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Let $f: \mathbb{R}^n \to \mathbb{R}$ be a smooth function and let's consider the conformally flat Riemannian metric $g = e^f \delta_{ij} dx^idx^j$ on $\mathbb{R}^n$.

Is it true that the Killing vector fields of the manifold $(\mathbb{R}^n, g)$ are exactly the Killing fields of the Euclidean space such that $f$ is constant along their flow? For sure if $K$ is an Euclidean Killing vector field with such property, then it must be a Killing field also for the manifold $(\mathbb{R}^n, g)$, but I'm not completely sure that those are exactly all the possible Killing fields of $(\mathbb{R}^n, g)$. I checked some simple cases and it seems to be the case.

Is it true in general? Or is it possible to find a counterexample where $(\mathbb{R}^n, g)$ has some symmetries that do not arise from symmetries of $f$?

Any help will be very appreciated! Thanks a lot!

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    $\begingroup$ Try the example (with $n=2$) where $$e^f = \frac{4}{(1+{x_1}^2+{x_2}^2)^2}.$$ You'll find that the space of Killing fields of $g$ has dimension $3$, but most of these Killing fields are not tangent to the level sets of $f$. $\endgroup$ – Robert Bryant Nov 22 '18 at 11:35
  • $\begingroup$ @RobertBryant Thanks a lot! So this answer negatively to my question. $\endgroup$ – Onil90 Nov 22 '18 at 12:42
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Hint
Assume that $h=e^fg$ is a conformal change of $g$ on a manifold $M$. Let $X$ be a vector field on $M$. Then
$${\cal{L}}_{X}h={\cal{L}}_{X}(e^fg)\\ ={\cal{L}}_{X}(e^f)g+e^f{\cal{L}}_{X}g\\ =e^fX(f)g+e^f{\cal{L}}_{X}g$$
where ${\cal{L}}_{X}$ denotes the Lie derivative in direction of $X$ on $M$. This leads to say "Killing fields of both metrics are the same if and only if $f$ is constant along the flow lines of these vector fields"

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  • $\begingroup$ Using this computation it is trivial to check that if $X \in \mathcal{Kill}(M, g)$ and if $f$ is constant along the flow of $X$, then $X \in \mathcal{Kill}(M, h)$. The problem is that in general the converse is not true. But it might be true in the case where $(M, g)$ is the standard Euclidean space. $\endgroup$ – Onil90 Nov 21 '18 at 16:22
  • $\begingroup$ @Onil90 Assume that $X$ is Killing on $(M,h)$, then it is conformal on $(M,g)$ with conformal factor $-X(f)$. It does not matter what is $g$. $\endgroup$ – Semsem Nov 21 '18 at 16:26
  • $\begingroup$ I am not interested in conformal Killing fields, but in Killing fields. $\endgroup$ – Onil90 Nov 21 '18 at 21:30
  • $\begingroup$ @Onil90 Killing vector fields are special cases of conformal vector fields when the factor vanishes. This will happen when $X(f)=0$ $\endgroup$ – Semsem Nov 21 '18 at 21:44
  • $\begingroup$ This leads to say "Killing fields of both metrics are the same if and only if $f$ is constant along the flow lines of these vector fields" $\endgroup$ – Semsem Nov 21 '18 at 21:49

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