Asymptotic Properties of Hilbert Geometry
aa r X i v : . [ m a t h . DG ] N ov Asymptotic Properties of Hilbert Geometry
Alexandr A. BORISENKO and Eugeny A. OLIN ∗
23 April 2007
Geometry Department, Mech.-Math. Faculty, Kharkov NationalUniversity, Pl. Svoboda 4, 310077-Kharkov, Ukraine.
E-mail: [email protected], [email protected]
Abstract
We show that the spheres in Hilbert geometry have the same vol-ume growth entropy as those in the Lobachevsky space. We give theasymptotic estimates for the ratio of the volume of metric ball to thearea of the metric sphere in Hilbert geometry. Derived estimates agreewith the well-known fact in the Lobachevsky space.
Key words:
Hilbert geometry, Finsler geometry, balls, spheres, volume,area, entropy.
Mathematical Subject Classifications (2000):
Hilbert geometry is the generalization of the Klein model of the Lobachevskyspace. The absolute there is an arbitrary convex hypersurface unlike anellipsoid in the Lobachevsky space. Hilbert geometries are simply connected,projectively flat, complete reversible Finsler spaces of constant negative flagcurvature − n + 1)-dimensional Hilbert geometry have the same volume growth entropy as thosein H n +1 , namely n . We obtain the analogous result for the spheres in Hilbertgeometry. Theorem 1.
Consider an ( n + 1) -dimensional Hilbert geometry asso-ciated with a bounded open convex domain U ⊂ R n +1 whose boundary is a C hypersurface with positive normal curvatures. Then we have lim t →∞ ln( Vol ( S nt )) t = n ∗ The second author was partially supported by the Akhiezer Foundation
1t is known [4, 5, 6, 7] that in the Lobachevsky space H n +1 of constantcurvature − { B n +1 t } t ∈ R + the following equalityholds lim ρ →∞ Vol ( B n +1 ρ ) Vol ( S nρ ) = 1 n Such a ratio in a more general case for λ - and h -convex hypersurfacesin Hadamard manifolds was considered in [4, 6, 7] by A. A. Borisenko, V.Miquel, A. Reventos and E. Gallego.Similar estimates in Finsler spaces were derived in [5] (see also [16]). Theorem [5].
Let ( M n +1 , F ) be an ( n +1) -dimensional Finsler-Hadamardmanifold that satisfies the following conditions:1. Flag curvature satisfies the inequalities − k K − k , k , k > ,2. S -curvature satisfies the inequalities nδ S nδ such that δ i < k i . Then for a family { B n +1 r ( p ) } r > we have n ( k − δ ) lim r →∞ inf Vol ( B n +1 r ( p )) Area ( S nr ( p )) lim r →∞ sup Vol ( B n +1 r ( p )) Area ( S nr ( p )) n ( k − δ ) . Our goal is to prove analogous result in Hilbert geometry for a family { B n +1 t } t ∈ R + . Applying the theorem from [5] is the rather difficult taskbecause the S -curvature in Hilbert geometry is difficult to calculate.As the result the following theorem is obtained. Theorem 2.
