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

[Updated on 5/10/2011: John Ma discovered that the result (re)discovered by him was proved by Lichnerowicz long time ago. $\underset{\sim}{>\;<}$]

This is a sequel to the previous post on a condition on a Riemmannian manifold to be isometric to a sphere. In that post, we proved that ${M}$ is a sphere if there is a nontrivial function ${\phi}$ such that

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

It is natural to ask if we can weaken the assumption to the existence of a nontrivial ${\phi}$ such that

$\displaystyle \Delta \phi =-n \phi.$

As it turns out, if we impose further condition on ${(M,g)}$, then it is true. More precisely, we can prove that

 Theorem 1 (Obata) For a compact Einstein manifold ${(M^n,g)}$ with positive scalar curvature ${n(n-1)}$, if there is a nonconstant function ${\phi}$ on ${M}$ such that $\displaystyle -\Delta \phi=n \phi, \ \ \ \ \ (1)$ then ${(M,g )}$ is isometric to the standard unit sphere.

Recall that ${M}$ is Einstein if its Ricci curvature ${Ric=c g}$ for some constant ${c}$, in this case we can let ${c=(n-1)k}$ and then the scalar curvature ${R=tr_g(Ric)}$ is ${n(n-1)k}$.

Actually Theorem 1 is the corollary of the following

 Theorem 2 (Obata, first eigenvalue estimate for laplacian) If ${(M,g)}$ is a connected compact Riemannian manifold, then if ${-\Delta \phi=n\lambda \phi}$ for some non-constant ${\phi}$ and nonzero ${\lambda}$, then $\displaystyle \lambda \geq \frac{(Rc(\nabla \phi), \nabla \phi)}{ (n-1) (\nabla \phi, \nabla \phi)}.$ The equality holds if and only if ${(M,g)}$ is isometric to the standard sphere of radius ${1/\sqrt{\lambda}}$ in the Euclidean space. In particular, if ${(M,g)}$ is Einstein with ${Ric = (n-1)g}$ and ${\lambda =1}$, i.e. there is ${\phi}$ with ${-\Delta \phi=n\phi}$, then ${(M,g)}$ is isometric to the standard unit sphere.

1. For an orientable compact ${M}$, for two tensors ${u,v}$ of the same type, we can define ${\langle u,v\rangle }$ at each point ${p\in M}$, we can then define

$\displaystyle (u,v)=\int_M \langle u,v \rangle dV.$

If ${M}$ is non-orientable, we then take its orientable double cover ${\widetilde M}$ and lifts ${u,v}$ to ${\widetilde M}$ (with the same notations ${u,v}$), we then define (naturally all tensors, and in particular ${g}$ and ${Rm}$ etc, can be lifted to ${\widetilde M}$. )

$\displaystyle (u,v)=\int_{\widetilde M}\langle u ,v\rangle d V.$

2. The $(0,2)$ tensor ${Ric=R_{ij}}$ can be raised to a $(1,1)$ tensor ${R_i^j}$ and thus can be treated as a linear map ${v^i\mapsto R_j^i v^j}$, we denote the later as ${Rc(v)\in TM}$. If ${M}$ is Einstein, i.e. ${Ric=(n-1)kg}$, then ${Rc}$ is just ${(n-1)k \;id}$.
3. We define ${\delta}$ as the differential operator which takes ${p}$-tensors to ${(p-1)}$-tensors by

$\displaystyle \delta u =-\nabla _i u^i \,_{i_{p-1}\cdots\, i_1}$

for ${u=u_{i_p \cdots i_1}}$. By Stokes theorem, ${\nabla }$ and ${\delta}$ are dual to each other:

$\displaystyle (\nabla u, v)=(u, \delta v)$

for ${(p-1)}$-tensor ${u}$ and ${p}$-tensor ${v}$. We define ${\Delta \phi=\nabla _i \nabla ^i \phi}$, in other words,

$\displaystyle \Delta =- \delta \nabla .$

Note that in this notations, ${-\Delta}$ is non-negative definite. (See Lemma 3 (2))

