Manifolds with many hyperbolic planes
aa r X i v : . [ m a t h . DG ] M a r Manifolds With Many Hyperbolic Planes
Samuel Lin and Benjamin Schmidt
Abstract.
We construct examples of complete Riemannian manifolds havingthe property that every geodesic lies in a totally geodesic hyperbolic plane.Despite the abundance of totally geodesic hyperbolic planes, these examplesare not locally homogenous.
1. Introduction
Hyperbolic n -space is the space H n = { ( x i ) ∈ R n | x n > } with the met-ric h n = x − n ( dx + · · · + dx n ) . Each tangent 2-plane to ( H n , h n ) exponentiatesto a totally-geodesic copy of ( H , h ). Similarly, hyperbolic surfaces abound in thenegatively curved locally symmetric manifolds. More precisely, these spaces satisfy: (P) : Each tangent vector to M is contained in a tangent -plane σ that exponenti-ates to an immersed totally-geodesic hyperbolic surface. We construct examples of manifolds with (P) that are not locally homogeneous.For n >
2, let M n = R × H n − = { ( t, ( x i )) | t ∈ R , ( x i ) ∈ H n − } . Main Construction.
For each ( a, b ) ∈ R × [ − , , M n admits a completeRiemannian metric h n,a,b satisfying (1) For fixed n > , the metrics h n,a,b depend smoothly on ( a, b ) . (2) There is a monomorphism
Isom( H n − , h n − ) → Isom( M n , h n,a,b ) . (3) Each tangent -plane to M n containing the tangent vector ∂∂t exponenti-ates to a totally-geodesic surface isometric to the hyperbolic plane. (4) The sectional curvature of a -plane σ at p = ( t, ( x i )) making angle θ ∈ [0 , π ] with ∂∂t is given by sec( σ ) = − − b − e a )((1 + b ) e t + (1 − b ) e − t ) sin ( θ ) . (5) For n > , ( M n , h n,a,b ) Riemannian covers a finite volume manifold ifand only if b + e a , or equivalently, if and only if ( M n , h n,a,b ) isisometric to ( H n , h n ) . Remark . Property (3) implies (P).
Remark . For n >
3, the Riemannian metric h n,a, has extremal sectionalcurvatures − − e a by (4); consequently, manifolds satisfying (P) can have cur-vatures pinched arbitrarily close to − rank rigidity [ ]-[ ], [ ]-[ ], [ ]-[ ], and in particular in manifolds of positive hyperbolic rank [
5, 6, 7, 8 ].A complete Riemannian manifold M has positive hyperbolic rank if along eachcomplete geodesic γ : R → M , there exists a parallel vector field V ( t ) such thatsec( ˙ γ, V )( t ) ≡ −
1. This hypothesis on geodesics is an infinitesimal analogue of (P).The negatively curved locally symmetric spaces, normalized so that − finite volume manifolds of higher hyper-bolic rank are all locally symmetric. There are no additional complete and finitevolume examples amongst manifolds having sec − ]. The same is known to betrue amongst the complete and finite volume non-positively curved Euclidean rankone manifolds with suitably pinched curvatures [ ]. Finally, complete and finitevolume three-dimensional manifolds of higher hyperbolic rank are real hyperbolicwithout any a priori sectional curvature bounds [ ].In contrast, it is known that complete infinite volume manifolds of higher hy-perbolic rank need not be locally symmetric . The negatively curved symmetricspaces are characterized amongst the homogeneous manifolds with positive hyper-bolic rank and sec − ]. Therein, a non-symmetric example is constructed.To our knowledge, the examples provided by the Main Construction are the firstthat are not locally homogenous.
