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This article is cited in 2 scientific papers (total in 2 papers) On volumes of hyperbolic right-angled polyhedra S. A. Alexandrova, N. V. Bogachevba, A. Yu. Vesnincde, A. A. Egorovd a Moscow Institute of Physics and Technology, Dolgoprudny, Moscow region, Russia b Institute for Information Transmission Problems of the Russian Academy of Sciences (Kharkevich Institute), Moscow, Russia c Tomsk State University, Tomsk, Russia d Novosibirsk State University, Novosibirsk, Russia e Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences, Novosibirsk, Russia
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     § 1. IntroductionStudying the volumes of hyperbolic polyhedra and hyperbolic manifolds is a fundamental problem in geometry and topology. We discuss finite-volume polyhedra with all dihedral angles equal to $\pi/2$ in hyperbolic $n$-space $\mathbb{H}^n$ and call them right-angled polyhedra. It is known that there are no compact right-angled polyhedra in $\mathbb{H}^n$ for $n>4$, and there are no finite-volume right-angled polyhedra in $\mathbb{H}^n$ for $n>12$ [8], with examples known only for $n<9$ (see [17]). It is worth mentioning that investigating the volumes of ideal right-angled polyhedra becomes increasingly more attractive and important in view of the theorem on the maximum volume of a generalized hyperbolic polyhedron with fixed 1-skeleton [4], the problem of a minimal ideal right-angled 4-dimensional hyperbolic polyhedron (see [15]), and the conjecture about hyperbolic right-angled knots (see [6]). In this paper we consider right-angled hyperbolic polyhedra in $\mathbb{H}^3$. An initial list of ideal right-angled polyhedra was presented in [11], and a list of compact ones in [12]. A detailed discussion of constructions of hyperbolic $3$-manifolds from right-angled polyhedra can be found in the recent survey [21]. In the compact case such manifolds are related to small covers (see [7] and [5], for example). Also, right-angled polyhedra are useful for the construction of hyperbolic 3-manifolds that bound geometrically [14]. Let us also mention several works on the interplay between the arithmeticity of hyperbolic reflection groups and the arithmeticity of hyperbolic links (see [13] and [16]). Here it turns out to be very useful that the fundamental groups of some hyperbolic link complements are commensurable with hyperbolic reflection groups since Vinberg’s theory of reflection groups [22] can be applied. In 1970 Andreev [1], [2] (also see [18]) obtained his famous characterization of hyperbolic acute-angled $3$-polyhedra (all of whose dihedral angles are at most $\pi/2$) of finite volume. For right-angled polyhedra (all of whose dihedral angles are equal to $\pi/2$) Andreev’s theorems provide simple necessary and sufficient conditions for realizing a particular combinatorial type as a compact, an ideal, or a finite-volume polyhedron in $\mathbb{H}^3$. Such realizations are determined uniquely up to an isometry of $\mathbb{H}^3$. Thus one can expect that geometric invariants of these polyhedra can be estimated via combinatorics. Lower and upper bounds for volumes of right-angled hyperbolic polyhedra in terms of the number of vertices were obtained by Atkinson [3]. In this paper we obtain new upper bounds on the volumes of ideal right-angled polyhedra (see Theorem 1.1), compact right-angled polyhedra (see Theorem 1.2) and finite-volume right-angled polyhedra with both finite and ideal vertices (see Theorem 1.3). Recall (see [23]) that volumes of hyperbolic $3$-polyhedra can usually be expressed in terms of the Lobachevsky function In order to formulate the main results more conveniently, we define two constants, which depend only on the values of the Lobachevsky function at certain points. The first constant, $v_8 = 8\Lambda(\pi/4)$, equals the volume of the regular ideal hyperbolic octahedron. Up to six decimal places, $v_8 = 3.663862$. The second constant, ${v_3 = 3\Lambda(\pi/3)}$, equals the volume of the regular ideal hyperbolic tetrahedron. Up to six decimal places, $v_3 = 1.014941$.1.1. Ideal right-angled hyperbolic $3$-polyhedraRecall that if $P \subset \mathbb{H}^3$ is an ideal right-angled polyhedron with $V$ vertices, then $V \geqslant 6$. Moreover, $V = 6$ if and only if $P$ is an octahedron, which can be described as the antiprism $A(3)$ with triangular bases. Thus, $\operatorname{Vol}(A(3)) = v_8$. The volume formula for antiprisms $A(n)$ (that is, ideal right-angled polyhedra with $V = 2n$ vertices, two $n$-gonal bases and $2n$ lateral triangles) for $n \geqslant 3$, was obtained by Thurston (see [19], Chs. 6 and 7):
$$
\begin{equation*}
\operatorname{Vol} (A(n))=2n \biggl[ \Lambda \biggl(\frac{\pi}{4}+\frac{\pi}{2n} \biggr)+\Lambda \biggl(\frac{\pi}{4}-\frac{\pi}{2n} \biggr) \biggr].
\end{equation*}
\notag
$$
For example, up to six decimal places, $\operatorname{Vol}(A(4)) = 6.023046$. In 2009, Atkinson obtained (see [3], Theorem 2.2) the following upper and lower bounds for volumes in terms of the number of vertices. Let $P$ be an ideal right-angled hyperbolic $3$-polyhedron with $V \geqslant 6$ vertices; then
$$
\begin{equation*}
\frac{v_8}{4} V-\frac{v_8}{2} \leqslant\operatorname{Vol} (P) \leqslant\frac{v_{8}}{2} V -2 v_{8}.
