= Solution
The product rule for the <Lie derivative> and the <Killing equation> give
$$
\mathcal L_K\widetilde g
=\mathcal L_K(\Lambda^2g)
=K(\Lambda^2)g+\Lambda^2\mathcal L_Kg
=2\frac{K(\Lambda)}{\Lambda}\widetilde g.
$$
Consequently $K$ is a <Conformal Killing vector field> for the rescaled <metric tensor>, with smooth <conformal Killing factor>
$$
\boxed{\sigma=K(\log|\Lambda|)=\frac{K^a\partial_a\Lambda}{\Lambda}.}
$$
It remains an actual <Killing vector field> precisely when $K(\Lambda)=0$: the <conformal factor> must be constant along its <integral curves of a vector field>.
Back to article page
