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Nov 4, 2017 at 20:20 comment added Jaap Eldering I just noticed that in my notes I've made the same mistake: the choice of atlas on the manifold should be part of the definition of uniformity, instead of saying "there exists an atlas ..."
Nov 4, 2017 at 14:41 comment added Francesco Polizzi ok, thank you again. This is very useful.
Nov 4, 2017 at 13:25 comment added Jaap Eldering Yes, any compatible atlas. Also, as a side remark: an atlas satisfying your condition can never be maximal, see also my remark under definition 4 in my notes.
Nov 4, 2017 at 12:55 comment added Francesco Polizzi Do you mean any atlas compatible with the original one, right? Of course, I want that $M$ remains the same embedded submanifold of $\mathbb{R}^N$ as before.
Nov 4, 2017 at 12:23 comment added Jaap Eldering @FrancescoPolizzi: indeed. So if you're ok with any atlas on $M$, then your condition can always be satisfied. Using bounded geometry essentially means that you're restricting to atlases generated by exponential maps. I also looked a bit into an alternative definition of uniformity that's only expressed in terms of the atlas, see my notes jaapeldering.nl/files/uniform-mflds.pdf. This definition is essentially equivalent to bounded geometry wrt. a class of metrics on $M$.
Nov 3, 2017 at 22:08 comment added Francesco Polizzi Thank you for your answer. When you write "all embedded manifolds $M \subset \mathbb{R}^N$ satisfy it", you mean (for instance) that if we have $$||i_{\alpha} \circ \phi_{\alpha}^{-1}||_{\infty} \leq M_{\alpha} \quad \mathrm{on} \; \; V_{\alpha}$$ then we can replace every $\phi_{\alpha}$ with $M_{\alpha} \phi_{\alpha}$ in order to obtain the uniform bound $$||i_{\alpha} \circ \phi_{\alpha}^{-1}||_{\infty} \leq 1 ?$$
Nov 3, 2017 at 20:24 history answered Jaap Eldering CC BY-SA 3.0