## A note on Obata’s theorem

In this short note, we will prove some simple extensions of Obata-type theorems. Let ${(M,g)}$ be a Riemannian manifold. Let ${\mathcal T^k}$ be the space of ${(0,k)}$-tensors and we will denote the tensor product ${\mathcal T^k\times \mathcal T^l \rightarrow \mathcal T^{k+l}}$ by ${\otimes}$. We will simply denote ${\overbrace{g\otimes \cdots \otimes g}^k}$ by ${g^k}$.

1. Condition to be a sphere

The classical Obata’s theorem states that for a complete Riemannian manifold ${(M,g)}$ such that there exists a non-trivial function ${f}$ with ${\nabla ^2 f =-fg}$, then ${(M,g)}$ is isometric to the standard unit sphere. It is interesting to see if there are other equations for which the existence of a non-trivial solution will ensure ${(M,g)}$ to be a sphere. E.g. if ${f}$ satisfies ${\nabla ^2 f =-f g}$, then obviously ${f}$ also satisfies ${\nabla ^4 f = fg^2}$. Suppose ${f}$ satisfies the latter equation, is ${(M,g)}$ necessarily a sphere? In this section I will answer this question (affirmatively), with the assumption of a compactness condition.

 Theorem 1 Suppose ${(M,g)}$ is an ${n}$-dimensional connected compact Riemannian manifold. Suppose there exists ${k\ge 1}$ and a non-trivial function ${f}$ on ${M}$ satisfying $\displaystyle \nabla ^{2k}f+a_{2k-2}(\nabla ^{2k-2}f)\otimes g + a_{2k-4}(\nabla^{2k-4}f)\otimes g^2\cdots + a_0 fg^{k}=0 \quad \textrm{for some }a_i\in \mathbb R \ \ \ \ \ (1)$ such that ${\pm\sqrt{-1}}$ are the only purely imaginary roots of the associated polynomial ${ x^{2k}+a_{2k-2}x^{2k-1}+\cdots +a_0}$. Then ${(M,g)}$ is isometric to the standard sphere. The converse also holds. If ${k=1}$, the compactness condition can be replaced by ${(M,g)}$ being complete, by the classical Obata’s theorem.

 Remark 1 We remark that the condition for ${M}$ to be compact cannot be omitted in general. For example, the function ${f=e^x}$ on the real line satisfies ${\nabla ^4 f = f g^2}$. On the other hand, if ${f}$ satisfies ${\nabla ^{2k}f+ (\nabla ^{2k-2}f)\otimes g=0}$ such that ${\Delta^{k-1}f=\overbrace {\Delta\cdots \Delta}^{k-1}f}$ is nontrivial, then by the classical Obata’s theorem applied to ${\Delta ^{k-1}f}$, the compactness assumption can be replaced by the completeness assumption.

 Remark 2 It is natural to ask if the assumption that ${\pm\sqrt{-1}}$ are the only purely imaginary roots is necessary. E.g. if the only purely imaginary roots are ${\pm \sqrt{-1}, \pm \sqrt{-1}k}$, ${k\in \mathbb R}$, can we still deduce that ${(M,g)}$ is a sphere (of certain radius)?

Proof of Theorem 1: The necessity is clear. 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

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

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

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

From this it is easy to see that (1) is true.

Conversely suppose ${M}$ has a function which satisfies (1). Take any ${q\in M}$, then on each geodesic ${l(s)}$ starting from ${q}$ parametrized by arclength, the function (when restricted to ${l}$) satisfies the linear ODE with constant coefficient

$\displaystyle f^{(2k)}(s) +a_{2k-2}f^{(2k-2)}(s)+\cdots + a_0f=0$

which has solution

$\displaystyle f=A\cos s+ B\sin s + \sum _{i=1}^{l}\sum_{j=0}^{m_i-1}(A_{i,j} s^je^{a_is}\cos (k_i s)+B_{i,j} s^je^{a_is}\sin (k_i s)) \ \ \ \ \ (2)$

for some ${l}$ and ${m_i}$ (multiplicity of the root ${e^{a_i+\sqrt{-1}k_i}}$ for the polynomial), and ${a_i\ne 0}$. As ${f}$ is bounded, it is easy to see that ${A_{i,j}=B_{i,j}=0}$ for all ${i,j}$. Thus

$\displaystyle f=A\cos s+ B\sin s =\sqrt{A^2 + B^2}\cos (s +\theta) \ \ \ \ \ (3)$

for some ${\theta}$, where ${A=f(q), B= \nabla _{l'(0)}f}$.

