 20 shows a particular case where U = V . Consider the loops l i , l j , l k and l m in A and A ′ . Their orientations based on the form ulas of Z T U and Z V W are mark ed by blue arro ws. In A , the blue arro w on l i o verlaps with that on l j after a clo c kwise rotation b y an angle 23 < π , whic h means     i x j x i y j y     > 0. Similarly , we ha ve     k x m x k y m y     > 0. In A ′ , one can chec k from the blue arro ws that      i ′ x k x i ′ y k y      < 0 and     j x m x j y m y     > 0. Therefore, O and O ′ are on differen t sides of σ i b y Lemma 5.3 . No w let O b e a space of Type 1, 2, 3 or 4 and O b e its closure. F or ev ery face σ of O , we will apply Lemma 5.2 and Lemma 5.3 to show that there exists a space O ′ of T yp e 1, 2, 3 or 4 whose closure O ′ is on the other side of σ with respect to O . T yp e 1 Spaces Without loss of generality , we can assume that O is O 1 . Due to the symmetry of A 1 , w e can also assume that the face σ is σ b . Lemma 5.4. The closur es O 1 and O 2 ar e on differ ent sides of σ b . Pr o of. Note that A 2 is obtained from A 1 b y moving the lab eled vertices 2 + , 2 − , 3 + , and 3 − in A 1 across the lo op l b sim ultaneously . Th us. similar to A and A ′ in Figure 11 , A 1 and A 2 differ only b y the lo op l b . Consider the lo cal frame on O 1 in Lemma 5.1 , which can b e extended contin uously to O 2 through the interior of σ b . In Figure 21 , we visualize the comp onent Z 2 + 3 + of this lo cal frame with A 1 and A 2 . Let ψ 1 b e defined as in Lemma 5.1 , and let ψ 2 b e the map from [0 , ∞ ) 8 to this lo cal frame via surfaces in O 2 . W e orien t the loops l b and l a in A 1 and A 2 according to the blue arro ws, and observ e from Figure 21 that there exist unit vectors  b x b y  , b ′ x b ′ y ! and  a x a y  suc h that Z 2 + 3 + = b  b x b y  + a  a x a y  in ψ 1 and Z 2 + 3 + = b b ′ x b ′ y ! + a  a x a y  in ψ 2 . Analogous to the first example in Figure 19 , from the directions of the blue arrows in Figure 21 , we see that     b x a x b y a y     < 0 and      b ′ x a x b ′ y a y      > 0. Thus, by Lemma 5.2 , O 1 and O 2 are on different sides of σ b . Corollary 5.5. Every T yp e 2 sp ac e is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . Pr o of. Without loss of generality , we prov e the result for O 2 . Let ψ 1 and ψ 2 b e the maps in the pro of of Lemma 5.4 . Referring to the pro of of Lemma 5.1 , we see that the key step is to show that ψ 2 is inv ertible. Let M 1 and M 2 b e the matrices representing ψ 1 and ψ 2 , resp ectively . By Lemma 5.1 , M 1 is in vertible. By Lemma 5.4 , the determinants of M 1 and M 2 ha ve opposite signs, so M 2 is inv ertible. The rest of pro of is iden tical to that of Lemma 5.1 in principle. T yp e 2 Spaces Without loss of generality , we can assume that O is O 2 . Observe from Figure 22 that there are some symmetries among lo ops in A 2 . A reflection ab out the axis on the left interc hanges l a and l b , l c and l g , l d and l h . A reflection ab out the axis on the right interc hanges l e and l f , l c and l h , l d and l g . Th us, in summary , w e can divide the 8 lo ops into three symmetry classes: 1) l a , l b ; 2) l e 24 Figure 21: Visualizing Z 2 + 3 + with A 1 and A 2 and l f ; 3) l c , l d , l g and l h . Therefore, it suffices to discuss the cases when the face σ is σ b , σ e or σ h . Since w e hav e sho wn that O 1 and O 2 are on differen t sides of σ b , w e only need to discuss the cases σ e and σ h . Figure 22: Two reflection symmetries of A 2 ab out the axes in dark blue. Lemma 5.6. The closur es O 2 and O 3 ar e on differ ent sides of σ h . Pr o of. Observe