Timeline for Geodesics on zero-curvature regions of closed surfaces of genus > 1 of non-positive curvature
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Sep 5, 2010 at 22:11 | answer | added | Sergei Ivanov | timeline score: 3 | |
| Aug 22, 2010 at 20:55 | answer | added | Sam Nead | timeline score: 0 | |
| Aug 4, 2010 at 10:15 | history | edited | BS. |
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| Aug 3, 2010 at 15:08 | comment | added | Nicolas Fernandez-Arias | Sergei: thank you, that clarifies a lot. Does this mean that there always is some way to identify D with the universal cover of M such that straight lines on zero curvature regions are Euclidean lines, anyways? That would be enough for my purposes. Also, the more important question is whether these geodesics on zero curvature regions are unaffected by negative curvature regions A to which they come close, even possibly tangent, as described in my question. It seems, from Willie Wong's post, that they would not be...is this correct? | |
| Aug 3, 2010 at 12:36 | history | edited | Charles Matthews | CC BY-SA 2.5 |
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| Aug 2, 2010 at 23:05 | comment | added | Sergei Ivanov | Rephrasing Willie Wong's comment: The issue is that the universal cover of $M$ is not $D$. You can identify it with $D$ in some way, but there is no "one true way" to do so. And what is a straight line after one identification is not straight after another - they differ by a map $\phi$ that "changes coordinates". Even if you restrict yourself to conformal coverings (as Tom Goodwillie suggested) $\phi$ can be any conformal bijection from $D$ to itself, and can transform any straight line to a circle arc. | |
| Aug 2, 2010 at 21:48 | comment | added | Tom Goodwillie | Maybe you mean to specify that the covering projection from the unit complex disk to the manifold is chosen to be conformal (w.r.t. the usual conformal structure of the complex numbers and the conformal structure determined by the given metric on the manifold). So that part of your question is, "If a region of the plane is given a metric in which right angles are what they appear to be and the curvature is zero, must geodesics be straight lines?" ? | |
| Aug 2, 2010 at 20:47 | history | edited | Nicolas Fernandez-Arias | CC BY-SA 2.5 |
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| Aug 2, 2010 at 20:46 | comment | added | Nicolas Fernandez-Arias | I edited my question -- I had forgotten to mention that D was supposed to be the universal cover of M. Willie: I am pretty sure I understand what your'e saying; does this change anything? Also, let me just check: even though the geodesics on regions of zero curvature may not look like straight lines, they will still not be affected in the scenario I describe, right? So if they happen to look like straight lines, they will remain so? | |
| Aug 2, 2010 at 20:41 | history | edited | Nicolas Fernandez-Arias | CC BY-SA 2.5 |
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| Aug 2, 2010 at 20:20 | comment | added | Willie Wong | Your question needs to be better formulated: let $\phi$ be an arbitrary smooth map from $\Omega\in\mathbb{R}^2$ to itself. Take the Euclidean metric $e$ on $\Omega$, and consider the pull-back metric $\phi^*e$. The curvature will still be zero, but geodesics no longer look like straight lines for $(\Omega,\phi^*e)$. The concept of Euclidean line really is chart dependent. Secondly, the condition of zero curvature is a closed condition. So $\gamma$ that is only tangent to $A$ in fact never passes through $A$. So geodesics should not be affected. | |
| Aug 2, 2010 at 20:16 | comment | added | Anton Petrunin | Could you formulate it clearly? | |
| Aug 2, 2010 at 19:11 | history | asked | Nicolas Fernandez-Arias | CC BY-SA 2.5 |