# Killing vector fields of a conformally flat Riemannian metric

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!

• 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 $f$. Nov 22, 2018 at 11:35
• @RobertBryant Thanks a lot! So this answer negatively to my question. Nov 22, 2018 at 12:42

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"
• 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. Nov 21, 2018 at 16:22
• @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 $g$. Nov 21, 2018 at 16:26
• @Onil90 Killing vector fields are special cases of conformal vector fields when the factor vanishes. This will happen when $X(f)=0$ Nov 21, 2018 at 21:44
• 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" Nov 21, 2018 at 21:49