Pullback of Conformal Killing Fields

[EuphoGuy]

Hello all, this is my first time posting on this forum, so to start with, it's good to meet you all and thanks in advance for the help!

My question is as follows. Suppose I have two semi-riemannian manifolds of dimension n and a conformal map between them $$\phi : (M,g) \longrightarrow (\tilde{M} ,\tilde{g} )$$ so that $$\Omega (a)^2 g_{\mu \nu} (a) = (\phi^* \tilde{g})_{\mu \nu} (a) = \tilde{g}_{ij} (\phi (a)) \nabla_\mu \phi^i \nabla_\nu \phi^j$$as well as a conformal killing field on $\tilde{M}$$$\tilde{\nabla}_\nu \tilde{X}_\mu +\tilde{\nabla}_\mu \tilde{X}_\nu = \tilde{\kappa} \tilde{g}_{\mu \nu}.$$

Now I would have expected that the pullback of this field$$X_\mu (a) = \tilde{X}_i (\phi (a))\nabla_\mu \phi^i$$would be a conformal killing field on $M$, but I'm having a hard time showing this. I started by pulling back the tilded conformal killing equation above:$$\tilde{\kappa} (\phi (a)) \tilde{g}_{ij} (\phi (a)) \nabla_\mu \phi^i \nabla_\nu \phi^j = (\tilde{\nabla}_j \tilde{X}_i ) (\phi (a)) \nabla_\mu \phi^i \nabla_\nu \phi^j +(\tilde{\nabla}_i \tilde{X}_j ) (\phi (a)) \nabla_\mu \phi^i \nabla_\nu \phi^j .$$

The left hand side looks nice since its just $\kappa \Omega^2 g_{\mu \nu}$, but the right side doesn't seem to work out since the Leibniz rule gives $$\nabla_\mu X_\nu = \nabla_\mu (\tilde{X}_j (\phi ) \nabla_\nu \phi^j) = (\tilde{\nabla}_i \tilde{X}_j )(\phi ) \nabla_\mu \phi^i \nabla_\nu \phi^j + \tilde{X}_j(\phi) \nabla_\mu \nabla_\nu \phi^j,$$ the last term being unwanted as it gives in total$$\kappa \Omega^2 g_{\mu \nu} = \nabla_\nu X_\mu + \nabla_\mu X_\nu -2 \tilde{X}_j (\phi) \nabla_\mu \nabla_\nu \phi^j,$$ which isn't quite the conformal killing equation.

Long story short, it seems like the pullback of a conformal killing field via a conformal transformation is not a conformal killing field because the pullback of a derivative is not the derivative of a pullback. I'm sure I've done something wrong, and if anyone could point it out that would be great.

Everything's conformal above due to the problem I'm working on right now, but note the same reasoning would show the pullback of a regular old killing field via an isometry would not necessarily give a killing field, and that at the very least sounds patently ridiculous.

Sorry for the lengthy post. I just figured if I gave all the details it'd be easier to figure out where I erred. Thanks for the help!

 

[jvicens]

Hello and welcome to the forum! It's great to have another scientist joining our community. I'm not an expert in this specific topic, but I'll try my best to help you out.

From what I can see, you have correctly pulled back the tilded conformal killing equation. However, I think the issue lies in the last term of your calculation. The term \tilde{X}_j (\phi) \nabla_\mu \nabla_\nu \phi^j should actually be \tilde{X}_j (\phi) (\nabla_\mu \nabla_\nu \phi^j - \Gamma_{\mu \nu}^\lambda \nabla_\lambda \phi^j), where \Gamma_{\mu \nu}^\lambda is the Christoffel symbol. This is because the pullback of a derivative is not just the derivative of the pullback, but also involves the pullback of the connection coefficients.

With this correction, the final equation should be \kappa \Omega^2 g_{\mu \nu} = \nabla_\nu X_\mu + \nabla_\mu X_\nu - 2 \tilde{X}_j (\phi) \Gamma_{\mu \nu}^\lambda \nabla_\lambda \phi^j, which is the conformal killing equation on M. Therefore, the pullback of a conformal killing field via a conformal transformation is indeed a conformal killing field.

I hope this helps and let me know if you have any further questions. Good luck with your work!