Mass and polyhedra in asymptotically hyperbolic manifolds
aa r X i v : . [ m a t h . DG ] F e b MASS AND POLYHEDRA IN ASYMPTOTICALLY HYPERBOLICMANIFOLDS
XIAOXIANG CHAI
Abstract.
Using the upper half space model, we evaluate a component ofthe hyperbolic mass functional evaluated on a special family of polyhedraextending a formula of Miao-Piubello. Introduction
We say that a three dimensional manifold (
M, g ) is an asymptotically hyperbolicmanifold if outside a compact set M is diffeomorphic to the standard hyperbolicspace ( H , ¯ g ) minus a geodesic ball and | e | ¯ g + | ¯ ∇ e | ¯ g + | ¯ ∇ ¯ ∇ e | ¯ g = O (e − τr ) , where ¯ ∇ is the connection on H , r is the distance function to a fixed point o and τ > . We fix o to be the point (0 , ,
1) in the upper half space model¯ g = x ) ((d x ) + (d x ) + (d x ) ) . Then 2 cosh r = x (( x ) + ( x ) + ( x ) + 1) . See [BP92, Chapter A]. We assume in this article that M is diffeomorphic to H .Using the upper half space coordinates, we say that ∆ is a polyhedron if each ofits faces is a plane in the Euclidean sense whatever the metric the upper half spacecarries.Let V = x and F be a face of the polyhedron, let ¯ ν be the ¯ g -normal pointingoutward of ∆ := ∆ q , we see that ¯ ν = x n a i ∂ i where a i are constants and a is a vector of length one under the Euclidean metric.Due to conformality of ¯ g to the Euclidean metric δ , faces of ∆ meets at constantangles. Easily,(1) ∂ ¯ ν V = x a i ∂ i ( x n ) = − a x = − a V. Note that a ∈ [ − ,
1] and the case of a = ± | a | <
1, this faces lies in a so-calledequidistant hypersurface. Suppose that the face lies in a plane which intersect the x axis at x = z , then following [Cha21], the vector X = x − z ∂ n is tangent tothe face and div( x − z ∂ n ) = z V . We calculate the second fundamental form ofeach face under the metric b . Pick a local coordinate y α where α = 1 ,
2. Using the conformality to δ , ¯ A αβ = h ∂ α , ∂ β i δ x ∂ a x ) = − a x ) h ∂ α , ∂ β i δ = − a h ∂ α , ∂ β i b . (2)Each face F is umbilic and the mean curvature is then ¯ H = − a . Euclideanspheres are also umbilic in the hyperbolic metric, however, they do not satisfy thecondition (1). Obviously, these discussions works for higher dimensionThe mass integrand for an asymptotically hyperbolic manifold (See [CH03]) is U = V div e − V d(tr b e ) + tr b e d V − e ( ¯ ∇ V, · ) . We assume that for a family of polyhedra indexed by q the mass satisfies thefollowing M ( V ) = Z ∂ ∆ q U i ¯ ν i d¯ σ + o (1) , where ¯ ν is the b -normal to the face of ∆ q and d¯ σ is the two dimensional volumeelement.Such a family is easy to find. For example, according to [CH03] or [Mic11], ifeach polyhedron of the family is enclosed by a geodesic sphere and enclose anothergeodesic sphere, radius of each sphere goes to infinity as ∆ q exhaust the manifold M , then such a family provides an example. Let E q be the all edges of ∆ q , α bethe dihedral angle for by neighboring faces along E q . We denote by d v , d σ andd λ respectively the three, two and one dimensional volume element. We put a barover a letter to indicate the quantity is calculated with respect to the backgroundmetric ¯ g . Theorem 1.
