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Let $(M, g)$ be an odd-dimensional Riemannian manifold, and let $X$ be a Killing vector field on $M$ (i.e. $L_X g = 0$). Show that $X$ cannot have isolated zeros.

I know that if $X$ vanishes at a point $p \in M$, then $X$ is tangent to the small geodesic spheres around point $p$, and so it is normal along every radial geodesic starting from $p$. However, I do not how to to use these properties in order to show that there must be other zeros of $X$ in arbitrarily small geodesic balls centered at $p$. In particular, I do not know how to use the fact that the dimension of $M$ is odd.

S.T.
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    If the dimension of $M$ is odd, then the dimension of a geodesic sphere is even. What do you know about vector fields on even dimensional spheres? – Jason DeVito - on hiatus Nov 06 '21 at 18:23
  • Oh yes, I did not think of this: there are no no-where vanishing vector fields on even-dimensional spheres. Thank you! – S.T. Nov 06 '21 at 18:28
  • @JasonDeVito Could you please write your comment as an answer so I can accept it and close the question? – S.T. Nov 06 '21 at 18:28
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    I could write it up, but I am happily to let you write it up (with more details, of course) if you'd like. – Jason DeVito - on hiatus Nov 06 '21 at 18:29
  • @JasonDeVito I wrote it up, thank you again! – S.T. Nov 06 '21 at 18:45
  • @JasonDeVito: In the comment to the following answer it is shown that (if I've understand it correctly) integral curves of Killing vector fields are circular. is this correct in general for Killing vector fields? and because of this you commented about geodesic sphere at first glance? – C.F.G Nov 07 '21 at 11:53

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Using @JasonDeVito's comment, fix $p \in M$ and assume that $X_p = 0$. Then, we know that $X$ is tangent to the small geodesic spheres centered at $p$.

However, we know that these geodesic spheres are the level sets of the function $q \mapsto d_g(p, q)$ (which is smooth outside $p$ and has unit gradient vector field, so it is a submersion), and thus they are codimension-$1$ embedded submanifolds, hence, in particular, they are even-dimensional.

These geodesic spheres are also diffeomorphic to the standard unit sphere in their respective dimension; this is because $\mathrm{exp}_p$ is a local diffeomorphism at $0 \in T_pM$, because $g_p$ is the standard Euclidean inner product, and because $\mathrm{exp}_p$ maps the small Euclidean spheres centered at $0 \in T_pM$ to the small geodesic spheres centered at $p \in M$. So, by the hairy ball theorem, $X$ must vanish at some other point on every small geodesic sphere centered at $p$, so $p$ is not an isolated zero.

S.T.
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  • I confused a bit. You are restricting $X$ to spheres centered at $p$ then using hairy ball theorem? – C.F.G Nov 06 '21 at 20:06
  • Yes; the geodesic spheres are topologically spheres, and $X$ is a vector field on these geodesic spheres, which can then be pushed forward to a vector field on the standard unit sphere (and there is a correspondence between the zeros of $X$ and the zeros of its pushfoward). – S.T. Nov 06 '21 at 20:09
  • My problem is that why $X$ is a (tangent) vector field on these geodesic spheres? – C.F.G Nov 06 '21 at 20:19
  • Let $(p, t) \mapsto \phi^t(p)$ be the flow of $X$. Then, since $X_p = 0$, we have $\phi^t(p) = 0$ for every $t \in \mathbb{R}$. Moreover, for each fixed $t$, the map $\phi^t$ is an isometry (this is because $X$ is Killing). Thus, for $q$ near $p$, we have $$d_g(p, q) = d_g (\phi^t(p), \phi^t(q)) = d_g (p, \phi^t(q)), $$ so the integral curve of $X$ through $q$ has constant distance to $p$. – S.T. Nov 06 '21 at 20:29
  • in your last comment, $\phi^t(p) = 0$ or $\phi^t(p) = p$? Thanks. I learned from You and Jason. – C.F.G Nov 07 '21 at 06:41
  • Indeed, it should have been $\phi^t(p) = p$, not zero. Thank you for pointing that out. – S.T. Nov 07 '21 at 08:58