On the volume growth of the hyperbolic regular $n$-simplex

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by  Ákos G. Horváth

ON THE VOLUME GROWTH OF THE HYPERBOLIC REGULAR n-SIMPLEX
A´KOSG.HORVA´TH
Abstract. Inthispaperwegivelowerandupperboundsforthevolumegrowthofaregularhyperbolic
simplex,namelyfortheratioofthen-dimensionalvolumeofaregularsimplexandthe(n−1)-dimensional
volumeofitsfacets. InadditiontothemethodsofU.HaagerupandM.Munkholmweuseathirdvolume
6 formisbasedonthehyperbolicorthogonalcoordinatesofabody. Inthecaseoftheideal,regularsimplex
1 ourupperboundgivesthebestknownupperbound. Ontheotherhand,alsointheidealcaseourgeneral
0 lowerbound,improvedthebestknownoneforn=3.
2

n
a
J

5 1. Introduction
1
1.1. The problem. The volume growthof the Euclidean regular n-simplex inscribed in the unit sphere

] can be calculated by the volume form
G
m
M (1) V (S)= V (F),
n n 1
n −
.
h where V (S), V (F) and m means the n-dimensional volume of the simplex, the (n 1)-dimensional
n n 1
at volume of its fac−et F and the height of the simplex, respectively. Thus we have Vn−(S)/Vn 1(F) =
m (n+1)/n2. Ifr is the radiusofthe circumscribedsphere ofthe regularsimplex rS then by simil−aritythe
ratio V (rS)/V (rF) is equal to r(n+1)/n2. In summary, the Euclidean ”volume growth” is not an
[ n n 1
interesting quan−tity, it can be determined exactly and easily.
1
In contrast to the Euclidean case in hyperbolic n-space there is at least one kind of regular simplices
v
9 whichvolumegrowthisveryimportant. Thisistheregular,idealsimplex. Inhyperbolicn-spaceasimplex
3 is calledregular if any permutationofits vertices canbe induced by anisometryof the space. Asimplex
9 ideal if all the vertices are on the sphere at infinity. There is only, up to isometry, only one ideal regular
3
simplex in hyperbolic n-space. It is known that the volume of a hyperbolic simplex is finite also if some
0
of the vertices are on the sphere at infinity. Contrary the case of the plane, there is non-congruent ideal
.
1
simplices arising an interesting question: Which one has maximal volume? The answer that a simplex
0
is of maximal volume if and only if it is ideal and regular was conjectured (for any n) by Thurston in
6
1 [15]. In three space it was proved by Milnor (see in [11] or [12]). In higher dimension spaces this result
: was published by U. Haagerup and H. Munkholm in [8]. Their prove based on the ”volume growth” of
v
an ideal, regular simplex, Proposition 2 in [8] says that (n 2)/(n 1)2 V (S)/V (F) 1/(n 1).
i n n 1
X Theseboundsreliesonaninterplaybetweenthevolumefor−msofthe−Poin≤care’shalf-sp−acem≤odeland−the
r volumeformoftheCayley-Klein’sprojectivemodelusingalsoGauss’divergenceformula. Unfortunately,
a
in hyperbolic n-space there is no formula analogous to (1), hence to compare the n-volume of a regular
simplex and the (n 1)-volume of its facets is a non-trivial exercise.
−
In recent decades the analytic investigations in hyperbolic space and even more the examination of
computational methods of the n-dimensional hyperbolic volume appreciated. Without mentioning the
exhaustive list I suggest to study the references [1, 10, 9, 2, 4, 5, 6, 7, 3, 14]. In this paper we give
estimatesforthe volumegrowthofaregularhyperbolicsimplex inthe abovedetailedsense. Inadditions
the methods of Haagerup and Munkholm we use a third volume form based on hyperbolic orthogonal
coordinates. Inthe caseoftheideal,regularsimplexourupperboundgivesthebestknownupperbound
provedby Haagerupand Munkholm in 1981. On the other hand, also in the ideal case our generallower
bound, improved the best known one for n=3.
1.2. Notation. We use the following notation in this paper:
Date:Jan. 15,2016.
2010 Mathematics Subject Classification. 51F10,52B10.
Key words and phrases. half-space model, hyperbolic orthogonal coordinates, hyperbolic volume inequality, projective
model.
1
2 A´KOSG.HORVA´TH
R,En and Hn: the set of real numbers, the Euclidean n-space and the hyperbolic n-spaces,
•
respectively,
,ρ(, ): the Euclidean length and the hyperbolic distance function, respectively,
• |·| · ·
V(): the volume function of the hyperbolic space,
• ·
Pn, h : Hn Pn: the half-space model of Hn (Poincare’s second model) and the standard
• →
mapping sending the hyperbolic space onto the model, respectively,
H: the boundary hyperplane of Pn,
•
CKn and p:Hn CKn: the projective model of Hn (Cayley-Klein’s model) and the standard
• →
mapping on the hyperbolic space to the model
x ,...,x : the Euclidean coordinates of the embedding Euclidean space En with respect to an
1 n
•
orthonormed basis in En,
S(n)=conv E ,...,E : the n-dimensional regular simplex inscribed in the unit sphere,
1 n+1
• { }
τ[n,t] = conv p 1(E ),...,p 1(E ) : the hyperbolic regular n-simplex with hyperbolic cir-
− 1 − n+1
• { }
cumradius r(t)=tanh−1(sint),
τ [n,t] = conv p 1(E ),...,p 1(E ),p 1(E ),...p 1(E ) : a facet of τ[n,t], it is a hy-
i − 1 − i 1 − i+1 − n+1
• { − }
perbolic, regular simplex of dimension n 1.
−
r : the hyperbolic radius of the circumscribed sphere of the k-dimensional faces of τ[n,t],
k
•
d : the hyperbolic distanceofthe circumcenterofτ[n,t]anda(k 1)-dimensionalface ofτ[n,t].
k
• −
2. The theorem
Denote by τ[n,t] and τ [n,t] the n-dimensional regular simplex inscribed in the sphere of radius
i
rn :=tanh−1(sint) andits ith facet, whichisa regularsimplex ofdimension(n 1)inscribedina sphere
−
of radius
sint 1 1
rn−1 :=tanh−1 1 qsin2−t n21 ,
− n2
q 
respectively. In this paper we prove the following theorem: (cid:0) (cid:1)
Theorem 1. We have the following two inequalities
n 1 1 sin2tn 1 sint 1 n 1
(2) (n+1)(cid:18)12(cid:19) − 1q− n2−−n1 n2−2−nsin2t(cost)tanh−1 1−qsin−2tn2−2n−n1≤ VV((ττi[[nn,,tt]]))
q p q 
and
n 1
V(τ[n,t]) 1 n2(1 sint)2(1+sint) −
(3) 1 − .
V(τi[n,t]) ≤ n 1 − s(n+sint)2(1+sint) (n2 1)sin2t(1 sint)2! 
− − − −
 
