A condition for a Riemannian manifold to be isometric to a sphere

Here is a theorem by Obata which gives a necessary and sufficient conditon for a Riemannian manifold (not necessarily compact) to be a standard sphere, which I think is quite interesting. (In Riemannian geometry, “to be a sphere” actually means “isometric to a sphere”. ) The proof is not very involved and I try to sketch it here.

Theorem 1 For a connected complete Riemannian manifold {(M^n,g)}, {n\geq 2}, it is isometric to the standard unit sphere if and only if there is a non-constant function {\phi} such that

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

(Of course, this means {\nabla^2 \phi(u,v)=-\phi g(u,v)} for all vector {u,v}.)  

Proof: The necessary condition is easy. Indeed, for {\mathbb{S}^n=\{(x_0, \cdots, x_n): \sum x_i^2=1\}}, the height function {x_0} when restricted to {\mathbb{S}^n} satisfies the given condition: it is known that

\displaystyle \nabla ^2 X(u,v)=- h(u,v)n=-g(u,v)X

where {X=(x_0,\cdots, x_n)} is the position function, {h} is the second fundamental form and {n} is the unit outward normal of {\mathbb{S}^n}. Thus for {\phi=x_0}, we have

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

Conversely suppose {M} has a function which satisfies the above equation. Take any {Q\in M} such that {\nabla \phi(Q)\neq 0}, then on each geodesic {l(s)} starting from {Q} parametrized by arclength, the function satisfies the ODE

\displaystyle \phi''(s) +\phi(s)=0

which has solution

\displaystyle  \phi=A\cos s+ B\sin s \ \ \ \ \ (2)

where {A=\phi(Q), B= \nabla _{l'(0)}\phi}. Take the geodesic {l} starting from {Q} such that {l'(0)= \nabla \phi(Q)}, then from (2), we know that there is {P_+} on {l} such that {\phi} takes the maximum on {M} and there is a {P_-} on {l} which {\phi} takes its minimum on {M}. We can further assume {\phi(P_+)=1} and thus {\phi(P_-)=-1}.

Now, for any geodesic on {P_+}, {\phi} is of the form

\displaystyle  \phi(s)=\cos s \ \ \ \ \ (3)

on it, where {s} is the arclength. Let {M_s} to be the set of points on {M} such that its distance form {P_+} is {s}. We have

Lemma 2 {M_\pi} consists of the point {P_-} only, and it is the only point with value of {\phi} equals {-1}.  

Proof: From the equation {\nabla^2 \phi=-\phi g}, {\nabla^2\phi} is nondegenerate when {\phi=-1}, and observe that {\phi=-1} on {M_\pi}, we know that {M_\pi} consists of non-singular critical points of {\phi}, which must be discrete by Morse lemma. On the other hand, since {M_\pi} is the image under exponential map of the unit sphere in {T_{P_+}M}, it is connected. We conclude that {M_\pi} consists of one point {Q} only, i.e. all geodesics starting from {P_+} must meets at {Q}. Thus {Q} is a conjugate point and therefore no geodesic from {P_+} which passes {Q} would remain minimizing. In particular, we must have {Q=P_-}, since {\cos s>-1} for {s<\pi}. \Box

In the same spirit, we also have

Lemma 3 {P_+} is the only point such that its {\phi} value is {1}.  

We can also see from (3) that there is no critical points except {P_\pm}. Furthermore, we have

Lemma 4 For {s\in (0, \pi)}, {M_s} is homeomorphic to an {(n-1)}-sphere.  

Proof: We claim that

\displaystyle \exp_{P_+}: S(s)\subset T_{P_+}M\rightarrow M_s

is one-one, and is thus a homeomorphism, where {S(s)} is the sphere with radius {s}.

Now, for any point {P\neq P_{\pm}}, since {P} is not a critical point, it can be seen from (2) that if a geodesic {\gamma} starting from {P} reaches {P_+} within an arclength of {\pi}, {\gamma'(0)= \nabla \phi(P)}, and so there is only one minimal geodesic joining {P_+} to {P}. The claim is proved. \Box

So now every point on {M_s} is determined by {(s,v)} where {v\in T_{P_+}M} is of unit length. For the standard sphere {\mathbb{S}^n} in the Euclidean space, fix {\overline P_+} to be its north pole, and we identify an orthnormal frame of {T_{P_+}M} to an orthnormal frame in {T_{\overline {P}_+}\mathbb{S}^n}. We then denote by {\overline v} the corresponding vector on {T_{\overline P_+}\mathbb{S}^n} with {v\in T_{P_+}M}, under this identification. We will use the geodesic polar coordinates at {\overline P_+} on {\mathbb{S}^n} and denote a point on it simply by {(s,\overline v)} (resp. a point in {M} by {(s,v)}).

