# If $\nabla X=h \cdot \text{Id}_{TM}$ for a vector field $X$ and $h \in C^{\infty}(M)$, is $h$ constant?

$\newcommand{\id}{\operatorname{Id}_{TM}}$ Let $M$ be a smooth connected closed Riemannian manifold.

Let $X$ be a vector field on $M$, and suppose that $\nabla X=h \cdot \id$, for some $h \in C^{\infty}(M)$.

Is it true that $h$ is constant? ($\nabla$ is the Levi-Civita connection on $M$).

This is true when the curvature of $M$ is zero. Indeed, we can consider $\nabla X$ as a one form on $M$ with values in $TM$: $\nabla X \in \Omega^1(M,TM)$. Then

$$R_{\nabla} (\cdot,\cdot)X=d_{\nabla}^2X=d_{\nabla}\nabla X =d_{\nabla}(h \cdot \id). \tag{1}$$

Using Leibnitz rule, $$d_{\nabla}(h \cdot \id)=h d_{\nabla} \id-dh \wedge \id=-dh \wedge \id. \tag{2}$$

(We used here the fact $\nabla$ is torsion-free and $d_{\nabla} \id$ is the torsion).

Combining equations $(1),(2)$, we obtain

$$R_{\nabla} (\cdot,\cdot)X=-dh \wedge \id.$$ When $R=0$, this means $dh \wedge \id=0$, i.e $$dh(v_1)v_2-dh(v_2)v_1=0$$ for all $v_1,v_2 \in \Gamma(TM)$. This implies $dh=0$, so $h$ is constant.

(In fact, so far we have only used torsion free and $R=0$, with no use of metricity).

$$\newcommand{\id}{\operatorname{Id}_{TM}}$$No. As a counterexample we can take the vector field $$X = \sin(\theta) \partial_\theta$$ on the two-sphere with the standard metric $$g = d \theta^2 + \sin(\theta)^2 d\varphi^2.$$ Near the north pole this looks like the homothetic vector field $$X(x) = x$$ on $$\mathbb R^2$$, while near the south pole it looks like $$X(x) = -x.$$

A coordinate calculation shows that $$\nabla X = \cos(\theta)\id,$$ since the only non-zero Christoffel symbols are $$\Gamma^\theta_{\varphi\varphi} = -\sin \theta \cos \theta$$ and $$\Gamma_{\theta\varphi}^\varphi = \cos \theta / \sin \theta.$$

• Thanks, your solution is very nice, and also easily visualisable. Do you think this example can be generalized to $\mathbb{S}^3$ (or to any other closed manifold of dimension $\ge 3$)? A naive attempt to take $X \oplus X$ on the product $\mathbb{S}^3 \otimes \mathbb{S}^3$ fails... Jan 22 '18 at 13:48
• I meant to take the product $X \oplus X$ on $\mathbb{S}^2 \otimes \mathbb{S}^2$ of course...when $X$ is your vector field. Jan 22 '18 at 14:10
• @AsafShachar: this example generalizes nicely to all spheres. To avoid hyperspherical coordinates you can do the calculation extrinsically - note that my $X$ is just the orthogonal projection of $-\partial_z$ on to the tangent bundle of $S^2 \subset \mathbb R^3$. You should be able to show that the obvious generalization of this vector field is an example on $S^n \subset \mathbb R^{n+1}.$ Jan 22 '18 at 14:28

Here is a generalization of Anthony Carapetis's (based on his comments and on some comments by Amitai Yuval).

Consider $\mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, and define $Z \in \Gamma(T\mathbb{S}^n)$ by $Z:=P(e_{n+1})$, where $P=P_{T\mathbb{S}^n}$ is the orthogonal projection on the tangent bundle of $\mathbb{S}^n$.

Claim: $\nabla^{\mathbb{S}^n} Z=-x_{n+1}\text{Id}_{T\mathbb{S}^n}$, i.e. $\nabla_w^{\mathbb{S}^n} Z=-x_{n+1}w$ for every $w \in \Gamma(T\mathbb{S}^n)$.

Let $x \in \mathbb{S}^n$; note that $P_x(v)=v-\langle x,v\rangle x$. ($P_x$ is the projection on $\{x\}^{\perp}$). Hence,

$$Z(x)=e_{n+1}-\langle x,e_{n+1}\rangle x=e_{n+1}-x_{n+1}x$$

Given $w \in T_x \mathbb{S}^n \subseteq \mathbb{R}^{n+1}$, let's calculate $\nabla_w^{\mathbb{S}^n} Z$:

$$\nabla_w^{\mathbb{R}^{n+1}} Z=\nabla_w^{\mathbb{R}^{n+1}}(e_{n+1}-x_{n+1}x)=-\nabla_w^{\mathbb{R}^{n+1}}x_{n+1}x.$$

Since $dx_i(w)=\langle \text{grad} (x_i),w \rangle =\langle e_i,w \rangle=w_i$, we get

$$-\nabla_w^{\mathbb{R}^{n+1}} Z=dx_{n+1}(w) x+x_{n+1}\nabla_w^{\mathbb{R}^{n+1}} x=w_{n+1}x+x_{n+1}(dx_1(w),\dots,dx_{n+1}(w))=$$

$$w_{n+1}x+x_{n+1}w.$$

So,

$$-\nabla_w^{\mathbb{S}^n} Z=-P(\nabla_w^{\mathbb{R}^{n+1}} Z)=P(w_{n+1}x+x_{n+1}w)=x_{n+1}w.$$

(Note $P(x)=0$, $P(w)=w$ since $w \perp x$).