Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 24, 1999, 31–43

ASYMPTOTIC VOLUME FORMULAE AND HYPERBOLIC BALL PACKING

T.H. Marshall

University of Auckland, Department of Mathematics Private Bag 92019, Auckland, New Zealand; t marshall@math.auckland.ac.nz

Abstract. We prove that the volume of an n-dimensional regular spherical simplex of edge length r < \frac{1}{2} π is asymptotically e^{−x}x^{n/2}(n + 1)^{1/2}/n!, as n → ∞, where x = sec r − 1. The same is true for hyperbolic simplices if we set x = 1 − sech r and replace e^{−x} by e^{x}.

We obtain error bounds for this asymptotic, and apply it to find an upper bound for the density of packings of balls of a given radius in hyperbolic n-space, for all sufficiently large n.

1. Introduction

The volume of a polytope Δ in an n-dimensional space of constant curvature K = ±1 satisfies Schläfli’s differential equation,

(1)  d Volume = \frac{K}{n − 1} ∑(Vol. codimension 2 face) d (angle)

the sum being taken over all faces of codimension two and their corresponding dihedral angles [10, Chapter 7, 2.2].

From this, for example, we may derive the well-known area formulae for spherical and hyperbolic triangles. However, in higher dimensions, volume calculation remains a difficult problem, even with Schläfli’s formula, because it requires the volumes of the codimension two faces as a function of the dihedral angles of the original polytope.

In this paper we derive an asymptotic formula for the volume, in either hyperbolic or spherical space, of a regular simplex, that is a simplex whose dihedral angles (or equivalently, whose edge lengths) are all equal. We let S_{n}(r) and H_{n}(r) denote, respectively, the volume of a regular n-simplex with edge length r in spherical space, and in hyperbolic space. We prove

Theorem 1. The volume

(2)  S_{n}(r) = \frac{(n + 1)^{1/2}x^{n/2}e^{−x}}{n!} (1 + s_{n}(r)),

1991 Mathematics Subject Classification: Primary 30F40, 57N10; Secondary 20F55.

where x = sec r − 1 and, for each r_{0} < \frac{1}{2}π, there is a constant C for which |s_{n}(r)| < C/n, whenever r ≤ r_{0} , and,

(3)  H_{n}(r) = \frac{(n + 1)^{1/2}x^{n/2}e^{x}}{n!} (1 + h_{n}(r)),

where x = 1 − sech r, and there is a constant C for which |h_{n}(r)| < C/n, for all r ∈ [0, ∞].

The case r = ∞ (x = 1) of (3) corresponds to the regular ideal simplex. The asymptotic (3), in this case, is given (without proof) by Milnor [8] (see [4] for a proof).

Of course (2) and (3) are closely related, and we obtain both as a corollary of Theorem 3 below, which also provides a related asymptotic expansion.

We also apply Theorem 3 to prove a result about ball packing in hyperbolic space. The local density of a packing is defined to be the ratio Vol.(B)/Vol.(D), where B is a ball in the packing, and D is the set of all points closer to B than to any other ball in the packing. Böröczky [1, Theorems 1 and 4] has shown that the local density of a packing by balls of radius \frac{1}{2}r is bounded above by the d_{n}(r), defined as follows.

Let Δ be a regular simplex in hyperbolic n-space, with edge length r, and let P be the union of n + 1 mutually tangent balls of radius \frac{1}{2}r with centres at each of the vertices of Δ. We define,

d_{n}(r) = Vol.(P ∩ Δ)/Vol.(Δ).

The limiting value d_{n}(0) = lim_{r→0} d_{n}(r) is obtained by doing the same construction in Euclidean space, in which case the dimensions of Δ are immaterial.

The limiting version of Böröczky’s result is that the density D_{n} of an ndimensional ball packing in Euclidean space does not exceed d_{n}(0). This result was proved by Rogers [9]. Asymptotically better estimates have since been found (see e.g. [5]).

The function d_{n}(r) has been proved to be strictly increasing, for n = 2 by Krammer (cited in [3, §37]), and for n = 3 by Böröczky and Florian [2]. Kellerhals [6] raises the question of whether this monotonicity also occurs in higher dimensions. For sufficiently high dimension we answer this question affirmatively.

