Asymptotic Properties of Hilbert Geometry
It is shown that the spheres in Hilbert geometry have the same volume growth entropy as those in the Lobachevsky space. Asymptotic estimates for the ratio of the volume of metric ball to the area of the metric sphere in Hilbert geometry are given. Derived estimates agree with the well-known fact in...
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nasplib_isofts_kiev_ua-123456789-1065102025-02-09T16:39:09Z Asymptotic Properties of Hilbert Geometry Borisenko, A.A. Olin, E.A. It is shown that the spheres in Hilbert geometry have the same volume growth entropy as those in the Lobachevsky space. Asymptotic estimates for the ratio of the volume of metric ball to the area of the metric sphere in Hilbert geometry are given. Derived estimates agree with the well-known fact in the Lobachevsky space. 2008 Article Asymptotic Properties of Hilbert Geometry / A.A. Borisenko, E.A. Olin // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 327-345. — Бібліогр.: 16 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106510 en Журнал математической физики, анализа, геометрии application/pdf Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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It is shown that the spheres in Hilbert geometry have the same volume growth entropy as those in the Lobachevsky space. Asymptotic estimates for the ratio of the volume of metric ball to the area of the metric sphere in Hilbert geometry are given. Derived estimates agree with the well-known fact in the Lobachevsky space. |
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Borisenko, A.A. Olin, E.A. |
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Borisenko, A.A. Olin, E.A. Asymptotic Properties of Hilbert Geometry Журнал математической физики, анализа, геометрии |
| author_facet |
Borisenko, A.A. Olin, E.A. |
| author_sort |
Borisenko, A.A. |
| title |
Asymptotic Properties of Hilbert Geometry |
| title_short |
Asymptotic Properties of Hilbert Geometry |
| title_full |
Asymptotic Properties of Hilbert Geometry |
| title_fullStr |
Asymptotic Properties of Hilbert Geometry |
| title_full_unstemmed |
Asymptotic Properties of Hilbert Geometry |
| title_sort |
asymptotic properties of hilbert geometry |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2008 |
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Asymptotic Properties of Hilbert Geometry / A.A. Borisenko, E.A. Olin // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 3. — С. 327-345. — Бібліогр.: 16 назв. — англ. |
| series |
Журнал математической физики, анализа, геометрии |
| work_keys_str_mv |
AT borisenkoaa asymptoticpropertiesofhilbertgeometry AT olinea asymptoticpropertiesofhilbertgeometry |
| first_indexed |
2025-11-28T01:07:42Z |
| last_indexed |
2025-11-28T01:07:42Z |
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1849994318491680768 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 3, pp. 327�345
Asymptotic Properties of Hilbert Geometry
A.A. Borisenko and E.A. Olin
Department of Mechanics and Mathematics, V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:borisenk@univer.kharkov.ua
evolin@mail.ru
Received November 2, 2007
It is shown that the spheres in Hilbert geometry have the same volume
growth entropy as those in the Lobachevsky space. Asymptotic estimates
for the ratio of the volume of metric ball to the area of the metric sphere
in Hilbert geometry are given. Derived estimates agree with the well-known
fact in the Lobachevsky space.
Key words: Hilbert geometry, Finsler geometry, balls, spheres, volume,
area, entropy.
Mathematics Subject Classi�cation 2000: 53C60, 58B20, 52A20.
1. Introduction
Hilbert geometry is a generalization of the Klein model of the Lobachevsky
space. The absolute there is an arbitrary convex hypersurface unlike an ellipsoid
in the Lobachevsky space. Hilbert geometries are simply connected, projectively
�at, complete reversible Finsler spaces of the constant negative �ag curvature �1.
In [12], B. Colbois and P. Verovic proved that the balls in an (n+1)-dimensional
Hilbert geometry have the same volume growth entropy as those in H n+1 , namely
n. We obtain an analogous result for the spheres in Hilbert geometry.
Theorem 1. Consider an (n+1)-dimensional Hilbert geometry associated with
a bounded open convex domain U � R
n+1 whose boundary is a C3 hypersurface
with positive normal curvatures. Then we have
lim
t!1
ln(Vol(Snt ))
t
= n:
The second author was partially supported by the Akhiezer Foundation.
c
A.A. Borisenko and E.A. Olin, 2008
A.A. Borisenko and E.A. Olin
It is known [4�7] that in the Lobachevsky space H n+1 of constant curvature
�1 for a family of metric balls fBn+1
t
gt2R+ the following equality holds:
lim
�!1
Vol(Bn+1
� )
Vol(Sn� )
=
1
n
:
The same ratio in a more general case for �- and h-convex hypersurfaces in
Hadamard manifolds was considered in [4, 6, 7] by A.A. Borisenko, V. Miquel,
A. Reventos and E. Gallego.
The similar estimates in Finsler spaces were derived in [5] (see also [16]).
