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ЧЕБЫШЕВСКИЙ СБОРНИК
Том 21. Выпуск 2.
УДК 515.162.8 + 514.132 DOI 10.22405/2226-8383-2020-21-2-65-83
Идеальные прямоугольные многогранники в пространстве Лобачевского1
А. Ю. Веснин, А. А. Егоров
Веснин Андрей Юрьевич — доктор физико-математических наук, член-корреспондент РАН, профессор, Новосибирский государственный университет; главный научный сотрудник, Институт математики им. С. Л. Соболева СО РАН, Новосибирск; главный научный сотрудник, Томский государственный университет, Томск (г. Новосибирск). e-mail: vesnin@math.nsc.ru
Егоров Андрей Александрович — магистрант кафедры геометрии и топологии, Новосибирский государственный университет, Новосибирск; лаборант-исследователь, Томский государственный университет, Томск (г. Новосибирск). e-mail: a.egorov2<Sg.nsu.ru
Аннотация
В работе рассматривается класс прямоугольных многогранников в трехмерном пространстве Лобачевского, все вершины которых лежат на абсолюте. Получены новые верхние оценки объемов через число граней многогранника. Вычислены объемы многогранников, имеющих не более, чем 23 граней. Показано, что наименьшие объемы реализуются на антипризмах и скрученных антипризмах. Установлены первые 248 значений объемов идеальных прямоугольных многогранников. Введен класс многогранников с изолированными треугольниками, получены комбинаторные оценки на существование и приведены минимальные примеры таких многогранников.
Ключевые слова: Пространство Лобачевского, идеальный многогранник, прямоугольный многогранник, антипризма
Библиография: 24 названия. Для цитирования:
А. Ю. Веснин, А. А. Егоров. Идеальные прямоугольные многогранники в пространстве Лобачевского // Чебышевский сборник, 2020, т. 21, вып. 2, с. 65-83.
1 Работа выполнена при частичной поддержке РФФИ (проект 19-01-00569)
CHEBYSHEVSKII SBORNIK Vol. 21. No. 2.
UDC 515.162.8 + 514.132 DOI 10.22405/2226-8383-2020-21-2-65-83
Ideal right-angled polyhedra in Lobachevsky space
A. Yu. Vesnin, A. A. Egorov
Vesnin Andrei Yurievich — Doctor of Physics and Mathematics, Corresponding member of RAS, Professor, Novosibirsk State University, Chief Researcher, Sobolev Institute of Mathematics, Novosibirsk; Chief Researcher, Tomsk State University, Tomsk (Novosibirsk). e-mail: vesnin@math.nsc.ru
Egorov Andrey Alexandrovich — Master student of the Department of Geometry and Topology,
Novosibirsk State University, Novosibirsk; Laboratory Assistant, Tomsk State University, Tomsk
(Novosibirsk).
e-mail: a. egorov2@g. nsu.ru
Abstract
In this paper we consider a class of right-angled polyhedra in three-dimensional Lobachevsky space, all vertices of which lie on the absolute. New upper bounds on volumes in terms the number of faces of the polyhedron are obtained. Volumes of polyhedra with at most 23 faces are computed. It is shown that the minimum volumes are realized on antiprisms and twisted antiprisms. The first 248 values of volumes of ideal right-angled polyhedra are presented. Moreover, the class of polyhedra with isolated triangles is introduces and there are obtained combinatorial bounds on their existence as well as minimal examples of such polyhedra are given.
Keywords: Hyperbolic 3-space, ideal polyhedron, right-angled polyhedron, antiprism Bibliography: 24 titles.
For citation:
A. Yu. Vesnin, A. A. Egorov, 2020, "Ideal right-angled polyhedra in Lobachevsky space" , Cheby-shevskii sbornik, vol. 21, no. 2, pp. 65-83.
To the 75th anniversary of academician Anatoly Timofeevich Fomenko.
Introduction
Applying of computer methods is a powerful tool for the study of three-dimensional hyperbolic manifolds. For example, the tabulation of manifolds obtained by Dehn surgery on manifolds with cusps led by S. V. Matveev and A.T. Fomenko [5] and independently J. Weeks [24] to recognizing the smallest volume closed orientable three-dimensional hyperbolic manifold. Todays it is known as the Weeks - Matveev - Fomenko manifold. Recall that it can be obtained by surgery on the Whitehead link and its volume is approximately equal to 0.942707.
In recent years many results appeared on the enumeration and classification of three-dimensional hyperbolic manifolds which admit decompositions into polyhedra with prescribed properties.