Theorem 2. For all sufficiently large n, d_{n}(r) is a strictly increasing function of r.

We conjecture that this result holds in all dimensions greater than one. The methods given here allow this to be tested in principle, but only through lengthy and tedious calculations. We therefore prove only the asymptotic result.

Since, for all r, d_{n}(r) < 1, monotonicity implies that, for each n, d_{n}(r) approaches a limit d_{n}(∞) as r → ∞. The local density of a packing by balls of any given radius is then bounded above by d_{n}(∞). This result has been used, when n = 3, to obtain improved lower bounds for the volumes of hyperbolic 3-manifolds and orbifolds. Details may be found in [7].

We use ⌊x⌋ to denote the greatest integer not exceeding x. Given two functions f(n) and g(n), defined on the natural numbers, we use f(n) ∼ g(n) to denote asymptotic equality. For taking half integer powers of negative numbers we adopt the convention that (−1)^{1/2} = i, whence (−1)^{(2n+1)/2} is i or −i, according as n is even or odd.

I would like to thank G.J. Martin for reading and correcting an earlier draft of this paper.

2. Simplex volumes

The edge length r and the dihedral angle θ of a regular n-dimensional spherical simplex are related by the well-known formula,

(4)  sec θ = sec r + (n − 1)

The same equation, with sech r replacing sec r, holds for the regular hyperbolic n-simplex. Thus, in the spherical case,

(5)

\frac{dr}{dθ} = \frac{dr}{d sec r} \frac{d sec r}{d sec θ} \frac{d sec θ}{dθ} = \frac{sec θ tan θ}{sec r tan r}

= \frac{(sec r + (n − 1))\sqrt{(sec r + (n − 1))^{2} − 1}}{sec r\sqrt{sec^{2}r − 1}}.

The same equation holds in the hyperbolic case, with sech r replacing sec r and a sign change on the right hand side and under the square root in the denominator.

The faces of regular simplex all have the same edge length as the original simplex. There are \binom{n+1}{2} faces of codimension two, so that, for the regular spherical simplex with dihedral angle θ, Schläfli’s equation may be written

(6)  \frac{dS_{n}(r)}{dθ} = \frac{1}{n − 1}\binom{n + 1}{2}S_{n−2}(r).

Similarly we have, in the hyperbolic case,

(7)  \frac{dH_{n}(r)}{dθ} = \frac{−1}{n − 1}\binom{n + 1}{2}H_{n−2}(r).

If z = sec r − 1, then (6) gives,

(8)

\frac{dS_{n}(r)}{dz} = \frac{dS_{n}(r)}{dθ} \frac{dθ}{dr} \frac{dr}{dz}

= \frac{n(n + 1)}{2(n − 1)} \frac{S_{n−2}(r)}{(n + z)((n + z)^{2} − 1)^{1/2}}

= \frac{(n + 1)^{1/2}}{2(n − 1)^{3/2}} (1 + \frac{z}{n})^{−1}(1 + \frac{z}{n + 1})^{−1/2}

× (1 + \frac{z}{n − 1})^{−1/2}S_{n−2}(r).

We obtain the corresponding equation for hyperbolic volumes by substituting H_{n}(r) and H_{n−2}(r) for S_{n}(r) and S_{n−2}(r), respectively, setting z = sech r − 1, and multiplying the right hand side by −1. Thus if we define a family of functions V_{n}(z) by

V_{0}(z) = 1, V_{1}(z) = ArcSec (1 + z)

and, for n ≥ 2

(9) V′_{n}(z) = \frac{(n + 1)^{1/2}}{2(n − 1)^{3/2}} (1 + \frac{z}{n})^{−1}(1 + \frac{z}{n + 1})^{−1/2}(1 + \frac{z}{n − 1})^{−1/2}V_{n−2}(z),

with V_{n}(0) = 0, then S_{n}(r) = V_{n}(sec r − 1), and H_{n}(r) = (−1)^{n/2}V_{n}(sech r − 1). Rather than using the functions V_{n}(z), it will be more convenient to work with the related family Y_{n}(z), given by

Y_{n}(z) = \frac{n!}{(\frac{1}{2}n)!(n + 1)^{1/2}} V_{n}(z).