Consider an ( n + 1) -dimensional Hilbert geometry asso-ciated with a bounded open convex domain U ∈ R n +1 whose boundary is a C hypersurface with positive normal curvatures. Fix a point o ∈ U , wewill consider this point as the origin and the center of all the consideredballs. Denote by ω ( u ) : S n → R + the radial function for ∂U , i. e. themapping ω ( u ) u , u ∈ S n is a parametrization of ∂U , and by ι : R n +1 → S n the mapping such that ι ( p ) = u p || u p || , u p is the radius-vector of a point p .Denote by K and k the maximum and minimum normal curvature of ∂U , c = max u ∈ S n ω ( u ) ω ( − u ) , ω = min u ∈ S n ω ( u ) , ω = max u ∈ S n ω ( u ) . Then wehave lim ρ →∞ sup Vol ( B n +1 ρ ) Vol ( S nρ ) n c n (cid:18) Kk (cid:19) n kω ) n +1 R S n ω ( u ) n du R ∂U ω ( ι ( p )) − n dp lim ρ →∞ inf Vol ( B n +1 ρ ) Vol ( S nρ ) > n c n (cid:18) kK (cid:19) n ( kω ) n R S n ω ( u ) n du R ∂U ω ( ι ( p )) − n dp or, more simple expression ρ →∞ sup Vol ( B n +1 ρ ) Vol ( S nρ ) n (cid:18) Kk (cid:19) n (cid:18) ω ω (cid:19) n +1 (cid:16) ω k (cid:17) n kω Vol E ( S n ) Vol E ( ∂U )lim ρ →∞ inf Vol ( B n +1 ρ ) Vol ( S nρ ) > n (cid:18) kK (cid:19) n (cid:18) ω ω (cid:19) n ω n ( kω ) n Vol E ( S n ) Vol E ( ∂U ) If U is a symmetric domain with respect to o then we have lim ρ →∞ sup Vol ( B n +1 ρ ) Vol ( S nρ ) n c n (cid:18) Kk (cid:19) n ω n ( kω ) n +1 Vol E ( S n ) Vol E ( ∂U )lim ρ →∞ inf Vol ( B n +1 ρ ) Vol ( S nρ ) > n c n (cid:18) kK (cid:19) n ( kω ) n ω n Vol E ( S n ) Vol E ( ∂U )Notice that in this theorem the ratio of the volume of the ball to the internal volume of the sphere is considered, unlike theorem [5], where the induced volume is used. In this section we recall some basic facts and theorems from Finsler geometrythat we need. See [16] for details.Let M n be an n -dimensional connected C ∞ -manifold. Denote by T M n = F x ∈ M n T x M n the tangent bundle of M n , where T x M n is the tangent spaceat x . A Finsler metric on M n is a function F : T M n → [0 , ∞ ) with thefollowing properties:1. F ∈ C ∞ ( T M n \{ } );2. F is positively homogeneous of degree one, i. e. for any pair ( x, y ) ∈ T M n and any λ > F ( x, λy ) = λF ( x, y );3. For any pair ( x, y ) ∈ T M n the following bilinear symmetric form g y : T x M n × T x M n → R is positively definite, g y ( u, v ) := 12 ∂ ∂t∂s [ F ( x, y + su + tv )] | s = t =0 The pair ( M n , F ) is called a Finsler manifold .If we denote by g ij ( x, y ) = 12 ∂ ∂y i ∂y j [ F ( x, y )] , g y ( u, v ) as g y ( u, v ) = g ij ( x, y ) u i v j For any fixed vector field Y defined on the subset U ⊂ M n , g Y ( u, v ) isa Riemannian metric on U .Given a Finsler metric F on a manifold M n . For a smooth curve c :[ a, b ] → M n the length is defined by the integral L F ( c ) = Z ba F ( c ( t ) , ˙ c ( t )) dt = Z ba q g ˙ c ( t ) ( ˙ c ( t ) , ˙ c ( t )) dt. Let { e i } ni =1 be an arbitrary basis for T x M n and { θ i } ni =1 the dual basisfor T ∗ x M n . Consider the set B nF ( x ) = (cid:8) ( y i ) ∈ R n : F ( x, y i e i ) < (cid:9) ⊂ T x M n .Denote by Vol E ( A ) the Euclidean volume of A . Then define the form dV F = σ F ( x ) θ ∧ ... ∧ θ n , here σ F ( x ) := Vol E ( B n ) Vol E ( B nF ( x ) ) . (1)and B n is the unit ball in R n .The volume form dV F determines a regular measure Vol F = R dV F andis called the Busemann-Hausdorff