Timeline for Is there a notion of a connection for which the horizontal lift of a curve depends on its orientation?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2019 at 9:44 | comment | added | Praphulla Koushik | I had written that as an answer and then I was not sure as question is not clear.. so deleted my answer... So, left it as a comment... | |
| Feb 2, 2019 at 8:19 | comment | added | Michael Bächtold | I agree with @PraphullaKoushik. Simply put: he lifts of the two curves $t\mapsto \gamma(t)$ and $t\mapsto \gamma(1-t)$ are not the same (their images are), so the usual lift is already orientation dependent. | |
| Jan 31, 2019 at 8:05 | comment | added | Praphulla Koushik | Given a curve $\gamma$ on $M$ and fixing a point $x$ in fibre of $\gamma(0)$, there exists a curve that starts at $x$. So, lift has starting point as $x$. Suppose you choose $t\mapsto \gamma(1-t)$, you fix a point $y$ in fibre of $\gamma(1)=\gamma(1-0)$, you get a lift whose starting point is $y$. This does not say horizantal lift of $\gamma(t)$ and $\gamma(1-t)$ are same if you are looking from orientation perspective. What is that I am misunderstanding in your question? | |
| Jan 31, 2019 at 8:00 | answer | added | alvarezpaiva | timeline score: 2 | |
| Jan 31, 2019 at 6:39 | answer | added | Praphulla Koushik | timeline score: 0 | |
| Jan 31, 2019 at 0:14 | history | edited | Josh Kirklin | CC BY-SA 4.0 |
added 64 characters in body
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| Jan 31, 2019 at 0:04 | history | asked | Josh Kirklin | CC BY-SA 4.0 |