These satisfy the simpler equation

(10)  Y′_{n}(z) = (1 + \frac{z}{n})^{−1}(1 + \frac{z}{n + 1})^{−1/2}(1 + \frac{z}{n − 1})^{−1/2}Y_{n−2}(z)

for n ≥ 2 with initial condition Y_{n}(0) = 0. Together with the initial functions,

Y_{0}(z) = 1, Y_{1}(z) = (\frac{2}{π} )^{1/2}ArcSec (1 + z) = \frac{z^{1/2}}{(\frac{1}{2})!} e^{−z}(1 + \frac{7}{12}z + ⋯),

(10) can be taken as defining the family Y_{n}(z). Now we have

(11)  S_{n}(r) = \frac{(\frac{1}{2}n)!(n + 1)^{1/2}}{n!} Y_{n}(sec r − 1)

and

(12)  H_{n}(r) = (−1)^{−n/2} \frac{(\frac{1}{2}n)!(n + 1)^{1/2}}{n!} Y_{n}(sech r − 1).

Clearly for n even, Y_{n} can be defined on the complex plane with the interval (−∞, −1) deleted, and is analytic in the interior of this region. For n odd, Y_{n}(z)/\sqrt{z} has the same properties.

Theorem 3. For each positive integer k,

(13)  Y_{n}(z) = \frac{z^{n/2}}{(\frac{1}{2}n)!}e^{−z}[1 + ∑_{i=1}^{k−1} r_{i}^{(n)}z^{i} + z^{k}ε_{k}^{(n)}(z)]

where (setting r_{0}^{(n)} = 1) we have,

(14)  r_{i}^{(n)} ∼ λ_{i}n^{−⌊(i+1)/2⌋}  as n → ∞,

the λ_{i} being defined recursively by λ_{0} = 1, λ_{1} = \frac{3}{2} , and,

(15)  λ_{i} = \frac{3λ_{i−2}}{i}

for even i ≥ 2 and,

(16)  λ_{i} = \frac{[−4λ_{i−3} + 3λ_{i−2}]}{(i − 1)} − 2λ_{i−1}

for odd i ≥ 3. For each compact region in C − (−∞, −1),

(17)  |ε_{k}^{(n)}(z)| ≤ Kn^{−⌊(k+1)/2⌋},

where K depends only on the region chosen.

Clearly Theorem 1 follows from Theorem 3 (with k = 1), using (11) and (12). Setting, for n ≥ 2,

(18)  \begin{aligned}h_{n}(z) & = (1 + \frac{z}{n})^{−1}(1 + \frac{z}{n + 1})^{−1/2}(1 + \frac{z}{n − 1})^{−1/2} \\ & = ∑_{i=0}^{∞} t_{i}^{(n)}z^{i} = ∑_{i=0}^{k−1} t_{i}^{(n)}z^{i} + s_{k}^{(n)}(z), \end{aligned}

we have t_{0}^{(n)} = 1,

(19)  t_{1}^{(n)} = −(\frac{2}{n} + \frac{1}{n(n^{2} − 1)}),

and

(20)  t_{i}^{(n)} ∼ \frac{(−1)^{i}(i + 1)}{n^{i}}  as n → ∞.

Also

(21)  \frac{s_{k}^{(n)}(z)}{z^{k}} = O(n^{−k})  (n → ∞),

uniformly in compact subsets of C − (−∞, −1).

3. Proofs of theorems

Proof of Theorem 3. Clearly we may write Y_{n}(z) in the form (13). We must prove (14)–(17). Setting ϱ_{k}^{(n)}(z) = e^{−z}z^{k+n/2}ε_{k}^{(n)}(z)/(\frac{1}{2}n)!, we have,

(22)  Y_{n}(z) = \frac{z^{n/2}}{(\frac{1}{2}n)!}e^{−z}[1 + ∑_{i=1}^{k−1} r_{i}^{(n)}z^{i}] + ϱ_{k}^{(n)}(z).

