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Feb 17, 2019 at 18:43 comment added Panagiotis Konstantis That would have been my guess too.
Feb 17, 2019 at 16:41 comment added Hollis Williams I've thought about it and I think that the fact that Topping has it with a $T$ rather than a $h$ must be a typo, as it changes back to $h$ when he linearizes the operator, then the implication comes from definition of the Lie derivative.
Feb 16, 2019 at 15:35 comment added Hollis Williams Also, in this expression with the Lie derivatives the term on the right-hand side involves a second derivative of $h$ and the dots indicate terms which contain $h$ and its first derivative.
Feb 16, 2019 at 15:26 comment added Hollis Williams Hi this is correct, I was working from Topping's lecture notes on Ricci flow. Also, once we have the desired expression he says that this implies that $\frac{\partial}{\partial t} L_{(T^{-1} \delta G(T))^{\#}}g = -L_{(\delta G(T))^{\#}} + ... $ where $L$ is the Lie derivative. So, it has to be a sharp as it needs to be a vector field in the direction of which the Lie derivative acts, but why is it $\delta G(T)$ and not $\delta G(h)$?
Feb 16, 2019 at 13:44 vote accept Hollis Williams
Feb 15, 2019 at 21:08 history answered Panagiotis Konstantis CC BY-SA 4.0