We now define the map

\displaystyle h :M \rightarrow \mathbb{S}^n


\displaystyle h(P_{\pm})=\overline P_{\pm },\quad h(s,v)=(s,\overline v).

As this map is a smooth bijection from a compact space to a Hausdorff space and its inverse is also smooth, we have

Lemma 5 {h} is a diffeomorphism.  

To show that it is also an isometry, we just have to prove that for any tangent vector {u} at {(s,v)}, the vector {dh(u)} at {(s, \overline v)} has the same length. By Gauss lemma, we can further assume that {u} is tangential to {M_s}, i.e. it is perpendicular to {v} when parallel translated by to {P_+}. Let us denote the parallel translation of {u} along the geodesic {s\mapsto (s,v)} by {u_s}, and let {\overline u_0=dh(u_0)}, we then parallel translate {\overline u_0} in the same way and denote it by {\overline u_s}. Obviously we then only have to prove that {dh(u_s)=\overline u_s}, as they both have the same length (why). We need the following

Lemma 6 For a function {f(v)} on {M_{\pi/2}}, if {F(s,v)= f(v)}, then

\displaystyle  u_0 (sF)= \sin s \;u_s(F)

on {M\setminus P_-}.  

Proof: As {\nabla _{\nabla s}u_s=0}, and by (3),

\displaystyle [ \nabla s, u_s ]= \nabla _{\nabla s}u_s- \nabla _{u_s}\nabla s= \nabla _{u_s} (\frac{1}{\sin s} \nabla \phi).

Since {\nabla _{u_s} s=\langle u_s , \nabla s\rangle=0}, so {\nabla _{u_s} \frac{1}{\sin s}=0}. Also, {\nabla _{u_s}\nabla \phi=-\phi u_s} by (1), so

\displaystyle [\nabla s, u_s]= -\cot s \;u_s.

On the other hand,

\displaystyle [\nabla s, u_s]F= \frac{d}{ds} u_s (F)

as {\frac{d}{ds}F=0}. We thus have

\displaystyle  \frac{d}{ds} u_s (F) = -\cot s u_s(F).

Multiply both sides by {\sin s} and integrating, we have

\displaystyle  \sin s \;u_s(F)= const= u_{\pi/2}(F).

Then {u_s(sF)= \frac{s}{\sin s}u_{\pi/2}(F)}. Taking {s\rightarrow 0}, and using the above equation, we can get the result. \Box

Note that on {\mathbb{S}^n}, we have exactly the same result, since it also satisfies (1), i.e.

\displaystyle \overline u_0 (s\overline F)= \sin s \;\overline u_s(\overline F).

({\overline F} as exactly defined in {M}, of course. )

We can now complete the proof of the theorem, it suffices to show that { dh(u_s)(\overline F)= \overline u_s (\overline F)} (why?). Note that by the above lemma,

\displaystyle  \begin{array}{rcl}  dh(u_s )\overline F= u_s (\overline F\circ h)= \frac{1}{\sin s } u_0 (s \overline F\circ h)=\frac{1}{\sin s } dh(u_0) (s \overline F)&=&\frac{1}{\sin s } \overline u_0 (s \overline F)\\ &=& \overline u_s (\overline F). \end{array}


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7 Responses to A condition for a Riemannian manifold to be isometric to a sphere

  1. johnmamanshun says:

    That’s a very intuitive proof! But seems that the condition is hard to check (Except in the Einstein case as in II)……

  2. KKK says:

    Yes indeed, in the PDE point of view, this corresponds to solving a system of n(n+1)/2 equations, which is much harder than solving the laplace type equation. Originally I want to know if merely a function satisfying \Delta \phi=-n\phi suffices to show that it is a sphere. I don’t know if the Einstein condition is really needed. I also want to do a similar thing in the hyperbolic setting. It seems promising to me, but I haven’t carried it out yet. [BTW, have you received my email? ]

    • johnmamanshun says:

      As least you have to add the curvature assuption. Or it is easy to construct a 2-torus (spanned by (1,0) and (0,\sqrt 2 \pi)) which has first nontrivial eigenvalue equals 2.

  3. KKK says:

    Ah… yes. Silly me :-P

  4. KKK says:

    Imposing a lower bound on Ricci curvature should be enough, I have included the result suggested by you in the note, see here.

    • johnmamanshun says:

      I check the note by Peter Li on geometric analysis. I think you have proved in II a theorem of Lichnerowicz, which states that if M is a compact manifold with Rc_g \geq (n-1)kg where k > 0. then the first eigenvalue satisfies \lambda \geq nk. Obata’s theorem corresponds to the equality case.

      Lichnerowicz theorem can be proved using Bochner-formula for 1 form (anyway it is more or less the same as your calculation).

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