from Figure 23 that A 3 is obtained from A 2 b y mo ving the lab eled vertices 1 + , 2 + , 1 − and 2 − across l h . Th us, A 2 and A 3 differ only b y l h , so σ h is a face of b oth O 2 and O 3 . W e consider the consisten t w ay of cutting and unfolding surfaces in O 2 demonstrated in the first picture of Figure 23 . The green segmen ts connecting 5 + to 2 − and 2 − to 3 − are not fully visible, but their antipo dal images are drawn. This unfolding giv es rise to a lo cal frame ( Z 5 − 2 + , Z 2 + 3 + , Z 3 + 4 + , Z 4 + 5 + ) on O 2 , and can b e extended to O 3 through the interior of σ h . Let ψ 2 and ψ 3 b e the maps from [0 , ∞ ) 8 to this local frame via surfaces in O 2 and O 3 , resp ectiv ely . In the second and the third pictures of Figure 23 , w e visualize the comp onen ts Z 5 − 2 + and Z 2 + 3 + in this lo cal frame with A 2 and A 3 . By orienting the lo ops l h , l a , l g and l b in A 2 and A 3 according 25 Figure 23: The unfolding demonstrated in the first picture gives rise to a lo cal frame, whose comp onen ts Z 5 − 2 + and Z 2 + 3 + are visualized with A 2 and A 3 . to the blue arrows, we observe that there exist unit v ectors  h x h y  , h ′ x h ′ y ! ,  a x a y  ,  g x g y  and  b x b y  suc h that Z 5 − 2 + = h  h x h y  + g  g x g y  + b  b x b y  and Z 2 + 3 + = a  a x a y  − b  b x b y  in ψ 2 . In addition, Z 5 − 2 + = g  g x g y  + b  b x b y  and Z 2 + 3 + = h h ′ x h ′ y ! + a  a x a y  − b  b x b y  in ψ 3 . Thus, we can apply Lemma 5.3 b y taking T = 5 − , U = V = 2 + , W = 3 + , i = h , j = a , k = g , and m = b . W e can compute the products of the determinants in Lemma 5.3 b y the directions of the blue arro ws in Figure 23 . But w e can also compare Figure 23 with Figure 20 directly to see that they are in the same situation. Therefore, O 2 and O 3 are on differen t sides of σ h b y Lemma 5.3 . Corollary 5.7. The interior of a T yp e 3 sp ac e is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . Pr o of. Without loss of generality , w e pro ve the result for O 3 . Consider the maps ψ 2 and ψ 3 in the pro of of Lemma 5.6 . Let M 2 and M 3 b e the matrices that represen t ψ 2 and ψ 3 , resp ectively . By Corollary 5.5 , O 2 is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ), so M 2 is inv ertible. By Lemma 5.6 , the determinan ts of M 2 and M 3 ha ve opp osite signs, so M 3 is inv ertible. The rest of the pro of is iden tical to that of Lemma 5.1 in principle. Lemma 5.8. The closur es O 2 and O 4 ar e on differ ent sides of σ e . Pr o of. Observe from Figure 24 that A 4 is obtained from A 2 b y mo ving the lab eled vertices 4 + , 4 − , 5 + , and 5 − across the lo op l e . There fore, A 2 and A 4 differ only by l e , so σ e is a face of b oth O 2 and O 4 . T o see that O 4 are on different sides of σ e , we use the same lo cal frame on O 2 in the pro of of Lemma 5.6 , and extend it to O 4 through the in terior of σ e . Let ψ 2 and ψ 4 b e the maps from [0 , ∞ ) 8 to this local frame via surfaces in O 2 and O 4 , resp ectiv ely . By visualizing the comp onen t Z 4 + 5 + of this lo cal frame with A 2 and A 4 in Figure 24 , we can find unit vectors  e x e y  , e ′ x e ′ y ! and  f x f y  suc h that Z 4 + 5 + = e  e x e y  + f  f x f y  in ψ 2 and Z 4 + 5 + = e e ′ x e ′ y ! + f  f x f y  in ψ 4 . 