\end{equation*}
\notag
$$
It is worth mentioning that both inequalities are sharp when $P$ is a regular ideal octahedron (that is, for $V=6$). Moreover, the upper bound is asymptotically sharp in the following sense: there exists a sequence of ideal right-angled polyhedra $P_{i}$ with $V_{i}$ vertices such that $\operatorname{Vol}(P_i) / V_i \to v_8/2$ as $i \to +\infty$. There are no ideal right-angled polyhedra for $V=7$, and $V=8$ if and only if $P$ is the antiprism $A(4)$ with quadrilateral bases. The following upper bound was obtained in [9], Theorem 2.2. Let $P$ be an ideal right-angled hyperbolic $3$-polyhedron with $V \geqslant 9$ vertices. Then
$$
\begin{equation*}
\operatorname{Vol} (P) \leqslant\frac{v_{8}}{2}V-\frac{5v_{8}}{2}.
\end{equation*}
\notag
$$
This inequality is sharp when $P$ is the double of a regular ideal octahedron along a face (that is, for $V=9$). The graphs of the above lower and upper bounds in comparison to the volumes of ideal right-angled polyhedra with up to $21$ vertices can be found in [9], Figure 1. Theorem 1.1. Let $P$ be an ideal right-angled hyperbolic $3$-polyhedron with $V$ vertices. Then the following inequalities hold. The proof of Theorem 1.1 is given in § 3. 1.2. Compact right-angled hyperbolic $3$-polyhedraIt is well known that, given a compact right-angled polyhedron in $\mathbb{H}^3$ with $V$ vertices, we have either $V = 20$ or $V \geqslant 24$ and $V$ is even. Moreover, $V = 20$ if and only if $P$ is a regular right-angled dodecahedron. The volume formula is known for an infinite series of compact right-angled Löbell polyhedra $L(n)$, $n \geqslant 5$, with $V = 4n$, two $n$-gonal bases (an upper and a lower one) and $2n$ lateral pentagonal faces. In particular, $L(5)$ is a regular dodecahedron. By [20], the volume of $L(n)$ is
$$
\begin{equation*}
\operatorname{Vol}(L(n))=\frac{n}{2} \biggl[ 2 \Lambda (\theta)+\Lambda \biggl( \theta+\frac{\pi}{n} \biggr)+\Lambda \biggl( \theta-\frac{\pi}{n} \biggr)-\Lambda \biggl( 2 \theta-\frac{\pi}{2} \biggr) \biggr],
\end{equation*}
\notag
$$
where $\theta={\pi}/{2}-\arccos \bigl({1}/(2 \cos (\pi/n))\bigr)$. Two-sided bounds for volumes of compact right-angled hyperbolic $3$-polyhedra were obtained by Atkinson (see [3], Theorem 2.3). Namely, if $P$ is a compact right-angled hyperbolic $3$-polyhedron with $V$ vertices, then
$$
\begin{equation*}
\frac{v_8}{32} V-\frac{v_8}{4} \leqslant \operatorname{Vol}(P)<\frac{5v_3}{8} V -\frac{25}{4} v_3.
\end{equation*}
\notag
$$
There exists a sequence of compact right-angled 3-polyhedra $P_i$ with $V_i$ vertices such that $\operatorname{Vol}(P_i)/V_i \to 5v_3/8$ as $i \to +\infty$. The upper bound can be improved if we exclude the case of a dodecahedron. Indeed, by Theorem 2.4 in [9], if $P$ is a compact right-angled hyperbolic $3$-polyhedron with $V \geqslant 24$ vertices, then
$$
\begin{equation*}
\operatorname{Vol}(P) \leqslant\frac{5v_3}{8} V-\frac{35}{4} v_3.
\end{equation*}
\notag
$$
Theorem 1.2. Let $P$ be a compact right-angled hyperbolic polyhedron with $V$ vertices. Then the following inequalities hold. The proof of Theorem 1.2 is given in § 4. 1.3. Right-angled hyperbolic $3$-polyhedra with finite and ideal verticesApart from the ideal and compact cases, Atkinson obtained volume bounds for right-angled hyperbolic polyhedra having both finite and ideal vertices (see [3], Theorem 2.4). If $P$ is a finite-volume right-angled hyperbolic polyhedron with $V_{\infty} > 0$ ideal vertices and $V_F$ finite vertices, then
$$
\begin{equation}
\frac{v_8}{4} V_\infty+\frac{v_8}{32} V_F-\frac{v_8}{4} \leqslant \operatorname{Vol}(P)<\frac{v_8}{2} V_\infty+\frac{5 v_3}{8} V_F-\frac{v_8}{2}.
\end{equation}
\tag{1.1}
$$
When more combinatorial information ia available about $P$, we are able to improve the upper bound as follows. Theorem 1.3. Let $P$ be a finite-volume right-angled hyperbolic $3$-polyhedron with $V_\infty > 0$ ideal vertices and $V_F$ finite vertices. If $V_\infty+V_F>17$, then The proof of Theorem 1.3 is given in § 5. § 2. Combinatorics of hyperbolic right-angled polyhedra2.1. Hyperbolic polyhedraLet $\mathbb{R}^{n,1}$ be the vector space $\mathbb{R}^{n+1}$ equipped with a scalar product $\langle \,\cdot\,{,}\,\cdot\, \rangle$ of signature $(n,1)$, and let $f_n$ be the associated quadratic form. The coordinate representation of $f_n$ with respect to an appropriate basis of $\mathbb{R}^{n,1}$ is Hyperbolic $n$-space $\mathbb{H}^{n}$ is the upper sheet (connected component) of the hyperboloid $f_n (x) = -1$:
$$
\begin{equation*}
\mathbb{H}^{n}=\bigl\{ x \in \mathbb{R}^{n, 1} \mid f_n (x)=-1,\,x_{0}>0 \bigr\}.
\end{equation*}
\notag
$$
In this model points in the ideal hyperbolic boundary correspond to isotropic vectors:
$$
\begin{equation*}
\partial \mathbb{H}^n=\bigl\{x \in \mathbb{R}^{n, 1} \mid f_n (x)=0,\,x_{0}>0\bigr\} / \mathbb R_+.