Assuming each dihedral angle satisfies the bound sin ¯ α > c > , thenthe mass M ( V ) is M ( V )= − Z ∂ ∆ q V ( H − ¯ H )d¯ σ + 2 Z E q V ( α − ¯ α )d¯ λ + Z ∂ ∆ q O (cosh − τ +1 r )d¯ σ + Z E q O (cosh − τ +1 r )d¯ λ + o (1) . (3)As suggested by Miao [Mia20], this type of formula may be used to promote theGromov dihedral rigidity [Gro18] to an integrated form. However, in the asymp-totically flat case, one can perturb graphically of a face and use Taylor expansionto the third order to see a counterexample.Also, the Miao-Piubello type formula (3) leads us to do comparison between aRiemannian polyhedron with the polyhedron whose faces are realized as Euclideanplanes in the upper half space model. In particular, one could consider the Gromovtype dihedral rigidity. The case when the model is a cone type polyhedron withbase faces lying on a horosphere is possible using the method developed by [Li20].Instead of minimal surface with capillary angle condition, one uses constant meancurvature two surfaces with capillary angle condition. ASS AND POLYHEDRA IN ASYMPTOTICALLY HYPERBOLIC MANIFOLDS 3 Example and proof
Before we show the proof of Theorem 1, we calculate an easy example otherthan the parabolic cylinder [JM21] to illustrate that the terms R ∂ ∆ q cosh − τ +1 r d¯ σ and R E q cosh − τ +1 r d¯ λ can be o (1). We assume the base face B is a regular n -sidepolygon lies at the horosphere { x = ε } and centered at (0 , , ε ) for ε small, theapex is (0 , , ε − ). The distance between the vertex of the polygon to the x -axisis ρ ( ε ). We assume that ρ ( ε ) = o ( ε − τ )and ρ ( ε ) → ∞ as ε →
0. We use aliases x = x , y = x , z = x for the coordinates.First, we deal with integrals on the edges. There are two type edges. By rota-tional symmetry, we can assume one edge is E = { (0 , − εz − ε ρ, z ) : z ∈ [ ε, ε − ] } . The length element d¯ λ on E is z − s ε ρ (1 − ε ) d z √ z − max { Cερ, } since we are take ε →
0. So Z E cosh − τ +1 r d¯ λ C max { ερ, } Z ε − ε z − τ ( z + 1 + ( − εz − ε ρ ) ) − τ +1 d z. We divide the integral into three parts. On [ ε, { Cερ, } Z ε z − τ ( z + 1 + ( − εz − ε ρ ) ) − τ +1 d z max { Cερ, } ( ε ρ ) − τ +2 Z ε z − τ d z = o (1)(4)since τ > . On [1 , ε − ], for 1 < a < τ − { Cερ, } Z ε − / z − τ ( z + 1 + ( − εz − ε ρ ) ) − τ +1 d z = max { Cερ, } Z ε − / z − τ − (2 τ − a ) ( z + 1 + ( − εz − ε ρ ) ) − τ +1+ 12 (2 τ − a ) d z C max { Cερ, } ρ − τ +2+2 τ − a Z ε − / z − a d z C max { Cερ, } ρ − τ + a (5) XIAOXIANG CHAI which is o (1) as ρ ( ε ) → ∞ . On [ ε − , ε − ], we have thatmax { Cερ, } Z ε − ε − / z − τ ( z + 1 + ( − εz − ε ρ ) ) − τ +1 d z max { Cερ, } Z ε − ε − / z − τ − τ +2 d z max { Cερ, } ε τ − . Requiring that ρ ( ε ) = o ( ε − τ ) this term is o (1). By rotational symmetry again, theother type edges we can just consider E = { ( ρ cos πn , y, ε ) : y ∈ [ − ρ sin πn , ρ sin πn ] } . The integral of cosh − τ +1 r on E is Z E cosh − τ +1 r d¯ λ C Z ρ sin πn − ρ sin πn ε − · ε − τ ( y + ρ cos πn + ε + 1) − τ +1 d y Cε − τ Z R ( x + 1) − τ +1 d y = o (1)as ε → F of ∆ lying on { z = ε } , we have Z F cosh − τ +1 r d¯ v C Z F ε − − τ ( x + y + ε + 1) − τ +1 d x d y Cε τ − Z ρ ( ε )0 ( s + 1) − τ +1 s d s = o (1)as ε →