Note 1. For n=3 and t=π/2, (2) is stronger than the left hand side of (3.1) in [8]. In fact,
sint 1 n 1
t→limπ/2(cost)tanh−1 1−qsin−2tn2−2n−n1=t→limπ/2(cost)tanh−1(sint)=1,
q 
hence we have for n=3 that
n 2 1 4 n+1 1 1
− = = = .
(n 1)2 4 ≤ 4√8 2n 1√n2 1 ≤ n 1 2
− − − −
Consequently, in dimension three our result is a generalization of the inequalities of [8].
Note 2. In the case of t=π/2, (3) gives the same bound as we saw in (3.1) in [8]. Really,
n 1
1 n2(1 sint)2(1+sint) −
1 −
n 1 − s(n+sint)2(1+sint) (n2 1)sin2t(1 sint)2! 
− − − −
 
in t=π/2 is equal to 1 (in all dimensions).
n 1
−
ON THE VOLUME GROWTH OF THE HYPERBOLIC REGULAR n-SIMPLEX 3
E
n+1
O
E
1
K
K n-1
n-2
E
i
Figure 1. Regular simplex in the projective model.
3. The proof of the theorem
The proof can be divided into three steps.
The first step uses that calculations which can be got from the Cayley-Klein (or projective) model.
We determine certain metric properties of the regular hyperbolic simplex of circumradius r .
n
In the second step using hyperbolic orthogonal coordinates we prove the general lower bound.
The third step contains the proof of the upper bound. In this section we use Poincare’s half-space
model.
For the good readability these steps can be found in three subsection and the result is the union of
the statements of Lemma 1 and Lemma 2.
3.1. Calculations in the projective model. We consider the map p : Hn CKn sending the
→
regularidealsimplexτ[n,π/2]intotheEuclideanregularsimplexp(τ[n,π/2])inscribedintheunitsphere.
(Clearly,themappcanbegivenconcretelybutitisnotimportantforourpurpose.) Foreveryt [0,π/2]
∈
p(τ[n,t]) denotes the regular simplex concentric with p(τ[n,π/2]) and with circumradius r = sint. For
simplicity S(n) denotes the regular simplex p(τ[n,π/2]). Observe that the volume of τ[n,t] can be
calculated from the volume form of CKn (see in [4])
n
dV = 1−r2 −(n+1)/2dx1...dxn, r2 := x2i,
i=1
(cid:0) (cid:1) X
and we get that
n+1 n+1
(4) V(τ[n,t])= 1 r2 − 2 dr =sinnt 1 (sin2t)r2 − 2 dr.
− −
Z Z
(sint)S(n) (cid:0) (cid:1) S(n) (cid:0) (cid:1)
Bysphericalsymmetrywecanchooseacoordinatesystemsuchaway,thatE =(0,...,0,sint)T. De-
n+1
notebyOandK thecenterofthesimplexS(n)andthecenteritsfacetconv E ,...,E ,respectively.
n 1 1 n
− { }
Then OK =sint/n and the hyperbolic distance of the corresponding points is
n 1
| − |
1 n+sint
ρ p−1(O),p−1(Kn 1) = ln =tanh−1(sint).
− 2 n sint
−
(cid:0) (cid:1)
Immediate calculation with cross-ratio gives the circumradius of the facets τ [n,t] (i = 1,...,n+1) of
i
τ[n,t]:
n2 sin2t+sint√n2 1
ρ p−1(E1),p−1(Kn 1) =ln − − .
− p ncost
Denote by K the center(cid:0)of the face conv E ,(cid:1)E ,...,E for i = 1,...,n 1. Then the fundamental
i 1 2 i+1
{ } −
orthosceme of the simplex τ[n,t] is the convex hull of the points E ,K ,K ,...,K ,O . Clearly
1 1 2 n 1
O { − }
τ[n,t] is disjointunion of (n+1)!congruentcopies of hence V(τ[n,t])=(n+1)!V( ). The successive
O O
edge lengths of can be determined by induction. As we saw
O
sint
rn :=ρ p−1(O),p−1(E1) =tanh−1(sint) and dn :=ρ p−1(O),p−1(Kn 1) =tanh−1
− n
(cid:18) (cid:19)
(cid:0) (cid:1) (cid:0) (cid:1)
4 A´KOSG.HORVA´TH
moreover
n2 sin2t+sint√n2 1
rn 1 :=ρ p−1(Kn 1),p−1(E1) =ln − − =
− − p ncost
(cid:0) (cid:1)
1 sin2t 1 +sint 1 1 sint 1 1