In [11] there are described hyperbolic three-dimensional manifolds that can be decomposed into regular ideal tetrahedra (up to 25 tetrahedra in the oriented case and up to 21 tetrahedrons in the non-oriented case). Three-dimensional hyperbolic manifolds that can be subdivided into Platonic polyhedra are listed in [13]. In [12] all three-dimensional orientable manifolds that can be obtained
from various realizations of an octahedron were constructed and classified. The paper [14] contains an initial list of 825 bounded right-angled hyperbolic polvhedra.
In this paper, the objects of our study are polvhedra which can be realized with right, ^/2, dihedral angles in a three-dimensional space of constant negative curvature H3, known as hyperbolic space or Lobachevskv space. Namely, we will consider only ideal right-angled hyperbolic polvhedra, that is, those for which all vertices lie on the absolute of Lobachevskv space. We will denote by X~R the class of ideal right-angled three-dimensional hyperbolic polvhedra. Recent results on the theory of right-angled polvhedra in Lobachevskv space and using them for constructing three-dimensional hyperbolic manifolds are given in the survey [2], see also [4]. The main attention in the survey was given to bounded right-angled polvhedra, while in this paper we will consider the case of ideal polvhedra. We will follow the standard terminology of the theory of hyperbolic manifolds; see, for example, [19].
Hyperbolic three-dimensional manifolds of finite volume, which can be decomposed into ideal right-angled polvhedra, have been intensively studied in last decade. In particular, due to their close relationship with the right-angled Coxeter groups. It is known that their fundamental groups have the LERF property, i.e. they are locally extended residuallv finite groups (each finitely generated subgroup is separable) [21]. Several types of hyperbolic three-dimensional manifolds admitting decomposition into ideal right-angled polvhedra are presented in [9]. Since the volume of a manifold is the sum of the volumes of the polvhedra into which it is decomposed, a description of the volumes of ideal right-angled polvhedra is interesting from this point of view.
The paper has the following structure. In Section 1 we recall some facts about the existence of ideal polvhedra in Lobachevskv spaces, in particular, Andreev's theorem (Theorem 1) and Rivin's theorem (Theorem 2), which give necessary and sufficient conditions for their existence in dimension three. In Section 2 the notion of twisted antiprism is introduced and a formula for volumes of right-angled twisted antiprisms is given (Theorem 7). The Table 2 provides information on the number of ideal right-angled polvhedra in Lobachevskv space with at most 23 faces and indicates the minimum and maximum volume values for each number of faces. The carried out calculations allow to propose a conjecture which polvhedra are of smallest volumes for an arbitrary number of faces (Conjecture 1). In Section 3 we obtain new upper bounds on the volume of an ideal right-angled polyhedron in terms of the number of its faces (Theorems 9 and 10). Also, we present ideal right-
23
first 248 values of volumes of ideal right-angled polvhedra (see Table 6). In Section 4 we introduce the notion of polvhedra with isolated triangles and give a lower bound on the number of faces of such polvhedra (Proposition 3). For the minimum possible number of faces equals to 26, two examples of polvhedra with isolated triangles are given.
1. Existence
1.1. Dimension and number of cusps
It was shown in [10] that right-angled polvhedra of finite volume can exist in Hra only for dimensions n < 13. Such polvhedra can have both finite vertices and cusps. At the same time [3], in dimensions n > 4 there are no bounded right-angled polvhedra. Thus, a right-angled polyhedron in Hra, n > 5, has at least one cusp. It was shown in [17] that in high dimensions right-angled polvhedra of finite volume should have a lot of cusps. Namely, the lower bounds c(n) of the number of cusps for dimensions n < 13 are given in Table 1.
We will be interested in right-angled hyperbolic polvhedra in which all vertices are cusps. Such polvhedra are called ideal. It was shown in [15] that in H™ n > 7, there are no ideal right-angled polvhedra. Examples of three-dimensional and four-dimensional ideal right-angled hyperbolic polvhedra will be given below.
Table 1: Lower bounds for the number of cusps.
n 6 7 8 9 10 11 12
c(n) 3 17 36 91 254 741 2200
1.2. Three-dimensional case
Necessary and sufficient conditions for a combinatorial polyhedron P to belong to the class XR can be obtained as a very special case of E. M. Andreev's theorem fl] on acute-angled polvhedra of finite volume.
Theorem 1. [1] Let, P be an abstract three-dimensional polyhedron with three or four faces meeting at each vertex, and P is not a simplex. The following conditions are necessary and sufficient conditions for the existence in H3 of a convex polyhedron of finite volume of a combinatorial type P with angles aij < ft/2:
0.0.< aij < ft/2.
