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 , , it is isometric to the standard unit sphere if and only if there is a non-constant function such that
(Of course, this means for all vector .)
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 .
on it, where is the arclength. Let to be the set of points on such that its distance form is . We have
Lemma 2 consists 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 3 is 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 4 For , 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
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 is 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 6 For a function on , if , then
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,