Under an infinitesimal diffeomorphism the Riemann metric changes by the Lie derivative $$ \delta g_{\mu\nu} = ({\mathcal L}_\xi G)_{\mu\nu}=\nabla_\mu \xi_\nu+\nabla_\nu \xi_\mu $$ and under a change of metric the LeviCivita Christoffel symbol changes by $$ \delta {\Gamma^{\alpha}}_{\beta \mu}= \frac 12 g^{\alpha \lambda} (\nabla_{\beta} \delta g_{\lambda \mu}+ \nabla_{\mu} \delta g_{\beta \lambda} \nabla_{\lambda} \delta g_{\beta \mu}) $$ If one plugs the variation of the metric into the variation of the Christoffel sysmbol there are many terms, but in Bardeen and Zumino's famous paper on anomalies ((Nuc Phys B 244 (1984) 421) they assert (eq 4.10, but in my notation) that $$ \delta {\Gamma^{\alpha}}_{\beta \mu}=\xi^\tau \partial_\tau {\Gamma^{\alpha}}_{\beta \mu}+ \partial_\beta \xi^\sigma {\Gamma^{\alpha}}_{\sigma \mu}+ \partial_\mu \xi^\sigma {\Gamma^{\alpha}}_{\beta \sigma}  \partial_\sigma \xi^\alpha {\Gamma^{\sigma}}_{\beta \mu}\partial^2_{\beta\mu}\xi^\alpha $$ This last equation makes sense as an infintesimal coordinate transformation or Lie derivative and greatly simplifies their later algebra. Is there a straightforward way to see the equivalence of the two computations? It's not plug and chug, but seems to lead into wasteland of Bianchi identities.

$\begingroup$ Ive posted an answer to to my question on physics stack exhange under the heading ``Variation of Christoffel symbol and Lie derivative'' $\endgroup$– mike stoneMay 4, 2017 at 15:00
1 Answer
If you consult Sec 3.1 of Wald's book on GR, you will see how Christoffel symbols can be interpreted as tensors parametrizing the difference between the coordinate connection $\partial$ and the LeviCivita connection $\nabla$, symbolically $\Gamma=\nabla\partial$ (cf. Wald's (3.1.7)). You are interested in $\Gamma'=\nabla\partial'$, where $\partial'$ is the new coordinate connection. Writing this in another way, $\Gamma' = (\nabla  \partial) + (\partial  \partial') = \Gamma + (\partial  \partial')$. Both terms in the last equality are tensors, so once you know them in one coordinate system, you know them in all coordinate systems, following the usual tensor transformation properties. The last formula in your question is essentially $\Gamma'$ written in the primed coordinates, to first order in the difference between the original and primed coordinates. The $\Gamma$dependent terms are just the tensor transformation rule for $\Gamma$ interpreted as a tensor, while the $\partial^2\xi$ term is precisely the $(\partial  \partial')$ difference. I'll leave checking that last claim as an exercise.

$\begingroup$ Thanks a ton! I did eventually figure it out on my own. The problem is that it is easy to get signs wrong when using the general argument  and I think that B and Z did exactly that as there should be no minus sign before the $\partial^2_{\mu\nu}\xi^\alpha)$ $\endgroup$ Jun 28, 2015 at 14:56

$\begingroup$ If you are happy with the answer, you may want to click the check mark next to it, to mark it as correct. Otherwise, the question will periodically appear on the front page as unanswered. $\endgroup$ Jun 28, 2015 at 19:56