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 1For a connected complete Riemannian manifold , , it is isometric to the standard unit sphere if and only if there is a non-constant function such that

*Proof:* The necessary condition is easy. Indeed, for , the height function when restricted to satisfies the given condition: it is known that

where is the position function, is the second fundamental form and is the unit outward normal of . Thus for , we have

Conversely suppose has a function which satisfies the above equation. Take any such that , then on each geodesic starting from parametrized by arclength, the function satisfies the ODE

where . Take the geodesic starting from such that , then from (2), we know that there is on such that takes the maximum on and there is a on which takes its minimum on . We can further assume and thus .

Now, for any geodesic on , is of the form

on it, where is the arclength. Let to be the set of points on such that its distance form is . We have

Lemma 2consists of the point only, and it is the only point with value of equals .

*Proof:* From the equation , is nondegenerate when , and observe that on , we know that consists of non-singular critical points of , which must be discrete by Morse lemma. On the other hand, since is the image under exponential map of the unit sphere in , it is connected. We conclude that consists of one point only, i.e. all geodesics starting from must meets at . Thus is a conjugate point and therefore no geodesic from which passes would remain minimizing. In particular, we must have , since for .

In the same spirit, we also have

Lemma 3is the only point such that its value is .

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

Lemma 4For , is homeomorphic to an -sphere.

*Proof:* We claim that

is one-one, and is thus a homeomorphism, where is the sphere with radius .

Now, for any point , since is not a critical point, it can be seen from (2) that if a geodesic starting from reaches within an arclength of , , and so there is only one minimal geodesic joining to . The claim is proved.

So now every point on is determined by where is of unit length. For the standard sphere in the Euclidean space, fix to be its north pole, and we identify an orthnormal frame of to an orthnormal frame in . We then denote by the corresponding vector on with , under this identification. We will use the geodesic polar coordinates at on and denote a point on it simply by (resp. a point in by ).

We now define the map

by

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

Lemma 5is a diffeomorphism.

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

Lemma 6For a function on , if , then

on .

*Proof:* As , and by (3),

Since , so . Also, by (1), so

On the other hand,

as . We thus have

Multiply both sides by and integrating, we have

Then . Taking , and using the above equation, we can get the result.

Note that on , we have exactly the same result, since it also satisfies (1), i.e.

( as exactly defined in , of course. )

We can now complete the proof of the theorem, it suffices to show that (why?). Note that by the above lemma,

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

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 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? ]

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

Ah… yes. Silly me :-P

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

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 where . then the first eigenvalue satisfies . 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).

Oh really? Seems I am too ignorant about the results in geometric analysis…