0. Now we consider a side face, we use the symbol S . We use the Euclideandistance ξ of a point in side edges to the z -axis and z to parametrized S . First, ξ = − εz − ε ρ , then S = { ( ξ cos πn , y, z ) : y ∈ [ − ξ sin πn , ξ sin πn ] , z ∈ [ ε, ε − ] } . It is easy to show that Z S cosh − τ +1 r d¯ σ = q ε ρ (1 − ε ) τ − Z ε − ε Z ξ sin πn − ξ sin πn z τ − ( ξ cos πn + y + 1 + z ) − τ +1 d y d z max { Cερ, C } Z ε − ε ξ sin πn z τ − ( ξ cos πn + 1 + z ) − τ +1 d z Similar to (4) and (5), we have thatmax { Cερ, C } ( Z ε + Z ε − / ) = o (1) . ASS AND POLYHEDRA IN ASYMPTOTICALLY HYPERBOLIC MANIFOLDS 5
For the integral over [ ε − / , ε − ], we absorb ξ sin πn by ξ cos πn + z + 1 andmax { Cερ, C } Z ε − ε − / ξ sin πn z τ − ( ξ cos πn + 1 + z ) − τ +1 d z max { Cερ, C } Z ε − ε − / z τ − ( ξ cos πn + 1 + z ) − τ +1+ 12 d z max { Cερ, C } Z ε − ε − / z τ − − τ +2+1 d z max { Cερ, C } ε τ − = o (1) . where we have required that ρ ( ε ) = o ( ε − τ ). Proof of Theorem 1.
We have on each face F of ∆ q ,(6) 2 V ( H − ¯ H ) = − U i ¯ ν i − div F ( V X ) + O (e − τr + r )where X is the vector field dual to the 1-form e (¯ ν, · ) with respect to the metric b | F .This is an easy consequence of [JM21, (2.5)] that U (¯ ν )=2 V ( ¯ H − H ) − div F ( V X ) + [(tr b e − e (¯ ν, ¯ ν )) h d V, ¯ ν i − V h ¯ A, e i b ] + O (e − τr + r ) . From (1) and (2), we obtain the desired formula (6). From (6), we have that Z ∂ ∆ q U i ¯ ν i d¯ σ = Z ∂ ∆ q [ − V ( H − ¯ H ) − div ∂ ∆ q ( V X )]d¯ σ + o (1)On each face F , using divergence theorem Z F div F ( V X )d¯ σ = Z ∂F V e (¯ ν, ¯ n )d¯ λ, where ¯ n is the b -normal to ∂F in F . On the edge F A ∩ F B , the contribution is Z F A ∩ F B V [ e (¯ ν A , ¯ n A ) + e (¯ ν B , ¯ n B )]d¯ λ. Let g ij = g ( ∂ i , ∂ j ), we have that ε ij := ( x n ) e ij = ( x n ) g ij − ( x n ) ¯ g ij = ( x n ) ¯ g ij − δ ij = O (e − τr )The g -normal to the face F A is then expressed as ν A = g ij a j ∂ j p g kl a k a l . We write cos θ = g ( ν A , ν B )=( a i b j g ij )( g kl a k a l ) − ( g pq a p a q ) − = a i b j g ij ( x n ) ( g kl ( x n ) a k a l ) − ( g pq ( x n ) a p a q ) − . XIAOXIANG CHAI
Up to here, it follows from the same lines as in [MP21, (3.9)-(3.30)] to show that Z F A ∩ F B V [ e (¯ ν A , ¯ n A ) + e (¯ ν B , ¯ n B )]d¯ λ = Z F A ∩ F B V (¯ α − α ) + Z F A ∩ F B V cosh − τ r d¯ λ. Therefore, M ( V )= − Z ∂ ∆ q V ( H − ¯ H )d¯ σ + 2 Z E q V ( α − ¯ α )d¯ λ + Z ∂ ∆ q cosh − τ +1 r d¯ σ + Z E q cosh − τ +1 r d¯ λ + o (1) , obtaining the theorem. (cid:3) References [BP92] Riccardo Benedetti and Carlo Petronio.
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