Timeline for Surfaces of constant Gauss curvature K spanned by two helices and two straight lines
Current License: CC BY-SA 3.0
22 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jun 30, 2016 at 8:46 | history | edited | Narasimham | CC BY-SA 3.0 |
related question (for defining a semi infinite surface between given boundaries)
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| Jun 17, 2016 at 10:40 | history | edited | Narasimham | CC BY-SA 3.0 |
SE Math related post mentioned
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| May 14, 2016 at 16:35 | comment | added | Narasimham | To me it appears the surface need not be smooth. For example geodesic polar coordinates centred around a point in the neighbourhood of an unsmooth $ K \equiv + $ surface/manifold can be continuously mapped/drawn or visualised across the cuspy edge, not only surviving but fully accommodating a sharp change of the orientation of the surface normal vector by $\pi$. I mention this even though this sounds as just a statement of belief.. | |
| May 13, 2016 at 4:16 | history | edited | Narasimham | CC BY-SA 3.0 |
discussion comment reg K singularities.
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| May 12, 2016 at 23:54 | history | edited | Narasimham | CC BY-SA 3.0 |
Cuspidal "gorges" possibility indicated
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| May 12, 2016 at 9:17 | history | edited | Narasimham | CC BY-SA 3.0 |
An explaining word added
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| May 12, 2016 at 7:06 | history | edited | Narasimham | CC BY-SA 3.0 |
Approx image for 3 signs of K
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| May 12, 2016 at 6:59 | history | edited | Narasimham | CC BY-SA 3.0 |
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| May 12, 2016 at 6:52 | history | edited | Narasimham | CC BY-SA 3.0 |
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| May 11, 2016 at 21:58 | comment | added | Robert Bryant | I'm not convinced that relaxation of boundary regularity will help. After all the surfaces with $K\equiv1$ are known to be real-analytic, and they may satisfy something similar to a Schwartz Reflection Principle (the same way minimal surfaces ($H\equiv0$) do), so that having a continuous boundary that contains a line segment would imply that the surface extends smoothly across the line line segment as a surface that satisfies $K\equiv1$, which, of course, cannot happen in the case $K\equiv1$. For example, it could happen that having a real-analytic boundary implies extendability across it. | |
| May 11, 2016 at 20:57 | comment | added | Narasimham | I forgot to mention that we are not talking about smooth continuously differentiable case but the cuspidal edge of an elliptic type sphere,which has to be twisted physically like a rubber sheet and placed on a straight line and by means of isometry formulation with its deformation after the twist around this straight cuspidal line. | |
| May 11, 2016 at 17:33 | comment | added | Robert Bryant | I don't understand your comment: A surface in $\mathbb{R}^3$ with positive Gauss curvature (in particular, when $K\equiv1$) cannot contain any straight lines, for, if it did, then the tangent plane at a point on such a line would contain the line, but we know that when $K>0$, the surface is locally strictly convex, so it cannot contain any lines. | |
| May 11, 2016 at 16:15 | comment | added | Narasimham | Afraid am not able to agree in both situations.For K>0, imagine a tightly wound hollow tube spiral wound on a cylinder.It produces approximately two spirals of both signs of $K$.If you accept negative then you must also accept positive K. The cuspidaledge can be imagined isometrically deformed so the cuspidaledge edhe can be placed on a straight line in case K=0. | |
| May 11, 2016 at 13:50 | history | edited | Narasimham | CC BY-SA 3.0 |
Typo in first line corrected.
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| May 11, 2016 at 13:19 | comment | added | Robert Bryant | What does 'it is required to find' mean in this case? Is this a homework problem? (If so, it is a very hard one.) In fact, I find it hard to believe that you will find a surface with this boundary that has $K\equiv1$. Certainly, if you do, it will not be smooth across the boundary because no surface with $K\equiv1$ can contain a straight line as a geodesic (for simple reasons). Similarly, I think you can rule out $K=0$ because you know that the surface will be ruled and it should be easy to show that no ruling can have lines that meet the helices in two points and sweep out a flat surface. | |
| May 11, 2016 at 13:00 | answer | added | Joseph O'Rourke | timeline score: 2 | |
| May 11, 2016 at 12:35 | history | edited | Narasimham | CC BY-SA 3.0 |
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| May 11, 2016 at 12:27 | history | edited | Narasimham | CC BY-SA 3.0 |
grammar
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| May 11, 2016 at 12:21 | history | edited | Narasimham | CC BY-SA 3.0 |
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| May 11, 2016 at 12:12 | history | edited | Narasimham | CC BY-SA 3.0 |
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| May 11, 2016 at 12:05 | history | asked | Narasimham | CC BY-SA 3.0 |