When n ≥ 2, substituting this form of Y_{n}(z) into (10), dividing through by e^{−z}z^{(n−2)/2}/(\frac{1}{2}(n − 2))!, and equating in turn the coefficients of z^{i} (0 ≤ i ≤ k−1), and the remainder gives,

(23)  r_{i}^{(n)}(1 + \frac{2i}{n}) − \frac{2}{n}r_{i−1}^{(n)} = ∑_{j=0}^{i} t_{i−j}^{(n)}r_{j}^{(n−2)}  (1 ≤ i ≤ k − 1)

and,

(24)  \begin{matrix}− \frac{z^{(n−2)/2}}{(\frac{1}{2}(n − 2))!}e^{−z}[\frac{2r_{k−1}^{(n)}}{n} z^{k}] + ϱ_{k}^{(n)}′(z) \\ = \frac{z^{(n−2)/2}}{(\frac{1}{2}(n − 2))!}e^{−z}(∑_{i=0}^{k−1} r_{i}^{(n−2)}s_{k−i}^{(n)}(z)z^{i}) + h_{n}(z)ϱ_{k}^{(n−2)}(z). \end{matrix}

whence, using (19),

(25)  \begin{aligned}ϱ_{k}^{(n)}′(z) & = \frac{z^{(n−2)/2}}{((n − 2)/2)!}e^{−z}[∑_{i=0}^{k−2} r_{i}^{(n−2)}s_{k−i}^{(n)}(z)z^{i} + r_{k−1}^{(n−2)}s_{2}^{(n)}(z)z^{k−1} \\ & + \frac{2}{n}(r_{k−1}^{(n)} − r_{k−1}^{(n−2)})z^{k} − \frac{r_{k−1}^{(n−2)}z^{k}}{n(n^{2} − 1)}] + h_{n}(z)ϱ_{k}^{(n−2)}(z). \end{aligned}

If we define,

R_{i}^{(n)} = \binom{i + \frac{1}{2}n}{i}r_{i}^{(n)},

(23) gives

(26) \begin{aligned}R_{i}^{(n)} & − R_{i}^{(n−2)} = \binom{i − 1 + \frac{1}{2}n}{i}[∑_{j=0}^{i−1} t_{i−j}^{(n)}r_{j}^{(n−2)} + \frac{2}{n}r_{i−1}^{(n)}] \\ & = \binom{i − 1 + \frac{1}{2}n}{i}[∑_{j=0}^{i−2} t_{i−j}^{(n)}r_{j}^{(n−2)} − \frac{r_{i−1}^{(n−2)}}{n(n^{2} − 1)} + \frac{2}{n}(r_{i−1}^{(n)} − r_{i−1}^{(n−2)})]. \end{aligned}

Also from (23), we have

(27) r_{i}^{(n)} − r_{i}^{(n−2)} = ∑_{j=0}^{i−2} t_{i−j}^{(n)}r_{j}^{(n−2)} − \frac{r_{i−1}^{(n−2)}}{n(n^{2} − 1)} + \frac{2}{n}(r_{i−1}^{(n)} − r_{i−1}^{(n−2)}) − \frac{2i}{n}r_{i}^{(n)}.

We prove (14) by induction on i, simultaneously with,

(28)  r_{i}^{(n)} − r_{i}^{(n−2)} ∼ −2⌊\frac{1}{2}(i + 1)⌋λ_{i}(n^{−⌊(i+3)/2⌋})  as n → ∞.

Since r_{i}^{(0)} = 1, (14) and (28) hold for i = 0. For i = 1, (26) gives

R_{1}^{(n)} − R_{1}^{(n−2)} = \frac{n}{2}(\frac{−1}{n(n^{2} − 1)}) = −\frac{1}{4}(\frac{1}{n − 1} − \frac{1}{n + 1}).

Since R_{1}^{(0)} = 1 and R_{1}^{(1)} = \frac{7}{8} , this gives, for even and odd n, respectively,

R_{1}^{(n)} = 1 − \frac{1}{4} + \frac{1}{4(n + 1)}  and  R_{1}^{(n)} = \frac{7}{8} − \frac{1}{8} + \frac{1}{4(n + 1)},

so that, for all n,

R_{1}^{(n)} = \frac{3n + 4}{4(n + 1)}, r_{1}^{(n)} = \frac{3n + 4}{2(n + 1)(n + 2)}.

