3
$\begingroup$

$\newcommand{\id}{\operatorname{Id}_{TM}}$ 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).

$\endgroup$
3
$\begingroup$

$\newcommand{\id}{\operatorname{Id}_{TM}}$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 $\Gamma_{\theta\varphi}^\varphi = \cos \theta / \sin \theta.$

$\endgroup$
3
  • $\begingroup$ 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... $\endgroup$ Jan 22 '18 at 13:48
  • $\begingroup$ 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. $\endgroup$ Jan 22 '18 at 14:10
  • $\begingroup$ @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 $S^n \subset \mathbb R^{n+1}.$ $\endgroup$ Jan 22 '18 at 14:28
0
$\begingroup$

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 $w \perp x$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.