A condition for a Riemannian manifold to be a hyperbolic space

[Update: A counter example when \phi>|\nabla\phi| is dropped and another version of this theorem by Pong Ting Kei (15-10-2011)]

In this post I will give a characterization when a complete connected Riemannian manifolds {(M,g)} is isometric to the hyperbolic space. Of course this is the analogue of Obata’s theorem for the necessary and sufficient condition for a manifold to be a sphere.

Recall that {\mathbb{R}^{n,1}=\{(x_0,x_1,\cdots, x_n)\in \mathbb{R}^{n+1}\}} is the Minkowski spacetime with metric {-dx_0^2+dx_1^2+\cdots +dx_n^2} and the hyperbolic space is

\displaystyle \mathbb{H}^n = \{ (x_0,\cdots, x_n)\in \mathbb{R}^{n,1}: x_0 ^2 -x_1^2-\cdots -x_n^2 =1, x_0>0\}

with induced metric. In geodesic polar coordinates, the metric can be expressed as

\displaystyle ds^2 + \sinh^2 s\, d\theta^2

where {s} is the distance function from the origin and {d\theta^2} is the metric on the standard unit ball. It can be checked that {\mathbb{H}^n} has constant sectional curvature {-1}, i.e. a space form.

We can now state the Obata-type theorem for hyperbolic space:

Theorem 1 Let {(M^n,g)} be a complete connected Riemannian manifold. Then it is isometric to the hyperbolic space if and only if there is a non-constant smooth function {\phi} on {M} such that

\displaystyle  \nabla ^2 \phi = \phi g \ \ \ \ \ (1)

and

\displaystyle  \phi> |\nabla \phi|.

 

The proof is quite similar to Obata’s proof, with some modification.
Proof: The necessary condition is clear. Indeed, for the position function {X\in \mathbb{R}^{n,1}} of {\mathbb{H}^n}, note that {X} can be regarded as the normal {N} of the hypersurface {\mathbb{H}^n\subset \mathbb{R}^{n,1}}, using the fact that {\partial_u X=u} for any tangent vector {u} of {\mathbb{H}^n} , together with {\nabla_u v= (\partial _u v)^T= \frac{\partial ^2X}{\partial u\partial v}+\langle \frac{\partial ^2X}{\partial u\partial v}, X\rangle X} , it is easy to see that the second fundamental form {h(u,v)= \langle \partial _uX,v\rangle= g(u,v)} and so

\displaystyle \nabla ^2 X(u,v)= h(u,v)N= g(u,v)X.

Thus if we take {\phi=x_0}, then { \nabla ^2 \phi = \phi g} and {\phi^2 -|\nabla \phi|^2=\cosh ^2 s-\sinh ^2 s=1}.

For the converse, suppose there is such {\phi}. Take any {q} such that {\nabla\phi(q)\neq0}. Then for any arclength-parametrized geodesic {\gamma} starting from this point, the equation is reduced to

\displaystyle \phi''=\phi,

so the solution is

\displaystyle \phi =A\cosh s+ B \sinh s,

where {A= \phi(q)} and {B= \nabla _{\gamma'(0)}\phi}. Using the compound angle formula, it can also be written as

\displaystyle \phi (s) =\sqrt{\phi(q)^2-|\nabla _{\gamma'}\phi|^2} \cosh (s+\theta)

where {\tanh \theta =B/A}. [This is where we have to use the condition {\phi> |\nabla \phi|}. ] From this form we see that when we choose {\gamma'(0)=\frac{\nabla\phi(q)}{|\nabla\phi(q)|}} and when {s=s_0=-\tanh^{-1}(\frac{ |\nabla \phi(q)|}{\phi(q)})}, then {\phi(\gamma(s_0))} is actually the unique minimum of {\phi} over {M}. (The uniqueness of the minimum can also be seen from the fact that \phi is strictly convex in view of (1). )

Now let {p=\phi(\gamma (s_0))}, we can further rescale {\phi} such that for any arclength-parametrized geodesic {\gamma (s)} from this point

\displaystyle  \phi=\phi(s) = \cosh s. \ \ \ \ \ (2)

It is not hard to see from (2) that the level set {\phi^{-1}(\cosh s)} is exactly the geodesic sphere {S_p(s)} centered at {p} of radius {s}. Furthermore, we claim that there is only one geodesic from each point to {p}. For, if there are two distinct geodesics {\gamma,\beta} joining {p} to {q}, then they must be of the same length {l} (why?), i.e. {\gamma,\beta:[0,l]\rightarrow M} both have starting point {p} and endpoint {q}. As {(M,g)} is complete, we can extend {\gamma} on {[0, l+1]}. (2) then shows that {\phi(\gamma(l+1))=\cosh (l+1)}. But then since {\beta|_{[0,l]}\cup \gamma |_{[l, l+1]}} is a broken geodesic joining {p} and {\gamma (l+1)}, the distance {d(p, \gamma(l+1))} must be strictly smaller than {l+1}, but then (2) shows that {\phi(\gamma (l+1))<\cosh(l+1) }, a contradiction.

