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).  

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