It is thus clear that (14) and (28) hold for i = 1. For the induction step we suppose that i ≥ 2 and (14) and (28) hold for all i < m. We prove (14) for i = m. Using (20) and the induction hypothesis, the last term in the sum in the square brackets of (26) is asymptotically 3λ_{m−2}n^{−⌊(m+3)/2⌋}. If m is even then, again using the induction hypothesis, all the other terms in the square brackets diminish at least as rapidly as n^{−(2+m/2)} whence,

R_{m}^{(n)} − R_{m}^{(n−2)} ∼ (\frac{3λ_{m−2}}{2^{m}m!} )n^{m/2−1}

so that,

R_{m}^{(n)} ∼ (\frac{3λ_{m−2}}{2^{m}m!} )(\frac{n^{m/2}}{2(\frac{1}{2}m)})

and finally,

r_{m}^{(n)} ∼ (\frac{3λ_{m−2}}{m} )n^{−m/2}

so that λ_{m} = 3λ_{m−2}/m. When m is odd the induction step is similar. This time the expression in the square brackets of (26) is dominated by the terms

t_{3}^{(n)}r_{m−3}^{(n−2)} + t_{2}^{(n)}r_{m−2}^{(n−2)} + \frac{2}{n}(r_{m−1}^{(n)} − r_{m−1}^{(n−2)}),

which, using (20) and the induction hypothesis, is asymptotically,

(−4λ_{m−3} + 3λ_{m−2} − 2(m − 1)λ_{m−1})n^{−(m+3)/2}

and the argument continues, as for the even case.

A similar argument, using (27), is used for the induction step in the proof of (28). We omit details.

Finally we prove (17). For each compact M ⊆ C − (−∞, −1), we show that there are constants, K_{n} for which, for all z ∈ M,

(29)  |ϱ_{k}^{(n)}(z)| ≤ K_{n} \frac{e^{|z|}|z|^{k+n/2}}{n^{⌊(k+1)/2⌋}(\frac{1}{2}n)!},

where the sequence {K_{n}} is bounded above. The bound (17) then follows from (29). For n = 0, 1 it is easy to see that there exist constants K_{0} and K_{1} respectively, for which (29) holds. For n ≥ 2, (25), along with (14), (21) and (28), gives,

(30)  |ϱ_{k}^{(n)}′(z)| ≤ C_{1} \frac{e^{|z|}|z|^{k+(n−2)/2}}{n^{⌊(k+3)/2⌋}(\frac{1}{2}(n − 2))!} + |h_{n}(z)ϱ_{k}^{(n−2)}(z)|,

where C_{1} is a constant depending only on M. Using the induction hypothesis,

(31) |ϱ_{k}^{(n)}′(z)| ≤ \frac{e^{|z|}|z|^{k+(n−2)/2}}{n^{⌊(k+1)/2⌋}(\frac{1}{2}(n − 2))!}(\frac{C_{1}}{n} + |h_{n}(z)|K_{n−2}(\frac{n}{n − 2})^{⌊(k+1)/2⌋}).

Using (21), there is a constant C_{2} , depending only on M, for which,

|h_{n}(z)| ≤ (1 + \frac{|z|}{(k + \frac{1}{2}n)})(1 + \frac{C_{2}}{n^{2}} )

whence,

(32)  \begin{aligned}|ϱ_{k}^{(n)}′(z)| & ≤ \frac{e^{|z|}|z|^{k+(n−2)/2}[1 + |z|/(k + \frac{1}{2}n)]}{n^{⌊(k+1)/2⌋}(\frac{1}{2}(n − 2))!} \\ & × (\frac{C_{1}}{n} + (1 + \frac{C_{2}}{n^{2}} )K_{n−2}(\frac{n}{n − 2})^{⌊(k+1)/2⌋}). \end{aligned}

