International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014
Complete Set of CHC Tetrahedrons
Pranab Kalita Bichitra Kalita Gauhati University Assam Engineering College Department of Mathematics Department of Computer Applications (M.C.A) Guwahati-781014, Assam, India Guwahati-781013, Assam, India
ABSTRACT P and two vertices x, y V P are adjacent if and only if In this article, using K. W. Roeder’s Theorem, some xy, is an edge of P . properties of CHC (compact hyperbolic coxeter) tetrahedrons have been developed which are facilitated by the link of graph theory and combinatorics, and it has been found that there are Definition 2.2: If the dihedral angle of an edge of a compact exactly 3 CHC tetrahedrons upto symmetry. hyperbolic polytope is ,n is a positive number, then n is n General Terms said to be the order of the edge. A trivalent vertex is defined Three Dimensional Polytope, Hyperbolic Space, Roland K. to be of order l,, m n if the three edges at that vertex are of W. Roeder Theorem. order l,, m n . Keywords Definition 2.3: A compact polytope in hyperbolic space with Dihedral angles, Face angles, Planar graph, Coxeter. coxeter dihedral angles is called a CHC (compact hyperbolic 1. INTRODUCTION coxeter) polytope. The angle between two faces of a polytope, measured from Suppose the dihedral angles at the edges e1,,,,, e 2 e 3 e 4 e 5 e 6 of a perpendiculars to the edge created by the intersection of the coxeter tetrahedron are respectively ,,,,, as planes is called a dihedral angle. A simple polytope P in n - n1 n 2 n 3 n 4 n 5 n 6 n dimensional hyperbolic space H is said to be coxeter, if the shown in figure 1. dihedral angles of P are of the form where, n is a v1 n positive integer 2 . P. Kalita and B. Kalita [1] found that there are exactly one, four and thirty coxeter Andreev’s tetrahedrons having respectively two edges of order n 6 , e e e n3 one edge of order n 6 and no edge of order n 6 , n 1 n1 4 n4 3 upto symmetry. There is no complete classification of hyperbolic coxeter polytopes. Vinberg proved in [22] that e 5 v e6 there are no compact hyperbolic coxeter polytopes in H n 4 n5 e n6 when n 30 . Roland K. W. Roeder’s Theorem [10] provides 2 the classification of compact hyperbolic tetrahedron by v2 n2 v3 restricting to non-obtuse dihedral angles. In this article, using Figure 1: CHCT n1,,,,, n 2 n 3 n 4 n 5 n 6 . K. W. Roeder’s Theorem, some properties of CHC (compact hyperbolic coxeter) tetrahedrons have been developed which are facilitated by the link of graph theory and combinatorics, Then the CHC tetrahedron is denoted as and it has been found that there are exactly 3 CHC CHCT n1,,,,, n 2 n 3 n 4 n 5 n 6 . tetrahedrons upto symmetry. Definition 2.4: A cell complex C on S 2 is called trivalent if The paper is organised as follows: each vertex is the intersection of three faces. The section 1 includes introduction. The section 2 includes some basic terminologies from graph theory and geometry. Definition 2.5: A 3-dimensional combinatorial polytope is a The section 3 focuses some already existed results on graph cell complex on that satisfies the following conditions: theory and geometry. Main results and conclusions are included in the sections 4 and 5 respectively. (a) Each edge of is the intersection of exactly two faces 2. BASIC TERMINOLOGIES (b) A nonempty intersection of two faces is either an edge or a vertex. Definition 2.1: Let P be a polyhedron. The abstract graph of (c) Each face is enclosed by not less than 3 edges. P is denoted by GP and is defined as Any trivalent cell complex on that satisfies the GPVPEP , , where VP is the set of vertices of above three conditions is said to be abstract polyhedron.
