# Variation of Christoffel symbol and Lie derivative

I've also asked this question on Math Overflow; I hope that asking in two separate fora is not a solecism.

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 Levi-Civita 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 symbol there are many terms, but in Bardeen and Zumino's famous paper on anomalies (Nucl. 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 infinitesimal co-ordinate 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.

• (a) SE sites are Q&A sites, not fora, (b) cross-posting is generally frowned upon, especially when it's a copy-paste job & not tweaked for the audience. – Kyle Kanos Jun 26 '15 at 16:25
• A solecism is a grammatical error. Perhaps you meant a faux pas? Posting on multiple SE sites is a mild breach of etiquette though I doubt the lynch mob will be stirring just yet. And while I'm posting frivolous comments, neither this site not MO are fora. – John Rennie Jun 26 '15 at 16:25
• Thanks for the mild rebuke. I'll not sin again. I'm not sure about the gramma points: Is not the plural of forum "fora" forums sounds wrong somehow? Also the dictionay gives solecism as a breach of good manners; a piece of incorrect behavior wit faux pas, gaffe, impropriety, social indiscretion, etc as synoyms :) – mike stone Jun 26 '15 at 16:40
• Thanks for the mild rebuke. I'll not sin again! Also @JohnRennie. My dictionary defines a solecism as "a breach of good manners; a piece of incorrect behavior" with faux-pas or grammar error as synonyms. But using the latin plural "fora" was just pretension on my part --sorry ;) – mike stone Jun 26 '15 at 16:46
• I believe we are getting sidetracked from the actual question. – Horus Jun 26 '15 at 16:48

## 1 Answer

It took me a couple of days to figure it out, but I can now answer my own question:

The problem with directly substituting the expression for $\delta g_{\mu\nu}$ into that for ${\delta \Gamma^\alpha}_{\beta\mu}$ is that there is no obvious source for the term $\xi^\sigma \partial_\sigma { \Gamma^\alpha}_{\beta\mu}$ that has to be the leading term in the Lie derivative of the connection. We can however proceed as follows $$\delta {\Gamma^{\alpha}}_{\beta\mu}= \frac 12 g^{\alpha\lambda}(\nabla_\beta(\nabla_\lambda \xi_\mu+\nabla_\mu\xi_\lambda)+ \nabla_\mu(\nabla_\beta \xi_\lambda+\nabla_\lambda\xi_\beta) -\nabla_\lambda(\nabla_\beta \xi_\mu+\nabla_\mu\xi_\beta))$$ $$=\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 g^{\lambda\alpha}([\nabla_\beta,\nabla_\lambda]\xi_\mu + [\nabla_\mu,\nabla_\lambda]\xi_\beta)$$ $$=\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 g^{\lambda\alpha}(-{R^\sigma}_{\mu\beta\lambda}- {R^\sigma}_{\beta\mu\lambda})\xi_\sigma$$ $$=\frac 12 (\nabla_\beta\nabla_\mu+\nabla_\mu\nabla_\beta)\xi^\alpha +\frac 12 \xi^\sigma({R^\alpha}_{\beta\sigma \mu}+ {R^\alpha}_{\mu\sigma \beta} )$$

$$=\xi^\sigma \partial_\sigma {\Gamma^{\alpha}}_{\beta\mu} +(\partial_\beta \xi^\lambda ){\Gamma^\alpha}_{\lambda\mu}+(\partial_\mu\xi^\lambda) {\Gamma^\alpha}_{\beta \lambda}- (\partial_\lambda \xi^\alpha) {\Gamma^{\lambda }}_{\beta\mu}+ \partial^2_{\beta\mu} \xi^\alpha$$ $$\equiv {({\mathcal L}_\xi \Gamma)^{\alpha}}_{\beta\mu}$$ In passing from the third line to the fourth we have used some symmetries of the Riemann tensor. We have then (somewhat tediously) substituted the usual expression for the curvature and the covariant derivatives in the last line. I end up with almost the same expression as Bardeen and Zumino, but with a sign difference in the inhomogenous part of the Christoffel symbol transformation

The net result is that first transforming the metric under the diffeomorphism and then computing the connection gets the same result as first computing the connection and then transforming it under a diffeomorphism. This had to be true of course, but there are many places where one can get signs wrong...!