26 Figure 24: Visualizing the comp onen t Z 4 + 5 + with A 2 and A 4 . Then w e orient l e and l f based on the form ulas of Z 4 + 5 + ab o v e, and mark their orientations by blue arrows in Figure 24 . W e can see that     e x f x e y f y     > 0 and      e ′ x f x e ′ y f y      < 0, so the result follows from Lemma 5.2 b y taking T = 4 + , U = 5 + , i = e and j = f . By an argumen t similar to Corollary 5.5 and 5.7 , w e obtain the following result: Corollary 5.9. The interior of a T yp e 4 sp ac e is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . T yp e 3 Spaces Without loss of generality , we can assume that O is the space O 3 arising from A 3 . Symmetries among lo ops in A 3 are less obvious. In Figure 25 , we sk etch A 3 in the first picture. Then w e push all the labeled v ertices in A 3 to the equator of the sphere as the second picture sho ws. Figure 25: When pushing all the lab eled vertices to the equator, lo ops in A 3 are mov ed or distorted so that their homotopy classes remain unchanged. When pushing the v ertices, we also mo ve or distort some lo ops in A 3 so that their homotop y classes remain the same. After that, we draw the lo ops l b and l f in the second picture (in which 27 b oth lo ops are distorted), and the remaining lo ops in the third picture (in which l d is distorted). T o verify b oth pictures, it would b e helpful to consider which v ertices are on the same side of each lo op in A 3 . The second and the third pictures in Figure 25 has a reflection symmetry about the dark blue axis. W e can thus divide the lo ops in A 3 in to five symmetry classes: 1) l a ; 2) l b and l f ; 3) l c and l h ; 4) l d ; 5) l e and l g . Since O 2 is on the other side of σ h with resp ect to O 3 b y Lemma 5.8 , it suffices to discuss the cases when σ is σ a , σ b , σ d or σ e Lemma 5.10. Ther e is a T yp e 3 sp ac e whose closur e is on the other side of σ a with r esp e ct to O 3 . Pr o of. Consider the lo op arrangement A ′ in the third picture of Figure 26 . It is obtained from A 3 b y moving the loop l a and p erturbing the lo op l g with the lab eled vertices 5 + and 5 − . Note that this do es not change the homotopy class of l g , so A ′ and A 3 differ only by l a . Let O ′ b e the space arising from A ′ and O ′ b e its closure. Then σ a is a common face of O 3 and O ′ . Figure 26: The unfolding demonstrated in the first picture gives rise to a lo cal frame, whose comp onen ts Z 1 + 5 + and Z 5 + 4 + are visualized with A 3 and A ′ . Consider the wa y of cutting and unfolding surfaces in O 3 demonstrated in the first picture of Figure 26 . The green segmen ts connecting 4 + to 2 − , 2 − to 1 − , and 1 − to 5 − are not fully visible, but we can see their antipo dal images. This unfolding giv es rise to a lo cal frame ( Z 4 − 2 + , Z 2 + 1 + , Z 1 + 5 + , Z 5 + 4 + ) on O 3 , and is extended to O ′ through the in terior of σ a . In Figure 26 , w e visualize the comp onen ts Z 1 + 5 + and Z 5 + 4 + of this local frame with A 3 and A ′ . T o show that O 3 and O ′ are on different sides of σ 3 , w e orient the lo ops l a , l c , l f and l e in A 3 and A ′ according to the blue arrows in Figure 26 . Let ψ 3 and ψ ′ b e the maps from [0 , ∞ ) 8 to the lo cal frame ab o v e via surfaces in O 3 and O ′ , respectively . Then we can see that there exist unit v ectors  a x a y  , a ′ x a ′ y ! ,  c x c y  ,  f x f y  and  e x e y  suc h that Z 1 + 5 + = a  a x a y  + c  c x c y  + e  e x e y  and Z 5 + 4 + = f  f x f y  − e  e x e y  in ψ 3 . In addition, Z 1 + 5 + = c  c x c y  + e  e x e y  and Z 5 + 4 + = a a ′ x a ′ y ! + f  f x f y  − e  e x e y  in ψ ′ . Therefore, w e can apply Lemma 5.3 b y taking T = 1 + , U = V = 5 + , W = 4 + , l i = l a , l j = l c , l k = l f and l m = l e . This is exactly the same situation as Figure 20 , so O 3 and O ′ are on differen t sides of σ a . 