Theorem [5]. Let (Mn+1; F ) be an (n + 1)-dimensional Finsler�Hadamard
manifold that satis�es the following conditions:
1. the �ag curvature satis�es the inequalities �k22 6 K 6 �k21, k1; k2 > 0;
2. the S-curvature satis�es the inequalities nÆ1 6 S 6 nÆ2 such that Æi < ki:
Then for a family fBn+1
r (p)gr>0 we have
1
n(k2 � Æ2)
6 lim
r!1
inf
Vol(Bn+1
r (p))
Area(Snr (p))
6 lim
r!1
sup
Vol(Bn+1
r (p))
Area(Snr (p))
6
1
n(k1 � Æ1)
:
Our goal is to prove an analogous result in Hilbert geometry for a family
fBn+1
t
gt2R+ . We can not apply the theorem from [5] because the S-curvature in
Hilbert geometry is di�cult to calculate.
As a result the following theorem is obtained.
Theorem 2. Consider an (n+1)-dimensional Hilbert geometry associated with
a bounded open convex domain U 2 R
n+1 whose boundary is a C3 hypersurface
with positive normal curvatures. Fix a point o 2 U , we assume this point to be
the origin and center of all balls considered. Denote by !(u) : Sn! R+ the radial
function for @U , i.e., the mapping !(u)u, where u 2 S
n is a parametrization of
@U , and by � : Rn+1 ! S
n the mapping such that �(p) =
up
jjupjj
, where up is the
radius-vector of a point p.
Denote by K and k the maximum and minimum normal curvatures of @U ,
c = maxu2Sn
!(u)
!(�u)
, !0 = minu2Sn !(u), !1 = maxu2Sn!(u). Then we have
lim
�!1
sup
Vol(Bn+1
� )
Vol(Sn� )
6
1
n
c
n
2
�
K
k
�n
2 1
(k!0)
n
2
+1
R
Sn
!(u)
n
2 duR
@U
!(�(p))�
n
2 dp
;
lim
�!1
inf
Vol(Bn+1
� )
Vol(Sn� )
>
1
n
1
c
n
2
�
k
K
�n
2
(k!0)
n
2
R
Sn
!(u)
n
2 duR
@U
!(�(p))�
n
2 dp
;
328 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
or, more simple expressions
lim
�!1
sup
Vol(Bn+1
� )
Vol(Sn� )
6
1
n
�
K
k
�n
2
�
!1
!0
�n+1 �!1
k
�n
2 1
k!1
VolE(S
n)
VolE(@U)
;
lim
�!1
inf
Vol(Bn+1
� )
Vol(Sn� )
>
1
n
�
k
K
�n
2
�
!0
!1
�n
2
!n0 (k!0)
n
2
VolE(S
n)
VolE(@U)
:
If U is a symmetric domain with respect to o, then we have
lim
�!1
sup
Vol(Bn+1
� )
Vol(Sn� )
6
1
n
c
n
2
�
K
k
�n
2 !n1
(k!0)
n
2
+1
VolE(S
n)
VolE(@U)
;
lim
�!1
inf
Vol(Bn+1
� )
Vol(Sn� )
>
1
n
1
c
n
2
�
k
K
�n
2
(k!0)
n
2 !n0
VolE(S
n)
VolE(@U)
:
Notice that in this theorem the ratio of the volume of the ball to the internal
volume of the sphere is considered, unlike in theorem [5], where the induced volume
is used.
2. Preliminaries
2.1. Finsler geometry
In this section we recall some basic facts and theorems from Finsler geometry
that we need. See [16] for details.
Let Mn be an n-dimensional connected C1-manifold. Denote by TMn =F
x2Mn TxM
n the tangent bundle of Mn, where TxM
n is the tangent space at x.
The Finsler metric on Mn is a function F : TMn ! [0;1) with the following
properties:
1. F 2 C1(TMnnf0g);
2. F is positively homogeneous of degree one, i.e., for any pair (x; y) 2 TMn
and any � > 0, F (x; �y) = �F (x; y);
3. for any pair (x; y) 2 TMn the following bilinear symmetric form gy : TxM
n�
TxM
n ! R is positively de�nite
gy(u; v) :=
1
2
@2
@t@s
[F 2(x; y + su+ tv)]js=t=0:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 329
A.A. Borisenko and E.A. Olin
The pair (Mn; F ) is called a Finsler manifold.
If we introduce the functions
gij(x; y) =
1
2
@2
@yi@yj
[F 2(x; y)];
then one can rewrite the form gy(u; v) as
gy(u; v) = gij(x; y)u
ivj :
For any �xed vector �eld Y de�ned on the subset U �Mn, gY (u; v) is a Rie-
mannian metric on U .