1. If Fijk is a vertex of P, then aij + ajk + aki > ft, if Fijki is a vertex, then aij + +ajk + aki + an = 2ft.
2. If Fi, Fj, Fk is a triangular prismatic element, then aij + ajk + aki < ft.
3. If Fi, Fj, Fk, Fi is a quadrilateral prismatic element, then aij + ajk + aki + < 2ft.
4- If P is a triangular prism with bases F1 and F2, then a13 + a14 + a15 + a23 + a24 + a25 < 3ft.
5. If among the faces Fi, Fj, Fk there are adjacent Fi and Fj, Fj and Fk, but Fi and Fk are not adjacent, but meet at a common vertex and all three faces don't meet at one vertex, then aij + ajk < ft.
For the case of an ideal right-angled polyhedron the conditions are significantly simplified since
ft/2
We also recall a result of I. Rivin [20], concerning arbitrary convex ideal hyperbolic polvhedra, which is formulated in terms of the dual graph.
Theorem 2. [20] Let, P be a plane polyhedral graph with a weight w(e) assigned to each edge e. Let P* be the dual graph of P and assume that the weight w*(e*) = ft — w(e) is assigned to the edge e* dual to e. Then P can be realized as a convex ideal polyhedron in H3 with dihedral angles w(e) at its edges if and only if the following conditions are satisfied:
(1) 0 < w*(e*) < ft for all e;
(2) if edges e\, e2,..., e^ bound a face in P*, then
w*(e*) + w*(e*2) + ... + w*(e*k) = 2ft;
(3) if edges e\, e*,..., e*k form a cycle in P*, that does not bound a face, then
w*(e\) + w*(e*2) + ... + w*(e*k) > 2ft.
Ideal hyperbolic polvhedra are also interesting from the point of views of Euclidean geometry, since they are exactly those polvhedra that can be inscribed into the ball. This correspondence and the Rivin's theorem made it possible to solve the problem of .Jacob Steiner on describing a sphere around a polyhedron.
In the case of a right-angled polyhedron, the weights of edges and the weights of dual edges in the Rivin's theorem are k/2 and all faces of P* are quadrilateral. Accordingly, all vertices of the polyhedron P £ IR are 4-valent.
It is easy to see that from the classical Euler formula for a polyhedron, V — E + F = 2, where V is the number of vertices, E is the number of edges, and F is the number of faces of a polyhedron P, and from the fact that each vertex of an ideal right-angled polyhedron is incident to exactly four edges, it follows that 2E = 4V. Thus, the relation F = V + 2 holds. Denote bv pu the number of &-gonal faces (k > 3) of a polyhedron of class IR. Then
P3 = 8 + ^ pk (k — 4). (1)
k> 5
Thus, each polyhedron from IR has at least 8 triangular faces. It is easy to see that the octahedron shown in the Figure 1 has the minimum number of faces among the polvhedra in IR.
Figure 1: The octahedron.
At the same time, as we will see below, the octahedron is also minimal in volume among all ideal right-angled polvhedra.
1.3. Four-dimensional case
Recall [8, Table I] that the only regular four-dimensional polyhedron for which each vertex has a type of the cone over a cube is a 24-eell. In particular, it is realized as an ideal right-angled polyhedron in H4.
Let P C H4 be an ideal right-angled polyhedron with the face vector f (P) = (/o,/1 ,f2,f3), where fi is number of its ¿-faces. Since P is a convex four-dimensional polyhedron, then its surface dP is homeomorphic to S3, that means that its Euler characteristic turns to zero. Thus, f0 — f1 + f2 — f3 = 0. The volume of the polyhedron can expressed in terms of the components of the face vector, namely, the following statement holds.
Lemma 1. [15] Let, P C H4 be an ideal right-angled polyhedron with a face vector f (P) = (f0, f\, f2, f3). Then its volume is equal to
, p /0 — /3 + 4 2 vol P =---n .
3
Since for a 24-cell, the face vector has the form (24, 96, 96, 24), its volume is 4-k2/3. It is shown
H4
2. Enumeration of polyhedra and their volumes
2.1. Antiprisms and the edge-twist operation
Considering the octahedron as a triangular antiprism, that is, a polyhedron with triangular top and bottom and with a lateral surface formed by two levels of triangles, it can be naturally generalized to the next infinite family of polyhedra. By an n-antiprism A(n), n > 3, we mean (2n + 2)-hedron with n-gonal top and bottom and with a lateral surface formed by two levels of n triangles in each; and with each vertex incident to four edges. Schlegel diagrams of polyhedra A(3) and A(4) are presented in Figure 3.