= 21lnq1−sin2t(cid:0)n12(cid:1) sintq1− n12 =tanh−1 1 qsin2−t n21 
− n2 − − n2 − n2
q q q 
(cid:0) (cid:1) (cid:0) (cid:1)
and
1 sin2t 1 + 1 sint 1 1
dn−1 :=ρ p−1(Kn−1),p−1(Kn−2) = 21lnq1−sin2t(cid:0)n12(cid:1) (n−11)sintq1− n12 =
(cid:0) (cid:1) − n2 − (n−1) − n2
q q
(cid:0) (cid:1)
sint 1 1
=tanh−1(n 1) q1 −sinn22t 1 
− − n2
 q 
(cid:0) (cid:1)
it can be proved by induction that
sint 1 k
(5) rn−k :=ρ p−1(Kn−k),p−1(E1) =tanh−1 q − n(n−k+1) 
1 sin2t k
(cid:0) (cid:1)  − n(n k+1) 
r − 
 (cid:16) (cid:17)
holds for k =1,...,(n 1) and
−
sint 1 k
(6) dn k :=ρ p−1(Kn k),p−1(Kn k 1) =tanh−1 q − n(n−k+1) 
− − − −
(n k) 1 sin2t k
(cid:0) (cid:1)  − − n(n k+1) 
 r − 
 (cid:16) (cid:17)
holds also for k=1,...,(n 1) if K means E .
0 1
−
With respect to an (n 1)-dimensional projective model we can determine the volume of a facet of
−
the regular n-simplex with circumradius r . In fact, the circumradius of its facets is r hence using
n n 1
−
the formula (4) we get
n
n 1
(7) V(τi[n,t])= 1 r2 −2 dr = dr.
− √1 r2
(tanhrn−Z1)S(n−1) (cid:0) (cid:1) sint√1−Zn12 S(n 1) (cid:18) − (cid:19)
r1−sin2t(n12) −
3.2. Hyperbolic orthogonal coordinates and the lower bound. Take a hyperbolic orthogonal
coordinatesystemwithoriginp 1(E )onthefollowingway. Letthex axisbethelinethroughthepoints
− 1 n
p 1(E )andp 1(E )containingthepointp 1(K ),too. Letthex axisthelineofthehyperbolic2-plane
− 1 − 2 − 1 1
containingthepointsp 1(E ),p 1(K )andp 1(K );anddirectedsuchthatthecoordinatesofthepoints
− 1 − 1 − 2
of the rectangular triangle p 1(E ),p 1(K ),p 1(K ) will be non-negative real numbers. Continue this
− 1 − 1 − 2
processandchoosethe axisx suchthatthe k 1subspaceis spannedby the axislin x ,x ,x ,...,x
k n 1 2 k
− { }
containsthe(k 1)-dimensionalorthoscemeisspannedbythepoints p 1(E ),p 1(K ),...,p 1(K )
− 1 − 1 − n 1
− { − }
in its positive orthant(hyperoctant). Then we canapply the following volume form(see paragraph3.3.3
in [4])
αnα1 αn−1
V( )= (coshn−1xn 1)(coshn−2xn 2) (coshx1)dxn 1 dx1dxn,
O ··· − − ··· − ···
Z Z Z
0 0 0
ON THE VOLUME GROWTH OF THE HYPERBOLIC REGULAR n-SIMPLEX 5
where
tanhd
αn 1 = tanh−1 n sinhxn 2
− (cid:18)sinhdn−1 − (cid:19)
tanhd
αn 2 = tanh−1 n−1 sinhxn 3
− (cid:18)sinhdn−2 − (cid:19)
.
.
.
tanhd
α2 = tanh−1 3 sinhx1
sinhd
(cid:18) 2 (cid:19)
tanhd
α1 = tanh−1 2 sinhxn
sinhd
(cid:18) 1 (cid:19)
α = d .
n 1
Observe that the volume of the facet :=conv p 1(E ),p 1(K ),p 1(K )...p 1(K ) is
n 1 − 1 − 1 − 2 − n 1
O − { − }
αnα1 αn−2
V( n 1)= (coshn−2xn 2) (coshx1)dxn 2 dx1dxn,
O − ··· − ··· − ···
Z Z Z
0 0 0
with the same α ’s. Consequently these formulas leads to a connection between V( ) and V( ) and
i n 1
O O −
betweenV(τ[n,t])=(n+1)!V( )andV(τ [n,t])=n!V( ). IntheproofweuseChebyshev’sintegral
i n 1
inequality (see in [13]), saying tOhat if f,g :[a,b] R havOe t−he same monotony then
→
b b b
1
f(x)g(x)dx f(x)dx g(x)dx,
≥ b a
Za − Za Za
and if f;g have opposite monotony, then the inequality should be reversed.
Lemma 1. For n 3 we have
≥
(8) V(τ[n,t])≥(n+1)(cid:18)21(cid:19)n−1 1q−1n2−−n1sinn22tn−2−ns1in2t(cost)tanh−1s1in−tqsin12−tnn2−2n−n11V(τi[n,t])
q p  
Proof. We compare the following two integrals
αnα1 αn−2 αn−1
V( )= coshn−1xn 1dxn 1 coshn−2xn 2 coshx1dxn 2 dx1dxn,
O ···  − −  − ··· − ···
Z Z Z Z
0 0 0 0
 