Given a Finsler metric F on a manifold Mn. For a smooth curve c : [a; b] !
Mn the length is de�ned by the integral
LF (c) =
bZ
a
F (c(t); _c(t))dt =
bZ
a
q
g _c(t)( _c(t); _c(t))dt:
Let feigni=1 be an arbitrary basis for TxM
n and f�ign
i=1 be a dual basis for
T �xM
n. Consider the set Bn
F (x)
=
�
(yi) 2 R
n : F (x; yiei) < 1
� TxM
n. Denote
by VolE(A) the Euclidean volume of A. Then de�ne the form
dVF = �F (x)�
1 ^ : : : ^ �n;
here
�F (x) :=
VolE(B
n)
VolE(B
n
F (x)
)
; (1)
and B
n is the unit ball in R
n .
The volume form dVF determines a regular measure VolF =
R
dVF and is
called the Busemann�Hausdor� volume form.
For any Riemannian metric g(u; v) = gij(x)u
ivj the Busemann�Hausdor�
volume form is the standard Riemannian volume form
dVg =
q
det(gij)�
1 ^ : : : ^ �n:
In [9] it was proved that the Busemann�Hausdor� measure for reversible met-
ric coincides with the n-dimensional outer Hausdor� measure. Recall that
the n-dimensional outer Hausdor� measure of a set A is de�ned by
�n = lim
r!0
�n;r;
�n;r = VolE(B
n) inf
X
i
�ni : 2�i < r;A �
[
i
B[xi; �i]; xi 2 A
!
:
330 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
It should be noticed here that if we calculate the Hausdor� measure for the
submanifold in a Finsler manifold with the symmetric metric, then we will obtain
the internal volume on submanifold in the metric induced from the ambient space.
But, unfortunately, the using of this volume implies certain di�culties. In our
case, when we consider the sphere as a submanifold, the following claim does not
hold
Vol(Bn
r ) =
rZ
0
Vol(Sn�1
t
)dt
if we use the internal volume. For details, see [16].
2.2. Hilbert geometry
Consider a bounded open convex domain U � R
n+1 whose boundary is a C3
hypersurface with positive normal curvatures in R
n equipped with a Euclidean
norm k � k.
For given two distinct points p and q in U , let p1 and q1 be the corresponding
intersection points of the hal�ines p + R�(q � p) and p + R+(q � p) with @U
(Fig. 1).
P1
Q1
P
Q
Fig. 1: Hilbert metric
Then consider the following distance function:
dU (p; q) =
1
2
ln
kq � q1k
kq � p1k
�
kp� p1k
kp� q1k
; (2)
dU (p; p) = 0:
The obtained metric space (U; dU ) is called Hilbert geometry and is a complete
noncompact geodesic metric space with the Rn -topology and in which the a�ne
open segments joining two points are geodesics [10].
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 331
A.A. Borisenko and E.A. Olin
The distance function is naturally associated with the Finsler metric FU on U .
For a point p 2 U and a tangent vector v 2 TpU = R
n
FU (p; v) =
1
2
kvk
�
1
kp� p�k
+
1
kp� p+k
�
; (3)
where p� and p+ are the intersection points of the half-lines p+R�v and p+R+v
with @U .
Then dU (p; q) = inf
R
I
FU (c(t); _c(t))dt when c(t) ranges over all smooth curves
joining p and q.
In is known (see for example [16]) that Hilbert metrics are the metrics of
constant �ag curvature �1.
When U = Bn
r , then we obtain the Klein model of the n-dimensional
Lobachevsky space H n , and the Finsler metric has the explicit expression
FBn
r
(p; v) =
s
kvk2
r � kpk2
+
< v; p >2
(r2 � kpk2)2
: (4)
In [10] it is proved that the balls of arbitrary radii are convex sets in Hilbert
geometry.
The asymptotic properties of Hilbert geometry have been obtained lately. All
these properties mean that Hilbert geometry is "almost" Riemannian at in�nity.
It is proved in [12] that Hilbert metric "tends" to Riemannian metric as follows.
Theorem [12]. Let C 2 R
n be a bounded open convex domain whose boundary
@C is a hypersurface of class C3 that is strictly convex. For any p 2 C, let Æ(p) > 0
be the Euclidean distance from p to @C. Then there exists a family (~lp)p2C of linear
transformations in R
n such that
lim
Æ(p)!0
FC(p; v)
k~lp(v)k
= 1
uniformly in v 2 R
nnf0g:
This means that the unit sphere in the tangent space of given Hilbert metric
tends to ellipsoid in continuous topology as the tangent point goes to the absolute.
