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Volume of Hyperbolic Octahedron with 3-Symmetry Volume of hyperbolic octahedron with 3-symmetry Nikolay Abrosimov (joint work with Ekaterina Kudina, Alexander Mednykh) Sobolev Institute of Mathematics Novosibirsk State University Novosibirsk, Russia BIRS Workshop on Discrete Geometry and Symmetry February 10, 2015 Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 1/16 Introduction Calculating volumes of polyhedra is a classical problem, that has been well known since Euclid and remains relevant nowadays. This is partly due to the fact that the volume of a fundamental polyhedron is one of the main geometrical invariants for a 3-dimensional manifold. Every 3-manifold can be presented by a fundamental polyhedron. That means we can pair-wise identify the faces of some polyhedron to obtain a 3-manifold. Thus the volume of 3-manifold is the volume of prescribed fundamental polyhedron. It is known that regular hyperbolic octahedron with all vertices at infinity is a fundamental polyhedron for Whitehead link manifold. On the other hand, the minimal volume hyperbolic manifold and many others can be obtained by Dehn surgery along Whitehead link. Thus, hyperbolic octahedra can serve as fundamental polyhedra for a wide class of 3-manifolds, including hyperbolic manifolds of small volume. The latter seems especially interesting if we arrange hyperbolic manifolds in order of volume increasing. Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 2/16 Introduction It is difficult problem to find the exact volume formulas for hyperbolic polyhedra of prescribed combinatorial type. It was done for hyperbolic tetrahedron of general type, but for general hyperbolic octahedron it is an open problem. Nevertheless, if we know that a polyhedron has a symmetry, then the volume calculation is essentially simplified. Firstly this effect was shown by Lobachevskij. He found the volume of an ideal tetrahedron, which is symmetric by definition. R.V. Galiulin, S.N. Mikhalev and I.Kh. Sabitov found the volumes of Euclidean octahedra with all possible types of symmetry, except the trivial one. The volumes of spherical octahedron with mmm or 2 m-symmetry | were given by N. Abrosimov, M. Godoy and A. Mednykh. The volume of hyperbolic octahedron with mmm-symmetry was obtained by N. Abrosimov and G. Baigonakova. Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 3/16 Definition An octahedron has 3-symmetry if it admits the antipodal involution and order 3 rotation z 2p v 3 2 a y a a v1 c c c v3 O c c c x a v a6 v 4 a v5 3 by Hermann Mauguin notation; S6 by Sch¨onflies notation; [2+, 6+] by Coxeter notation; 3 by orbifold notation. × Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 4/16 2π 2π cos sin 0 1 0 0 3 − 3 − R 2π 2π S = sin cos 0 , = 0 1 0 . 3 3 − 0 0 1 0 0 1 − Since the symmetry group acts transitively on the set of vertices of octahedron , then we fix one vertex v = (r, 0, h) and consider its orbit O 1 under action of R and S: r √3 r √3 v = (r, 0, h), v = , , h , v = , , h , 1 2 −2 2 3 −2 − 2 (1) r √3 r √3 v = ( r, 0, h), v = , , h , v = , , h . 4 − − 5 2 − 2 − 6 2 2 − Now we know the coordinates of vertices of octahedron . O Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 5/16 Using the coordinates of vertices, we compute edge lengths. Then we calculate normal vectors for faces and find the cosines of dihedral angles. r a a2 = 3 r 2, cos A = − = − , √r 2 + 16 h2 3 (4 c2 a2) − r 2 8 h2 ap2 2 c2 c2 = 4 h2 + r 2, cos C = − = − . r 2 + 16 h2 −a2 4 c2 − Remark Let = (a, c) is Euclidean octahedron with 3-symmetry. Then O O a2 (i) edge lengths of satisfy the condition 0 < < 3 ; O c2 C (ii) dihedral angles of are related by equation 2 sin = √3 sin A, O 2 π 2 π where A ,π , C 0, . ∈ 2 ∈ 3 Converse is also true. a2 The volume of such an octahedron is V = 3 c2 a2 3 p − Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 6/16 Caley-Klein