and
αnα1 αn−2
V( n 1)= coshn−2xn 2 coshx1dxn 2 dx1dxn.
O − ··· − ··· − ···
Z Z Z
0 0 0
Since
k 1
1 2k 1 − 2k
cos2k(x) = + cos(2(k l)x)
4k k 22k 1 l −
(cid:18) (cid:19) − l=0(cid:18) (cid:19)
X
k
1 2k+1
cos2k+1x = cos((2(k l)+1)x)
4k l −
l=0(cid:18) (cid:19)
X
then using the identity coshx=cos(ix) we have
k 1
1 2k 1 − 2k
cosh2kx = + cosh(2(k l)x)
4k k 22k 1 l −
(cid:18) (cid:19) − l=0(cid:18) (cid:19)
X
k
1 2k+1
cosh2k+1x = cosh((2(k l)+1)x).
4k l −
l=0(cid:18) (cid:19)
X
6 A´KOSG.HORVA´TH
Let denote Fn 1(x) the antiderivative of the functions coshn−1x. Then we have
−
k 1
1 2k x+ 1 − 2k 1 sinh(2(k l)x) if n=2k
4k k 22k−1 l 2(k l) −
(9) F (x)= l=0 −
n−1  41k(cid:0)k(cid:1)2kl+1 (2(kP1l)+(cid:0)1)s(cid:1)inh((2(k−l)+1)x) if n=2k+1.
l=0 −
Since Fn 1(0)=0 we get P(cid:0) (cid:1)
− αn−1
fn 1(xn 2):= coshn−1(xn 1)dxn 1 =
− − − −
Z
0
k 1
1 2k α (x )+ 1 − 2k 1 sinh(2(k l)α (x )) if n=2k
= 4k k n−1 n−2 22k−1 l=0 l 2(k−l) − n−1 n−2
 41k(cid:0)l=k0(cid:1)2kl+1 (2(k−1l)+1) sinh(P(2((cid:0)k−(cid:1)l)+1)αn−1(xn−2)) if n=2k+1
where  P(cid:0) (cid:1)
tanhd n 1 2sin2t
αn 1(xn 2)=tanh−1 n sinhxn 2 =tanh−1 − 1 sinhxn 2 .
− − (cid:18)sinhdn−1 − (cid:19) rn+1s − n(n−1) − 
 