By checking the conditions of Andreev's theorem or Rivin's theorem, it is easy to see that the antiprisms A(n), n > 3, can be realized as ideal right-angled polyhedra in H3.
A. Kolpakov demonstrated in [15] that the polyhedra A(n) are minimal in the following sense
Theorem 3. [15] For n > 3 the antiprism, A(n) has the smallest number of faces, equal to 2n + 2, among all the ideal right-angled hyperbolic polyhedra with at least one n-gonal face.
As will be clear below, antiprisms play an important role in understanding the structure of the set X~R of all ideal right-angled hyperbolic polyhedra.
Theorem 2 admits to characterize the class X~R in terms of the dual graphs of the polyhedral graphs. The dual graph must be a quadrangulation of the sphere, that is, a finite graphs with quadrilateral faces on a 2-sphere. Furthermore, the dual graph cannot contain a cycle of length four that separates two faces. The class of graphs with these properties was considered in [7], where it was denoted by Q4. Using the results of [7] on graphs from the class Q4 and passing from dual graphs to the original graphs, one can state the following result.
Theorem 4. ¡7] The class of 4-valent 3-connected and cyclically 6-connected planar graphs is generated by 1-skeletons of antiprisms A(n) and by edge-twist moves.
Recall that a graph is said to be cyclically k-connected if k is the smallest number of edges such that removing them decomposes the graph into two components each of which contains a cycle.
The edge-twist move is defined as follows. Let P £ X~R and assume that some face of the polyhedron has four distinct ideal vertices that split in pairs of connected by edges e\ and e2.
P P*
Figure 2: An edge-twist move.
Then the transformation involves removing e\ and e2, creating a new vertex v and connecting it by edges with the above four vertices. We denote the resulting polyhedron by P* and say that P* is obtained from P by an edge-twist move. For example, A(4)* in Figure 3 is obtained from A(4) by the edge-twist move. The edges ei, e2 and the new vertex v are indicated in the figure.
Theorems 2 and 4 lead to the following result.
Theorem 5. Each ideal right-angled hyperbolic polyhedron is an antiprism or can be obtained from some antiprism by a finite number of edge-twist moves.
Let us introduce a class of polyhedra obtained from antiprisms. Let A(n), n > 4, be an antiprism, and ei and e2 be two edges that belong to one of n-gonal faces, such that there is a third edge on the same face to which they are both adjacent, let us denote it by e3. In other words, e\ and e2
Figure 3: Polvhedra A(3), A(4) and A(4)*.
are adjacent through an edge. We apply the edge-twist move to the edges e1 and e2. As one can
see from Figure 4, illustrating the case of the antiprism A(6), when we apply edge-twist move to
the edges adjacent through an edge, the combinatorial structure changes as follows. The antiprism
A(n) had 2n + 2 faces: two n-gonal feces and 2n triangular faces. The new polyhedron A(n)* has
2n + 3 faces: one n-gonal face, one (n — 1)-gonal face, two quadrilateral faces and (2n — 1) triangular
faces. The polyhedron A(n)* will be referred to as a twisted antiprism. Observe that the number
of faces of the antiprism is always even, but the number of faces of the twisted antiprism is always
odd. As well as the antiprism, by virtue of Andreev's or Rivin's theorems, any twisted antiprism
H3
Figure 4: Polvhedra A(6) and A(6)*.
Denote bv v8 the volume of an ideal right-angled hyperbolic octahedron. The numerical value of this quantity will be given in the next section.
Lemma 2. Let A(n), n > 4, be an ideal right-angled antiprism in H3 having 2n + 2 faces and A(n)* be an ideal right-angled twisted antiprism obtained from A(n) and having 2n + 3 faces. Then for the volume of the twisted antiprism the following equality holds:
vol(A(n)*) = vol(A(n — 1)) + v8.
Proof. Since the edge-twist move of edges adjacent through one edge is a local transformation of a polyhedron, we will illustrate the proof for the case n = 6. For an arbitrarv n > 4, the proof is analogous.
Let us consider the antiprism A(n — 1). The left side of Figure 5 presents the antiprism A(5). Put an ideal right-angled octahedron on one of its faces to the triangular face ABC. Since the triangular faces of the antiprism and the triangular faces of the octahedron are ideal triangles, these faces are pairwise isometric. The remaining seven faces of the octahedron are drawn inside the triangle ABC in the middle of Figure 5. Recall that the dihedral angles at the edges of the antiprism and at the edges of the octahedron are equal to ^/2. Therefore, when we combine the antiprism and the
octahedron along the triangular face ABC in the resulting polyhedron, the angles at the edges AB, BC and AC will be equal to ft and the corresponding faces will belong to the same plane in pairs. Thus the polyhedron obtained by combining the antiprism and the octahedron will have
A B B C A C
absent.
