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Jan 19, 2021 at 10:51 comment added Igor Khavkine @Andresito Yes, $g^{jk}\nabla_k$ in what I wrote is the covariant derivative wrt to $y$ with tensor index raised by the metric $g^{jk}(y)$ at $y$. It is easy to see that $\sigma^j(x,y) = O(x-y)$ in any coordinate system $(y^j)$, hence good enough for a Taylor expansion. Even better, $\sigma^j(x,y) = z^j$ when $(z^j)$ are normal coordinates centered at $x$ (compare their definitions!).
Jan 19, 2021 at 6:26 comment added Pether Cll_ Thanks for your answer! but why you say a good substitute for the difference of coordinates? And second, what is $\nabla_k \sigma(x,y)$ I understand this is a covariant derivative, so ¿ 'k' is the coordinate index?. Thanks!
Jan 11, 2021 at 18:07 history answered Igor Khavkine CC BY-SA 4.0