Without loss of generality we can assume that n = 2k+1, the other case when n = 2k can be proved
analogouslyleading to the same result. Since fn 1(xn 2) is strictly increasingand sinhx,tanh−1(x) x
− − ≥
for all positive x, we have
k
1 2k+1 1
f (x ) (2(k l)+1)α (x ))=α (x ))=
n−1 n−2 ≥ 4k l (2(k l)+1) − n−1 n−2 n−1 n−2
l=0(cid:18) (cid:19) −
X
tanhd tanhd
=tanh−1 n sinhxn 2 n sinhxn 2,
(cid:18)sinhdn−1 − (cid:19)≥ sinhdn−1 −
implying that
αnα1 αn−2 αn−1
V( )= coshn−1xn 1dxn 1 coshn−2xn 2 coshx1dxn 2 dx1dxn
O ···  − −  − ··· − ··· ≥
Z Z Z Z
0 0 0 0
αnα1 αn−2 
tanhd
n sinhxn 2coshn−2xn 2dxn 2coshn−3xn 3 coshx1dxn 3 dx1dxn.
≥ sinhdn−1 Z0 Z0 ··· Z0 − − − − ··· − ···
Using the Chebysev’s inequality we get that
αn−2 αn−2
coshα 1
sinhxn 2coshn−2xn 2dxn 2 n−2− coshn−2xn 2dxn 2
Z0 − − − ≥ αn−2 Z0 − − ≥
αn−2 αn−2
α 1tanhd
n−2 coshn−2xn 2dxn 2 n−1 sinhxn 3 coshn−2xn 2dxn 2.
≥ 2 Z0 − − ≥ 2 sinhdn−2 − Z0 − −
So
αnα1 αn−3 αn−2
1 tanhd tanhd
V( ) n n−1 sinhxn 3 coshn−2xn 2dxn 2
O ≥ 2sinhdn 1 sinhdn 2 ··· −  − − ×
− − Z0 Z0 Z0 Z0
coshn−3xn 3dxn 3coshn−4xn 4 coshx1dxn 4 dx1dxn 
× − − − ··· − ···
From (9) can be seen easily that the function
αn−2
coshn−2xn 2dxn 2 coshn−3xn 3
 − −  −
Z
0
 