By (3), we can deduce that ${\nabla ^2 f =-fg}$. Indeed, if ${\lbrace e_i\rbrace_{i=1}^n}$ is an orthonormal basis which diagonalizes ${\nabla ^2 f(p)}$, then for fixed ${i}$, for the geodesic ${\gamma}$ emanating from ${p}$ such that ${\gamma'(0)=e_i}$, we have, by (3),

$\displaystyle \nabla ^2 f (e_i, e_i)= (f\circ \gamma)''(0)= - f (p).$

Since ${p}$ and ${\lbrace e_i\rbrace}$ are arbitrary, we have

$\displaystyle \nabla ^2 f =-fg.$

We then deduce by Obata’s theorem that ${(M,g)}$ is the standard unit sphere.

$\Box$

2. A splitting result

In this section, I will show a simple splitting result. The proof is quite simple, and I suspect that it should be known to the experts.

 Theorem 2 Suppose ${(M,g)}$ is a connected complete Riemannian manifold such that there exists a nontrivial function ${f}$ on ${M}$ satisfying $\displaystyle \nabla ^2 f=0. \ \ \ \ \ (4)$ Then there is a totally geodesic connected complete hypersurface ${M_0}$ such that ${(M,g)}$ is isometric to ${M_0\times \mathbb R}$. The converse also holds.

Proof: It is easy to see that ${\nabla f}$ is a parallel Killing vector field. In particular, ${|\nabla f|}$ is a non-zero constant everywhere. In particular, ${M_0:=f^{-1}(0)}$ is a smooth hypersurface. We claim that ${M_0}$ is connected and is totally geodesic. Let ${p, q\in M_0}$ and let ${\gamma}$ be a geodesic in ${(M,g)}$ with ${\gamma(0)=p}$, ${\gamma(l)=q}$. Then on ${\gamma}$, ${f}$ satisfies ${f''=0}$, ${f(0)=0}$ and ${f(l)=0}$. In particular, this implies ${\gamma\subset M_0}$ and so ${M_0}$ is connected. On the other hand, if ${\alpha}$ is a geodesic in ${(M,g)}$ such that ${\alpha'(0)\in T_pM_0}$. Then we have ${(f\circ \alpha)'(0)= \langle \nabla f, \alpha'(0)\rangle=0}$. Therefore ${\alpha\subset M_0}$ and so ${M_0}$ is totally geodesic. By Hopf-Rinow theorem, ${M_0}$ is also complete. We now claim that ${M=M_0\times \mathbb R}$. Let ${\phi_t}$ be the one-parameter family of diffeomorphism generated by the vector field ${\nabla f}$. Define ${\Phi: M_0 \times \mathbb R\rightarrow M}$ by ${\Phi(x, t)= \phi_t(x)}$. It is easy to see that ${\Phi}$ is smooth bijection, with inverse given by ${\Phi(y)= (\phi_{-f(x)}(x), f(x))}$, so ${\Phi}$ is a diffeomorphism. Since ${\nabla f}$ is a Killing vector field, it follows that ${(M, g)}$ is isometric to ${M_0 \times \mathbb R }$. $\Box$

 Corollary 3 Let ${(M,g)}$ be an ${n}$-dimensional connected Riemannian manifold. Let ${V=\lbrace \nabla f: f\textrm{ satisfies \;}}$ (4)${\rbrace}$. Suppose ${\dim V=k}$. Then there is a complete connected totally geodesic submanifold ${N}$ of dimension ${n-k}$ such that ${M}$ is isometric to ${N\times \mathbb R^k}$. The converse also holds.

Proof: The case for ${k=0}$ is trivial and the ${k=1}$ case is shown by Theorem 2. Suppose ${\lbrace X_i=\nabla f_i\rbrace\subset V}$ are independent. By the proof of Theorem 2, we have ${M= M_1\times \mathbb R}$, where ${M_1=f_1^{-1}(0)}$. By the total geodesic-ness of ${M_1}$, the function ${f_2}$ when restricted to ${(M_1, g_{M_1})}$ satisfies (4) and is non-trivial. Thus by induction, it is easy to see that the assertion is true. $\Box$

 Remark 3 After some googling, I found that Corollary 3 has already been proved in a paper of Wu and Ye (Theorem 5.2).