Figure 5: A(5) and its union with an octahedron.
It is easy to see that the resulting polyhedron coincides combinatoriallv with the polyhedron A(6)* and is also right-angled. By virtue of Andreev's uniqueness theorem, these polvhedra are isometric, which implies equality of volumes. □
2.2. The Lobachevsky function
Traditionally, volumes of polvhedra in three-dimensional hyperbolic space are computed in terms of function
e
A(9) = - j log I2sin(t)l dt, 0
which J. Milnor introduced in the survey [16] and called it the Lobachevsky function. He demonstrated that the volume of an ideal tetrahedron T(a, f, 7) in a three-dimensional hyperbolic space with dihedral angles a, f and 7 (meaning that at one of the vertices dihedral angles are a, f 7, where a + f + 7 = ft, and dihedral angles at the opposite edges of the tetrahedron coincide) is calculated by the formula:
vol(T(a, p, 7)) = A(a) + A(f) + A(j).
Splitting an ideal right-angled octahedron into four ideal tetrahedra with dihedral angles a = ft/2, f = ft/4, 7 = ft/4 and using A(n/2) = 0, we obtain
v8 = 8A^ ^ = 3.663862376708876.....
A formula for the volume of a right-angled antiprism was presented by W. Thurston in his well-known lectures [23, Chapter 6.8], where the antiprism was called by a drum with triangular sides, and its volume was used to calculate the volume of the complement to some link in a three-dimensional sphere.
Theorem 6. [23] For n > 3 the volume of a right-angled n-ant,iprism, is given by
v°UA(„))=2„ [A(¡ + ) +A (\ - 2n)] •
where A(x) is the Lobachevsky function.
Theorem 6 implies that vol(A(n)) is asymptotically equivalent to ^fn when n -particular, the above formula gives the volume of an ideal right-angled octahedron A(3):
oo. In
vol(A(3)) = 8A(^/4) = v8, where we used properties of the Lobachevskv function [19].
Theorem 7. For the volume of the twisted antiprism A(n)*, n > 4; the following formula holds:
vol(A(n)*) = 2(n - 1)
4 4++Л ( 4- 2ПГ-Г))} +8Л (4)
proof. It follows from Lemma 2 and Theorem 6. □
2.3. Volumes of polyhedra with at most 23 faces
As we noted above, the Euler's formula implies that any ideal right-angled polyhedron has at 8
For each n = 8,10,11,..., 23 Table 2 gives the number of ideal right-angled polyhedra in H3 n
Table 2: Ideal right-angled polyhedra.
# of faces # of polyhedra # of volumes min volume max volume
8 1 1 3.663863 3.663863
9 0 0 - -
10 1 1 6.023046 6.023046
11 1 1 7.327725 7.327725
12 2 2 8.137885 8.612415
13 2 2 9.686908 10.149416
14 9 7 10.149416 12.046092
15 11 7 11.801747 13.350771
16 37 17 12.106298 14.832681
17 79 31 13.813278 16.331571
18 249 79 14.030461 18.069138
19 671 172 15.770160 19.523353
20 2182 495 15.933385 21.241543
21 6692 1359 17.694323 22.894415
22 22131 4276 17.821704 24.599233
23 72405 13031 19.597248 26.228126
Combinatorial enumeration of polyhedra was done with using the computer program plan-tri [18]. Volumes of hyperbolic polyhedra were calculated with the applying some modification of the computer program SnapPea [22]. Values of volumes are given up to 10-6.
Denote bv P^'^d P,max polyhedra which realize minimal and maximal values of volumes in
n
Table 3.
The volume calculations showed that the following fact holds for n < 23. If n is even, then the smallest volume is achieved on the antiprism, that is, P™in = A(k), where n = 2k + 2. If n is odd, then the smallest volume is achieved on the twisted antiprism, that is, P™in = A(k)*, where n = 2k + 3. We formulate this observation in the following form.
Table 3: The first seven ideal right-angled polvhedra
(1) vo\(P8) = 3.663863 (2) vo\(Pw) = 6.023046 (3) vol(Pn) = 7.327725
(4) vol(P$m) = 8.137885 (5) vol(P^ax) = 8.612415
(6) vol(Pf3in) = 9.686908 (7) vol(Pf3ax) = 10.149416
Proposition 1. For ideal right-angled hyperbolic polyhedra with at •most 23 faces, the. minimum value of volumes is achieved on antiprisms and twisted antiprisms. The minimum and maximum volumes are. presented in Tables 2, 4 and 5.
n > 8
is, P™in = A(k), where n = 2k + 2. If n > 11 is odd, then the minimum volume is achieved on the twisted antiprism,, that is, P™in = A(k)*, where n = 2k + 3.