3. Calculating the Volume Growth Entropy of Spheres
In this section we will prove that for an (n+1)-dimensional Hilbert geometry
lim
t!1
ln(Vol(Snt ))
t
= n:
Consider a bounded open convex domain U � R
n+1 whose boundary is a C3
hypersurface with positive normal curvatures in Rn .
332 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
Fix a point o 2 U , we assume this point to be the origin and center of all
balls considered. Denote by !(u) : Sn ! R+ the radial function for @U , i.e., the
mapping !(u)u, where u 2 S
n is a parametrization of @U . Let Bn+1
r (o) be the
metric ball of radius r centered at a point o, Snr (o) = @Bn+1
r (o) be the metric
sphere.
We will use the following lemma that shows the order of growth of Hilbert
distance from the sphere to @U in terms of Euclidean distance. We also estimate
the deviation of tangent and normal vectors to sphere from those to @U .
Lemma 1. Let !(u)u : Sn! R+ be the parametrization of @U , �t(u) : S
n!
R+ be the parametrization of the sphere of radius t.
Then, as t!1:
1. !(u)� �t(u) = �(u)e�2t + �o(e�2t);
�(u) = !(u)
�
!(u)
!(�u)
+ 1
�
;
2. !0
i
(u)� �0
t;i
(u) = �i(u)e
�2t + �o(e�2t);
�i(u) =
"
!0i(u)
�
2
!(u)
!(�u)
+ 1
�
+
�
!(u)
!(�u)
�2
!0i(�u)
#
;
3. !00
ij
(u)� �00
t;ij
(u) = �ij(u)e
�2t + �o(e�2t);
!(u)3�ij(u) = !(u)2[2!0i(�u)!
0
j(�u)� !(�u)!00ij(�u)]
+!(�u)2[2!0i(u)!
0
j(u) + !(�u)!00ij(u)]
+2!(�u)!(u)[!0j(�u)!
0
i(u) + !0i(�u)!
0
j(u)]:
P r o o f o f L e m m a 1. We are going to obtain the explicit expression
for �t(u). Let q = 0 be the center of the sphere, p be a point on the sphere. Using
formula (3), we obtain the equation on the function �t(u)
1
2
ln
�
!(u)
!(�u)
�
!(�u) + �t(u)
!(u)� �t(u)
�
= t:
By direct computation we have
�t(u) =
!(�u)!(u)(e2t � 1)
!(u) + !(�u)e2t
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 333
A.A. Borisenko and E.A. Olin
1. Consider the di�erence
!(u)� �t(u) = !(u)�
!(�u)!(u)(e2t � 1)
!(u) + !(�u)e2t
=
!2(u) + !(�u)!(u)
!(u) + !(�u)e2t
= !(u)
�
!(u)
!(�u)
+ 1
�
e�2t + �o(e�2t); t!1:
2. Analogously, we obtain
!0(u)� �0t;i(u)
=
!0
i
(u)!(�u)2e2t + 2e2t!(u)!(�u)!0
i
(u) + !(u)2(!0
i
(u) + !0
i
(�u)(e2t � 1))
(!(u) + !(�u)e2t)2
=
"
!0i(u)
�
2
!(u)
!(�u)
+ 1
�
+
�
!(u)
!(�u)
�2
!0i(�u)
#
e�2t + �o(e�2t); t!1:
3. It can be proved in the same manner.
Denote the minimum and maximum Euclidean normal curvatures of @U by k
and K, respectively.
We also use the notation !0 = minu2Sn !(u), !1 = maxu2Sn !(u).
The following lemma gives the estimates on the angle between the radial and
normal directions at points from @U .
Lemma 2. For a given point m = !(um)um 2 @U denote by N(m) the
normal vector at m. Then
cos\(um; N(m)) >
!0
R
:
P r o o f o f L e m m a 2. This lemma follows from a more general
theorem.
Theorem [4, 6, 7]. Let N be a hypersurface in a Riemannian manifold M .
Consider N as de�ned by the equation t = �(�) of class C2, where �(�) is the
distance to point o. N can be seen as the 0-level set of the function F = t � �.
For given point P 2 N we consider all the vectors to be attached at P . Consider
Y = gradN�
kgradN�k
. Let x, that is orthogonal to the radial direction, be the unit vector
in the plane spanned on y and on the radial direction. Let ' be the angle between
the normal direction and the radial direction at point P 2 N .
If kn is the normal curvature at P in the direction given by Y , �n is the normal
curvature in the direction of x of the sphere centered at o of radius �, and d'
ds
is
the derivative of ' with respect to the arc parameter of the integral curve of Y
by P , then
kn = �n cos'+
d'
ds
:
334 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
Now we can prove Lem. 2.
