Complete Set of CHC Tetrahedrons
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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 . 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.