2. Warping hyperbolic -space over the Euclidean line. Let t denote the Euclidean coordinate on R and r > H .Consider the foliation of H by the family of (Euclidean) upper-half semicir-cles with common center the origin in R . Each such semicircle, parameterizedappropriately, is an h -geodesic: For each r >
0, the map F r : R → H defined by F r ( t ) = ( r tanh( t ) , r sech( t )) parameterizes the semicircle through (0 , r )in the clockwise fashion as a unit-speed h -geodesic with the initial point (0 , r ).Define a diffeomorphism F : R × H → H by F ( t, r ) = F r ( t ). In the coordinates ( t, r ) on R × H , the metric F ∗ ( h ) is givenby F ∗ ( h ) = dt + cosh ( t ) h = dt + e t )) h . Hence, ( H , h ) is isometric to a warped product of the hyperbolic line over theEuclidean line.In fact, the hyperbolic plane H is isometric to many different warpings of H over R . The warping functions are described in terms of solutions to the secondorder differential equation(2.1) φ ′′ ( t ) + ( φ ′ ( t )) − . For a ∈ R and − b
1, the solution to (2.1) determined by the initialconditions φ (0) = − a and φ ′ (0) = b is given by(2.2) φ a,b ( t ) = ln( (1 + b ) e t + (1 − b ) e − t e a ) . Let M = R × H and let π : M → R be the first coordinate projection. Ignorethe slight abuse of notation in letting φ a,b also denote the function π ∗ ( φ a,b ) on M .For a and b as above, consider the warped product Riemannian metric h ,a,b = dt + e φ a,b h on M . In this notation, h , , = F ∗ ( h ) . Lemma . For each ( a, b ) ∈ R × [ − , , ( M , h ,a,b ) is isometric to thehyperbolic plane ( H , h ) . Proof.
The proof only uses the fact that φ a,b is a solution to (2.1). For thisreason, and to simplify notation, set h = h ,a,b and φ = φ a,b in the remainder ofthis proof. As ( R , dt ) and ( H , h ) are complete, so too is their warped product( M , h ) [ , Lemma 40, pg. 209]. Therefore, it suffices to prove that h has constantcurvature − − ,Proposition 42 (4), pg. 210], the sectional curvature at a point is given by − φ ′′ − ( φ ′ ) . (cid:3)
3. Main Construction: Warping hyperbolic ( n − -space over theEuclidean line. Fix n > a, b ) ∈ R × [ − , φ a,b be as in (2 . h n,a,b denotethe warped product Riemannian metric on M n = R × H n − defined by h n,a,b = dt + e φ a,b h n − . The metric h n,a,b is complete by [ , Lemma 40, pg. 209] We now consider theproperties (1)-(5) stated in the Main Construction. SAMUEL LIN AND BENJAMIN SCHMIDT
Property (1):
Property (1) is immediate since the warping functions φ a,b dependsmoothly on ( a, b ) ∈ R × [ − , Property (2):
Given F ∈ Isom( H n − , h n − ), define ¯ F : M n → M n by ¯ F ( t, ( x i )) =( t, F ( x i )) . Then ¯ F ∈ Isom( M n , h n,a,b ) and the mapIsom( H n − , h n − ) → Isom( M n , h n,a,b )defined by F ¯ F is a monomorphism. Property (3):
To find one such totally-geodesic hyperbolic plane, define R ∈ Isom( H n − , h n − ) by R ( x , . . . , x n − , x n − ) = ( − x , . . . , − x n − , x n − ) . As ¯ R is an isometry, its fixed point setΣ = { ( t, (0 , . . . , , x n − )) } ⊂ M n is a totally-geodesic surface. The map ( t, (0 , . . . , , x n − )) ( t, x n − ) defines anisometry between the induced metric on Σ and ( M , h ,a,b ). By Lemma 2.1, Σ isisometric to the hyperbolic plane. We conclude by showing that given an arbitrarytangent plane σ as in Property (3) there is an isometry of M n that carries a tangentplane to Σ to σ .Let π : M n → R denote the first coordinate projection and F ¯ t the fiber above¯ t ∈ R . For each p ∈ F ¯ t , let G p denote the set of tangent 2-planes to M n at p thatcontain the vector ∂∂t ( p ). Let X ¯ t = ∪ p ∈ F ¯ t G p . Let σ denote the tangent space to Σ at the point (¯ t, (0 , . . . , , σ ∈ X ¯ t . As Isom( H n − , h n − ) acts transitively on unit-tangent vectors to H n − ,Isom( M n , h n,a,b ) acts transitively on the set X ¯ t . Hence, there exists an isometry ¯ I of M n that carries σ to σ .As exp( σ ) = exp( d ¯ I ( σ )) = ¯ I (exp( σ )) = ¯ I (Σ) , exp( σ ) is a totally geodesic surface isometric to the hyperbolic plane. Property (4):