Consider any integral curve
of y
kyk
. Since the angle ' takes its value in the
interval [0; �=2], then there is a supremum '0 of it. If at some point
(s0) the
value '0 is achieved, then at this point we have '0 = 0 and
cos' =
kn
�n
:
The minimum possible value of kn is equal to k = 1
R
, and the maximum
possible value of �n is equal to 1
!0
. Hence we have
cos' =
kn
�n
>
!0
R
:
And Lemma 2 follows.
P r o o f o f T h e o r e m 1. Now we are going to estimate the volume
of a sphere Snt in Hilbert geometry. The idea of proof is to obtain the Hausdor�
measure of this sphere. It follows from the reversibility of Hilbert metrics that
the Hausdor� measure coincides with the Finslerian Busemann�Hausdor� volume
[9].
Fix a point p on the sphere Snr . Since the spheres are convex, we can choose
the vector u 2 S
n such that p = �t(u). More generally, for a given origin
o 2 R
n+1 denote the corresponding radius vector by up and consider the function
� : Rn+1 ! S
n such that �(p) =
up
jjupjj
. Then we can write that p = �t(�(u)).
Denote the point !(�(p))�(p) 2 @U by m. Consider the vector vm which is
tangent to @U at m, the vector nm which is orthogonal to vm with respect to the
Euclidean inner product such that the point o belongs to the plane P spanned on
vm and nm. Let km be the curvature of the section of @U by P at m. Consider
a special coordinate system in the plane P such that the axis z is directed as
nm and the axis x is directed as vm. Then, in this special coordinate system the
section of @U can be locally expressed as
z(x) =
1
2
kmx
2 + �o(x2); x! 0:
Later on we will work with this section.
Draw the secant of the sphere that is parallel to the tangent vector at p (Fig. 2).
Put d = jja1�a2jj, Æ(p) = jjm�pjj, Æ1 = jjb1�f jj, Æ2 = jjf�b2jj, h = jjp�f jj.
Let us estimate the function Æ(p). From Lemma 1 we have
Æ(p) 6 !(u)
�
!(u)
!(�u)
+ 1
�
e�2t + �o(e�2t); t!1: (5)
From the triangle pm1m we have that Æ(p) t cos\(um; vm)(!(u) � �t(u)).
Finally, using Lem. 2, we obtain Æ(p) > !0
R
(!(u)� �t(u)):
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 335
A.A. Borisenko and E.A. Olin
b1
a1
a2
b2
f f1
p
mm1 x
z
O
SrU
Fig. 2: Proof of Theorem 1
Consequently,
Æ(p) >
!0
R
!(u)
�
!(u)
!(�u)
+ 1
�
e�2t + �o(e�2t); t!1: (6)
Then we estimate Æ1 and Æ2. Let
z = a(p)x+ Æ(p) + h
be the equation of the secant in a special coordinates system. We suppose h to
decrease faster than Æ(p). Thus in further computations h will be neglected.
We �nd the intersection points of this line with the boundary @U . From the
expression for the boundary we have
a(p)x+ Æ(p) =
1
2
kmx
2:
Thus
x1;2 =
a(p)�
p
a(p)2 + 2kmÆ(p)
km
:
It follows from Lem. 1 that a(p) = a0(u)e
�2t; t!1, for some function a0(u)
and, consequently, a(p) = O(Æ(p)); Æ(p) ! 0. Therefore we have
x1;2 = �
s
2Æ(p)
km
+ �o(
p
Æ(p)); Æ(p) ! 0;
z1;2 =
1
2
kmx
2 + �o(x2)jx=x1;2 =
1
2
Æ(p) + �o(Æ(p)); Æ(p) ! 0
336 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
and
Æ1 =
q
x21 + (z1 � Æ(p))2 =
s
2Æ(p)
km
+ �o(Æ(p)) = Æ2; Æ(p) ! 0:
Therefore, the turning of the tangent, as the point goes to @U , does not
in�uence the asymptotic behavior of Æi.
Compute the Hilbert length of segment a1a2. Denote it by dU . Then, as
h! 0,
dU t
1
2
ln
�
d+ Æ1
Æ2
�
d+ Æ2
Æ1
�
=
1
2
ln
��
d
Æ2
+
Æ1
Æ2
�
�
�
d
Æ1
+
Æ2
Æ1
��
t
1
2
�
d
Æ1
+
d
Æ2
�
t
d
p
kmp
2(Æ(p) + h)
+ �o(
p
1=Æ(p)); Æ(p) ! 0:
We are showing now that the limit of ratio of dU to the Finslerian length
~dU of the geodesic arc a1a2 is equal to 1 as the arc is subtended to a point.