holds the assumption of Chebysev’s inequality and hence we can continue the process.
2 αnα1 αn−4
1 tanhd tanhd tanhd
n n 1 n 2
V( ) − − sinhxn 4
O ≥(cid:18)2(cid:19) sinhdn−1sinhdn−2sinhdn−3 Z0 Z0 ··· Z0 − ×
ON THE VOLUME GROWTH OF THE HYPERBOLIC REGULAR n-SIMPLEX 7
~
E
n+1
~
~O
K
n-1
~
E
i O
P
Figure 2. Regular simplex in the half-space model.
αn−3αn−2
coshn−2xn 2coshn−3xn 3dxn 2dxn 3 coshn−4xn 4dxn 4
× − − − −  − − ×
Z Z
0 0
 coshn−5xn 5 coshx1dxn 5 dx1dxn
× − ··· − ···
and so on... By induction we get the inequality
1 1 n−1 (tanhdn tanhd2)d1
V(τ[n,t])=V( ) ··· V( )=
n 1
(n+1)! O ≥(cid:18)2(cid:19) sinhdn−1···sinhd1 O −
1 n−1 (sinhdn)d1 1 n−1 (sinhdncoshd1)d1V(τi[n,t])
= V( )= .
n 1
(cid:18)2(cid:19) coshdn···coshd2sinhd1 O − (cid:18)2(cid:19) coshrnsinhd1 n!
But
sint 1
sinhd = , coshr = ,
n n
n2 sin2t cost
−
sint 1 n 1 p sint 1 n 1 1 sin2tn 1
d1 =tanh−1 1 qsin2−tn2−n1 , sinhd1 = qcos−t 2−n coshd1 = q −cost 2−n ,
− 2−n
we get the inequality 
n 1 1 sin2tn 1 sint 1 n 1
V(τ[n,t])≥(n+1)(cid:18)21(cid:19) − 1q− n2−−n1 n2−2−nsin2t(cost)tanh−1 1−qsin2−tn2−2n−n1 V(τi[n,t]),
as we stated. q p   ✷
3.3. Half-space model and the upper bound. In the half-space model we consider an Euclidean
orthogonalcoordinate system with origin O lying on the boundary (horizontal) hyperplane H. For the
P
sake of simplicity denote by P the point h(p 1(P)) Pn where P CKn is a point of the projective
−
^ ] ∈ ∈
model. AssumethatO ,K ,OandE lyingonaverticalline,whichisthex axisofourcoordinate
P n 1 n+1 n
−e
system(seeinFig. 2). Fromtheregularityofτ[n,t]followsthattheEuclideansimplexconv E ,...,E
1 n
{ }
ofdimension(n 1)isalsoregulaerinEuclideansenseandlyinginahorizontal(paralleltoH)hyperplane.
−
Set O K^ = 1 then O O = n+sint and O E] = (n+sint)(1+sint). If α is thefEuclidfean
| P n−1| | P | n−sint | P n+1| (n−sint)(1−sint)
angle E O O∠ (1 i n 1) then wqe have: q
i P
≤ ≤ − e
1+sinα
f e ρ(E1K)=ln
cosα
and the equation between α and t is
e
n2 sin2t+sint√n2 1 1+sinα
− − =
ncost cosα
p
8 A´KOSG.HORVA´TH
implying that
sint= nsinα or sinα= √n2−1sint = sint 1− n12.
√n2 cos2α n2 sin2t 1q sin2t
− − − n2
p q
The Euclidean coordinates of the points E =(xi), j =1,...,n, i=1,...,n+1 hold the equalities:
i j
n−1 f (n2 1)sin2t
(xi)2 =sin2α= − for all i=1...n,
j n2 sin2t
Xj=1 −
n2cos2t
(xi)2 =cos2α= for all i=1...n
n n2 sin2t
−
n 1
− (n+sint)(1+sint)
(xn+1)2 =0, (xn+1)2 = .
j n (n sint)(1 sint)
j=1 − −
X
Let denote by C = yi,...,yi ,0 T and γ the center and the radius of the sphere through the points
i 1 n 1 i
] ] ] −
E ,...,E ,E ,...,E . This sphere contains a facet of h(τ[n,t]). Hence for all i = 1,...,n we
1 i 1 i+1 (cid:0) n+1 (cid:1)
{ − }
have (n 1) equalities holding for the coordinates above. These are
−
f
n 1
− (n+1)sint
xkyi = − =:c where k =i, k =1,...,n.
j j (n sint)(1 sint) 6
j=1 − −
X
wInetrgoedtucethenotationXi =:[xkj]kj==11,,......,,nn−;k16=i ∈R(n−1)×(n−1) andyi =[y1i,...,yni−1]T,respectively. Then
(yi)TXi =[c,...,c] or equivalently (Xi)Tyi =[c,...,c]T =c[1,...,1]T.
Since conv E ,...,E is a regular simplex we get
1 n
{ }
f f 1 1 ... 1 1
−n 1 −n 1 −n 1
1 1− 1 ..−. −1
R(n−1):=(Xi)T(Xi)= (nn22−1s)insi2nt2t −n...−1 ... −n...−1 ... −n...−1 .
−  1 ... 1 1 1 
 −n 1 −n 1 −n 1 
 −1 1 ..−. 1 1− 
 −n 1 −n 1 −n 1 
 − − − 
It follows that
(Xi)T =R(n 1)(Xi) 1 and so ((Xi)T) 1 =(Xi)(R(n 1)) 1.
− − −
− −
Since the inverse of R(n 1) is equal to
−
2 1 1 ... 1
1 2 1 ... 1
(R(n−1))−1 = n(nn2+−1s)insi2nt2t ... ... ... ... ... 
 1 ... 1 2 1 
 
