In this note, we will prove some simple extensions of Obata-type theorems.
Let (M, g) be a Riemannian manifold. Let Tk be the space of (0, k)- tensors and we will denote the tensor product Tk × Tl → Tk+l by ⊗. We will simply denote
k
z }| {
g ⊗ · · · ⊗ g by gk.
1 Condition to be a sphere
The classical Obata’s theorem states that for a complete Riemannian mani- fold (M, g) such that there exists a non-trivial function f with ∇2f = −f g, 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 ∇2f = −f g, then ob- viously f also satisfies ∇4f = f g2. 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 1.1. Suppose (M, g) is an n-dimensional connected compact Rie- mannian manifold. Suppose there exists k ≥ 1 and a non-trivial analytic function f on M satisfying
∇2kf + a2k−2(∇2k−2f ) ⊗ g + · · · + a0f g2k = 0 for some ai ∈ R (1.1) such that ±√
−1 are the only purely imaginary roots of the associated poly- nomial x2k+ a2k−2x2k−1+ · · · + a0. 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 com- plete, 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 = ex on the real line satisfies
∇4f = f g2. On the other hand, if f satisfies ∇2kf + (∇2k−2f ) ⊗ g = 0 such that ∆k−1f =
k−1
z }| {
∆ · · · ∆ f is nontrivial (this is automatic is M is compact), then by the classical Obata’s theorem applied to ∆k−1f , the compactness as- sumption can be replaced by the completeness assumption.
Remark 2. It is natural to ask if the assumption that ±√
−1 are the only purely imaginary roots is necessary. E.g. if the only purely imaginary roots
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are ±√
−1, ±√
−1k, k ∈ R, can we still deduce that (M, g) is a sphere (of certain radius)?
Proof of Theorem 1.1. Theorem 1.1 The necessity is clear. Indeed, for Sn= {(x0, · · · , xn) : P x2i = 1}, the height function x0 when restricted to Sn satisfies the given condition: it is known that
∇2X(u, v) = −h(u, v)ν = −g(u, v)X
where X = (x0, · · · , xn) is the position function, h is the second fundamental form and ν is the unit outward normal of Sn. Thus for f = x0, we have
∇2f = −f g.
From this it is easy to see that (1.1) is true.
Conversely suppose M has a function which satisfies (1.1). Take any q ∈ 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
f(2k)(s) + a2k−2f(2k−2)(s) + · · · + a0f = 0 which has solution
f = A cos s + B sin s +
l
X
i=1 mi−1
X
j=0
(Ai,jsjeaiscos(kis) + Bi,jsjeaissin(kis)) (1.2)
for some l and mi (multiplicity of the root eai+
√−1ki for the polynomial), and ai 6= 0. As f is bounded, it is easy to see that Ai,j = Bi,j = 0 for all i, j.
Thus
f = A cos s + B sin s =√
A2+ B2cos(s + θ) (1.3) for some θ, where A = f (q), B = ∇l0(0)f .
By (1.3), we can deduce that ∇2f = −f g. Indeed, if {ei}ni=1 is an or- thonormal basis which diagonalizes ∇2f (p), then for fixed i, for the geodesic γ emanating from p such that γ0(0) = ei, we have, by (1.3),
∇2f (ei, ei) = (f ◦ γ)00(0) = −f (p).
Since p and {ei} are arbitrary, we have
∇2f = −f g.
We then deduce by Obata’s theorem that (M, g) is the standard unit sphere.
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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 2.1. Suppose (M, g) is a connected complete Riemannian manifold such that there exists a nontrivial function f on M satisfying
∇2f = 0. (2.1)
Then there is a totally geodesic connected complete hypersurface M0 such that (M, g) is isometric to M0× R. The converse also holds.
Proof. It is easy to see that ∇f is a parallel Killing vector field. In particular,
|∇f | is a non-zero constant everywhere. In particular, M0 := f−1(0) is a smooth hypersurface. We claim that M0 is connected and is totally geodesic.
Let p, q ∈ M0 and let γ be a geodesic in (M, g) with γ(0) = p, γ(l) = q. Then on γ, f satisfies f00 = 0, f (0) = 0 and f (l) = 0. In particular, this implies γ ⊂ M0 and so M0 is connected. On the other hand, if α is a geodesic in (M, g) such that α0(0) ∈ TpM0. Then we have (f ◦ α)0(0) = h∇f, α0(0)i = 0.
Therefore α ⊂ M0 and so M0 is totally geodesic. By Hopf-Rinow theorem, M0 is also complete.
We now claim that M = M0× R. Let φt be the one-parameter family of diffeomorphism generated by the vector field ∇f . Define Φ : M0 × R → M by Φ(x, t) = φt(x). It is easy to see that Φ is smooth bijection, with inverse given by Φ(y) = (φ−f (x)(x), f (x)), so Φ is a diffeomorphism. Since ∇f is a Killing vector field, it follows that (M, g) is isometric to M0× R.
Corollary 2.1. Let (M, g) be an n-dimensional connected Riemannian man- ifold. Let V = {∇f : f satisfies (2.1)}. 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 × Rk. The converse also holds.
Proof. The case for k = 0 is trivial and the k = 1 case is shown by Theorem 2.1. Suppose {Xi = ∇fi} ⊂ V are independent. By the proof of Theorem 2.1, we have M = M1 × R, where M1 = f1−1(0). By the total geodesic- ness of M1, the function f2 when restricted to (M1, gM1) satisfies (2.1) and is non-trivial. Thus by induction, it is easy to see that the assertion is true.
Remark 3. After some googling I’ve found that Corollary 2.1 has already been proved in a paper of Wu and Ye: A Note On Obata’s Rigidity Theorem I (Theorem 5.2).
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