The above argument shows that {\exp_p : T_pM \rightarrow M} is a bijection, and clearly {\exp_p} and its inverse (if exists) are smooth, thus {\exp _p} is actually a diffeomorphism.

We will show that then {g} is pulled back to {T_pM\cong \mathbb{R}^n} under this map and using the usual spherical coordinates (i.e. normal coordinates around {p}), the pulled back metric is {ds^2 +\sinh ^2 s \,d\theta^2}, and thus {(M,g)} is isometric to the hyperbolic space. For this purpose we make use of the Jacobi field.

First of all, fixed any {v\in T_pM\cong \mathbb{R}^n} with length {1}. Let {\gamma (s)=\exp_p(sv)}. Fix {s_0} and let {E\in T_{\gamma(s_0)}M} such that {\gamma'(s_0)\perp E} is orthonormal. Parallel transport {E} along {\gamma }, which we denote by {E(s)}. We then claim that

\displaystyle  J(s)=\sinh s \,E(s) \ \ \ \ \ (3)

is an Jacobi field with {J(0)=0}.

As

\displaystyle \nabla _{\gamma'}\nabla _{\gamma'}J= \sinh s \,E(s)=J(s).

By the Jacobi field equation

\displaystyle \nabla _{\gamma'}\nabla _{\gamma'}J+ R(J, \gamma')\gamma'=0,

we only have to prove that

\displaystyle  R(J, \gamma')\gamma'=-J. \ \ \ \ \ (4)

To see this, note that (1) implies

\displaystyle \nabla _X (\nabla \phi)=\phi X.

Also (2) implies {\gamma'(s)= \frac{\nabla \phi}{|\nabla \phi|}}, so

\displaystyle  |\nabla \phi|^2R(J,\gamma')\gamma'= \nabla _J \nabla_{\nabla \phi}\nabla \phi-\nabla _{\nabla \phi} \nabla_J\nabla \phi- \nabla _{[J,\nabla \phi]}\nabla \phi. \ \ \ \ \ (5)

By {\nabla _X (\nabla \phi)=\phi X},

\displaystyle  \nabla _J\nabla _{\nabla \phi}(\nabla \phi)= \nabla _J(\phi\nabla \phi)= J(\phi)\nabla \phi+\nabla _J(\nabla \phi)=J(\phi)\nabla \phi+\phi^2 J=\phi^2 J \ \ \ \ \ (6)

noting that {J(\phi)= g(\nabla \phi, J)=0}. We also have

\displaystyle  \nabla _{\nabla \phi}\nabla _J(\nabla \phi)= \nabla_{\nabla\phi} (\phi J)= |\nabla \phi|^2 J+ \phi \nabla _{\nabla \phi}J \ \ \ \ \ (7)

and

\displaystyle  \nabla _{[J,\nabla \phi]}( \nabla \phi)= \phi[J, \nabla \phi]= \phi(\nabla _J \nabla \phi- \nabla _{\nabla \phi}J)= \phi^2J- \phi \nabla _{\nabla \phi}J. \ \ \ \ \ (8)

(6)(7)(8) then gives the RHS of (5):

\displaystyle |\nabla \phi|^2R(J,\gamma')\gamma'= -|\nabla \phi|^2J, \quad\text{i.e. }R(J,\gamma')\gamma'= -J.

This proves (4). To finish the proof, for any vector {X\in T_{\gamma(s_0)}M}, {X= X^\perp+ X^T}, where {X^T} is tangential to the sphere {S_p(s_0)=\{s=s_0\}} and {X^\perp} is a multiple of {\frac{\partial}{\partial s}}. By Gauss lemma, {X^\perp\perp_g X^T} and so {|X|_g^2= |X^\perp|^2+|X^T|^2}. If we denote the Euclidean metric on {T_pM} by {\overline g} and push forward it to {M} by {\exp}, which we still denote by {\overline g}, then by Gauss lemma, {|X^\perp|_{\overline g}=|X^\perp|_{g}}. Let {x^i} be the normal coordinates around {p} and so in this coordinates, {X= \sum_i X^i \frac{\partial}{\partial x^i}}. Take a Jacobi field {J(s)} on {\gamma(s)} such that {J(s_0)=X^T} and {J(0)=0} (it exists because there is no pair of conjugate points on M (why?)). Then on one hand, it is well known that

\displaystyle  J(s)= \frac{s}{s_0} \sum_iX^i \frac{\partial}{\partial x^i} \ \ \ \ \ (9)

On the other hand, we know from (3) that {J(s)= \sinh s E(s)} where {E(s)} is parallel along {\gamma}, this implies {\nabla _{\gamma'(0)}J=E(0)}, thus

\displaystyle |X^T|^2= |J(s_0)|^2 = \sinh ^2 s_0\, |E(s)|^2= \sinh ^2 s_0 |E(0)|^2 =\sinh ^2 s_0\,|\nabla _{\gamma'(0)}J|^2.

