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Let $M$ be a Riemannianiam manifold with Levi-Civita connection and $X $ be a smooth vector field on $M$. Let $\phi : (-\epsilon, \epsilon) × V \to M$ be the local flow of $X$ in $M$.

Problem is- if $X$ is a killing field on $M $, $p \in M$ and $U$ be a normal neighborhood of $p$. If $p$ is the unique point satisfying $X(p)= 0$ then in $U$, $X$ is tangent to the geodesic spheres centred at $p$.

My attempt: With some hints, I've got from here and there, I know that first, we need to have that $\phi(t,p) = p \forall t\in(-\epsilon, \epsilon)$, which I am able to show.

Next, I think to show the required, it suffices to show that for any $q \in exp_{p}(S(0,\epsilon) = S_({p}(\epsilon))$, a geodesic sphere contained in $U$ which is of the form $q= exp_{p}(v)$ for some $v$ such that $|v|=\epsilon$, the radial vector joining $p$ to $q$ is orthogonal to $X(q)$.

But I don't know how to proceed to show this. Also, where do we use the uniqueness of point $p$ as the only zero of $X$?

It would be lovely if anyone could provide me some hint to solve this. Thank you!!

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It's a standard fact that, for any $t \in (-\varepsilon,\varepsilon)$, the flow $\phi_t(\cdot) = \phi(t,\cdot)$ along $X$ for time $t$ is an isometry. Since each $\phi_t$ fixes $p$, the result follows immediately from the first variation formula.

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    $\begingroup$ I'm sorry but we haven't done first variation formula in class, it's a problem in do Carmo's chapter 3 while he introduces curvature and such not before chapter 4 so I believe we could do it without using that. $\endgroup$
    – Shreya
    Commented Apr 5, 2019 at 3:48
  • $\begingroup$ Also, do you mind telling where exactly uniqueness of point $p$ is being used? $\endgroup$
    – Shreya
    Commented Apr 5, 2019 at 5:33
  • $\begingroup$ The argument doesn't require that $p$ be the unique zero of $X$. It's easy to see that this isn't essential; for example, consider the Killing field $X$ obtained as the variational derivative of rotation around a line in $\mathbb{R}^3$. Every point along the line will be a zero of $X$, and each orbit circle is equidistant from each point on the line. $\endgroup$
    – James Dibble
    Commented Apr 5, 2019 at 19:55

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