\end{equation*}
\notag
$$
A convex hyperbolic $n$-polyhedron is the intersection, with non-empty interior, of a finite family of closed half-spaces in hyperbolic $n$-space $\mathbb{H}^n$. A hyperbolic Coxeter $n$-polyhedron is a convex hyperbolic $n$-polyhedron $P$ all of whose dihedral angles are integer sub-multiples of $\pi$, that is, have the form $\pi/m$ for some integers $m\geqslant 2$. A hyperbolic Coxeter polyhedron is called right-angled if all of its dihedral angles are $\pi/2$. A generalized1 convex polyhedron is said to be acute-angled if all of its dihedral angles do not exceed $\pi/2$. It is known that generalized Coxeter polyhedra are the natural fundamental domains of discrete reflection groups in spaces of constant curvature (see [22]). A convex $n$-polyhedron has a finite volume if and only if it is the convex hull of finitely many points in the closure $\overline{\mathbb{H}^n} = \mathbb{H}^n \cup \partial \mathbb{H}^n$. If a convex polyhedron is compact, then it is a convex hull of finitely many proper points of $\mathbb{H}^n$, which are its proper (or finite) vertices. Finally, a convex polyhedron is called ideal if all of its vertices are placed on the ideal hyperbolic boundary $\partial \mathbb{H}^n$ (such vertices are also called ideal). It is known that compact acute-angled polyhedra (in particular, compact Coxeter polyhedra) in $\mathbb{H}^n$ are simple, that is, every vertex belongs to exactly $n$ facets (and therefore $n$ edges). Two compact polyhedra $P$ and $P'$ in Euclidean space $\mathbb{E}^n$ are combinatorially equivalent if there is a bijection between their faces that preserves the inclusion relation. A combinatorial equivalence class is called a combinatorial polyhedron. Note that if a hyperbolic polyhedron $P \subset \mathbb{H}^n$ is of finite volume, then the closure $\overline{P}$ of $P$ in $\overline{\mathbb{H}^n}$ is combinatorially equivalent to a compact polyhedron in $\mathbb{E}^n$. The following result is a special case of Andreev’s theorem (see [1] and [2]). Theorem 2.1. Let $\mathcal{P}$ be a combinatorial $3$-polyhedron. Then a finite-volume right-angled hyperbolic $3$-polyhedron $P \subset \overline{\mathbb{H}^3}$ that realizes $\mathcal{P}$ exists if and only if: A vertex $v$ of a right-angled polyhedron $P \subset \overline{\mathbb{H}^3}$ lies in $\mathbb{H}^3$ (that is, is finite) if and only if it belongs to exactly three faces of $P$. If a vertex $v$ is contained in four faces of $P$, then $v \in \partial\mathbb{H}^3$. 2.2. The combinatorics of ideal right-angled hyperbolic $3$-polyhedraLet $P$ be an ideal hyperbolic right-angled $3$-polyhedron. Let $V$ be the number of vertices, $E$ be the number of edges and $F$ be the number of faces of $P$. Then the Euler characteristic of $P$ is $V - E + F = 2$. Every ideal vertex of a finite-volume right-angled hyperbolic $3$-polyhedron is contained in exactly four edges, which implies that $4V = 2E$, and so $F = V + 2$. Let $p_k$ denote the number of $k$-gonal faces of $P$. Then the previous equalities yield
$$
\begin{equation*}
\sum_{k \geqslant 3} p_k=F, \qquad \sum_{k \geqslant 3} k p_k=4V \quad\text{and}\quad p_3=8+ \sum_{k \geqslant 5} (k-4) p_k.
\end{equation*}
\notag
$$
We say that two vertices of $P$ are adjacent if they are connected by an edge. Two vertices are quasi-adjacent if they belong to the same face but are not adjacent. Example 2.1. Two ideal right-angled polyhedra with $24$ vertices from [11] are shown in Figure 1. Each vertex of these polyhedra belongs to exactly one triangular face and three quadrilateral faces. Thus, each vertex of those polyhedra has exactly three quasi-adjacent vertices. Lemma 2.1. Let $P$ be an ideal right-angled hyperbolic $3$-polyhedron with $V > 24$ vertices. Then there is a vertex that has at least four quasi-adjacent vertices. Proof. Let $q(v)$ be the number of vertices that are quasi-adjacent to $v$. Then the average number of quasi-adjacent vertices in $P$ is
$$
\begin{equation*}
\begin{aligned} \, \frac{\sum_v q(v)}{V} &=\frac{1}{V} \sum_{k \geqslant 3} k (k-3) p_k =\frac{1}{V} \sum_{k \geqslant 3} k^2 p_k-\frac{3}{V} \sum_{k \geqslant 3} k p_k \\ &=\frac{1}{V} \sum_{k \geqslant 3} k^2 p_k-12 =\frac{1}{V} \biggl[3^2\,p_3+4^2\biggl(F- p_3-\sum_{k \geqslant 5} p_k\biggr) +\sum_{k \geqslant 5} k^2 p_k\biggr]-12 \\ &=\frac{1}{V} \biggl[3^2\biggl(8+\sum_{k \geqslant 5} (k-4)p_k\biggr) +4^2\biggl(F-p_3- \sum_{k \geqslant 5} p_k\biggr)+\sum_{k \geqslant 5} k^2 p_k\biggr]-12 \\ &=\frac{1}{V}\biggl[3^2\biggl(8+\sum_{k \geqslant 5} (k-4)p_k\biggr) +4^2\biggl(\!V-6\,{-} \sum_{k \geqslant 5} (k-3) p_k\biggr){+}\sum_{k \geqslant 5} k^2 p_k\biggr]{-}\,12 \\ &=4-\frac{24}{V}+\frac{1}{V} \sum_{k \geqslant 5} (k^2-7k+12) p_k \geqslant 4-\frac{24}{V}> 3. \end{aligned}
\end{equation*}
\notag
$$
Therefore, there is a vertex in $P$ that has at least four quasi-adjacent vertices. The proof is complete. Remark 2.1. Each vertex of a $k$-gonal face has at least $k-3$ quasi-adjacent vertices. So if a polyhedron has a $k$-gonal face for $k \geqslant 7$, then it has a vertex with at least four quasi-adjacent ones. We say that a face $f$ and a vertex $v$ are incident if $v$ belongs to $f$. A face $f$ and a vertex $v$ are quasi-incident if they are not incident, but $v$ has an incident face $f'$ such that $f'$ shares an edge with $f$. Proposition 2.1. Let $P$ be an ideal right-angled hyperbolic $3$-polyhedron with ${V \!\geqslant\! 73}$ vertices and only triangular and quadrilateral faces. Then there is a vertex without incident and quasi-incident triangular faces (see Figure 2). Proof. Let $F$ denote the number of faces of $P$. Because there are only triangular and quadrilateral faces, the set of faces of $P$ consists of $8$ triangles and $F-8$ quadrilaterals. Every triangular face is incident or quasi-incident to at most nine vertices (see Figure 3, where incident vertices are marked black and quasi-incident ones are marked white). Therefore, at most $72$ vertices of $P$ can be incident or quasi-incident to triangular faces. The proposition is proved.