In this work, the main concentration is on 3-dimensional compact hyperbolic polytopes whose base spaces are
18 International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014 homeomorphic to a convex polyhedron and whose sides are Hence the face angles are calculated from the dihedral angles. silvered. The compact hyperbolic polyhedron is simple, They are non-obtuse, since the right-hand side of the equation therefore, the combinatorial polyhedron of a compact is positive for i,, j k are non-obtuse. hyperbolic polyhedron can be known from 3-connected planar graph of the polyhedron. In case of compact hyperbolic Theorem 3.5: (Roland K. W. Roeder, 2013, [10]) Let tetrahedron, the corresponding 3-connected planar graph has 4 16,, be a set of proposed non-obtuse dihedral angles and vertices, 6 edges, 4 faces (excluding exterior one). let 1 1,,,,,, 6 12 1 6 be the face angles given Hyperbolic geometry is difficult to visualize as many of by equation (1), corresponding to these proposed dihedral its theorems are contradictory to similar theorems of angles. Euclidean geometry which are very similar to us. Therefore There is a compact hyperbolic tetrahedron with dihedral technology has been used in creation of geometric models in Euclidean space to visualize Hyperbolic geometry. Some angles if and only if developed models are The Klein Model, The Poincare Model (1) For each edge e , 0 . etc. Hyperbolic [10] planes in these models correspond to i i 2 Euclidean hemispheres and Euclidean planes that intersect the (2) Whenever 3 distinct edges e,, e e meet at a vertex, boundary perpendicularly. Furthermore, these models are i j k correct conformally. That is, the hyperbolic angle between a i j k . pair of such intersecting hyperbolic planes is exactly the (3) For each face the sum of the face angles satisfies Euclidean angle between the corresponding spares or planes. i j k . Furthermore this tetrahedron is unique. 3. KNOWN RESULTS Theorem 3.1: (Blind and Mani) If P is a convex In this article, the dihedral angles of a tetrahedron are polyhedron, then the graph GP determines the entire restricted to be coxeter and hence non-obtuse as well. combinatorial structure of P . In other words, if two simple Therefore, the concentration is only on the CHC tetrahedrons polyhedral have isomorphic graphs, then their combinatorial given by the Theorem 3.5. polyhedral are also isomorphic. 4. MAIN RESULTS Theorem 3.2: (Ernst Steinitz) A graph GP is a graph of a Now the main results will be established below. Theorem 4.1: In a CHC tetrahedron T , there is no vertex of 3-dimensional polytope P if and only if it is simple, planar order of the form 2,2,n 2 . and 3-connected.
Lemma 3.3: (E. M. Andreev [23]) Suppose that three planes Proof: Suppose there exists at least one vertex of order PPP,, intersect pair-wise in H 3 with non-obtuse . Therefore the dihedral angles are v1 v 2 v 3 dihedral angles i,, j k . Then, PPPv,, v v intersect at a 1 2 3 i, j , k , , ;n 2 . The corresponding face 22n vertex in 3 if and only if . The planes H i j k angles are calculated by using the equation (1): intersect in if and only if the inequality is strict (that is cosi cos j cos k cos i j k ). i sinjk sin Lemma 3.4: (E. M. Andreev [23]) Given a trivalent vertex of cos cos cos a hyperbolic polyhedron, we can compute the angles of the 22n faces in terms of the dihedral angles. If the dihedral angles sin sin 2 n are non-obtuse, these angles are also . 2 0 By lemma 3.4, 0 , therefore . Similarly it can Proof: Let v be a finite trivalent vertex of a polyhedron P . i 2 i 2 After an appropriate isometry, we can assume that is the origin in the Poincare ball model, so that the faces at are be deduced that j and k . Now, 2 n subsets of Euclidean planes through the origin. A small sphere centered at the origin will intersect in a spherical triangle i j k Q whose angles are the dihedral angles between faces. Call 22nn That is the Roeder’s 3rd condition (of Theorem 3.5) is not these angles ,, . 1 2 3 satisfied. Hence, in a CHC tetrahedron T , there is no vertex The edge lengths of are precisely the angles in the faces at of order of the form 2,2,n 2 . the origin. Suppose that has edge length 1,, 2 3 with Theorem 4.2: In a CHC tetrahedron T , there is no vertex of the edge i opposite the angle i for each i 1,2,3 , the law order of the form 2,3,3 . of cosines in spherical geometry states that:
cosi cos j cos k cosi ...(1) sinjk sin
19 International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014
Proof: Suppose there exists at least one vertex of order Corollary 4.5: In a CHC tetrahedron , the edges of order 2 2,3,3 . Therefore the dihedral angles are must be disjoint.
Proof: Suppose there exists at least two adjacent edges of i,,,, j k and the face angles given by (1) are: 233 order 2. Then the order of the vertex at which the two edges of order 2 are adjacent becomes 2,2,n 2 . This is not 11 1 1 1 1 ,i j , k cos ,cos ,cos 3 33 possible by Theorem 4.1. Therefore the edges of order 2 must be disjoint. But , That is the Roeder’s 3rd condition (of i j k Theorem 3.5) is not satisfied. Hence, in a CHC tetrahedron T The following Theorem 4.6 and corollary 4.7 have been taken , there is no vertex of order of the form 2,3,3 . from [1] which will be used in this present work.