With these notations in mind, we have

 Lemma 3 Suppose ${-\Delta \phi=n\lambda \phi}$ , we have $\displaystyle (\nabla \phi, \nabla \phi)= -(\Delta \phi, \phi)= n\lambda(\phi,\phi).$ $\displaystyle (\nabla ^2 \phi, \phi g)= (\Delta \phi , \phi)= -(\nabla \phi,\nabla \phi).\ \ \ \ \ (2)$ $\displaystyle \delta \nabla ^2 \phi=n\lambda \nabla \phi -Rc(\nabla \phi).\ \ \ \ \ (3)$ $\displaystyle (\nabla ^2 \phi, \nabla ^2 \phi)= n\lambda (\nabla\phi , \nabla \phi)-(Rc(\nabla \phi), \nabla \phi). \ \ \ \ \ (4)$ For ${v= \nabla ^2 \phi +\lambda \phi g}$, $\displaystyle (v,v) =(n-1) \lambda (\nabla \phi,\nabla \phi) -(Rc(\nabla \phi),\nabla \phi).\ \ \ \ \ (5)$

Proof: The first two are easy. For (3), consider

$\displaystyle \begin{array}{rcl} -g^{ik}\nabla _i \nabla _k \nabla _j \phi&=&-g^{ik}\nabla _i \nabla _j \nabla _k \phi\quad (\nabla _i \nabla _j f= \nabla _j \nabla _i f)\\ &=&- g^{ik} \nabla _j \nabla_i \nabla _k \phi -R_j ^i \nabla _i \phi\quad (\text{Ricci identity})\\ &=& \nabla _j (-g^{ik} \nabla _i \nabla _k \phi)-R_j^i \nabla _i \phi\\&=& (n\lambda \nabla \phi-Rc(\nabla \phi))_j.\end{array}$

For (4), ${ (\nabla ^2 \phi,\nabla ^2\phi)= (\nabla \phi,\delta \nabla ^2 \phi)=(\nabla \phi, n\lambda \nabla \phi-Rc(\nabla \phi))}$. The last formula is the consequence of the above. $\Box$

As a corollary, as ${(v,v)\geq 0}$ and ${(\nabla \phi,\nabla \phi)>0}$, we have

$\displaystyle \lambda \geq \frac{ (Rc(\nabla \phi),\nabla \phi)}{(n-1) (\nabla \phi,\nabla \phi)}.$

This is the eigenvalue estimate of Theorem 2. In the case where ${M}$ is Einstein with ${Ric=(n-1)kg}$, the condition is reduced to

$\displaystyle \lambda \geq k.$

Also, note that the equality holds if and only if ${\nabla ^2 \phi=- \lambda \phi g}$, thus by Obata’s theorem, ${(M,g)}$ is isometric to a standard sphere of radius ${1/\sqrt{\lambda}}$ in the Euclidean space if ${M}$ is orientable. Actually ${M}$ can’t be non-orientable, otherwise, since ${\nabla ^2 \phi=- \lambda \phi g}$ on ${\widetilde M}$, we also have ${\nabla ^2 \phi=- \lambda \phi g}$ on ${M}$, and thus both are isometric to a standard sphere of the same radius, which is impossible (as one is orientable and the other is not, note that there is no orientation assumption in Obata’s theorem).

As remarked by John Ma, we actually can weaken the assumption of being Einstein by imposing some conditions on its curvature in Theorem 1 [Updated: this result was discovered by Lichnerowicz and rediscovered by John Ma]:

 Theorem 4 (Obata-Lichnerowicz-John Ma’s theorem, 2011) For a compact ${(M^n,g)}$ with a lower bound ${1}$ on its (normalized) Ricci curvature in the sense that $\displaystyle Ric\geq (n-1)g,$ if there is a nonconstant function ${\phi}$ on ${M}$ such that $\displaystyle -\Delta \phi=n \phi, \ \ \ \ \ (6)$ then ${(M,g )}$ is isometric to the standard unit sphere.

Proof: Indeed, the above proof shows that

$\displaystyle \lambda \geq \frac{ (Rc(\nabla \phi),\nabla \phi)}{(n-1) (\nabla \phi,\nabla \phi)}\geq \frac{ (n-1)(\nabla \phi,\nabla \phi)}{(n-1) (\nabla \phi,\nabla \phi)}=1.$

Since the equality is attained, ${v=\nabla^2\phi+\phi g=0}$ and we can apply Obata’s theorem. $\Box$