volume form .For any Riemannian metric g ( u, v ) = g ij ( x ) u i v j the Busemann-Hausdorffvolume form is the standard Riemannian volume form dV g = q det( g ij ) θ ∧ ... ∧ θ n . It was proved in [9] that the Busemann-Hausdorff measure for reversiblemetric coincides with the n -dimensional outer Hausdorff measure. Recallthat the n -dimensional outer Hausdorff measure of a set A is defined by ν n = lim r → ν n,r ,ν n,r = Vol E ( B n ) inf X i ρ ni : 2 ρ i < r, A ⊆ [ i B [ x i , ρ i ] , x i ∈ A ! It should be noticed here that if we calculate the Hausdorff measure forthe submanifold in a Finsler manifold with the symmetric metric then wewill obtain the internal volume on submanifold in the metric induced fromthe ambient space. But unfortunately using of this volume implies certaindifficulties. In our case when we consider a sphere as the submanifold thefollowing claim does not hold
Vol ( B nr ) = Z r Vol ( S n − t ) dt, if we use the internal volume. For details, see [16].4 .2 Hilbert geometry Consider a bounded open convex domain U ⊂ R n +1 whose boundary isa C hypersurface with positive normal curvatures in R n equipped with aEuclidean norm k · k .For given two distinct points p and q in U , let p and q be the corre-sponding intersection point of the half line p + R − ( q − p ) and p + R + ( q − p )with ∂U (Fig. 1). Figure 1: Hilbert metricThen consider the following distance function. d U ( p, q ) = 12 ln k q − q kk q − p k × k p − p kk p − q k (2) d U ( p, p ) = 0The obtained metric space ( U, d U ) is called Hilbert geometry and is a com-plete noncompact geodesic metric space with the R n -topology and in whichthe affine open segments joining two points are geodesics [10].The distance function is associated in a natural way with the Finslermetric F U on U . For a point p ∈ U and a tangent vector v ∈ T p U = R n F U ( p, v ) = 12 k v k (cid:18) k p − p − k + 1 k p − p + k (cid:19) (3)where p − and p + is the intersection point of the half-lines p + R − v and p + R + v with ∂U .Then d U ( p, q ) = inf R I F U ( c ( t ) , ˙ c ( t )) dt when c ( t ) ranges over all smoothcurves joining p to q .In is known (see for example [16]) that Hilbert metrics are the metricsof constant flag curvature −
1. 5hen U = B nr then we obtain the Klein model of the n -dimensionalLobachevsky space H n and the Finsler metric has the explicit expression F B nr ( p, v ) = s k v k r − k p k + < v, p > ( r − k p k ) (4)It is proved in [10] that the balls of arbitrary radii are convex sets inHilbert geometry.The asymptotic properties of Hilbert geometry have been obtained lately.All this properties mean that Hilbert geometry is ”almost” Riemannian atinfinity. It is proved in [12] that Hilbert metric ”tends” to riemannian metricas follows. Theorem [12].
Let
C ∈ R n be a bounded open convex domain whoseboundary ∂ C is a hypersurface of class C that is strictly convex. For any p ∈ C let δ ( p ) > be the Euclidean distance from p to ∂ C . Then there existsa family ( ~l p ) p ∈C of linear transformations in R n such that lim δ ( p ) → F C ( p, v ) k ~l p ( v ) k = 1 uniformly in v ∈ R n \{ } This means that the unit sphere in the tangent space of given Hilbertmetric tends to ellipsoid in continuous topology as the tangent point goesto the absolute.