Theorem 4.6: In a tetrahedron , if any three vertices are of Theorem 4.3: In a CHC tetrahedron T , the order of each of same order, then the fourth vertex is also of same order. the vertices is one of the forms: 2,3,4 , 2,3,5 . Corollary 4.7: In a tetrahedron , the number of same order
Proof: Let ei,, e j e k be three distinct edges of orders ni,, n j n k vertices can be either 2 or 4. meet at one vertex v in T . The respective dihedral angles of Theorem 4.8: Let be a CHC tetrahedron with at least one at v are ,, , with positive integers vertex is of order 2,3,4 . Then there are exactly 2 such ni n j n k nd upto symmetry. ni, n j , n k 2 . By Roeder’s 2 condition (of Theorem 3.5): 1 1 1 Proof: Let be a CHC tetrahedron with at least one vertex 1 i j k is of order . By corollary 4.7, in a CHC tetrahedron ni n j n k n i n j n k , the number of same order vertices can be either 2 or 4. So, upto permutations, the triples ni,, n j n k are respectively Therefore, there will be three cases: Case 1: all (four) the 2,2,n 2 , 2,3,3 , 2,3,4 , 2,3,5 . The vertices of order vertices are of order , Case 2: two vertices are of 2,2,n 2 are not possible by Theorem 4.1 and the vertices order and Case 3: one vertex is of order . of order 2,3,3 are not possible by Theorem 4.2. For the Case 1: All the vertices of are of order . vertices of order ni, n j , n k 2,3,4 and In this case, there will be exactly 1 of this type as shown in figure 2 upto symmetry. ni, n j , n k 2,3,5 ; the corresponding dihedral angles are: v1 i,,,,,,,,, j k i j k 2 3 4 2 3 5 For i,,,, j k , the face angles given by (1) are: 234 e e 1 2 4 3 e3 4 1112 ,i j , k cos , ,cos 43 e 3 5 v4 e6 rd 4 2 And i j k , That is the Roeder’s 3 condition (of e2 Theorem 3.5) is satisfied. Hence, in a CHC tetrahedron T , v2 3 v3 there is vertex of order of the form 2,3,4 . Figure 2: CHCT-1 2,3,4,3,4,2 For i,,,, j k , the face angles given by (1) are: 2 3 5 Case 2: Two vertices are of order . 1 1 1 ,i j , k cos 0.7947 ,cos 0.8506295 ,cos 0.93415 Obviously the two vertices of order with either rd disjoint edges of order 2 or common edge of order 2. And i j k , That is the Roeder’s 3 condition (of Case 2.1: Two vertices of order with disjoint edges Theorem 3.5) is satisfied. Hence, in a CHC tetrahedron , there is vertex of order of the form 2,3,5 . Therefore in a of order 2. Suppose the vertices v and v are of order with CHC tetrahedron , the order of the edges at one vertex is 1 3 one of the forms: 2,3,4 , 2,3,5 . disjoint edges e1 and e6 of order 2.
Corollary 4.4: In a CHC tetrahedron , the number of edges of order 2 at one vertex is exactly one.
Proof: In a CHC tetrahedron , the order of the edges at one vertex is one of the forms: 2,3,4 , 2,3,5 . Therefore the number of edges of order 2 at one vertex is exactly one.
20 International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014
v1 v1
e e 1 2 e4 3 e3 4 1 2 e4 3 e3 4
e5 v4 e6 e5 v4 e6 m m 2 e2 2 4 e2
v2 3 v3 v2 3 v3 Figure 3: CHC Tetrahedron with Two vertices of order Figure 6: Another CHC Tetrahedron with Two vertices of 2,3,4 having disjoint edges of order 2. order having common edge of order 2.
