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What is the variation of Gauss-Bonnet term total derivative of?

i.e. Variation of Gauss-Bonnet combination $= \nabla_{\mu} C^{\mu}$.

What's $C^{\mu}$ in 4-dimensions?

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  • $\begingroup$ A fairly careful and recent discussion of this can be found here: arxiv.org/abs/1008.5154v1. They discuss this using the language of forms (where the result has been known for a long time), and using just functions of the metric. $\endgroup$ – Lloyd Sep 19 '17 at 14:17
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If you just want to know why the Gauss-Bonnet Term is topological, you should take a look at the generalized gauss bonet theorem.

The integral over the gauss-bonet term is proportional to the euler-characteristic, which is a topological invariant, so it can't contribute to the dynamics.

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According to this website, for a four dimensional manifold, $$ G = \nabla_{\alpha}J^{\alpha}, $$ where $$ G = R^2 -4 R_{\alpha \beta} R^{\alpha \beta} + R_{\alpha \beta \gamma \delta}R^{\alpha \beta \gamma \delta}, $$ and $$ J^{\alpha} = \epsilon^{\alpha \beta \gamma \delta} \epsilon_{\rho \sigma}^{\;\;\; \mu \nu} \Gamma^{\rho}_{\;\; \mu \beta} \left[ \frac{1}{2} R^{\sigma}_{\;\; \nu \gamma \delta} + \frac{1}{3} \Gamma^{\sigma}_{\;\; \lambda \gamma} \Gamma^{\lambda}_{\;\; \nu \sigma} \right]. $$ So $G$ becomes a topological term in the action, which does not contribute to the dynamics. However, I have yet to check it myself...

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  • $\begingroup$ The link is now broken, and I've been unable to find the Living Review that mentions this result. Could you update the link? Thanks! $\endgroup$ – Lloyd Sep 14 '17 at 14:46
  • $\begingroup$ A nice discussion of writing the Lanczos-Lovelock Lagrangians as total derivatives can be found here: arxiv.org/abs/1008.5154v1. In this article they also provide a fairly careful discussion as to why the equation you're quoting is incorrect for general metrics. $\endgroup$ – Lloyd Sep 19 '17 at 14:13

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