In this section we will prove that for an ( n +1)-dimensional Hilbert geometrylim t →∞ ln( Vol ( S nt )) t = n, as it is in H n +1 .Consider a bounded open convex domain U ⊂ R n +1 whose boundary isa C hypersurface with positive normal curvatures in R n .Fix a point o ∈ U , we will consider this point as the origin and thecenter of all the considered balls. Denote by ω ( u ) : S n → R + the radialfunction for ∂U , i. e. the mapping ω ( u ) u , u ∈ S n is a parametrizationof ∂U . Let B n +1 r ( o ) be the metric ball of radius r centered at a point o , S nr ( o ) = ∂B n +1 r ( o ) be the metric sphere.We will use the following lemma that shows the order of growth of theHilbert distance from the sphere to ∂U in terms of the Euclidean distance.We also estimate the deviation of the tangent and normal vectors to spherefrom those to ∂U . 6 emma 1. Let ω ( u ) u : S n → R + be the parametrization of ∂U , ρ t ( u ) : S n → R + – the parametrization of the sphere of radius t .Then, as t → ∞ : ω ( u ) − ρ t ( u ) = ∆( u ) e − t + ¯ o ( e − t );∆( u ) = ω ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) ω ′ i ( u ) − ρ ′ t,i ( u ) = ∆ i ( u ) e − t + ¯ o ( e − t );∆ i ( u ) = " ω ′ i ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) + (cid:18) ω ( u ) ω ( − u ) (cid:19) ω ′ i ( − u ) ω ′′ ij ( u ) − ρ ′′ t,ij ( u ) = ∆ ij ( u ) e − t + ¯ o ( e − t ) ,ω ( u ) ∆ ij ( u ) = ω ( u ) [2 ω ′ i ( − u ) ω ′ j ( − u ) − ω ( − u ) ω ′′ ij ( − u )]+ ω ( − u ) [2 ω ′ i ( u ) ω ′ j ( u )+ ω ( − u ) ω ′′ ij ( u )]++2 ω ( − u ) ω ( u )[ ω ′ j ( − u ) ω ′ i ( u ) + ω ′ i ( − u ) ω ′ j ( u )]P r o o f o f l e m m a 1. We are going to obtain the explicit expressionfor ρ 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 ρ t ( u )12 ln (cid:20) ω ( u ) ω ( − u ) × ω ( − u ) + ρ t ( u ) ω ( u ) − ρ t ( u ) (cid:21) = t By the direct computation we have ρ t ( u ) = ω ( − u ) ω ( u )( e t − ω ( u ) + ω ( − u ) e t
1. Consider the difference ω ( u ) − ρ t ( u ) = ω ( u ) − ω ( − u ) ω ( u )( e t − ω ( u ) + ω ( − u ) e t == ω ( u ) + ω ( − u ) ω ( u ) ω ( u ) + ω ( − u ) e t = ω ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) e − t + ¯ o ( e − t ) , t → ∞
2. We obtain analogously ω ′ ( u ) − ρ ′ t,i ( u ) == ω ′ i ( u ) ω ( − u ) e t + 2 e t ω ( u ) ω ( − u ) ω ′ i ( u ) + ω ( u ) ( ω ′ i ( u ) + ω ′ i ( − u )( e t − ω ( u ) + ω ( − u ) e t ) == " ω ′ i ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) + (cid:18) ω ( u ) ω ( − u ) (cid:19) ω ′ i ( − u ) e − t + ¯ o ( e − t ) , t → ∞
7. It can be proved in the same manner. (cid:3)
Denote by k and K the minimum and maximum Euclidean normal cur-vatures of ∂U .We also use the notations ω = min u ∈ S n ω ( u ), ω = max u ∈ S n ω ( u ).The following lemma gives the estimates on the angle between the radialand normal directions at the points from ∂U . Lemma 2.
For a given point m = ω ( u m ) u m ∈ ∂U denote by N ( m ) thenormal vector at m . Then cos ∠ ( u m , N ( m )) > ω R P r o o f o f l e m m a 2. This lemma follows from the more generaltheorem.
Theorem [4, 6, 7].