28 Finally , we sho w that O ′ is a Type 3 space by sho wing that A ′ is combinatorically equiv alent to A 3 . W e mo ve the lab eled vertices in A ′ to the equator of the sphere as Figure 27 shows, and mov e or distort lo ops if necessary so that their homotopy classes remain unchanged. One can verify the second and the third pictures of Figure 27 by comparing the vertices on the same side of each lo op in A ′ . Then we compare these tw o pictures with the last tw o pictures in Figure 25 , we see that they differ b y a rotation by π when w e ignore all the lab els. Figure 27: The loop arrangemen t A ′ is combinatorically equiv alent to A 3 . Lemma 5.11. Ther e is a T yp e 1 sp ac e whose closur e is on the other side of σ b with r esp e ct to O 3 . Pr o of. Consider the lo op arrangement A ′ obtained from A 3 b y moving the lo ops l b , l g and the lab eled v ertices 2 + and 2 − to their p ositions in the second picture of Figure 28 . Note that this only c hanges the homotopy class of l b . Let O ′ b e the space arising from A ′ and O ′ b e its closure. Then σ b is a common face of O 3 and O ′ . Figure 28: Visualizing Z 2 + 5 − on A 3 and A ′ , which is combinatorically equiv alen t to A 3 . Consider the lo cal frame ( Z 4 − 2 + , Z 2 + 1 + , Z 1 + 5 + , Z 5 + 4 + ) on O 3 in Lemma 5.10 . If we replace the comp onen t Z 5 + 4 + b y Z 5 − 4 − and then by − ( Z 5 − 4 − + Z 4 − 2 + ) = Z 2 + 5 − , we get another frame on O 3 . W e extend this lo cal frame to O ′ through the interior of σ b , and visualize the comp onen t Z 2 + 5 − with A 3 and A ′ in Figure 28 . 29 T o see that O 3 and O ′ are on differen t sides of σ b , w e orient the lo ops l b and l g in A 3 and A ′ according to the blue arrows in Figure 28 . Let ψ 3 and ψ ′ b e the maps from [0 , ∞ ) 8 to the ab ov e lo cal frame via surfaces in A 3 and A ′ , resp ectiv ely . Then w e can observe from Figure 28 that there exist unit vectors  b x b y  , b ′ x b ′ y ! and  g x g y  suc h that Z 2 + 5 − = b  b x b y  + g  g x g y  in ψ 3 and Z 2 + 5 − = b b ′ x b ′ y ! + g  g x g y  in ψ ′ . Thus, we can apply Lemma 5.2 b y taking U = 2 + , V = 5 − , i = b and j = g . According to the directions of the blue arrows in Figure 28 , we can see that     b x g x b y g y     < 0 and      b ′ x g x b ′ y g y      > 0, so O 3 and O ′ are on differen t sides of σ b . Finally , w e show that O ′ is a Type 1 space. W e just p erturb some lo ops and lab eled vertices in A ′ to their p ositions in the third picture of Figure 28 . This do es not change the homotopy class of an y loop, so A ′ is combinatorically equiv alent to A 1 . Lemma 5.12. Ther e is a T yp e 4 sp ac e whose closur e is on the other side of σ d with r esp e ct to O 3 . Pr o of. Consider the lo op arrangement A ′ obtained by moving the lo op l d to its p osition in the second picture of Figure 29 . Let O ′ b e the space arising from A ′ and O ′ b e its closure. Then σ d is a common face of O 3 and O ′ . Figure 29: Visualizing Z 2 + 1 + and Z 5 + 4 + with A 3 and A ′ (com binatorically equiv alent to A 4 ). Consider the lo cal frame in Lemma 5.10 on O 3 (and extended to O ′ through the interior of σ d ). The comp onents Z 2 + 1 + and Z 5 + 4 + of this lo cal frame are visualized with A 3 and A ′ in Figure 29 . Then we orient the lo ops l d , l h , l e and l f according to the blue arrows. Let ψ 3 and ψ ′ b e the maps from [0 , ∞ ) 8 to the ab o ve lo cal frame via surfaces in A 3 and A ′ , resp ectiv ely . F rom Figure 29 , we can see that there exist unit vectors  d x d y  , d ′ x d ′ y ! ,  h x h y  ,  e x e y  and  f x f y  suc h that Z 2 + 1 + = d  d x d y  + h  h x h y  + f  f x f y  and Z 5 + 4 + = e  e x e y  − f  f x f y  in ψ 3 . In addition, Z 2 + 1 + = h  h x h y  + f  f x f y  and Z 5 + 4 + = d d ′ x d ′ y ! + e  