Integrating then gives, (33)

\begin{aligned}|ϱ_{k}^{(n)}(z)| & ≤ \frac{e^{|z|}|z|^{k+n/2}}{n^{⌊(k+1)/2⌋}(\frac{1}{2}n)!}(\frac{n}{2k + n})(\frac{C_{1}}{n} + (1 + \frac{C_{2}}{n^{2}} )K_{n−2}(\frac{n}{n − 2})^{⌊(k+1)/2⌋}) \\ & = K_{n} \frac{e^{|z|}|z|^{k+n/2}}{n^{⌊(k+1)/2⌋}(\frac{1}{2}n)!}, \end{aligned}

where

K_{n} = (\frac{n}{2k + n})(\frac{C_{1}}{n} + (1 + \frac{C_{2}}{n^{2}} )K_{n−2}(\frac{n}{n − 2})^{⌊(k+1)/2⌋}).

Since, as n → ∞, n/(2k+n) and (n/(n − 2))^{⌊(k+1)/2⌋} are asymptotically 1−2k/n and 1 + (2/n)⌊(k + 1)/n⌋, respectively, one can readily verify that the sequence {K_{n}} approaches a finite limit.

Proof of Theorem 2. Let Δ be a regular hyperbolic simplex Δ with edge length r and dihedral angle θ. The link of Δ is the (n − 1)-dimensional spherical simplex obtained by intersecting it with a small sphere centred at one of its vertices, and scaling to obtain a simplex in the unit sphere. The link of Δ thus has the same dihedral angle as Δ, but a different edge length, which we denote by r̃.

Now let B be a hyperbolic ball of radius \frac{1}{2}r centred at a vertex of Δ. In terms of volume, the intersection B ∩Δ is the same proportion of B as the link of Δ is of the unit (n − 1)-sphere, S^{n−1} . Recalling that the volume of a hyperbolic n-ball of radius x is Volume (S^{n−1}) ∫_{0}^{x} sinh^{n−1} t dt, the volume of B ∩ Δ is given by

S_{n−1}(r̃) ∫_{0}^{r/2} sinh^{n−1} t dt,

whence, adding n+1 disjoint regions with this volume, and dividing by the volume of Δ, we obtain

(34)  d_{n}(r) = (n + 1)\frac{S_{n−1}(r̃)}{H_{n}(r)} ∫_{0}^{r/2} sinh^{n−1} t dt.

We set,

x = sech r = sec r̃ − 1,

and

y = 1 − sech r = 1 − x,

and also let

c = cosh(\frac{1}{2}r) = \sqrt{\frac{1}{2}(x^{−1} + 1)},
s = sinh(\frac{1}{2}r) = \sqrt{\frac{1}{2}(x^{−1} − 1)}.

The change of variable, u = sinh t, and integration by parts gives

(35)  \begin{aligned}∫_{0}^{r/2} sinh^{n−1} t dt & = ∫_{0}^{s} u^{n−1}(1 + u^{2})^{−1/2} du \\ & = \frac{1}{n}[\frac{s^{n}}{c} + ∫_{0}^{s} u^{n+1}(1 + u^{2})^{−3/2} du] \\ & ≤ \frac{s^{n}}{nc}[1 + \frac{cs^{2}}{(n + 2)}]. \end{aligned}

Also, for n ≥ 6, (36)

∫_{0}^{r/2} sinh^{n−1} t dt = ∫_{0}^{s} u^{n−2}(1 + u^{−2})^{−1/2} du

= \frac{1}{n − 1}[\frac{s^{n}}{c} − ∫_{0}^{s} u^{n−4}(1 + u^{−2})^{−3/2} du]

= \frac{1}{n − 1}[\frac{s^{n}}{c} − \frac{s^{n}}{(n − 3)c^{3}} + \frac{3}{(n − 3)} ∫_{0}^{s} u^{n−6}(1 + u^{−2})^{−5/2} du]

≤ \frac{s^{n}}{(n − 1)c}[1 − \frac{1}{(n − 3)c^{2}} + \frac{3}{(n − 3)(n − 5)c^{4}} ].