Let N be a hypersurface in a Riemannian manifold M . Consider N as defined by the the equation t = ρ ( θ ) of class C , where ρ ( θ ) is the distance to a point o . N can be seen as the 0-level set of thefunction F = t − ρ . For given point P ∈ N we consider all the vectors tobe attached at P . Denote by Y = grad N ρ k grad N ρ k . Let x be a unit vector in theplane spanned on y and the radial direction that is orthogonal to the radialdirection . Let ϕ be the angle between the normal direction and the radialdirection at the point P ∈ N .If k n is the normal curvature at P in the direction given by Y , µ n is thenormal curvature in the direction of x of the sphere centered at o of radius ρ and dϕds is the derivative of ϕ with respect to the arc parameter of the integralcurve of Y by P then k n = µ n cos ϕ + dϕds Now we can prove lemma 2.Consider any integral curve γ of y k y k . Since the angle ϕ takes its valuein the interval [0 , π/
2] then there is a supremum ϕ of it. If at some point γ ( s ) the value ϕ is achieved then we have at this point ϕ ′ = 0 andcos ϕ = k n µ n The minimum possible value of k n is equal to k = R , and the maximumpossible value of µ n is equal to ω . Hence we havecos ϕ = k n µ n > ω R And lemma 2 follows. (cid:3)
P r o o f o f t h e o r e m 1. Now we are going to estimate thevolume of a sphere S nt in Hilbert geometry. The idea of proof is to obtainthe Hausdorff measure of this sphere. It follows from the reversibility of8ilbert metrics that the Hausdorff measure coincides with the FinslerianBusemann-Hausdorff volume [9].Fix the point p on the sphere S nr . Since the spheres are convex we canchoose the vector u ∈ S n such that p = ρ t ( u ). More generally, for a givenorigin o ∈ R n +1 denote by u p the corresponding radius vector and considerthe function ι : R n +1 → S n such that ι ( p ) = u p || u p || . Then we can write that p = ρ t ( ι ( u )).Denote by m the point ω ( ι ( p )) ι ( p ) ∈ ∂U . Consider the vector v m whichis tangent to ∂U at m , the vector n m which is orthogonal to v m with respectto the Euclidean inner product such that the point o belongs to the plane P spanned on v m and n m . Let k m be the curvature of the section of ∂U by P at m . Consider the special coordinate system in the plane P : let the axe z be directed as n m , and the axe x be directed as v m . Then in this specialcoordinate system the section of ∂U can be locally expressed as z ( x ) = 12 k m x + ¯ o ( x ) , x → p (Fig. 2). Figure 2:Put d = || a − a || , δ ( p ) = || m − p || , δ = || b − f || , δ = || f − b || , h = || p − f || . 9et us estimate the function δ ( p ). From lemma 1 we have δ ( p ) ω ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) e − t + ¯ o ( e − t ) , t → ∞ (5)From the triangle pm m we have that δ ( p ) ≈ cos ∠ ( u m , v m )( ω ( u ) − ρ t ( u )).Finally, using lemma 2 we obtain δ ( p ) > ω R ( ω ( u ) − ρ t ( u ))Consequently, δ ( p ) > ω R ω ( u ) (cid:18) ω ( u ) ω ( − u ) + 1 (cid:19) e − t + ¯ o ( e − t ) , t → ∞ (6)Then we estimate δ and δ . Let z = a ( p ) x + δ ( p ) + h be the equation of the secant in the special coordinates system. We will thinkat once that h decreases faster than δ ( p ). Thus in the further computationswe will neglect h .We find the intersection points of this line with the boundary ∂U . Fromthe expression for the boundary we have a ( p ) x + δ ( p ) = 12 k m x . Thus x , = a ( p ) ± p a ( p ) + 2 k m δ ( p ) k m It follows from lemma 1 that a ( p ) = a ( u ) e − t , t → ∞ , for some function a ( u ) and consequently a ( p ) = O( δ ( p )) , δ ( p ) →
