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Feb 8, 2016 at 14:26 comment added Robert Bryant @Deane: Yes, the exponential map provides a $C^1$ set of 'normal coordinates'. To see this, assume that $M=\mathbb{R}^n$. Then using the notation in my answer, there exists a smooth map $E_2:U\to\mathbb{R^n}$ so that $\gamma_u(t) = tu + t^2E_2(u,t)$ for all $(u,t)\in U\subset\Sigma\times\mathbb{R}$. From this and the compactness of $\Sigma_p$, one sees that, for all $\delta>0$ sufficiently small, there is a constant $K>0$ such that $$|\,\exp_0(v)-v\,|\le K\,|v|^2$$ for all $v$ with $|v|<\delta$. Of course, this implies that $\exp_0$ is $C^1$ at $0\in\mathbb{R}^n$.
Feb 5, 2016 at 22:33 comment added Deane Yang So there are $C^1$ normal coordinates?
Feb 5, 2016 at 18:18 review Low quality posts
Feb 5, 2016 at 19:05
Feb 5, 2016 at 18:16 comment added ABIM Oh so it fails to be $C^{\infty}$, can you explain why? Or provide a simple example?
Feb 5, 2016 at 18:03 history answered alvarezpaiva CC BY-SA 3.0