e x e y  − f  f x f y  in ψ ′ . Therefore, we can apply 30 Lemma 5.3 b y taking T = 2 + , U = 1 + , V = 5 + , W = 4 + , i = d , j = h , k = e and m = f . According to the directions of the blue arro ws in Figure 29 , we conclude that     d x h x d y h y     ·     e x f x e y f y     > 0 and      d ′ x e x d ′ y e y      ·     h x f x h y f y     < 0. Thus O 3 and O ′ are on differen t sides of σ d . Finally , A ′ is com binatorically equiv alent to A 4 . W e ma y rotate the sphere so that the v ertex 5 + is at the center of our view, and then p erturb the loops in A ′ to their p ositions in the third picture of Figure 29 . In fact, this do es not change the homotop y class of each lo op. Although we ha ve not found a wa y to demonstrate this easily without a 3D mo del, it is still helpful to compare the v ertices on the same side of each lo op in the second and the third pictures. Finally , the third picture is equiv alent to A 4 up to a rotation if w e ignore all the labels. Lemma 5.13. Ther e is a T yp e 3 sp ac e whose closur e is on the other side of σ e with r esp e ct to O 3 . Pr o of. Consider the lo op arrangement A ′ obtained by moving the lo ops in A 3 to their p ositions in the second picture of Figure 30 . Note that this does not c hange the homotop y classes of an y loop except for l e . Let O ′ b e the space arising from A ′ and O ′ b e its closure. Then σ e is a common face of O 3 and O ′ . Figure 30: Visualizing Z 5 + 4 + with A 3 and A ′ , which is combinatorically equiv alen t to A 3 . Consider the local frame in Lemma 5.10 on O 3 , and extend it to O ′ through the in terior of σ e . In Figure 30 , we visualize the comp onent Z 5 + 4 + of this lo cal frame with A 3 and A ′ , and orient the lo ops l e and l g according to the blue arrows. Let ψ 3 and ψ ′ b e the maps from [0 , ∞ ) 8 to this local frame via surfaces in A 3 and A ′ , resp ectively . Then there exist unit v ectors  e x e y  , e ′ x e ′ y ! ,  f x f y  suc h that Z 5 + 4 + = e  e x e y  + f  f x f y  in ψ 3 and Z 5 + 4 + = e e ′ x e ′ y ! + f  f x f y  in ψ ′ . Th us, we can apply Lemma 5.2 by taking T = 4 + , U = 5 + , i = e and j = f . Based on the directions of the blue arro ws in Figure 30 , we can see that     e x f x e y f y     > 0 and      e ′ x f x e ′ y f y      < 0. Th us, O 3 and O ′ are on differen t sides of σ e . Finally , A ′ is combinatorically equiv alen t to A 3 . T o see this, we rotate the second picture of A ′ b y π to obtain the third picture in Figure 30 , which is equiv alent to A 3 if w e ignore the lab els. 31 W e hav e thus finished the discussion with Type 3 spaces. T yp e 4 Spaces In the picture of A 4 in Figure 17 , we observe that there are essentially t wo symmetry classes of lo ops: 1) l a , l b , l e and l f ; 2) l c , l d , l g and l h . In Lemma 5.8 , w e hav e shown that there exists a T yp e 2 space whose closure is on the other side of σ e with resp ect to O 4 . In Lemma 5.12 , we hav e sho wn that there exists a Type 3 space whose closure is on the other side of σ d with resp ect to O 4 . This completes the discussion with T yp e 4 spaces. Let us summarize the results from the lemmas and the corollaries in this section so far: Lemma 5.14. L et O b e a sp ac e of T yp e 1, 2, 3 or 4 and O b e its closur e. Then O is op en in C 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . In addition, for any fac e σ of O , ther e exists a sp ac e of T yp e 1, 2, 3 or 4 whose closur e is on the other side of σ with r esp e ct to O . Analogous to Section 4, we consider the mo duli space ∆ of O , the space of surfaces in O with area 1. Let ∆ b e the completion (with resp ect