Specialize the coordinate system on R
n+1 so as a1 = 0. Let w(t) : [0; T ] ! U
be a parametrization of the arc. Then the segment from point a1 = 0 to point
a2 = w(t) can be parameterized by v(s) = s
t
w(t) : [0; t] ! U . Calculate the
lengths of v and w
~dU =
tZ
0
FU (w(s); _w(s))ds;
dU =
tZ
0
FU (v(s); _v(s))ds =
tZ
0
FU
�
s
t
w(t);
1
t
w(t)
�
ds:
From the intermediate-value theorem for integrals we have
~dU =
tZ
0
FU (w(s); _w(s))ds = tFU (w(s0); _w(s0)); s0 2 [0; t];
dU =
tZ
0
FU
�
s
t
w(t);
1
t
w(t)
�
ds = tFU
�
s1
t
w(t);
1
t
w(t)
�
; s1 2 [0; t]:
Now we subtend the arc to a point, i.e., let t ! 0. Then s0; s1 ! 0, and we
obtain
~dU
dU
=
tFU (w(s0); _w(s0))
tFU
�
s1
t
w(t); 1
t
w(t)
� �! FU (0; _w(0))
FU (0; _w(0))
= 1:
And the statement is proved.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 337
A.A. Borisenko and E.A. Olin
Now our goal is to calculate the Hausdor� measure of sphere Snr . Denote
by Æ0(r) the Hausdor� distance from the points of sphere to the absolute @U .
Consider a covering fBig of the sphere Snr by balls with diameters ~di centered at
points pi 2 Snr . Denote by ki the normal curvature of @U that corresponds to the
i-th sphere from the covering (as above). As we saw, we can replace ~di by the
lengths of corresponding chords di of the sphere S
n
r .
Then the Hausdor� measure and, consequently, the Finslerian Busemann�
Hausdor� measure are given by
Vol(Snt ) = VolE(B
n) inf
dBi
X
i
di
p
kip
2(Æ(pi) + h)
!n
+ �o(
p
1=Æ0(t)n); Æ0(t)! 0;
where in�mum is calculated over all coverings of the sphere Snr .
Our metric sphere Snr is su�ciently smooth, so we can proceed to the integral
over Snr
Vol(Snt ) = VolE(B
n) inf
dBi
X
i
di
p
kip
2(Æ(pi) + h)
!n
+ �o(
p
1=Æ0(t)n)
= inf
dBi
X
i
p
kip
2(Æ(pi) + h)
!n
VolE(Bi) + �o(
p
1=Æ0(t)n); Æ0(t)! 0:
Denote by dp the area element of Snt . Proceeding to integral and estimating
lead to
Vol(Snt ) > k
n
2
Z
Sn
t
�
1
2Æ(p)
�n
2
dp+ �o(
p
1=Æ0(t)n); Æ0(t)! 0;
Vol(Snt ) 6 K
n
2
Z
Sn
t
�
1
2Æ(p)
�n
2
dp+ �o(
p
1=Æ0(t)n); Æ0(t)! 0:
The using of the explicit estimates (5), (6) for Æ(p) results
Vol(Snt ) > k
n
2
Z
Sn
t
�
2!(�(p))
�
!(�(p))
!(��(p))
+ 1
���n
2
dp � ent + �o(ent); t!1; (7)
Vol(Snt ) 6 K
n
2
Z
Sn
t
�
2
!0
R
!(�(p))
�
!(�(p))
!(��(p))
+ 1
���n
2
dp�ent+�o(ent); t!1: (8)
And Theorem 1 follows. �
338 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
4. Estimation of Ratio of the Volume of the Ball to the Volume
of the Sphere
Here we will �nd the asymptotic behavior of volume of the metric ball Bn+1
�
in Hilbert geometry. Further there will be used the method introduced in [13]
where we improve some necessary estimates.
The volume of a metric ball is given by the integral
Vol(Bn+1
� ) =
Z
B
n+1
�
�(p)dp:
Here �(p) is the Busemann�Hausdor� volume form. And the volume estimating
problem is reduced to the estimating of the volume form. Recall (1) that
�(p) := �FU (p) =
VolE(B
n)
VolE(B
n
FU (p)
)
:
Thus we have to estimate the volume of the unit sphere in the tangent space at
point p 2 U .
We will use the following simple lemma.
Lemma 3. There exists a value �0 such that for any points p 2 U in the
neighborhood d(p; @U) 6 �0 there exists a unique point �(p) 2 @U : d(p; �(p)) =
d(p; @U)
Put m = �(p) 2 @U . The minimum and maximum Euclidean normal curva-
tures of @U denote by k and K, respectively. Then, at any point m 2 @U the
tangent sphere of radius R := 1
k
contains U , the tangent sphere of radius r := 1
K
is contained in U [2]. On two tangent spheres of the radii r and R at this point
we construct corresponding Klein metrics Fr and FR. We can give the explicit
expressions (4) for them.