0. Therefore we have x , = ± s δ ( p ) k m + ¯ o ( p δ ( p )) , δ ( p ) → z , = 12 k m x + ¯ o ( x ) | x = x , = 12 δ ( p ) + ¯ o ( δ ( p )) , δ ( p ) → δ = q x + ( z − δ ( p )) = s δ ( p ) k m + ¯ o ( δ ( p )) = δ , δ ( p ) → δ i .10ompute the Hilbert length of the segment a a . Denote it by d U . Thenas h → d U ≈
12 ln (cid:20) d + δ δ × d + δ δ (cid:21) = 12 ln (cid:20)(cid:18) dδ + δ δ (cid:19) × (cid:18) dδ + δ δ (cid:19)(cid:21) ≈≈ (cid:18) dδ + dδ (cid:19) ≈ d √ k m p δ ( p ) + h ) + ¯ o ( p /δ ( p )) , δ ( p ) → d U to the Finslerianlength ˜ d U of the geodesic arc a a is equal to 1 as the arc is subtendedto a point. Specialize the coordinate system on R n +1 so as a = 0. Let w ( t ) : [0 , T ] → U be a parametrization of the arc. Then the segment fromthe point a = 0 to the point a = w ( t ) can be parameterized by v ( s ) = st w ( t ) : [0 , t ] → U . Calculate lengths of v and w .˜ d U = Z t F U ( w ( s ) , ˙ w ( s )) dsd U = Z t F U ( v ( s ) , ˙ v ( s )) ds = Z t F U (cid:18) st w ( t ) , t w ( t ) (cid:19) ds. From the intermediate-value theorem for integrals we have˜ d U = Z t F U ( w ( s ) , ˙ w ( s )) ds = tF U ( w ( s ) , ˙ w ( s )) , s ∈ [0 , t ] d U = Z t F U (cid:18) st w ( t ) , t w ( t ) (cid:19) ds = tF U (cid:18) s t w ( t ) , t w ( t ) (cid:19) , s ∈ [0 , t ]Now we are subtending the arc to a point, i. e. let t →
0. Then s , s →
0, and we obtain:˜ d U d U = tF U ( w ( s ) , ˙ w ( s )) tF U (cid:0) s t w ( t ) , t w ( t ) (cid:1) −→ F U (0 , ˙ w (0)) F U (0 , ˙ w (0)) = 1And the statement is proved.Now our goal is to calculate the Hausdorff measure of the sphere S nr .Denote by δ ( r ) the Hausdorff distance from the points of the sphere tothe absolute ∂U . Consider a covering { B i } of the sphere S nr by balls ofdiameters ˜ d i centered at points p i ∈ S nr . Denote by k i the normal curvatureof ∂U that corresponds to the i -th sphere from the covering (as above). Aswe saw, we can replace ˜ d i by the lengths of the corresponding chords d i ofthe sphere S nr .Then the Hausdorff measure, and, consequently, the Finslerian Busemann-Hausdorff measure is given by Vol ( S nt ) = Vol E ( B n ) inf d Bi X i d i √ k i p δ ( p i ) + h ) ! n + ¯ o ( p /δ ( t ) n ) , δ ( t ) → , S nr .Our metric sphere S nr is sufficiently smooth, so we can proceed to theintegral over S nr . Vol ( S nt ) = Vol E ( B n ) inf d Bi X i d i √ k i p δ ( p i ) + h ) ! n + ¯ o ( p /δ ( t ) n ) == inf d Bi X i √ k i p δ ( p i ) + h ) ! n Vol E ( B i ) + ¯ o ( p /δ ( t ) n ) , δ ( t ) → , Denote by dp the area element of S nt . Proceeding to integral and esti-mating leads to Vol ( S nt ) > k n Z S nt (cid:18) δ ( p ) (cid:19) n dp + ¯ o ( p /δ ( t ) n ) , δ ( t ) → Vol ( S nt ) K n Z S nt (cid:18) δ ( p ) (cid:19) n dp + ¯ o ( p /δ ( t ) n ) , δ ( t ) → δ ( p ) as the result we have Vol ( S nt ) > k n Z S nt (cid:18) ω ( ι ( p )) (cid:18) ω ( ι ( p )) ω ( − ι ( p )) + 1 (cid:19)(cid:19) − n dp · e nt + ¯ o ( e nt ) , t → ∞ (7) Vol ( S nt ) K n Z S nt (cid:18) ω R ω ( ι ( p )) (cid:18) ω ( ι ( p )) ω ( − ι ( p )) + 1 (cid:19)(cid:19) − n dp · e nt +¯ o ( e nt ) , t → ∞ (8)And theorem 1 follows. (cid:4) Here we will find the asymptotic behavior of the volume of the metric ball B n +1 ρ in Hilbert geometry. We will use the method introduced in [13] inwhich some necessary estimates were improved.The volume of a metric ball is given by the integral Vol ( B n +1 ρ ) = Z B n +1 ρ σ ( p ) dp Here σ ( p ) is the Busemann-Hausdorff volume form. And the volume esti-mating problem is reduced to the estimating of the volume form. Recall (1)that σ ( p ) := σ F U ( p ) = Vol E ( B n ) Vol E ( B nF U ( p ) ) . p ∈ U .We will use the following simple lemma. Lemma 3.