model 4 Consider Minkowski space R1 with scalar product X , Y = x y x y x y + x y . h i − 1 1 − 2 2 − 3 3 4 4 The Caley-Klein model of hyperbolic space is the set of vectors K = (x , x , x , 1) : x2 + x2 + x2 < 1 forming the unit 3-ball in the { 1 2 3 1 2 3 } hyperplane x4 = 1. The lines and planes in K are just the intersections of ball K with Euclidean lines and planes in the hyperplane x4 = 1. Let V , W K. Assume V = (v, 1), W = (w, 1), where v, w R3. Then ∈ ∈ the scalar product of V and W in Minkowski space is expressed via the Euclidean scalar product in R3 by the formula V , W = 1 v, w E . h i − h i The distance between vectors V and W in Caley-Klein model is defined by equation V , W ch ρ (V , W )= h i . (2) V , V W , W h i h i A plane in K is a set = V K :pV , N = 0 , where N = (n, 1), P { ∈ h i } n, n E > 0 is a normal vector to the plane . h i P Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 7/16 Caley-Klein model In model K consider two planes , with normal vectors N, M P Q correspondingly. Then every of four dihedral angles between the planes , is defined by relation P Q \ N, M cos ( , )= h i . (3) P Q ± N, N M, M h i h i p Now let V1 = (v1, 1), V2 = (v2, 1), V3 = (v3, 1) are three non-coplanar vectors in K. Then there is a unique plane passes through them: = V K : V , N = 0 with normal vector N = (n, 1), where the P { ∈ h i } coordinates of vector n R3 are uniquely determined as the solution of a ∈ system of linear equations v , n E 1 = 0, h 1 i − v , n E 1 = 0, h 2 i − v , n E 1 = 0. h 3 i − Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 8/16 z 2p v 3 2 a y a a v1 c c c v3 O c c c x a v a6 v 4 a v5 2π − 2π − cos sin 0 0 1 0 0 0 3 3 2π 2π − sin cos 0 0 0 1 0 0 Vi =(vi , 1), R = , S = . 3 3 − 0 0 10 0 0 1 0 0 0 01 0 0 0 1 Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 9/16 In Caley-Klein model K consider vectors Vi = (vi , 1), i = 1,..., 6, where vi are the vertices of Euclidean octahedron with 3-symmetry, studied before. The isometries R, S of Euclidean 3-space are naturally extended to isometries of hyperbolic space K: R : (v, 1) (Rv, 1), S : (v, 1) (Sv, 1). → → As before, R is an order 3 rotation around the axe Ox3, and S is an antipodal involution at point (0,0,0,1), which is the centre of K. In contrast to Euclidean case, we have one additional condition r 2 + h2 < 1. Or equivalently, all the vertices Vi K. Without lost of generality, assume ∈ that r > 0, h > 0. Then the vectors Vi = (vi , 1), i = 1,..., 6, are determine the vertices of a hyperbolic octahedron with 3-symmetry. O Using the coordinates of vertices we compute the edge lengths by (2): 2+ r 2 2 h2 2 r 2 + 2 h2 ch a = − , ch c = − . (4) 2 (1 r 2 h2) 2 (1 r 2 h2) − − − − Solving (7) with respect to r 2, h2 we get 4 (ch a 1) 3 ch c ch a 2 r 2 = − , h2 = − − . (5) 3 (ch a + ch c) 3 (ch a + ch c) Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 10/16 Proposition A hyperbolic octahedron = (a, c), admitting 3-symmetry, with edge lengths a, c is exist if andO onlyO if 3 ch c ch a 2 > 0 . − − Using coordinates of vertices Vi , i = 1,..., 6, by formula (3) we get the cosines of dihedral angles up to choice of signs. But in particular case of regular hyperbolic octahedron we already know the answer. This allows us to choose the signs correctly. Proposition (ch c ch a 1)√ch a 1 cos A = − − − , (1 + 2 ch a)(2 ch2 c ch a 1) q − − (6) 1 ch c + ch a ch c ch2 c cos C = − − . 2 ch2 c ch a 1 − − Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 11/16 We differentiate the volume as a composite function ∂V ∂V ∂A ∂V ∂C = + , ∂a ∂A ∂a ∂C ∂a (7) ∂V ∂V ∂A ∂V ∂C = + . ∂c ∂A ∂c ∂C ∂c By Schl¨afli formula we have ℓθ dV = dθ = 3 adA 3 c dC, − 2 − − Xθ ∂V ∂V hence = 3 a, = 3 c. ∂A − ∂C − Abrosimov-Kudina-Mednykh (IM & NSU) Hyperbolic octahedron with 3-symmetry Feb. 10,2015 12/16 We substitute all the expressions into formulas (7).
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