Timeline for Submanifolds of $\mathbb{R}^N$ whose local charts have uniformly bounded derivatives
Current License: CC BY-SA 3.0
7 events
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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 |