There exists a value ρ such that for any points p ∈ U in the neighborhood d ( p, ∂U ) ρ there exist a unique point π ( p ) ∈ ∂U : d ( p, π ( p )) = d ( p, ∂U )Put m = π ( p ) ∈ ∂U . Denote by k and K the minimum and maximumEuclidean normal curvatures of ∂U . Then at any point m ∈ ∂U the tangentsphere of radius R := k contains U , the tangent sphere of radius r := K is contained in U [2]. On two tangent spheres of the radii r and R at thispoint we construct corresponding Klein metrics F r and F R . We can give theexplicit expressions (4) for them.Then the following inequalities hold Vol E (cid:16) B n +1 F r ( p ) (cid:17) Vol E (cid:16) B n +1 F U ( p ) (cid:17) Vol E (cid:16) B n +1 F R ( p ) (cid:17) (9)As it was shown in [13]: Vol E (cid:16) B n +1 F R ( p ) (cid:17) = Vol E ( B n +1 ) R n +1 ( − (cid:18) − d ( p, m ) R (cid:19) ) n +22 Vol E (cid:16) B n +1 F r ( p ) (cid:17) = Vol E ( B n +1 ) r n +1 ( − (cid:18) − d ( p, m ) r (cid:19) ) n +22 Thus, we have1 R n +1 (cid:26) − (cid:16) − d ( p,m ) R (cid:17) (cid:27) n +22 σ ( p ) r n +1 (cid:26) − (cid:16) − d ( p,m ) r (cid:17) (cid:27) n +22 (10)Consider the mappingΦ( u, s ) = tanh( s ) ω ( u ) u : S n × R −→ U It was shown in [13] that the mapping Φ( u, s ) satisfies the following proper-ties1. Φ( S n , [0 , ρ − c ]) ⊆ B n +1 ρ ⊆ Φ( S n , [0 , ρ + 1]) where c = sup u ∈ S n ω ( u ) ω ( − u ) Hence,
Vol (Φ( S n , [0 , ρ − c ])) Vol ( B n +1 ρ ) Vol (Φ( S n , [0 , ρ + 1]))2. | Jac (Φ( u, s )) | = ω ( u ) n +1 tanh n ( s )(1 − tanh ( s ))13e improve the first property.Fix d >
0. Consider the difference ρ t ( u ) − ω ( u ) tanh( t + d ) = ω ( u ) (cid:18) − ω ( u ) + ω ( − u ) e t ω ( − u ) + ω ( u ) − tanh( t + d ) (cid:19) == ω ( u ) (cid:18) e t + d ) + 1 − ω ( u ) + ω ( − u ) e t ω ( − u ) + ω ( u ) (cid:19) == ρ t ( u ) − ω ( u ) tanh( t + d ) = ω ( u ) e − t (cid:18) e − d − − ω ( u ) ω ( − u ) (cid:19) +¯ o ( e − t ) , t → ∞ Thus B n +1 ρ ⊆ Φ( S n , [0 , ρ + d ]) for sufficiently large ρ if2 e − d − − ω ( u ) ω ( − u ) d > −
12 ln (cid:20) (cid:18) c (cid:19)(cid:21) := d and B n +1 ρ ⊇ Φ( S n , [0 , ρ + d ]) for sufficiently large finite ρ if d −
12 ln (cid:20)
12 (1 + c ) (cid:21) := d Fix the values d and d and choose sufficiently large ρ .Then Vol (Φ( S n , [0 , ρ + d ])) Vol ( B n +1 ρ ) Vol (Φ( S n , [0 , ρ + d ])) (11)Notice that if the domain U is centrally-symmetric then d = d = 0.In the worst case when c → ∞ we have d → ln √ ≈ . <
1. Inclusion(11) is more precise than it was obtained in [13]. It will be essentially usedin the proof of theorem 2.The volume of the set Φ( S n , [ ρ , ρ ]) is given by. Vol (Φ( S n , [ ρ , ρ ])) = Vol E ( B n ) Z S n Z ρρ σ (Φ( u, s )) | Jac (Φ( u, s )) | dsdu It is known [13] that | J ac (Φ( u, s )) | = ω ( u ) n +1 tanh n ( s )(1 − tanh ( s )) = ω ( u ) n +1 e s (cid:16) e s − e s +1 (cid:17) n +1 e s − Z S n Z ρρ ω ( u ) n +1 e s “ e s − e s +1 ” n +1 e s − R n +1 (cid:18) − (cid:16) − d (Φ( u,s ) ,∂U ) R (cid:17) (cid:19) n +22 dsdu Vol (Φ( S n , [ ρ , ρ ]))14 ol (Φ( S n , [ ρ , ρ ])) Z S n Z ρρ ω ( u ) n +1 e s “ e s − e s +1 ” n +1 e s − r n +1 (cid:18) − (cid:16) − d (Φ( u,s ) ,∂U ) r (cid:17) (cid:19) n +22 dsdu Out next task is to find the asymptotic behavior of the integral Z r e s “ e s − e s +1 ” n +1 e s − (1 − (1 − Ce − s ) ) n +22 ds After the changing of the variable y = e − s , we obtain the integral Z e − r − y − „ y − − y − « n +1 y − − (1 − (1 − Cy ) ) n +22 dy = Z e − r − y ) n +1 (1 + y ) n +1 ( y − Cy (2 − Cy )) n +22 dy == Z e − r C n +22 y n +22 n − · (1 − y ) − n n +22 (1 + y ) n +1 ( y − − Cy ) n +22 dy Notice that lim y → " (1 − y ) − n n +22 (1 + y ) n +1 ( y − − Cy ) n +22 = − Z r e s “ e s − e s +1 ” n +1 e s − (1 − (1 − Ce − s ) ) n +22 ds = 1 nC n +22 n − e nr + ¯ o ( e nr ) , r → ∞ (12)The expression for Vol E (cid:16) B n +1 F R ( p ) (cid:17) includes the quantity d ( p, m ) = d ( p, ∂U ).Thus we need the estimates of d ( p, m ) for the point p = Φ( u, s ). So, d (Φ( u, s ) , ω ( u ) u ) = ω ( u ) − tanh( s ) ω ( u ) = ω ( u ) − e s − e s + 1 ω ( u ) = 2 ω ( u )1 + e s Finally, d (Φ( u, s ) , ∂U ) ω ( u ) e − s + ¯ o ( e − s ) (13)On the other hand analogously as formula (6) we get d (Φ( u, s ) , ∂U ) > ω R ω ( u ) e − s + ¯ o ( e − s ) (14)Using (12), (13), (14), one can compute that1 n C e nρ + ¯ o ( e nρ ) Vol (Φ( S n , [ ρ , ρ ])) n C e nρ + ¯ o ( e nρ ) , ρ → ∞ (15)15 = 12 n Z S n (cid:18) ω ( u ) R (cid:19) n du C = 12 n R n +22 ω n +22 Z S n (cid:18) ω ( u ) r (cid:19) n du And, taking into account (11), (15), we have1 n C e nd e nρ + ¯ o ( e nρ ) Vol ( B n +1 ρ ) n C e nρ e nd + ¯ o ( e nρ ) , ρ → ∞ (16)P r o o f o f t h e o r e m 2. It follows from (6), (7), (16) that:lim ρ →∞ sup Vol ( B n +1 ρ ) Vol ( S nρ ) n n/ e nd R n +22 ω n +22 R S n (cid:16) ω ( u ) r (cid:17) n duk n R ∂U (cid:16) ω ( ι ( p )) (cid:16) ω ( ι ( p )) ω ( − ι ( p )) + 1 (cid:17)(cid:17) − n dp n c n (cid:18) Kk (cid:19) n kω ) n +1 R S n ω ( u ) n du R ∂U ω ( ι ( p )) − n dp n c n (cid:18) Kk (cid:19) n ω n ( kω ) n +1 Vol E ( S n ) Vol E ( ∂U )Note that c ω ω . Hencelim ρ →∞ sup Vol ( B n +1 ρ ) Vol ( S nρ ) n (cid:18) Kk (cid:19) n (cid:18) ω ω (cid:19) n +1 (cid:16) ω k (cid:17) n kω Vol E ( S n ) Vol E ( ∂U )lim ρ →∞ inf Vol ( B n +1 ρ ) Vol ( S nρ ) > n n/ e nd R S n (cid:16) ω ( u ) R (cid:17) n duK n R ∂U (cid:16) ω R ω ( ι ( p )) (cid:16) ω ( ι ( p )) ω ( − ι ( p )) + 1 (cid:17)(cid:17) − n dp > n c n (cid:18) kK (cid:19) n ( kω ) n R S n ω ( u ) n du R ∂U ω ( ι ( p )) − n dp > n c n (cid:18) kK (cid:19) n ω n ( kω ) n Vol E ( S n ) Vol E ( ∂U ) > n (cid:18) kK (cid:19) n (cid:18) ω ω (cid:19) n ω n ( kω ) n Vol E ( S n ) Vol E ( ∂U )And the theorem follows. (cid:4) E x a m p l e 1. Let U = B n +1 ρ . Then we get the Klein model of theLobachevsky space. Applying theorem 2 to this space implies ω ( u ) = 1 k = 1 K = r = R = ω = ρ = 1 Z ∂U du = ρ n Vol E ( S n )Therefore we have obtained the well-known resultlim ρ →∞ Vol ( B n +1 ρ ) Vol ( S nρ ) = 1 n E x a m p l e 2. One should not hope that for all metrics of negativecurvature such result holds.Let U be a open bounded strongly convex domain in R n , o = 0 ∈ R n .Given a point x ∈ U and a direction y ∈ T x U \{ } ≃ U \{ } . The Funkmetric F ( x, y ) is a Finsler metric that satisfies the following condition x + yF ( x, y ) ∈ ∂U. Then Hilbert metric is a symmetrized Funk metric F U ( x, y ) = 12 [ F ( x, y ) + F ( x, − y )]Funk metrics are of constant negative curvature − /
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