Then the following inequalities hold:
VolE
�
Bn+1
Fr(p)
�
6 VolE
�
Bn+1
FU (p)
�
6 VolE
�
Bn+1
FR(p)
�
: (9)
As it was shown in [13]
VolE
�
Bn+1
FR(p)
�
= VolE(B
n+1)Rn+1
(
1�
�
1�
d(p;m)
R
�2
)n+2
2
;
VolE
�
Bn+1
Fr(p)
�
= VolE(B
n+1)rn+1
(
1�
�
1�
d(p;m)
r
�2
)n+2
2
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 339
A.A. Borisenko and E.A. Olin
Thus, we have
1
Rn+1
�
1�
�
1� d(p;m)
R
�2�n+2
2
6 �(p) 6
1
rn+1
�
1�
�
1� d(p;m)
r
�2�n+2
2
: (10)
Consider the mapping
�(u; s) = tanh(s)!(u)u : Sn� R �! U:
In [13] it was shown that the mapping �(u; s) satis�es the following properties:
1. �(Sn; [0; � � c]) � Bn+1
� � �(Sn; [0; � + 1]), where c = supu2Sn
!(u)
!(�u)
:
Hence,
Vol(�(Sn; [0; � � c])) 6 Vol(Bn+1
� ) 6 Vol(�(Sn; [0; � + 1])):
2. jJac(�(u; s))j = !(u)n+1 tanhn(s)(1 � tanh2(s)):
We improve the �rst property.
Fix d > 0. Consider the di�erence
�t(u)� !(u) tanh(t+ d) = !(u)
�
1�
!(u) + !(�u)
e2t!(�u) + !(u)
� tanh(t+ d)
�
= !(u)
�
2
e2(t+d) + 1
�
!(u) + !(�u)
e2t!(�u) + !(u)
�
= �t(u)� !(u) tanh(t+ d) = !(u)e�2t
�
2e�2d � 1�
!(u)
!(�u)
�
+ �o(e�2t); t!1:
Thus Bn+1
� � �(Sn; [0; � + d]) for su�ciently large � if
2e�2d � 1�
!(u)
!(�u)
6 0;
d > �
1
2
ln
�
1
2
�
1 +
1
c
��
:= d1;
and Bn+1
� � �(Sn; [0; � + d]) for su�ciently large �nite � if
d 6 �
1
2
ln
�
1
2
(1 + c)
�
:= d2:
Fix the values d2, d1 and choose a su�ciently large �0.
340 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
Then
Vol(�(Sn; [0; � + d2])) 6 Vol(Bn+1
� ) 6 Vol(�(Sn; [0; � + d1])): (11)
Notice that if the domain U is centrally symmetric, then d1 = d2 = 0. In the
worst case, when c ! 1, we have d1 ! ln
p
2 � 0:347 < 1. Inclusion (11) is
more precise than it was obtained in [13]. It will be essentially used in the proof
of Th. 2.
The volume of the set �(Sn; [�0; �]) is given by
Vol(�(Sn; [�0; �])) = VolE(B
n)
Z
Sn
�Z
�0
�(�(u; s))jJac(�(u; s))jdsdu:
It is known [13] that
jJac(�(u; s))j = !(u)n+1 tanhn(s)(1 � tanh2(s)) = !(u)n+1
4e2s
�
e2s�1
e2s+1
�n+1
e4s � 1
:
And, using the estimates (10), we obtain
Z
Sn
�Z
�0
4!(u)n+1
e2s
�
e
2s
�1
e2s+1
�n+1
e4s�1
Rn+1
�
1�
�
1� d(�(u;s);@U)
R
�2�n+2
2
dsdu 6 Vol(�(Sn; [�0; �]));
Vol(�(Sn; [�0; �])) 6
Z
Sn
�Z
�0
4!(u)n+1
e
2s
�
e
2s
�1
e2s+1
�n+1
e4s�1
rn+1
�
1�
�
1� d(�(u;s);@U)
r
�2�n+2
2
dsdu:
Out next task is to �nd the asymptotic behavior of the integral
rZ
0
4e2s
�
e
2s
�1
e2s+1
�n+1
e4s�1
(1� (1� Ce�2s)2)
n+2
2
ds:
After the changing of the variable y = e�2s, we obtain the integral
1Z
e�2r
�8
y�2
�
y
�1
�1
y�1+1
�
n+1
y�2�1
(1� (1� Cy)2)
n+2
2
dy =
1Z
e�2r
8(1� y)n+1
(1 + y)n+1(y2 � 1)(Cy(2� Cy))
n+2
2
dy
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 341
A.A. Borisenko and E.A. Olin
=
1Z
e�2r
1
C
n+2
2 y
n+2
2 2
n
2
�2
�
(1� y)�
n
2 2
n+2
2
(1 + y)n+1(y2 � 1)(2� Cy)
n+2
2
dy:
Notice that
lim
y!0
"
(1� y)�
n
2 2
n+2
2
(1 + y)n+1(y2 � 1)(2 � Cy)
n+2
2
#
= �1:
Taking into account the above and making the inverse change of
variable, we get
rZ
0
4e2s
�
e
2s
�1
e2s+1
�n+1
e4s�1
(1� (1� Ce�2s)2)
n+2
2
ds =
1
nC
n+2
2 2
n�2
2
enr + �o(enr); r !1: (12)
The expression for VolE
�
Bn+1
FR(p)
�
includes the quantity d(p;m) = d(p; @U).