Then the possibilities of orders for e5 are m 2,3,4,5 . For In figure 5, there is no choice for m . In figure 6, only choice m 2 , the order of v2 becomes 2,2,3 which is not is m 2 which falls in case 1. possible by Theorem 4.1. For m 3 , the order of becomes Case 3: One vertex v1 is of order 2,3,4 . 2,3,3 which is not possible by Theorem 4.2. For m 4 , it By corollary 4.4, the number of edges of order 2 at one vertex falls in case 1. For m 5 , there is exactly 1 such T of this is exactly one and by corollary 4.5, the edges of order 2 must type as shown in figure 4 upto symmetry. be disjoint. Therefore if is of order with e1 is of
v1 order 2, then e6 must be of order 2. Suppose the order of the
edges e2 and e5 are m1 and m2 respectively.
v1 e 1 2 e4 3 e3 4
e1 2 e4 3 e 4 e5 v4 e6 3 5 e 2 2 e 5 v4 e6 v2 3 v3 m2 e 2 Figure 4: CHCT-2 2,3,4,3,5,2 2 v m v 2 1 3 Figure 7: CHC Tetrahedron with one vertex of order Case 2.2: Two vertices of order 2,3,4 with common edge of order 2. To avoid symmetry, assume mm12 . Therefore, m1 3 and Suppose the vertices v1 and v2 are of order with m2 3,4,5 upto symmetry. For m1 3 and m2 3 , the order common edge e1 of order 2. Then the possibilities are as of v becomes 2,3,3 , which is not possible by Theorem 4.2 shown in figures 5 and 6. 2
and m1 3 , m2 4 as well as , m2 5 lead to case 2. v1 Hence, there is no such T of this type.
Theorem 4.9: Let T be a CHC tetrahedron with at least one vertex is of order 2,3,5 and no vertex is of order of the e e 1 2 e4 3 3 4 forms 2,3,4 . Then there are exactly 1 such upto symmetry.
e5 v4 e6 m ? Proof: Let be a CHC tetrahedron with at least one vertex 3 e2 is of order and no vertex is of order of the forms v2 4 v3 Figure 5: CHC Tetrahedron with Two vertices of order . By corollary 4.7, in a CHC tetrahedron , the having common edge of order 2 number of same order vertices can be either 2 or 4. Therefore, there will be three cases: Case 1: all (four) the vertices are of
order , Case 2: two vertices are of order and Case 3: one vertex is of order . Case 1: All the vertices of are of order . In this case, there is exactly 1 as shown in figure 8 upto symmetry.
21 International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014
v1 v1
e e 1 2 e4 3 e3 5 1 2 e4 3 e3 5
e 5 v4 e6 e5 v4 e6 2 m 5 e2 5 e2 v2 3 v3 v2 3 v3 Figure 8: CHCT-3 2,3,5,3,5,2 Figure 10: CHC Tetrahedron with Two vertices of order having common edge of order 2. Case 2: Two vertices are of order 2,3,5 . Then the only possibility of order for e is m 2 and this Obviously the two vertices of order with either 6 falls in case 1. Hence, there is no such of this type. disjoint edges of order 2 or common edge of order 2 Case 3: One vertex v is of order . Case 2.1: Two vertices of order with disjoint edges 1 By corollary 4.4, the number of edges of order 2 at one vertex of order 2. is exactly one and by corollary 4.5, the edges of order 2 must Suppose the vertices v1 and v3 are of order with be disjoint. Therefore if is of order with e1 is of disjoint edges e and e of order 2. 1 6 order 2, then must be of order 2. Suppose the order of the
v1 edges e2 and e5 are m1 and m2 respectively.
v1
e 1 2 e4 3 e3 5 e e 1 2 4 3 e3 5
e5 v4 e6 m e5 v e e2 2 4 6 m2 e 2 v 3 v 2 2 3 v m v Figure 9: CHC Tetrahedron with Two vertices of order 2 1 3 2,3,5 having disjoint edges of order 2. Figure 11: CHC Tetrahedron with one vertex of order
To avoid symmetry, assume mm12 . Therefore, m1 3 and Then the possibilities of orders for e5 are m 2,3,4,5 .
m2 3,4,5 upto symmetry. For , m2 3 , the order of For m 2 , the order of v2 becomes 2,2,3 which is not possible by Theorem 4.1. v2 becomes 2,3,3 which is not possible by Theorem 4.2.