Let p = ( t, ( x i )) ∈ M and let v, w, ∂∂t be orthonormal vectors at p .Property (3) implies that(3.1) R ( ∂∂t , w, w, ∂∂t ) = − . By [ , Proposition 42 (5), pg. 210],(3.2) R ( v, w, w, v ) = − − b − e a )((1 + b ) e t + (1 − b ) e − t ) . By [ , Proposition 42 (3), pg. 210],(3.3) R ( ∂∂t , w ) v = 0 . ANIFOLDS WITH MANY HYPERBOLIC PLANES 5
Now fix a two dimensional subspace σ ⊂ T p M . Assume that σ makes angle θ ∈ [0 , π ] with ∂∂t . Then there exist unit length vectors v and w perpendicular to ∂∂t such that { ¯ v = cos( θ ) ∂∂t + sin( θ ) v, w } is an orthonormal basis of σ . By (3.1)-(3.3) and the symmetries of the curvaturetensor, sec( σ ) = R (¯ v, w, w, ¯ v )= cos ( θ ) R ( ∂∂t , w, w, ∂∂t ) + sin ( θ ) R ( v, w, w, v )= − − b − e a )((1 + b ) e t + (1 − b ) e − t ) sin ( θ ) . Property (5):
As ( M n , h n,a,b ) is simply connected and complete, it is isometricto ( H n , h n ) if and only if it has constant sectional curvatures −
1. By (4), this isequivalent to the parameters satisfying 1 = b + e a . As ( H n , h n ) Riemannian coversfinite volume manifolds, it remains to prove that when n > = b + e a ,( M n , h n,a,b ) does not Riemannian cover a finite volume manifold.Assume that 1 = b + e a . As isometries preserve sectional curvatures, Property(4) implies that each isometry F ∈ Isom( M n , h n,a,b ) has the form F (( t, x i )) =( F ( t ) , F ( x i )) for some F ∈ Isom( R , dt ) and F ∈ Isom( H n − , h n − ). If b = ± F is the identity. If b = ±
1, then F is either the identity or reflection aboutthe critical point t = ln( √ − b ) − ln( √ b ) of ((1 + b ) e t + (1 − b ) e − t ) .Let ( N, g ) be a Riemannian manifold with universal covering ( M n , h n,a,b ). Itsfundamental group π ( N ) is identified with a subgroup Γ of Isom( M, h n,a,b ) thatacts properly discontinuously and fixed point freely on M n . The association γ γ described above defines a homomorphism Φ : Γ → Isom( R , dt ) whose kernel hasindex at most two in Γ. Let ( ¯ N , ¯ g ) be the Riemannian covering of ( N, g ) associatedto the subgroup ker Φ of Γ.As each isometry in ker Φ acts trivially on the first factor of M n = R × H n − ,the metric ¯ g is also a warped product metric with the same warping function e φ n,a,b .By (4), ¯ g has negative sectional curvatures. By [ , Lemma 7.6], e φ n,a,b is a non-constant convex function on ( ¯ N , ¯ g ). By [ , Proposition 2.2], ( ¯ N , ¯ g ) has infinitevolume. Therefore, ( N, g ) has infinite volume, as required.
4. Concluding remarks and questions.
Remark . For n >
1, let g n denote the Euclidean metric on R n . An anal-ogous construction replacing ( H n − , h n − ) with ( R n − , g n − ) produces completenonpositively curved metrics on R n satisfying (P). In this setting, the parameters( a, b ) = (0 ,
1) correspond to the well-known warped product expression for H n aris-ing from normal coordinates about the horosphere through (0 , . . . , , ∈ H n − . Question . Are all complete Riemannian metrics on M that satisfy (P)isometric to a metric as described in the Main Construction or in Remark 4.1? SAMUEL LIN AND BENJAMIN SCHMIDT
Question . The constant curvature metric belongs to a one parameterfamily of pairwise non-isometric metrics on the manifold M n each of which satisfies(P) as in Remark 1.2. Are the symmetric metrics on the other negatively curvedsymmetric spaces similarly deformable through metrics satisfying (P)?In view of the existing hyperbolic rank-rigidity results for finite volume mani-folds, it seems reasonable to make the following conjecture: Conjecture . A finite volume Riemannian manifold that satisfies (P) islocally symmetric.
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Department of Mathematics, Michigan State university, East lansing, MI, 48823,
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