Using (9) and the fact that {\nabla \frac{\partial }{\partial x^i}|_p=0}, we know that {\nabla_{\gamma'(0)}J= \frac{1}{s_0}\sum _i X^i \frac{\partial}{\partial x^i}|_p}. So the above becomes

\displaystyle |X^T|_g^2= \frac{\sinh ^2 s_0}{s_0^2}\left|\sum_i X^i \left.\frac{\partial }{\partial x^i}\right|_p\right|_g^2=\frac{\sinh ^2 s_0}{s_0^2}|X|_{\overline g}^2

since {g} agrees with {\overline g} at {p}. We conclude that

\displaystyle |X|_g ^2 =|X^\perp|^2 + \frac{\sinh^2 s}{s^2}|X^T|_{\overline g}^2

for {X\in T_{\gamma(s)}M}.

Finally, we know that the Euclidean metric is given by

\displaystyle \overline g=ds^2 +s^2 \;d\theta^2

where {d\theta ^2} is the standard metric on the unit sphere. So we can transform the above to

\displaystyle g=ds^2 +\sinh ^2 s \;d\theta^2,

which is the hyperbolic metric. \Box

Remark 1 We will show by an example that the condition {\phi> |\nabla \phi|} cannot be dropped.

Consider the Riemannian manifold {(M,g)=(\mathbb{R}\times N, ds^2 + e^{2s} h)} where {(N, h)} is a complete Riemannian manifold. Let {\phi=e^s}. We first show that this function satisfies

\displaystyle \nabla ^2 \phi= \phi g.

Choose local coordinates {x^i, i=1,\cdots, n-1} on {N}. Denote {g} by {\langle\cdot,\cdot \rangle}. Then

\displaystyle  \begin{array}{rcl}  \nabla ^2\phi(\partial_i,\partial_j) = \partial_i\partial_j \phi- \nabla _i \partial _j (\phi) = -\langle \nabla _i \partial _j, \partial_s\rangle \phi' &=& \langle \partial _j , \nabla_s\partial _i\rangle \phi'\\ &=& \frac{1}{2}\partial_s\langle \partial _j , \partial _i\rangle \phi'\\ &=& e^{3s} h_{ij} \\ &=& \phi g(\partial_i,\partial_j), \end{array}

\displaystyle \nabla ^2\phi(\partial_s,\partial_s)=\phi''=e^ s g(\partial_s,\partial_s)=\phi g(\partial_s,\partial_s),

and

\displaystyle  \begin{array}{rcl}  \nabla ^2\phi(\partial_s,\partial_i) =\partial_s\partial_i \phi- \nabla _s \partial _i (\phi) &=& - \langle \nabla _s \partial _i,\partial_s\rangle \phi'\\ &=& - \frac{1}{2}\partial_i\langle \partial_s,\partial_s\rangle \phi'\\ &=&0\\ &=&\phi g(\partial _s , \partial_i). \end{array}

We now show that {(M,g)} is complete. We will denote {e^{2s}h} by {g_s} and the distance (resp. diameter) function corresponding to {g_s} by {d_{g_s}} (resp. {diam_{g_s}}), etc.

Let {K} be a closed and bounded subset in {(M,g)}, it suffices to show that it is compact. Let {p_n=(x_n,y_n)} be a sequence in {C}. It is easy to see that {\pi_1(C)} is bounded and thus by passing into subsequence, we can w.l.o.g assume that {x_n \rightarrow x\in \mathbb{R}}. Here {\pi_1: \mathbb{R}\times N\rightarrow \mathbb{R}} is the projection. Also this shows that {\pi_1(C)\subset [a,b]}. Obviously we have {e^{2a}h\leq g_s\leq e^{2b}h} for {s\in \pi_1(C)}. We now claim that {y_n} is bounded in {(N,h)}. Indeed, for {C_s= C\cap(\{s\}\times N)}, {e^a diam_h (C_s)=diam_{g_a}(C_s)\leq diam_{g_s}(C_s)\leq diam _{g}(C)}. So {d_h(y_m,y_n)\leq e^{-a} diam_g(C)}. As {(N,h)} is complete, we can pick a subsequence {y_{n_k}\rightarrow y\in (N,h)} as {k\rightarrow\infty}. It remains to show that {(x_{n_k}, y_{n_k})\rightarrow (x,y)} in {(M,g)} and {(x,y)\in C}. But the second assertion follows from the first because {C} is closed in {(M,g)}. To show the first claim, note that