 1 1 ... 1 2 
 
 
we get
2 1 1 ... 1 1
1 2 1 ... 1 1
yi =Xi(R(n 1)) 1[c,...,c]T = (n+sint) Xi .. .. .. .. ..  .. =
− . . . . . .
− −nsint(1 sint)   
−  1 ... 1 2 1  1 
  
 1 1 ... 1 2  1 
  
  
ON THE VOLUME GROWTH OF THE HYPERBOLIC REGULAR n-SIMPLEX 9
n
xk
1
k=1
 kPn6=i 
1  xk2 
= (n+sint) Xi 1.. = (n+sint)  kkP=6=1i.. = (n+sint) n v ,
.  .  k
−sint(1 sint)   −sint(1 sint)  −sint(1 sint)
−  1  −  n xk  − kXk==1i
 1   k=1 n−2  6
   kPn6=i 
 xk 
 n 1 
 k=1 − 
 kP6=i 
n
where v :=[xk,...,xk ]T. Since v =0, we get that
k 1 n 1 k
− k=1
P
(n+sint)
yi = v .
i
sint(1 sint)
−
Hence
(n+sint)2 (n2 1)(n+sint)
yi 2 = v 2 = − ,
| | sin2t(1 sint)2| i| (n sint)(1 sint)2
− − −
implying that
n+sint 2
γ2 = .
i 1 sint
(cid:18) − (cid:19)
Let denote by ǫ[n 1,t]=conv F ,...,F where F is the orthogonalprojection of E to H. ǫ[n 1,t]
1 n i i
− { } −
can be dissected into n congruent simplices with a common vertex O . Denote by ǫ [n 1,t] these
P i
−
simplices, concretely, we set ǫ [n 1,t] := α v α 0, α 1 . fIf a point z Pn
i − { j=i j j | j ≥ j=i j ≤ } ∈
is in h(τ[n,t]), then it is on the form z = v +z6 (v)e , where v ǫ([n6 1,t]). If z h(τ[n,t]) with
n n
P ∈ P − ∈
v ǫ [n 1,t] then z C 2 γ2 hence we have that
∈ n − k − nk ≤ n
2
n 1 n 1
− − α xj yn +z2 (γ )2.
 j i − i n ≤ n
i=1 j=1
X X
 
From this inequality we get
2
z2 n+sint 2 n−1 n−1α xj (n+sint) xn =
n ≤ 1 sint −  j i − sint(1 sint) i
(cid:18) − (cid:19) i=1 j=1 −
X X
 
2
2 n 1 n 1
= n+sint − − α + (n+sint) xj =
1 sint −  j sint(1 sint) i
(cid:18) − (cid:19) i=1 j=1(cid:18) − (cid:19)
X X
 