Thus we need the estimates of d(p;m) for point p = �(u; s). So,
d(�(u; s); !(u)u) = !(u)� tanh(s)!(u) = !(u)�
e2s � 1
e2s + 1
!(u) =
2!(u)
1 + e2s
:
Finally,
d(�(u; s); @U) 6 2!(u)e�2s + �o(e�2s): (13)
On the other hand, analogously as formula (6) we get
d(�(u; s); @U) > 2
!0
R
!(u)e�2s + �o(e�2s): (14)
Using (12)�(14), one can compute that
1
n
C1e
n� + �o(en�) 6 Vol(�(Sn; [�0; �])) 6
1
n
C2e
n� + �o(en�); �!1; (15)
C1 =
1
2n
Z
Sn
�
!(u)
R
�n
2
du;
C2 =
1
2n
R
n+2
2
!
n+2
2
0
Z
Sn
�
!(u)
r
�n
2
du:
And, taking into account (11), (15), we have
1
n
C1e
nd2en� + �o(en�) 6 Vol(Bn+1
� ) 6
1
n
C2e
n�end1 + �o(en�); �!1: (16)
342 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3
Asymptotic Properties of Hilbert Geometry
P r o o f o f T h e o r e m 2. It follows from (6), (7), (16) that
lim
�!1
sup
Vol(Bn+1
� )
Vol(Sn� )
6
1
n
1
2n=2
end1
R
n+2
2
!
n+2
2
0
R
Sn
�
!(u)
r
�n
2
du
k
n
2
R
@U
�
!(�(p))
�
!(�(p))
!(��(p))
+ 1
���n
2
dp
6
1
n
c
n
2
�
K
k
�n
2 1
(k!0)
n
2
+1
R
Sn
!(u)
n
2 duR
@U
!(�(p))�
n
2 dp
6
1
n
c
n
2
�
K
k
�n
2 !n1
(k!0)
n
2
+1
VolE(S
n)
VolE(@U)
:
Note that c 6 !1
!0
. Hence
lim
�!1
sup
Vol(Bn+1
� )
Vol(Sn� )
6
1
n
�
K
k
�n
2
�
!1
!0
�n+1 �!1
k
�n
2 1
k!1
VolE(S
n)
VolE(@U)
;
lim
�!1
inf
Vol(Bn+1
� )
Vol(Sn� )
>
1
n
1
2n=2
end2
R
Sn
�
!(u)
R
�n
2
du
K
n
2
R
@U
�
!0
R
!(�(p))
�
!(�(p))
!(��(p))
+ 1
���n
2
dp
>
1
n
1
c
n
2
�
k
K
�n
2
(k!0)
n
2
R
Sn
!(u)
n
2 duR
@U
!(�(p))�
n
2 dp
>
1
n
1
c
n
2
�
k
K
�n
2
!n0 (k!0)
n
2
VolE(S
n)
VolE(@U)
>
1
n
�
k
K
�n
2
�
!0
!1
�n
2
!n0 (k!0)
n
2
VolE(S
n)
VolE(@U)
:
And the theorem follows. �
E x a m p l e 1. Let U = B
n+1
� . Then we get the Klein model of Lobachevsky
space. Applying Theorem 2 to this space, we get
!(u) =
1
k
=
1
K
= r = R = !0 = �;
c = 1;Z
@U
du = �nVolE(S
n):
Therefore we have obtained the well-known result
lim
�!1
Vol(Bn+1
� )
Vol(Sn� )
=
1
n
:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 3 343
A.A. Borisenko and E.A. Olin
E x a m p l e 2. One should not hope that this result holds for all metrics
of negative curvature.
Let U be an open bounded strongly convex domain in Rn , o = 0 2 R
n . Given
a point x 2 U and a direction y 2 TxUnf0g ' Unf0g. The Funk metric F (x; y)
is a Finsler metric that satis�es the condition
x+
y
F (x; y)
2 @U:
Then Hilbert metric is a symmetrized Funk metric
FU (x; y) =
1
2
[F (x; y) + F (x;�y)] :
Funk metrics are of constant negative curvature �1=4, but for these metrics
lim
r!1
Vol(Bn+1
r )
Vol(Snr )
=1
holds ([5]).
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