2.3. The combinatorics of compact right-angled hyperbolic $3$-polyhedraLet $P$ be a compact right-angled hyperbolic $3$-polyhedron. Let $V$ denote the number of vertices, $E$ the number of edges and $F$ the number of faces. The Euler characteristic of $P$ is $V - E + F = 2$. Every vertex of a compact right-angled hyperbolic $3$-polyhedron is incident to three edges, which implies that $3V = 2E$ and $F = V/2 + 2$. Let $p_k$ denote the number of $k$-gonal faces of $P$. By Theorem 2.1, $p_3 = 0$ and $p_4 = 0$, and the previous equalities imply that
$$
\begin{equation*}
\sum_{k \geqslant 5} p_k=F, \qquad \sum_{k \geqslant 5} k p_k=3V \quad\text{and}\quad p_5=12+ \sum_{k \geqslant 7} (k-6) p_k.
\end{equation*}
\notag
$$
An edge $e$ and a vertex $v$ are incident if $v$ lies on $e$. We say that an edge $e$ and a vertex $v$ are quasi-incident if they are not incident, but at least one vertex of $e$ belongs to the same face as $v$. Since each vertex of a compact right-angled hyperbolic polyhedron $P$ is trivalent, four faces, $f_1$, $f_2$, $f_3$ and $f_4$, are arranged around each edge $e$ of $P$ as shown in Figure 4. If each $f_i$ is $k_i$-gonal, then the number of vertices quasi-incident to $e$ is equal to $\sum_{i=1}^4 k_i-10$ (note that Theorem 2.1 implies that $f_2\cap f_4=\varnothing$). Example 2.2. Consider the polyhedron shown in Figure 5. It has $80$ vertices and it is known as a fullerene C80 in structural chemistry. Each edge of C80 is quasi-incident to $13$ vertices. Indeed, every edge of C80 has a pentagon and three hexagons around it. Lemma 2.2. Let $P$ be a compact right-angled hyperbolic 3-polyhedron with $V > 80$ vertices. Then there is an edge with at least $14$ quasi-incident vertices. Proof. Let $q(e)$ be the number of vertices that are quasi-incident to an edge $e$. Then the average number of quasi-incident vertices in $P$ equals
$$
\begin{equation*}
\begin{aligned} \, &\frac{\sum_{e} q(e)}{E} =\frac{1}{E} \biggl[\sum_{k \geqslant 5} k (k-2) p_k+\sum_{k \geqslant 5} k (k-3) p_k\biggr] =\frac{2}{E} \sum_{k \geqslant 5} k^2 p_k-\frac{5}{E} \sum_{k \geqslant 5} k p_k \\ &\quad=\frac{2}{E} \sum_{k \geqslant 5} k^2 p_k-10 =\frac{2}{E} \biggl[5^2 p_5+6^2 p_6+\sum_{k \geqslant 7} k^2 p_k\biggr]-10 \\ &\quad=\frac{2}{E}\biggl[5^2 \biggl(12+\sum_{k \geqslant 7} (k-6) p_k\biggr) +6^2\biggl(F-p_5- \sum_{k \geqslant 7} p_k\biggr)+\sum_{k \geqslant 7} k^2 p_k\biggr]-10 \\ &\quad=\frac{2}{E}\biggl[5^2\biggl(12+\sum_{k \geqslant 7} (k-6) p_k\biggr) +6^2\biggl(\frac{E}{3}-10-\sum_{k \geqslant 7} (k-5) p_k\biggr)+\sum_{k \geqslant 7} k^2 p_k\biggr]-10 \\ &\quad=14-\frac{120}{E}+\frac{2}{E} \sum_{k \geqslant 7} (k^2-11 k+30) p_k \geqslant 14-\frac{120}{E}=14-\frac{80}{V}>13. \end{aligned}
\end{equation*}
\notag
$$
Therefore, there exists an edge with at least $14$ quasi-incident vertices. The lemma is proved. Corollary 2.1. Let $P$ be a compact right-angled hyperbolic $3$-polyhedron with $V> 80$ vertices. Then there is an edge with $k_i$-gonal faces around it, $i=1,\dots,4$, such that $\sum_{i=1}^4 k_i \geqslant 24$. 2.4. The combinatorics of right-angled hyperbolic $3$-polyhedra with both finite and ideal verticesLet $P$ be a right-angled hyperbolic $3$-polyhedron with $V_F$ finite and $V_{\infty}$ ideal vertices. Let $E$ denote the number of its edges, and $F$ be the number of its faces. Then the Euler characteristic of $P$ is Since every ideal vertex is incident to four edges and each finite vertex is incident to three edges, we obtain $3V_{F}+4V_{\infty} = 2E$. Hence $F=V_F/2+V_{\infty}+2$.We say that two faces are neighbours if they have a common vertex. Lemma 2.3. Let $P$ be a finite-volume right-angled hyperbolic $3$-polyhedron with $V_F$ finite and $V_\infty$ ideal vertices. Proof. Assume that $P$ has $F$ faces. Given a face $f_i \in P$, $i = 1, \dots, F$, denote the number of finite vertices in $f_i$ by $V_{F}^i$ and the number of ideal vertices in $f_i$ by $V^i_{\infty}$. Then the average number $A$ of neighbouring faces in $P$ is
