Timeline for Comparison theorem for Lambert quadrilateral
Current License: CC BY-SA 3.0
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| Feb 23, 2016 at 18:11 | comment | added | forevenone | @RobertBryant Thanks, you are right. For the application I have in mind, the two manifolds would be simply-connected. (In fact, I am considering $M$ here to be a universal cover of some compact surface with negative curvature bounded from above, so it's diffeomorphic to $\mathbb R^2$.) | |
| Feb 23, 2016 at 18:05 | comment | added | Robert Bryant | In order to have a hope of proving something like this, you are going to need to assume that the sides of the quadrilateral form a contractible loop in the surface. On a compact surface with $K\equiv -1$, you can have 'squares', i.e., a periodic sequence of 4 congruent geodesic segments such that each successive pair meet at right angles at their endpoints. (Of course, this can't happen in the hyperbolic plane, but the hyperbolic plane is simply-connected.) | |
| Feb 23, 2016 at 1:40 | history | edited | forevenone |
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| Feb 22, 2016 at 22:51 | history | asked | forevenone | CC BY-SA 3.0 |