Timeline for is the closure of a totally convex set again totally convex?
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2010 at 15:10 | comment | added | Sergei Ivanov | Suppose it does. Then the intersection points are not conjugate along the vertical segment, so you can vary it (as a geodesic) by varying the endpoints. Let the endpoints go to each other along this supposed second geodesic. The family of geodesics defined by these endpoints (started from the vertical segment) has decreasing length, because they form sharp angles with that second geodesic. Hence this family will be free of conjugate points for all time, and you'll get an arbitrarily short pair of distinct geodesics with the same endpoints, a contradiction. | |
| Dec 10, 2010 at 14:29 | comment | added | Luc | I'm sorry to ask. My intuition tells me it should be clear. I really tried to get the computation but I didn't succeed. Could you give another hint in how to conclude that no geodesic can intersect the $y$-axis at two points with distance less than $\pi$? | |
| Dec 6, 2010 at 11:29 | vote | accept | Luc | ||
| Dec 4, 2010 at 1:41 | history | edited | Sergei Ivanov | CC BY-SA 2.5 |
fixed typos
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| Dec 3, 2010 at 23:12 | history | answered | Sergei Ivanov | CC BY-SA 2.5 |