The statement 1 confirms the conjecture for n < 23.
3. Upper and lower volume bounds
Bilateral bounds for the volumes of ideal right-angled polyhedra in terms of the number of vertices were obtained by K. Atkinson in [6].
Theorem 8. ¡6] Let, P be an ideal right-angled polyhedron with N vertices, then
(N - 2) • V4 <vo 1(P) < (N - 4) • |.
Table 4: Ideal right-angled polvhedra with n faces having minimum and maximum volume, U<n< 18.
Both inequalities became equalities when P is the regular ideal hyperbolic octahedron. Moreover, there exists a sequence of ideal right-angled polyhedra Pi with N vertices such thatvol(Pi)/Ni tends to v8/2 as i —y ^o.
Figure 6 shows the graphs of the upper and lower bounds from Theorem 8, the set of volume values of ideal right-angled polvhedra with at most 23 faces, where volumes of antiprisms and
n
19<n< 23.
twisted antiprisms are separately highlighted.
The upper bound in Theorem 8 can be improved as follows.
Til P N
the octahedron. Let, F\ and F2 be two faces of P such that F\ is n\-gon, and F2 is n2-gon, where
Figure 6: The set of volumes and Atkinson bounds.
ni,n2 > 4. Then for its volume the following upper bound holds:
vol(P) < (N - £ -f ).*
P
is different from an octahedron, then it would have two faces that have at least four sides. For
Fi F2
Fi F2
Fi F2
induction, attaching at each step a copy of the polyhedron P. Put P1 = P. Define P2 = P1 U^ P1,
Pi Fi P2
polyhedron with the number of vertices N2 = 2N - n1 and the volume vol(P2) = 2vol(P). The
P2 F2 P
P2 along this face. We get P3 = P2 Uf2 P = P Uf1P Uf2 P. Obviously, P3 is an ideal right-angled polyhedron with the number of vertices N3 = 3N - n1 - n2 and the volu me vol(P3) = 3vol(P).
P F1
F2, we obtain the polyhedron P2k+1 = P2k-1 Uf1PUf2 P, which is an ideal right-angled polyhedron with N2k+1 = (2k + 1)N - k,n1 - kn2 vertices and of volume vol(P2^+1) = (2k + 1) vol(P). Now let us apply the upper bound from Theorem 8 to polyhedron P2k+i-
(2k + 1) vol(P) < ((2k + 1)N - km - kn2 - 4) -8.
Dividing both sides of the inequality by (2k + 1) and passing to the limit as k — to, we obtain the required inequality.
(2) Let the faces F1 and F2 be adjacent. Put P2 = P Ufi P. The constructed polyhedron P2 has N2 = 2N — n1 vertices and its volume is two times the volume of the polyhedron P. By construction, the polyhedron P2 has a face F21, which is a (2n2 — 2)-gon. Since the face of F1 has at least 4 edges, there is a face in P adjacent to Fi, but not adjacent to F2. As a result of attaching P along F1, this face will turn into a face F22 in a polyhedron P2 that has at 1 east 4 sides. Thus, in P2 there is a pair of non-adjacent faces F21 and F22, each of which has at least 4 sides. This situation corresponds to the already proved case (1). Thus, for the polyhedron P2 and its non-adjacent faces F11 and F12 we get:
(2n2 — 2) 4\ vs
2 vol( P) < N2 —
2 2 2
from where we receive
and therefore,
□
2 vol(P) < (2N — m — n2 — 1) y.
vol(P) < (N — ^ — .