to the real hyperb olic metric) of ∆. When O has Type 1, 2, 3 or 4, a similar argument to Lemma 4.3 shows that ∆ is a real hyperb olic ideal 7-simplex. Consider the moduli space M 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) and its metric completion M 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ). Then Lemma 5.14 implies the follo wing result analogous to Corollary 4.4 : Corollary 5.15. If O has T yp e 1, 2, 3 or 4, then every element in ∆ or in the interior of a fac e of ∆ has an op en neighb orho o d in M 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) . Finally , we are ready to pro ve the follo wing result, whic h is an analog of Theorem 4.5 : Theorem 5.16. Every p olyhe dr al surfac e in M 10 ( δ 1 , δ 2 , δ 3 , δ 4 , δ 5 ) c an b e de c omp ose d into at most 2  8 2  = 56 p ar al lelo gr ams, and the de c omp osition is invariant under the antip o dal map. Pr o of. The idea of pro of is essentially the same as that of Theorem 4.5 . W e construct infinitely man y copies of the moduli spaces of Type 1, 2, 3 or 4. Then w e glue their metric completions (which are 7-simplices) to form a space X such that ev ery face b elongs to exactly tw o 7-simplices, one on eac h side. By Lemma 5.14 , suc h X exists. Then we define the map ι analogously . Awa y from the co dimension-t w o b oundaries of X , the op enness of ι is guaranteed by Corollary 5.15 . Finally , there are 8 lo ops in ev ery lo op arrangement, each tw o of which intersect exactly twice as great circles, hence the n umber 2  8 2  . 6 Discussions and F uture W ork Our pro of of Theorem 5.16 is based on Corollary 5.15 . In fact, this metho d of pro of has the p oten tial to b e applied in higher dimensions as long as we can generalize Lemma 5.14 (and thus Corollary 5.15 ) to higher dimensions. T o ac hieve this, w e will need to construct differen t types of lo op arrangements. In addition, for each lo op arrangemen t A and an y lo op l i in A , we need to mo ve l i to an appropriate position to get an adjacen t lo op arrangement A ′ suc h that the closures of the spaces arising from A and A ′ are on differen t sides of the face σ i . The difficult y is that we cannot alwa ys find a p osition to mov e l i b y induction, even though we may remo ve another tw o lo ops l j and l k from A to obtain a lo op arrangement of a low er dimensional space. T his is b ecause 32 when we mov e l i b y the induction hypothesis, we ma y mov e some lab eled v ertices at the same time, making it imp ossible to add l j and l k bac k so that they remain great circles in their original homotop y classes. Without induction, it would b e tedious to enumerate and c heck all types of lo op arrangemen ts as we can expect from the w ork in Section 5. Nev ertheless, we b eliev e that Theorem 5.16 can b e generalized to higher dimensions. W e sum- marize it as the conjecture below to w ork on in the future: Conjecture. L et δ 1 , δ 2 , . . . , δ N b e N p ositive numb ers that sum up to 2 π . Then every surfac e in M 2 N ( δ 1 , δ 2 , . . . , δ N ) c an b e de c omp ose d into at most 2  2 N − 2 2  p ar al lelo gr ams, and the de c omp osition is invariant under the antip o dal map. Ac kno wledgmen ts W e are very grateful to Professor Richard Sc hw artz and Professor Peter Do yle for their very helpful commen ts and supp ort. References [1] Bav ard, C., & Ghys, ´ E. (1992). Polygones du plan et p oly edres hyperb oliques. Geometriae Dedicata, 43(2), 207-224. [2] Deligne, P ., & Mosto w, G. D. (1986). Mono drom y of hypergeometric functions and non-lattice integral mon- odromy . 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