do Carmo: near isolated zeros, killing field tangent to geodesic spheres

Question

Exercise 3.5b of do Carmo's Riemannian Geometry asks the reader to prove that given a Killing field $X$ on a manifold $M$, an isolated zero $p$ of $X$, and a normal neighborhood $U$ of $p$ in which $X$ has no other zeros, $X$ is tangent (in $U$) to the geodesic spheres centered at $p$.

I have a "proof", but it doesn't use the fact that $X_p=0$, so I must be missing something. Here's the "proof":

Let $\phi_t$ be the flow of $X$ on $U$, and let $q=\exp_p{v}\in U$. Since $\phi_t$ is an isometry, $\phi_t(q) = \exp_{\phi_t(p)} d(\phi_t)_p v$. So, denoting the radial distance function at $p$ by $r_p$, we get $$r_{\phi_t(p)}(\phi_t(q)) = \|d(\phi_t)_pv\| = \|v\|,$$ which means $\frac{d}{dt}\!\left[r_{\phi_t(p)}(\phi_t(q)) \right]=0$. But by the chain rule, $$\frac{d}{dt}\!\left[r_{\phi_t(p)}(\phi_t(q)) \right] = \begin{pmatrix} \text{stuff} &d(r_p)_q X_q \end{pmatrix},$$ where $\text{stuff}$ is an unimportant block matrix. Thus, $d(r_p)_q X_q=0$. Hence, $\left\langle \frac{\partial}{\partial r_p}, X\right\rangle\equiv 0$. Thus, $X$ is tangent to the geodesic spheres in $U$.

What did I miss?

Answer 1

Here is a solution that makes the assumption $X_p = 0$ very transparent.

The geodesic vector field has a flow $G\colon I\times TU\to TM$ (cf. do Carmo p. 65), where by definition, $$ \gamma(1,p,v) \equiv (\pi\circ G)(1,p,v) = \exp_pv, $$ where $\pi\colon TM\to M$ is the canonical projection. It will be convenient in what follows to use this $\gamma$. If $\theta\colon (-\varepsilon,\varepsilon)\times U\to M$ is the flow of $X$, we will also make use of the two maps \begin{align*} q&\mapsto \theta_t(q) \equiv \theta(t,q),\\ t&\mapsto \theta^{(q)}(t) \equiv \theta(t,q). \end{align*} For each fixed $q\in U$, $\big(\theta^{(q)}\big)'(0) = X_q$.

Let $S_\delta(p)\subset U$ be a geodesic sphere centered at $p$, and let $q\in S_\delta(p)$. Then $q$ is the endpoint $(s=\delta)$ of a certain unit-speed geodesic—concretely there is some $v\in S_1(0)\subset T_pM$ so that $q = \gamma(\delta,p,v)$. By the assumption that $X$ is Killing, for every fixed $t\in(-\varepsilon,\varepsilon)$, $\theta_t$ is an isometry, so $$ [0,\delta]\ni s\longmapsto \gamma\big(s,p,(d\theta_t)_p(v)\big) $$ is a unit-speed geodesic.

Consider the parametrized surface (cf. Definition 3.3 of do Carmo, p. 67) $$ f(s,t) \equiv (\theta_t\circ\gamma)(s,p,v). $$ For each $t$, the curve $s\mapsto f(s,t)$ is a unit-speed geodesic starting from $p$ because $\theta_t$ is an isometry. By the chain rule, $$ \frac{\partial f}{\partial s}(0,t) = (d\theta_t)_{p}\bigg(\frac{\partial}{\partial s}\gamma(0,p,v)\bigg) = (d\theta_t)_p(v). $$ By definition of $\gamma$, and our observation that $s\mapsto f(s,t)$ is a unit-speed geodesic with endpoint ($s=\delta$) in $S_\delta(p)$, we thus see that $$ f(s,t) = \gamma(s,p,(d\theta_t)_p(v)). $$ For each fixed $t$, the curve $s\mapsto f(s,t)$ is a geodesic by the discussion above, so we have for all $(s,t)$, $$ \frac{D}{\partial s}\frac{\partial f}{\partial s} = 0. $$ Moreover, $\frac{\partial f}{\partial s}(\delta,0) = \frac{\partial}{\partial r}\big|_q$. Looking at the definition, we see that $\frac{\partial f}{\partial t}(\delta,0) = X_q$. Now we see that we would like to show that $$ \left< \frac{\partial f}{\partial s}(\delta,0),\frac{\partial f}{\partial t}(\delta,0)\right> = 0. $$ For all $(s,t)$, by the lemma of symmetry (cf. do Carmo p. 68) \begin{align*} \frac{\partial}{\partial s}\left< \frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\right> &= \left< \underbrace{\frac{D}{\partial s}\frac{\partial f}{\partial s}}_{=\ 0},\frac{\partial f}{\partial t}\right> + \left< \frac{\partial f}{\partial s},\frac{D}{\partial s}\frac{\partial f}{\partial t}\right>\\ &= \left< \frac{\partial f}{\partial s}, \frac{D}{\partial t}\frac{\partial f}{\partial s}\right> & \text{(symmetry)}\\ &= \frac{1}{2}\frac{\partial}{\partial t}\left< \frac{\partial f}{\partial s},\frac{\partial f}{\partial s}\right>. \end{align*} Now, for any fixed $s$, $\left< \frac{\partial f}{\partial s},\frac{\partial f}{\partial s}\right>$ is constant as a function of $t$ because $\theta_t$ is an isometry, so the last expression in the above display is identically $0$.

We thus deduce in particular that $\left< \frac{\partial f}{\partial s}(s,0),\frac{\partial f}{\partial t}(s,0)\right>$ is constant as a function of $s\in[0,\delta]$. By smooth dependence on initial conditions, \begin{align*} \lim_{s\to 0} \frac{\partial f}{\partial t}(s,0) = \lim_{s\to 0}\big(\theta^{(\gamma(s,p,v))}\big)'(0) = \big(\theta^{(p)}\big)'(0) = X_p = 0, \end{align*} so we conclude that $\left< \frac{\partial f}{\partial s}(\delta,0),\frac{\partial f}{\partial t}(\delta,0)\right> = 0$, which was our goal.

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We essentially reproduced a large portion of the proof of Gauss' lemma (cf. do Carmo p. 69), so there may be a way to shorten this proof using Gauss' lemma.

Answer 2

Let $q \in U$ be arbitrary. There is a unique geodesic $\gamma$ in $U$ that joins $p$ and $q$. $\varphi_{t_0}$ is isometry, so $\varphi_{t_0} \circ \gamma$ is also a geodesic. Note that $X(p) = 0$, so $\varphi(t, p) = p$ for all $t$. Thus $\varphi_{t_0} \circ \gamma$ is geodesic with one endpoint at $p$ and of same length as $\gamma$, say $r$. Hence the other endpoint of $\varphi_{t_0} \circ \gamma$, i.e. $\varphi(t_0, q)$ is on the geodesic sphere $S_r(p)$ for all $t_0 \in (-\epsilon, \epsilon)$. Therefore, $X_q$ is tangent to $S_r(p)$.