For m 3 , the order of becomes 2,3,3 which is not For m1 3 , m2 4 , the order of v2 becomes 2,3,4 which possible by Theorem 4.2. cannot be taken by assumption. For m1 3 , m2 5 , it falls in For m 4 , the order of becomes 2,3,4 which cannot be case 1. Hence, there is no such of this type. taken by assumption. For m 5 , it falls in case 1. Hence, there is no such T of this From theorems 4.8 and 4.9, the total number of CHC type. tetrahedrons upto symmetry is: 2 1 3 . These are namely Case 2.2: Two vertices of order with common edge CHCT-1 2,3,4,3,4,2 , of order 2. CHCT-2 2,3,4,3,5,2 , , Suppose the vertices v and are of order with 1 CHCT-3 2,3,5,3,5,2 common edge e1 of order 2 upto symmetry. These 3 tetrahedrons can be realized uniquely [10] in Hyperbolic space and these are nothing but the 3 coxeter Andreev’s tetrahedrons found in theorems 3.20 and 3.21 in [1]. 5. CONCLUSIONS In this article, it is found that there are exactly 3 CHC (compact hyperbolic coxeter) tetrahedrons upto symmetry in real projective space. These 3 tetrahedrons can be realized uniquely [10] in Hyperbolic space and these are nothing but the 3 coxeter Andreev’s tetrahedrons found in theorems 3.20 and 3.21 in [1]. This research can be extended to other
22 International Journal of Computer Applications (0975 – 8887) Volume 87 – No.11, February 2014 compact as well as non-compact hyperbolic polytopes in [13] Pavel Tumarkin, Compact Hyperbolic Coxeter n spaces of different dimensions. development of the template. polytopes with n 3 facets, The Electronic Journal 6. REFERENCES of Combinatorics 14 (2007). [1] P. Kalita and B. Kalita, Properties of Coxeter Andreev’s [14] R.K.W. Roeder, Constructing hyperbolic polyhedral Tetrahedrons, IOSR Journal of Mathematics, Volume 9, using Newton’s method, Experiment. Math. 16, 463- Issue 6, pp 81-105, 2014. 492 (2007). [2] John G. Ratcliffe, Foundations of Hyperbolic Manifolds, [15] Roland K.W. Roeder, John H. Hubbard and William D. ©1994 by Springer-Verlag, New York, Inc. Dunbar, Andreev’s Theorem on Hyperbolic Polyhedra, Ann. Inst. Fourier, Grenoble 57, 3 (2007), [3] Chris Godsil, Gordon Royle, Algebraic Graph Theory, 825-882. Springer International Edition. [16] D. Cooper, D. Long and M. Thistlethwaite, Computing [4] Gil Kalai, Polytope Skeletons and Paths, ©1997 by CRC varieties of representations of hyperbolic 3-manifolds Press LLC. into SL4, , Experiment. Math. 15, 291305 [5] Dipankar Mondal, Introduction to Reflection Groups, (2006). April 26, 2013, Triangle Group (Course Project). [17] Yunhi Cho and Hyuk Kim. On the volume formula [6] Projective Linear Group, for hyperbolic tetrahedral. Discrete Comput. http://en.wikipedia.org/wiki/Projective_linear_group, Geom., 22 (3): 347-366, 1999. access in October, 2013. [18] Tomaz Pisanski, Milan Randic, Bridges between [7] Hyperbolic Tetrahedron, Geometry and Graph Theory, ISSN 1318-4865, Preprint http://mathworld.wolfram.com/Hyperbolic Series, Vol. 36 (1998), 595. Tetrahedron.html, access in October, 2013. [19] Raquel diaz, Non-convexity of the space of [8] J. Mcleod, Hyperbolic Coxeter Pyramids, Advances in dihedral angles of hyperbolic polyhedra. C. R. Acad. Sci. Pure Mathematics, Scientific Research, 2013, 3, 78- Paris Ser. I Math., 325 (9):993-998, 1997. 82. [20] E. B. Vinberg, Geometry II, Encyclopedia of Maths, Sc. [9] Tetrahedron, Wikipedia, the free encyclopedia, access in 29. Springer 1993. October, 2013. [21] E. B. Vinberg, Hyperbolic Reflection Groups, Uspekhi [10] [10] Roland K. W. Roeder, Compact hyperbolic Mat. Nauk 40, 29-66 (1985). tetrahedra with non-obtuse dihedral angles, August 10, 2013, arxiv.org/pdf/math/0601148. [22] E. B. Vinberg, The absence of crystallographic groups of reflections in Lobachevskij spaces of large dimensions, [11] Aleksandr Kolpakov, On extremal properties of Trans. Moscow Math. Soc. 47 (1985), 75-112. hyperbolic coxeter polytopes and their reflection groups, Thesis No: 1766, e-publi.de, 2012. [23] E. M. Andreev, On Convex Polyhedral of Finite Volume in Lobacevskii Space, Math. USSR Sbornik 10, 413-440 [12] Anna Felikson, Pavel Tumaarkin, Coxeter polytopes (1970). with a unique pair of non intersecting facets, Journal of Combinatorial Theory, Series A 116 (2009) 875- 902.
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