Now differentiating (34) with respect to θ, using (6) and (7), we obtain, when n ≥ 3 (37)

−\frac{d}{dθ} d_{n}(r) = \frac{−(n + 1)}{H_{n}(r)^{2}}[\frac{dS_{n−1}(r̃)}{dθ}(∫_{0}^{r/2} sinh^{n−1} t dt)H_{n}(r)

+ S_{n−1}(r̃)\frac{d}{dθ} (∫_{0}^{r/2} sinh^{n−1} t dt)H_{n}(r)

− \frac{dH_{n}(r)}{dθ}S_{n−1}(r̃) ∫_{0}^{r/2} sinh^{n−1} t dt]

= \frac{(n + 1)S_{n−1}(r̃) ∫_{0}^{r/2} sinh^{n−1} t dt}{2H_{n}(r)}

× [−\frac{n(n − 1)S_{n−3}(r̃)}{(n − 2)S_{n−1}(r̃)} − \frac{sinh^{n−1}(\frac{1}{2}r)}{∫_{0}^{r/2} sinh^{n−1} t dt} \frac{dr}{dθ} − \frac{n(n + 1)H_{n−2}(r)}{(n − 1)H_{n}(r)}].

We denote the expression in square brackets in (37) by X, and its three summands by X_{1} , X_{2} and X_{3} respectively. Since r is a strictly decreasing function of θ, we must prove that X is positive for all r > 0. Using the hyperbolic version of (11),

(12) and (4) we may rewrite X as

(38) \begin{matrix}− 2n^{1/2}(n − 1)(n − 2)^{1/2} \frac{Y_{n−3}(x)}{Y_{n−1}(x)} + \frac{s^{n−1}(x + n − 1)((x + n − 1)^{2} − 1)^{1/2}}{x(1 − x^{2})^{1/2} ∫_{0}^{r/2} sinh^{n−1}t dt} \\ + 2(n + 1)^{1/2}n(n − 1)^{1/2} \frac{Y_{n−2}(−y)}{Y_{n}(−y)}. \end{matrix}

Showing that X is positive requires some care, as a lot of cancellation occurs in the sum. For each n, the terms X_{1} and X_{2} both go to infinity as x → 0, as do X_{2} and X_{3} when y → 0, but the sum is bounded in (0, 1). We will also show that the sum grows like n^{2} as n → ∞, even though the individual terms grow like n^{3} .

We set

η_{n}(z) = [(2zY_{n−2}(z))/(nY_{n}(z))] − 1.

By Theorem 3 (with k = 3), η_{n}(t)/t diminishes like n^{−2} , uniformly for t ∈ [−1, 1]. We have,

(39)  X_{1} = −\frac{n^{1/2}(n − 1)^{2}(n − 2)^{1/2}}{x} (1 + η_{n−1}(x)),

and, using (36), for n ≥ 6, (40)

\begin{aligned}X_{2} & ≥ x^{−1}y^{−1}n^{1/2}(n − 1)^{2}(n − 2)^{1/2}(1 + \frac{x}{n − 1})(1 + \frac{x}{n})^{1/2}(1 + \frac{x}{n − 2})^{1/2} \\ & × [1 − \frac{1}{(n − 3)c^{2}} + \frac{3}{(n − 3)(n − 5)c^{4}} ]^{−1}, \end{aligned}

whence (41)

X_{1} + X_{2} ≥ \frac{1}{xy} n^{1/2}(n − 1)^{2}(n − 2)^{1/2}(1 − y + 3x/n + xy(1 + x)^{−1}/n + ε_{1}n^{−2}x)

= \frac{1}{y} n^{1/2}(n − 1)^{2}(n − 2)^{1/2}(1 + 3/n + y(1 + x)^{−1}/n + ε_{1}n^{−2})

= \frac{1}{y} n^{3}(1 + y(1 + x)^{−1}/n + ε_{3}n^{−2})

where terms of the form ε_{i} are bounded uniformly for x ∈ [0, 1], and for all n ≥ 6. Since,

(42)  X_{3} = −\frac{1}{y} n^{3}(1 − 1/n^{2})^{1/2}(1 + η_{n}(−y)),

for each x_{0} < 1, there is an n_{0} for which X is positive for all x ≤ x_{0} , and n ≥ n_{0} .