\displaystyle  \begin{array}{rcl}  d_g((x_{n_k}, y_{n_k}), (x,y)) &\leq& |x_{n_k}-x|+ \max_s d_{g_s}(y_{n_k},y)\\ &\leq& |x_{n_k}-x|+ e^{2b} d_{h}(y_{n_k}, y)\\ &\rightarrow & 0 \end{array}

as {k\rightarrow \infty}.  

Remark 2 As remarked by Pong Ting Kei, the condition {\phi >|\nabla \phi|} can also be replaced by the condition

\displaystyle \phi\geq c>0

for some constant {c}. Indeed, this condition together with {\nabla ^2\phi=\phi g} implies that {\nabla^2 \phi \geq cg}, i.e. {\phi} is strongly convex. By some convex analysis, this will show that {\phi} must have a (necessarily unique) minimum, as in the proof of Theorem 1, and the proof can proceed as before.

We now give some details of the convex analysis. First we can w.l.o.g. assume that {c=1}. Thus we have {\nabla^2\phi\geq g}. Fix any {p,q\in M}, let {l} be the distance between them and {\gamma:[0.l]\rightarrow M} be a unit speed geodesic joining {p} to {q}. We let {f(s)=\phi (\gamma(s))}. From {\nabla^2\phi\geq g}, we immediately have {f''\geq1} on {[0,l]}. By fundamental theorem of calculus, for {b\geq a}, f'(b)\geq f'(a)+b-a. So

\displaystyle  \begin{array}{rcl}  f(l)-f(\frac{l}{2})=\int_{\frac{l}{2}}^l f'(t)dt &\geq& \int_{\frac{l}{2}}^l( f'(\frac{l}{2})+ (t-\frac{l}{2}))dt\\ &=&f'(\frac{l}{2})\frac{l}{2}+ \int_0^\frac{l}{2}tdt\\ &=&f'(\frac{l}{2})\frac{l}{2}+ \frac{l^2}{8}. \end{array}

Similarly

\displaystyle  \begin{array}{rcl}  f(\frac{l}{2})-f(0)=\int_0^{\frac{l}{2}} f'(t)dt &\leq& \int_0^{\frac{l}{2}}( f'(\frac{l}{2})- (\frac{l}{2}-t))dt\\ &=&f'(\frac{l}{2})\frac{l}{2}- \int_0^\frac{l}{2}sds\\ &=&f'(\frac{l}{2})\frac{l}{2}- \frac{l^2}{8}. \end{array}

Subtract the second inequality from the first one, we have

\displaystyle  \frac{1}{2}(f(0)+f(l))-f(\frac{l}{2})\geq\frac{l^2}{8}.

We have showed that for any minimal geodesic {\gamma} joining {p,q},

\displaystyle \frac{1}{2}(\phi(p)+\phi(q))-\frac{d(p,q)^2}{8}\geq\phi(\gamma(\frac{d(p,q)}{2})).

We now prove the each sublevel set {\{x\in M: \phi(x)\leq c\}} is bounded, with the diameter depending on {c}. Indeed, if {f(p), f(q)\leq c}, join them by a geodesic {\gamma} of length {l=d(p,q)}, then

\displaystyle 0<\phi(\gamma(\frac{l}{2}))\leq \frac{1}{2}(\phi(p)+\phi(q))-\frac{d(p,q)^2}{8}\leq c-\frac{d(p,q)^2}{8}.

Thus {d(p,q)<\sqrt{8c}}. Therefore {diam\{\phi\leq c\}\leq \sqrt{8c}}. The sublevel sets are all closed and bounded, and so is compact, therefore the minimum of {\{\phi\leq a\}\neq \emptyset} must be the minimum of {\phi} in {M}.  

Therefore we have

Theorem 2 (Pong Ting Kei’s theorem, 2011) Let {(M^n,g)} be a complete connected Riemannian manifold. Then it is isometric to the hyperbolic space if and only if there is a smooth function {\phi} on {M} such that

\displaystyle  \nabla ^2 \phi = \phi g

and

\displaystyle  \phi\geq c>0.

 

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2 Responses to A condition for a Riemannian manifold to be a hyperbolic space

  1. gilbertweinstein says:

    Hi, right below equation 2, should that read \phi^{-1}(cosh(s)) is the sphere S_p(s)? [Oh, yes. Corrected, thanks! -KKK]

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