(1) Our aim is to show that which is equivalent to
$$
\begin{equation*}
8 V_\infty+3 V_F>5 V_\infty+\frac{5}{2} V_F+10, \quad \text{that is, } 3 V_\infty+\frac{1}{2} V_F> 10.
\end{equation*}
\notag
$$
Using the inequality $V_F + V_{\infty} > 15$ we obtain
$$
\begin{equation*}
3 V_\infty+\frac{1}{2} V_F> 3 V_\infty+\frac{15}{2}-\frac{1}{2} V_\infty=\frac{5}{2} V_\infty+\frac{15}{2} \geqslant 10
\end{equation*}
\notag
$$
if $V_\infty \geqslant 1$. Thus, there is a face $f \in P$ with at least six neighbouring faces.
Analogously, using the inequality $V_F + V_{\infty} \geqslant 15$, we obtain
$$
\begin{equation*}
3 V_\infty+\frac{1}{2} V_F \geqslant 3 V_\infty+\frac{15}{2}-\frac{1}{2} V_\infty=\frac{5}{2} V_\infty+\frac{15}{2}>10
\end{equation*}
\notag
$$
if $V_\infty > 1$. Thus, there is a face $f \in P$ with at least six neighbouring faces.
(2) Let us use formula (2.1) for the average number $A$ of neighbours:
$$
\begin{equation*}
\begin{aligned} \, A &=\frac{8V_\infty+3V_F}{V_\infty+(1/2)V_F+2} =\frac{6(V_\infty+(1/2) V_F+ 2)+2V_\infty-12}{V_\infty+(1/2) V_F+2} \\ &\geqslant\frac{6(V_\infty+(1/2) V_F+2)-6}{V_\infty+(1/2) V_F+2}=6-\frac{6}{V_\infty+(1/2) V_F+2}, \end{aligned}
\end{equation*}
\notag
$$
where we have used that $2 V_\infty - 6 \geqslant 0$. Since $V_\infty + V_F > 17$ and $V_\infty \geqslant 3$, we have
$$
\begin{equation*}
V_\infty+\frac{1}{2} V_F+2 \geqslant 11+\frac{1}{2} V_\infty \geqslant\frac{25}{2}.
\end{equation*}
\notag
$$
Hence the average number of neighbours satisfies the inequality
Since $V_\infty + V_F > 17$ and $V_\infty \geqslant 3$, the polyhedron $P$ has
$$
\begin{equation*}
F=\frac{1}{2} V_F+V_\infty+2 \geqslant 11+\frac{1}{2} V_\infty>12
\end{equation*}
\notag
$$
faces. But $F$ is an integer, so $F \geqslant 13$. Assume that $P$ has $k \leqslant 6$ faces with six neighbours. Then $P$ has $F-k$ faces with at most five neighbours, and the average number $A$ of neighbours in $P$ satisfies the inequality
$$
\begin{equation*}
A \leqslant\frac{6 k+5 (F-k)}{F}=5+\frac{k}{F} \leqslant 5+\frac{6}{13}=\frac{71}{13}.
\end{equation*}
\notag
$$
Since ${71}/{13}<{138}/{25}$, we obtain a contradiction. Hence $P$ has at least seven faces such that each of them has six neighbouring faces.
(3) By (2.1), using $V_\infty \geqslant 6$ we obtain the following inequality for the average number of neighbours: Since $A \geqslant 6$ and $f$ has at most five neighbours, there is a face $f'$ of $P$ with at least seven neighbours. The lemma is proved.§ 3. Proof of Theorem 1.1The Lobachevsky function is concave on the interval $[0,{\pi}/{2}]$, which implies that
$$
\begin{equation*}
\sum_{k=1}^m \Lambda(x_k) \leqslant m\Lambda\biggl(\frac{\sum_{k=1}^m x_k}{m}\biggr).