n1
2J2
Th eorem 10. Let P be an ideal right-angled hyperbolic polyhedron with N > 17 faces and having only triangular or quadrilateral faces. Then for its volume the following upper bound holds:
vol( P) < (N — 5) Is
proof. Observe that in the polyhedron P there are three quadrilateral faces F1, F2, F3 such that F2 is adjacent to F1 and F3 both. In fact, assume that there is no such triple of faces. Then each quadrilateral face is adjacent to at most one quadrangular face. If a quadrilateral face has no adjacent quadrilateral (we will say that it is isolated), then through its four sides it is adjacent to the triangular faces. If two quadrilaterals are adjacent to each other and none of them is adjacent to another quadrilateral (we will say that the faces form a pair), then their union is adjacent through
k1 2 quadrilateral faces, then through their sides they are adjacent to triangular faces through 4n1 + 6n2 sides. Since the polyhedron does not contain n-gonal faces for n > 5, it follows from Euler's formula that the number of triangles is 8. Their total number of sides is 24. If the number of faces N > 17 and eight of them are triangles, then n1 + 2n2 > 9 and S = 4n1 + 6n2 sides of triangles are required. Using the fact that 2n2 > 9 — n^, we obtain S > 4n1 + 3(9 — n1) = 27 + n1 > 24. This contradiction implies that there is a triple of sequentially adjacent quadrilateral faces F1, F2, F3, where F2 is adjacent to F1 and F3
Let us consider the union P2 = P U^2 P of two copies of P along F2. Then the doubled faces F1 and F3 of the polyhedron P will give two hexagonal faces in the polyhedron P2. The total number
P2 2 N — 4 P2
hexagonal faces:
2vol( P) < (2 N — 4 — 6 — 6)? ■
from where we receive
vol( P) < (N — 5) Is, □
The following statement describes the structure of the initial part of the set of volumes of ideal right-angled polvhedra.
Proposition 2. The volume values of ideal right-angled hyperbolic polyhedra not exceeding 5v8 are listed in Table 6. The number of such values is 248.
PROOF. By virtue of a lower bound from Theorem 8, if the number of faces of the ideal right-angled P F F - 2
( F - 4) • V4 < vol(P).
If F > 24 then this bound is at least 5v8- Thus, the volume of any polyhedron with at least 24 faces is bounded below by 5v8 = 18.319312, where the approximate value is indicated on the right-hand
23
values of volumes not exceeding 5^ is 248, All of them are listed in Table 6. □
Table 6: The first 248 values of volumes.
i 3,663863 51 15,46561 101 16,735095 151 17,477141 201 17,974896
2 6,023046 52 15,478658 102 16,744556 152 17,509421 202 17,98967
3 7,327725 53 15,495403 103 16,750301 153 17,516143 203 18,009307
4 8,137885 54 15,546518 104 16,755495 154 17,517167 204 18,026172
5 8,612415 55 15,654866 105 16,769779 155 17,52091 205 18,038106
6 9,686908 56 15,655017 106 16,780195 156 17,528985 206 18,045655
7 10,149416 57 15,709955 107 16,798534 157 17,530777 207 18,047625
8 10,806002 58 15,720116 108 16,805953 158 17,548392 208 18,058361
9 10,991587 59 15,77016 109 16,829048 159 17,55096 209 18,063652
10 11,136296 60 15,795313 110 16,83204 160 17,558575 210 18,063815
11 11,447207 61 15,803436 111 16,855785 161 17,571217 211 18,069138
12 11,801747 62 15,8569 112 16,864012 162 17,577434 212 18,084139
13 12,106298 63 15,85949 113 16,896062 163 17,5839 213 18,08961
14 12,276278 64 15,933385 114 16,961302 164 17,600432 214 18,092676
15 12,414155 65 15,94014 115 16,974442 165 17,615398 215 18,099757
16 12,46092 66 15,958101 116 16,98803 166 17,616542 216 18,109351
17 12,611908 67 15,959551 117 17,004375 167 17,633184 217 18,109786
18 12,854902 68 15,996629 118 17,014633 168 17,671046 218 18,128273
19 12,883862 69 16,049989 119 17,024507 169 17,694323 219 18,129371
20 13,020639 70 16,061517 120 17,061166 170 17,701559 220 18,133727
21 13,310579 71 16,078017 121 17,061237 171 17,70449 221 18,144299
22 13,350771 72 16,158579 122 17,061342 172 17,709902 222 18,152718
23 13,447108 73 16,172462 123 17,110971 173 17,712742 223 18,15859
24 13,677298 74 16,213678 124 17,140322 174 17,740113 224 18,167534
25 13,714015 75 16,27577 125 17,159342 175 17,751064 225 18,1677
26 13,813278 76 16,295989 126 17,165397 176 17,759743 226 18,173199
27 13,907355 77 16,324638 127 17,169868 177 17,766925 227 18,175729
28 14,030461 78 16,330917 128 17,174806 178 17,766983 228 18,180264
