Timeline for The logarith map as a contraction [closed]
Current License: CC BY-SA 3.0
16 events
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Oct 5, 2014 at 18:55 | vote | accept | the42nd | ||
Oct 5, 2014 at 15:49 | history | closed |
YCor Chris Godsil Suvrit Stefan Kohl♦ D.-C. Cisinski |
Needs details or clarity | |
Oct 5, 2014 at 8:02 | comment | added | YCor | Would you also define "contraction mapping"? this has several possible meanings. | |
Oct 5, 2014 at 1:29 | answer | added | dennis sullivan | timeline score: 8 | |
Oct 4, 2014 at 23:06 | comment | added | the42nd | Yes, I was hoping to gain some insight from what appears to be a bad analogy. I've deleted that part. | |
Oct 4, 2014 at 22:43 | history | edited | the42nd | CC BY-SA 3.0 |
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Oct 4, 2014 at 22:29 | comment | added | YCor | You still haven't corrected the wrong example noticed by Joonas. On $\mathbb{R}$ the logarithm is not contracting, you seem to be confusing between the "usual" logarithm and local inverses of the exponential map for Riemannian manifolds, which, in the case of $\mathbb{R}$, is not the logarithm map. | |
Oct 4, 2014 at 22:24 | history | edited | the42nd | CC BY-SA 3.0 |
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Oct 4, 2014 at 22:13 | comment | added | the42nd | Correction: "..we are looking at the logarithm map in a neighbourhood around $q_0$, where the exponential map is injective." | |
Oct 4, 2014 at 22:12 | history | edited | the42nd | CC BY-SA 3.0 |
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Oct 4, 2014 at 22:00 | comment | added | the42nd | Thanks for the feedback. Q is a Riemannian manifold and TQ it's tangent bundle. In particular, we are looking at the logarithm map in a neighborhood around $q_0$ where it is injective. I'll edit my post. | |
Oct 4, 2014 at 21:56 | history | edited | the42nd | CC BY-SA 3.0 |
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Oct 4, 2014 at 20:34 | review | Close votes | |||
Oct 5, 2014 at 15:49 | |||||
Oct 4, 2014 at 19:40 | comment | added | Joonas Ilmavirta | What are $TQ$ and $Q$? It seems to me that $Q$ is a Riemannian manifold and $TQ$ its tangent bundle. In this case your first example is wrong, since the exponential map on a manifold homeomorphic to $\mathbb R$ is always isometric. The logarithm is (typically) only locally defined and arbitrarily close to being an isometry when the domain is shrinked. If your logarithm means something else, these remarks don't apply, but you could make your question clearer. | |
Oct 4, 2014 at 19:09 | review | First posts | |||
Oct 4, 2014 at 19:33 | |||||
Oct 4, 2014 at 19:09 | history | asked | the42nd | CC BY-SA 3.0 |