\end{equation*}
\notag
$$
Let $P$ be an ideal right-angled hyperbolic polyhedron and $v$ be a vertex of $P$, which is called the apex in what follows. For every face $f$ there is a unique projection $u$ of $v$ onto $f$. This projection lies in the interior of $f$ unless $f$ meets one of the faces containing $v$. Projecting $u$ onto the edges of $f$ decomposes $P$ into tetrahedra known as orthoschemes: a hyperbolic tetrahedron with vertices $P_1$, $P_2$, $P_3$ and $P_4$ is said to be an orthoscheme if the edge $P_1 P_2$ is orthogonal to the plane $P_2 P_3 P_4$, and $P_1 P_2 P_3$ is orthogonal to $P_3 P_4$. For a face formed by vertices $v_1$, $v_2$, $v_3$ and $v_4$ such a decomposition is shown in Figure 6. Thus, we obtain eight tetrahedra, which have the common edge $vu$. Consider the tetrahedron formed by $v$, $u$, $w_4$ and $v_4$, where $v$ and $v_4$ are ideal vertices and $u$ and $w_4$ are finite vertices. Dihedral angles at the edges $v w_4$, $w_4 u$ and $u v_4$ are equal to $\pi/2$. If the dihedral angle at $vu$ is $\alpha$, then the dihedral angle at $v v_4$ is $\pi/2 - \alpha$ and the dihedral angle at $v_4 w_4$ is $\alpha$. Thus, this tetrahedron is determined by $\alpha$, and we call $\alpha$ the parameter of the tetrahedron. By Ch. 7 in [19], the volume of the tetrahedron formed by $v$, $u$, $w_4$ and $v_4$ is $\frac{1}{2} \Lambda(\alpha)$. Example 3.1. Let $P$ be the antiprism $A(4)$ with vertex set $V=\{v_1,v_2,\dots,v_8\}$. A decomposition of $A(4)$ with apex $v_1$ is shown in Figure 7. We define the tetrahedral cone $C(v)$ of the vertex $v$ as the union of the tetrahedra in the decomposition that contain $v$. Therefore, $A(4)$ decomposes into cones and $\operatorname{Vol}(A(4)) = \sum_{k=2}^8 \operatorname{Vol}(C(v_k))$. The vertices $v_2$, $v_4$, $v_5$ and $v_8$ are adjacent to $v_1$. The cone $C(v_4)$ consists of two tetrahedra with parameters $\alpha$ and $\beta$. Since the dihedral angle at the edge $v_1 v_4$ is $\pi/2$, the sum $(\pi/2 - \alpha) + (\pi/2 - \beta)$ equals $\pi/2$ and $\alpha + \beta$ equals $\pi/2$. The concavity of $\Lambda(x)$ implies that
$$
\begin{equation*}
\operatorname{Vol}(C(v_4))=\frac{1}{2} \Lambda(\alpha)+\frac{1}{2} \Lambda(\beta) \leqslant \Lambda\biggl(\frac{\pi}{4}\biggr).
\end{equation*}
\notag
$$
The same holds for every vertex adjacent to $v_1$. For more details, see Proposition 5.2 in [3]. A similar argument yields $\operatorname{Vol}(C(v)) \leqslant 2\Lambda(\pi/4)$ if $v$ is quasi-adjacent to $v_1$ (that is, $v=v_3$), and $\operatorname{Vol}(C(v)) \leqslant 4\Lambda(\pi/4)$ for every other vertex $v$ (that is, for $v_6$ and $v_7$). So
$$
\begin{equation*}
\operatorname{Vol}(A(4)) \leqslant 4 \Lambda\biggl(\frac{\pi}{4}\biggr)+1\cdot 2 \Lambda\biggl(\frac{\pi}{4}\biggr)+2\cdot 4 \Lambda\biggl(\frac{\pi}{4} \biggr)=14 \Lambda\biggl(\frac{\pi}{4}\biggr).
\end{equation*}
\notag
$$
Example 3.1 shows the idea of the proof of the following lemma. Lemma 3.1. Let $P$ be an ideal right-angled hyperbolic polyhedron with $V$ vertices. If there is a vertex with $m$ quasi-adjacent vertices, then Proof. Every vertex has exactly four adjacent vertices. This provides the inequality The lemma is proved. To prove part (3) of Theorem 1.1 we sum the volumes of the tetrahedra over the faces rather than over the vertices. The tetrahedral cone $C(f)$ of the face $f$ is the union of the tetrahedra in the decomposition that have a part of $f$ as a face. Such a cone is shown in Figure 6, c. Lemma 3.2. Let $P$ be a decomposed ideal right-angled hyperbolic polyhedron and $f$ be a $k$-gonal face of $P$ that does not contain the apex. Then the following hold. Proof. If $f$ is quasi-incident to the apex, then the projection of the apex does not lie in the interior of $f$, the cone $C(f)$ contains $2k - 2$ tetrahedra of the decomposition (like $v_3 v_4 v_7$ in Figure 7) and
If $f$ is not quasi-incident to the apex, then the projection of the apex lies in the interior of $f$, the cone $C(f)$ contains $2k$ tetrahedra of the decomposition (like $v_3 v_6 v_7$ in Figure 7) and The lemma is proved.Now we prove Theorem 1.1. Proof of Theorem 1.1. Lemma 3.1, Lemma 2.1 and Remark 2.1 provide items (1) and (2) of Theorem 1.1.