We now consider the behaviour of X, when x is bounded below, say x ≥ x_{1} > 0. We have, using (35) (43)

X_{2} = \frac{s^{n−1}(n − y)((n − y)^{2} − 1)^{1/2}}{x(1 − x^{2})^{1/2} ∫_{0}^{r/2} sinh^{n−1}t dt}

= \frac{n(n^{2} − 1)^{1/2}s^{n−1}}{x(1 − x^{2})^{1/2} ∫_{0}^{r/2} sinh^{n−1} t dt}(1 − \frac{y}{n})(1 − \frac{y}{n + 1})^{1/2}(1 − \frac{y}{n − 1})^{1/2}

≥ \frac{n^{2}(n^{2} − 1)^{1/2}}{xy} [1 + \frac{cs^{2}}{(n + 2)}]^{−1}[1 − \frac{2(2n^{2} − 1)y}{n(n^{2} − 1)} ]^{1/2}

= \frac{n^{2}(n^{2} − 1)^{1/2}}{xy} [1 + \frac{cy}{2(n + 2)x}]^{−1}[1 − \frac{2(2n^{2} − 1)y}{n(n^{2} − 1)} ]^{1/2}

= \frac{n^{2}(n^{2} − 1)^{1/2}}{xy} [1 − \frac{cy}{2nx} − \frac{2y}{n} + \frac{ε_{3}y}{n^{2}} ]

= \frac{n^{2}(n^{2} − 1)^{1/2}}{y} [1 + \frac{y}{x} − \frac{cy}{2nx^{2}} − \frac{2y}{nx} + \frac{ε_{3}y}{n^{2}x}],

Asymptotic volume formulae and hyperbolic ball packing 43

where here, and in the sequel, terms of the form ε_{i} are bounded uniformly in n and x ≥ x_{1} .

Hence

(44)  \begin{aligned}X_{2} + X_{3} & ≥ \frac{1}{y} n^{2}(n^{2} − 1)^{1/2}[\frac{y}{x} − \frac{cy}{2nx^{2}} − \frac{2y}{nx} + \frac{ε_{4}y}{n^{2}x}] \\ & = \frac{1}{x}n^{2}(n^{2} − 1)^{1/2}[1 − \frac{c}{2nx} − \frac{2}{n} + \frac{ε_{4}}{n^{2}} ] \\ & = \frac{1}{x}n^{3}[1 − \frac{c}{2nx} − \frac{2}{n} + \frac{ε_{5}}{n^{2}} ]. \end{aligned}

Since, from (39)

(45)  X_{1} = −\frac{1}{x}n^{3}(1 − 3/n + ε_{6}/n^{2}),

it follows that, when x_{1} is chosen so that c/x_{1} < 2, X is positive for all x ≥ x_{1} , and all sufficiently large n. We have thus shown that X is positive, so that d_{n}(r) is strictly increasing, for all r, when n is sufficiently large.

References

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[2] Böröczky, K., and A. Florian: Über die dichteste Kugelpackung im Hyperbolischen Raum. - Acta Math. Hungar. 15, 1964, 237–245.

[3] Fejes Tóth, L.: Regular Figures. - MacMillan, 1964.

[4] Haagerup, U., and H.J. Munkholm: Simplices of maximal volume in hyperbolic nspace. - Acta Math. 147, 1981, 1–11.

[5] Kabatiansky, G.A., and V.I. Levenshtein: Bounds for packings on a sphere and in space. - Problemy Peredachi Informatsii 1, 1978, 3–25.

[6] Kellerhals, R.: Regular simplices and lower volume bounds for hyperbolic n-manifolds. - Ann. Global Anal. Geom. 13, 1995, 377–392.

[7] Martin, G.J.: The volume of regular tetrahedra and sphere packing in hyperbolic 3space. - Math. Chronicle 20, 1991, 127–147.

[8] Milnor, J.W.: How to compute volume in hyperbolic space. - In: Collected Papers Vol. 1, Geometry, Publish or Perish, 1994.

[9] Rogers, C.A.: The packing of equal spheres. - Proc. London Math. Soc. 8, 1958, 609–620.

[10] Vinberg, E.B. (Ed.): Geometry II, Encyclopaedia of Mathematical Sciences Vol. 29, Springer-Verlag, 1993.

Received 24 February 1997