Timeline for Subdividing a Compact Bounded Curvature Manifold into Charts with Bounded Lipschitz Constant
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 2, 2019 at 7:22 | comment | added | macbeth | I don't see why the Lipschitz constant $L$ should depend on curvature: domain-rescaling an $L$-Lipschitz chart gives a new chart which is 1-Lipschitz. Indeed, let $U\subseteq\mathbb{R}^k$ be open and let $\phi:U\to M$ be a $L$-Lipschitz chart for $M$, i.e., $\forall x,y\in U, |\phi(x)-\phi(y)|\leq L|x-y|$. Then, defining $\hat\phi$ by $\hat\phi(x):=\phi(L^{-1}x)$, we have that $\forall x,y\in LU, |\hat\phi(x)-\hat\phi(y)|=|\phi(L^{-1}x)-\phi(L^{-1}y)|\leq |x-y|$. | |
| Jun 1, 2019 at 4:49 | comment | added | user141213 | Yes, I believe the Lipschitz-ness is important. The manifold is embedding in $\mathbb{R}^d$, it has some predefined geometry. I want to construct a set of coordinate charts on $M$ such that each chart is Lipschitz. This must depend on the curvature; as $L$ becomes smaller I should need more charts to ensure that each satisfy this Lipschitz property since the curvature is constant. | |
| May 29, 2019 at 17:26 | comment | added | macbeth | Is the Lipschitz-ness aspect of your question really relevant? The number of smooth charts needed to cover $M$ is finite and a topological invariant (see mathoverflow.net/questions/2615/… ), and the charts in a smooth atlas can be shrunk very slightly to give a new atlas in which all charts $\phi$ have $\sup |D\phi|<\infty$. Domain-rescaling these charts gives a third atlas in which all charts have $\sup |D\phi| = 1$. Now we have a 1-Lipschitz atlas whose size is a topological invariant. | |
| May 28, 2019 at 23:30 | review | First posts | |||
| May 28, 2019 at 23:38 | |||||
| May 28, 2019 at 23:28 | history | asked | user141213 | CC BY-SA 4.0 |