Let us prove (3). Let $P$ be an ideal right-angled hyperbolic polyhedron with $V \geqslant 73$ vertices and $F$ faces, which can only be triangular or quadrilateral. By Proposition 2.1 there is a vertex quasi-incident to eight pairwise different quadrilaterals. Then by Lemma 3.2
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Vol}(P) &\leqslant 8\cdot 3\Lambda\biggl(\frac{\pi}{3}\biggr)+8 (4- 1)\Lambda\biggl(\frac{\pi}{2\cdot 4-2}\biggr)+(F-20) 4\Lambda\biggl(\frac{\pi}{4}\biggr) \\ &= (V-18) 4\Lambda\biggl(\frac{\pi}{4}\biggr) +24\biggl[\Lambda\biggl(\frac{\pi}{3}\biggr)+ \Lambda\biggl(\frac{\pi}{6}\biggr)\biggr]. \end{aligned}
\end{equation*}
\notag
$$
To complete the proof we recall that $v_8=8 \Lambda( \pi/4)$ and $v_3=3 \Lambda( \pi/3 )=2 \Lambda( \pi/6)$ (since $\Lambda (x)$ is odd and $\pi$-periodic, and $\Lambda (2x) =2 \Lambda (x)+2 \Lambda (x+{\pi}/{2})$; see [19], Lemma 7.1.4). § 4. The proof of Theorem 1.24.1. Proof of statement (1) of Theorem 1.2We enumerate the faces as in Figure 4, so that $f_1$ and $f_3$ contain $e$. Since $V > 80$, by Corollary 2.1 there is an edge $e \in P$ such that for the $k_i$-gonal faces $f_i$, $i=1, \dots, 4$, we have $\sum_{i=1}^4 k_i \geqslant 24$. Let $P'$ be the polyhedron obtained by gluing $P$ with its mirror reflection in the plane passing through the face $f_1$. Then $P'$ has $V' = 2V - 2k_1$ vertices. Denote the $(2k_2 - 4)$-gonal face of $P'$ containing $f_2$ by $f_2'$, the $(2k_3 - 4)$-gonal face of $P'$ containing $f_3$ by $f_3'$, and the $(2k_4 - 4)$-gonal face of $P'$ containing $f_4$ by $f_4'$. Recall the following volume bound from [10]. Lemma 4.1 (see [10], Corollary 3.2). Let $P$ be a compact right-angled hyperbolic $3$-polyhedron with $V$ vertices. Let $f_1$, $f_2$ and $f_3$ be three faces of $P$ such that $f_2$ is adjacent to both $f_1$ and $f_3$, and $f_i$ is $k_i$-gonal for $i=1,2,3$. Then the following formula holds: Applying Lemma 4.1 to the polyhedron $P'$ we obtain
$$
\begin{equation*}
2\operatorname{Vol}(P)=\operatorname{Vol}(P') \leqslant \bigl(2V-2k_1-(2k_2-4)-(2k_3-4)-(2k_4-4)+ 4\bigr)\frac{5 v_3}{8},
\end{equation*}
\notag
$$
so that
$$
\begin{equation*}
\operatorname{Vol}(P) \leqslant (V-k_1-k_2-k_3-k_4+8)\frac{5 v_3}{8} \leqslant (V-16)\frac{5 v_3}{8}=\frac{5 v_3}{8} V-10 v_3,
\end{equation*}
\notag
$$
where we have used inequality $\sum_{i=1}^4 k_i \geqslant 24$. Statement (1) is proved. 4.2. The proof of statement (2) of Theorem 1.2Set $k_1 = k$. Since any face of $P$ has at least five sides, we have $k_i \geqslant 5$ for $i=2,3,4$. Then from the preceding inequality we obtain
$$
\begin{equation*}
\operatorname{Vol}(P) \leqslant (V-k-5-5-5+8)\frac{5 v_3}{8}=\frac{5 v_3}{8} V-\frac{5k+ 35}{8} v_3.
\end{equation*}
\notag
$$
Theorem 1.2 is proved.
§ 5. Proof of Theorem 1.3Observe that formula (1.1) implies that the following upper bound holds for a polyhedron $P$ with $V_\infty > 0$ ideal and $V_F$ finite vertices:
$$
\begin{equation*}
\operatorname{Vol}(P)<\frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-\frac{v_8}{2}.
\end{equation*}
\notag
$$
Let $P^1$ denote the polyhedron $P$, and let $k_\infty^1$ ($k_F^1$) be the number of ideal (finite, respectively) vertices of a face $f^1$ of $P^1$. Consider the union $P^2$ of $P^1$ with its mirror reflection in the plane containing $f^1$. Then $P^2$ is right-angled and has $V^2_\infty = 2V_\infty - k^1_\infty$ ideal vertices and $V^2_F = 2V_F - 2k^1_F$ finite vertices. Applying to $P^2$ the upper bound from (1.1) we obtain
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Vol}(P) &=\frac{\operatorname{Vol}(P^2)}{2}< \frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-\biggl(\frac{v_8}{4} k^1_\infty+\frac{5v_3}{8} k^1_F+\frac{v_8}{4} \biggr) \\ &=\frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-c_1-\frac{v_8}{4}, \end{aligned}
\end{equation*}
\notag
$$
where Let $k_\infty^2$ ($k_F^2$) be the number of ideal (finite, respectively) vertices of a face $f^2$ of $P^2$. Consider the union $P^3$ of $P^2$ with its mirror reflection in the plane containing the face $f^2$. Then $P^3$ is right angled and has $V^3_\infty = 4V_\infty -2 k^1_\infty - k^2_\infty$ ideal vertices and $V^3_F = 4V_F - 4k^1_F - 2k^2_F$ finite vertices. Applying to $P^3$ the upper bound from (1.1) we obtain
$$
\begin{equation*}
\operatorname{Vol}(P)=\frac{\operatorname{Vol}(P^3)}{4}<\frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-c_1-c_2-\frac{v_8}{8},
\end{equation*}
\notag
$$
where Continuing the process recursively, we obtain
$$
\begin{equation}
\operatorname{Vol}(P)<\frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-c_1-c_2-\dots-c_n-\frac{v_8}{2^{n+1}},
\end{equation}
\tag{5.1}
$$
where for $i=1, \dots, n$ we have
$$
\begin{equation*}
c_i=\frac{v_8}{2^{i+1}} k^i_\infty+\frac{5v_3}{2^{i+2}} k^i_F.
\end{equation*}
\notag
$$
Suppose that for every $i = 1,2,\dots$, the face $f^i$ is distinct from an ideal triangle and has at least six neighbouring faces. We show in Lemma 5.1 below that one can choose such a face $f^i$ of $P^i$ for every $i$. Then the quantity $c_i$ is smallest for $k^i_\infty=2$ and $k^i_F=2$. Thus, we get that
$$
\begin{equation*}
\sum_{i=1}^n c_i \geqslant\biggl(v_8+\frac{5v_3}{2}\biggr) \biggl(\sum_{i=1}^n\frac{1}{2^i}\biggr) =\biggl(v_8+\frac{5v_3}{2}\biggr) \biggl(1-\frac{1}{2^n}\biggr).
