## 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

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

by

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

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,

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

$\Box$

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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).

• KKK says:

Oh really? Seems I am too ignorant about the results in geometric analysis… ${\underset{\bigtriangledown}{\eqcirc\,\eqcirc}}$