29 14,103121 79 16,331571 129 17,19799 179 17,769525 229 18,180633
30 14,160931 80 16,339295 130 17,199831 180 17,773653 230 18,207313
31 14,171606 81 16,382246 131 17,201332 181 17,790452 231 18,21988
32 14,273414 82 16,39832 132 17,22483 182 17,812693 232 18,233526
33 14,469865 83 16,448631 133 17,233217 183 17,821704 233 18,234257
34 14,494727 84 16,465777 134 17,238195 184 17,824793 234 18,244844
35 14,635461 85 16,48952 135 17,280423 185 17,835469 235 18,247553
36 14,655449 86 16,49154 136 17,303311 186 17,83745 236 18,276848
37 14,766948 87 16,506891 137 17,324068 187 17,844054 237 18,281813
38 14,800159 88 16,518764 138 17,341161 188 17,845073 238 18,287301
39 14,832681 89 16,535273 139 17,342423 189 17,857212 239 18,28917
40 14,898794 90 16,538867 140 17,354288 190 17,860804 240 18,291323
41 15,031667 91 16,547725 141 17,354866 191 17,864013 241 18,292895
42 15,052463 92 16,575188 142 17,362724 192 17,864685 242 18,299323
43 15,07859 93 16,595363 143 17,377493 193 17,894018 243 18,300817
44 15,11107 94 16,605736 144 17,377877 194 17,899631 244 18,304268
45 15,126498 95 16,615815 145 17,38534 195 17,901906 245 18,307302
46 15,169623 96 16,627568 146 17,420943 196 17,907162 246 18,31334
47 15,253393 97 16,657287 147 17,429408 197 17,918936 247 18,316267
48 15,323216 98 16,678106 148 17,45025 198 17,922791 248 18,319312
49 15,350907 99 16,684502 149 17,470253 199 17,937276
50 15,367058 100 16,726449 150 17,470735 200 17,944583
4. Polyhedra with isolated triangles
Recall that an ideal right-angled polyhedron has at least eight triangles. In the case of an octahedron, each triangle is adjacent to three other triangles along sides. From Table 4 and 5 one can see that with the increasing the number of faces of polyhedra with maximum volume, the
triangles move away from each other more and more. From the presence of common sides, the situation changes towards the presence of common vertices. The question appears when polvhedra arise in which all triangular faces are isolated, that is, no two triangular faces have common vertices. In this case we will call the polyhedron ITR-polyhedron, emphasizing that it satisfies the isolated triangles rule.
Proposition 3. Let P be an ideal right-angled polyhedron in Lobachevsky space having N faces. Denote by p3 the number of its triangular faces. If N < 3p3 + 2, then P is not, ITR-polyhedron.
proof. Let n be the maximum number of edges in faces of the polyhedron P. Denote bv Pk, k = 3,..., n, the number of k-gonal faces in P. Then ^fc=3 Pfc = N. Recall that bv the formula (1), p3 = 8 + Y^,fc=4Pk(k — 4). Assume, on the contrary, that P is an ITR-polvhedron. So at each vertex of the triangle there are meet three more vertices related to the faces that are not triangular. The
3 3
be at least 9p3. Let us calculate this number:
n n n n
kpfc = 4p4 + ^ kpfc = 4p4 + ^ pk(k — 4) + 4 ^ pk
k=4 k=5 fc=5 fc=5
n
= p3 — 8 + 4 ^ pk = p3 — 8 + 4(N — p3) = 4N — 3p3 — 8.
k=4
Demanding inequality
4 N — 3p3 — 8 > 9p3,
we get
N > 3p3 + 2,
which contradicts the original condition. Therefore, for N < 3p3 + 2 the polyhedron P cannot have isolated triangles. □
3 8
polvhedra with at most 25 faces. But among the 26-faeed polvhedra there are two such examples, which are shown in Figure 7. The volume of the ITR-polyhedron shown on the left-hand side is
Figure 7: 26-gonal polvhedra with isolated triangles
31.0930375, and the volume of the ITR-polyhedron shown on the right-hand side is 31.1668675. Proposition 4. There are. infinitely many ITR-polyhedra.
PROOF. Let Pi be an ITR-polvhedron (for example, it can be taken any of the polvhedra shown in Figure 7) and let F1 be one of its triangular faces. Consider the polyhedron P2 = P1 U^ P1 obtained by gluing two copies of the polyhedron P1 along the face P1. Obviously, the polyhedron P2 will also be an ideal right-angled polyhedron with isolated triangles. Suppose, for certainty, that the faces of the polyhedron P1; which have common sides with P1, have n15 n2, n3 sides, where n > 4, f = 1,2,3, bv virtue of the triangle isolation property. When these faces are doubled, they will turn into faces of a polyhedron P2, having, respectively, 2n — 2 sides, i = 1, 2, 3. If P1 have
p3№) triangular faces, then P2 will have p3(P2) = 2p3(P^ — 2 = p3(P1) + X]i=1(2nj — 6) triangular □
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Получено 14.12.2019 г.
Принято в печать 11.03.2020 г.