\end{equation*}
\notag
$$
Taking the limit as $n \to \infty$ in (5.1) we obtain
$$
\begin{equation*}
\operatorname{Vol}(P) \leqslant\frac{v_8}{2} V_\infty+\frac{5v_3}{8} V_F-\biggl(v_8+\frac{5v_3}{2}\biggr).
\end{equation*}
\notag
$$
Thus, the bound required in Theorem 1.3 is obtained. It remains to prove Lemma 5.1. Let $N_6(P)$ denote the set of faces of a polyhedron $P$ that have the following property: $f \in N_6(P)$ if $f$ has at least six neighbouring faces. This set is not empty by part (1) of Lemma 2.3. Lemma 5.1. If $V_\infty + V_F > 17$, then for any $i=1, \dots, n$ the set $N_6(P^i)$ contains at least one face that is not an ideal triangle. Proof. Observe that for any $i \geqslant 1$ the polyhedron $P^i$ is not an octahedron. Indeed, this holds true for $P^1 = P$ since $V^1_{\infty} = V_{\infty}$, $V^1_F = V_F$ and $V_\infty^1 + V_F^1 > 17$. Suppose for a contradiction that $P^2$ is an octahedron. Then $2V_\infty^1\geqslant V_\infty^2=6$ and
$$
\begin{equation*}
\begin{aligned} \, \operatorname{Vol}(P^2) &=2 \operatorname{Vol}(P^1) \geqslant2\frac{4 V_\infty^1+V_F^1-8}{32} v_8 \\ &>\frac{4 V_\infty^1+17-V_\infty^1-8}{16} v_8 \geqslant\frac{18 v_8}{16}>v_8, \end{aligned}
\end{equation*}
\notag
$$
which is a contradiction. Finally, let us show that for any $i \geqslant 3$ the polyhedron $P^i$ is not an octahedron. If we suppose for a contradiction that $P^i$ is an octahedron, then $V_\infty^1 \geqslant 1$ and by inequality (1.1) which is a contradiction again.
Let $i = 1$ and suppose for a contradiction that all faces in $N_6(P^1)$ are ideal triangles. Then part (2) of Lemma 2.3 implies that $N_6(P^1)$ contains at least seven ideal triangles. Denote the set of faces of $P^1$ by $\mathcal{F}$ and the number of ideal vertices of a face $f$ by $I(f)$. Assume that $P^1$ contains at most five ideal vertices. Then we obtain a contradiction:
$$
\begin{equation*}
21=3\cdot 7 \leqslant \sum_{f \in \mathcal{F}} I(f) \leqslant 4\cdot 5=20.
\end{equation*}
\notag
$$
Thus, $P^1$ contains at least $k+4\geqslant 6$ ideal vertices.
Since $P^1$ is not an octahedron, there is a face $f'$ that is not an ideal triangle. Therefore, $f' \notin N_6(P^1)$, so that $f'$ has at most five neighbouring faces. Then by part (3) of Lemma 2.3 there is a face $f''$ with at least seven neighbouring faces. Therefore, $f''$ is not an ideal triangle. However $f'' \in N_6(P^1)$, which contradicts the assumption. Now let $i \geqslant 2$ and suppose for a contradiction that for some $i \geqslant 2$ each face in $N_6(P^i)$ is an ideal triangle. The polyhedron $P^{i}$ is the union of two copies of $P^{i-1}$ along the face $f^{i-1}$. Let $D^{i-1}$ be the set of faces of $P^{i-1}$ that have a common edge with $f^{i-1}$. Let $S^{i}$ denote the set of faces of $P^{i}$ that contain a face in $D^{i-1}$. That is, $S^{i}$ consists of all new faces that appear after gluing two copies of $P^{i-1}$ along $f^{i-1}$. By Theorem 2.1 each face of a right-angled polyhedron has at least five neighbours. Hence each face in $S^{i}$ has at least six neighbours and $S^i \subset N_6(P^i)$. Therefore, each face in $S^i$ is an ideal triangle by assumption. Then each face in $D^{i - 1}$ is a triangle with two ideal and one finite vertex. Moreover, $f^{i - 1}$ is a face with an even number of vertices, such that its ideal and finite vertices alternate. Namely, if $f^{i-1}$ has $2k$ vertices, then there are $k$ ideal and $k$ finite vertices, and $k \geqslant 2$. Observe that there are at least two ideal vertices in $P^{i - 1}$ that are not contained in $f^{i-1}$. Indeed, since all faces in $D^{i-1}$ are triangles with two ideal vertices, $P^{i-1}$ has at least one ideal vertex $v$ that is not contained in the face $f^{i-1}$. Suppose that such a vertex $v$ is unique. Then $v$ is incident to all of the vertices of $f^{i-1}$. If ${k \geqslant 3}$, then $v$ is a vertex of valency $2k$. The latter is impossible by Theorem 2.1. If $k = 2$, then $P^{i-1}$ is a quadrilateral pyramid with five faces, while by Theorem 2.1 the quadrilateral pyramid is not a right-angled polyhedron. Thus, $P^{i-1}$ must have at least two ideal vertices that are not contained in $f^{i-1}$. Hence $P^i$ has at least $k + 4 \geqslant 6$ ideal vertices. Since $P^i$ is not an octahedron, it has at least one face $f'$ that is not an ideal triangle. Using part (3) of Lemma 2.3 we obtain that there is a face $f''$ of $P^i$ that has at least seven neighbours and thus it cannot be an ideal triangle. This contradicts our assumption about $N_6(P^i)$, and the lemma is proved. AcknowledgementAuthors thank the Laboratory of Combinatorial and Geometric Structures at the Moscow Institute for Physics and Technology for organizing the Summer 2021 Research Program for Undergraduates where the present research started.
Citation in format AMSBIB
\Bibitem{AleBogVes23} |
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