2 n 1 n 1
= n+sint − − α + (n+sint) α + (n+sint) xjxk =
1 sint − j sint(1 sint) k sint(1 sint) i i
(cid:18) − (cid:19) i=1j,k=1(cid:18) − (cid:19)(cid:18) − (cid:19)
X X
2 n 1 n 1
= n+sint − α + (n+sint) α + (n+sint) − xjxk =
1 sint − j sint(1 sint) k sint(1 sint) i i
(cid:18) − (cid:19) j,k=1(cid:18) − (cid:19)(cid:18) − (cid:19)i=1
X X
n+sint 2 (n2 1)sin2t n−1 (n+sint) 2
= − α +
1 sint − n2 sin2t  j sint(1 sint) −
(cid:18) − (cid:19) − Xj=1(cid:18) − (cid:19)

n 1
1 − (n+sint) (n+sint)
α + α + =
j k
−n 1 sint(1 sint) sint(1 sint) 
− j=k=1(cid:18) − (cid:19)(cid:18) − (cid:19)
6X

n+sint 2 (n2 1)sin2t n−1 (n+sint)
= − α2+2α +
1 sint − n2 sin2t  j jsint(1 sint)
(cid:18) − (cid:19) − Xj=1(cid:18) −

10 A´KOSG.HORVA´TH
(n+sint)2 1 n−1 (n+sint) (n+sint)2
+ α α +(α +α ) + =
sin2t(1 sint)2 − n 1 j k j k sint(1 sint) sin2t(1 sint)2 
− (cid:19) − j6=Xk=1(cid:18) − − (cid:19)

n+sint 2 (n2 1)sin2t 1 (n+sint)2
= − (n 1) (n 1)(n 2)
1 sint − n2 sin2t − − n 1 − − sin2t(1 sint)2−
(cid:18) − (cid:19) − (cid:18) − (cid:19) −
(n2 1)sin2t n−1 1 n−1 (n+sint)
− 2 α (α +α )
− n2 sin2t  j − n 1 j k sint(1 sint)−
− Xj=1 − j6=Xk=1 −
 
(n2 1)sin2t n−1 1 n−1 (n+sint)(1+sint)
− α2 α α =
− n2 sin2t  j − n 1 j k (n sint)(1 sint)−
− Xj=1 − j6=Xk=1 − −
 
2(n+1)sint n−1 (n2 1)sin2t n−1 1 n−1
α − α2 α α .
−(n sint)(1 sint) j − n2 sin2t  j − n 1 j k
− − Xj=1 − Xj=1 − j6=Xk=1
On the other hand obviously z 2 1 and hence  
k k ≥
2
z2 1 n−1 n−1α xj =1 (n2−1)sin2t n−1α2 1 n−1 α α .
n ≥ −  j i − n2 sin2t  j − n 1 j k
Xi=1 Xj=1 − Xj=1 − j6=Xk=1
   
These inequalities imply the assumption
n 1
(n+sint)(1+sint) 2(n+1)sint −
(10) 1 v 2 z2 α v 2,
−| | ≤ n ≤ (n sint)(1 sint) − (n sint)(1 sint) j −| |
− − − − j=1
X
n 1 n 1
where|v|2 = (nn22−−1s)insi2nt2t j=−1α2j − n−11j=−k=1αjαk! is the squareofthe Euclideannormofthe vectorv.
6
n 1 P P
Denote the quantity − α by α(v). Then the volume is:
j
j=1
P
n q((nn−+ssiinntt))((11−+ssiinntt))−(n−2s(inn+t)1()1s−insitnt)α(v)−|v|2
V(τ[n,t])= zn−ndzndρ.
i=1 Z Z
Xv∈ǫi[n−1,t] zn=√1−|v|2
Sincethesetofthevectorsv belongingtomorethanoneǫ [n 1,t],hasmeasurezero,theaboveequality
i
−
can be written of the form:
n 1
1 1 −
V(τ[n,t])= dρ
n 1 1 ρ2! −
− ǫ[nZ1,t] −
−
p
n 1
−
1 1
(11) dρ,
−n−1ǫ[nZ1,t]  ((nn+ssiinntt))((11+ssiinntt)) − (n2s(inn+t)1()1sinsitnt)α(v)−ρ2
− q − − − − 
where ρ2 = v 2 and α(v) is uniquely determined by the vector v.
| |
By formula (3.3) we can get an upper bound on the volume of τ[n,t].
Lemma 2.
n 1
1 n2(1 sint)2(1+sint) −
(12) V(τ[n,t]) 1 − V(τ [n,t]).
≤ n 1 − s(n+sint)2(1+sint) (n2 1)sin2t(1 sint)2!  i
− − − −
 
Proof. Clearly
(n+sint)(1+sint) 2(n+1)sint (n+sint)(1+sint)
α(v) v 2 v 2
s(n sint)(1 sint) − (n sint)(1 sint) −| | ≤s(n sint)(1 sint) −| |
− − − − − −