In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2. --Dan ________________________________________________________________________________________ It goes without saying that .
Sounds like a fun puzzle, but I don't see how the resulting space can avoid having pinch-points that don't look like R^2 locally. The kind of space I'm picturing is what you get from shrinking the equator of a sphere to a point: the result is a bouquet of two spheres, and in the vicinity of the point where they meet, the space isn't a 2-manifold. How can a theta-curve on a torus avoid this fate? Am I totally on the wrong track, or does my puzzlement reflect what's cool about the puzzle? Jim Propp On 3/29/12, Dan Asimov <dasimov@earthlink.net> wrote:
In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2.
--Dan
________________________________________________________________________________________ It goes without saying that .
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On Thursday 29 March 2012 20:25:57 Dan Asimov wrote:
In the torus T^2, find an embedded theta curve C (i.e., 2 points joined by each of 3 otherwise disjoint arcs) such that if C is identified to a point, the quotient space is topologically a sphere S^2.
Blank space so that you don't have to read what follows if you don't want to ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more ... and more. OK, so why isn't this easy? Represent the torus as a square with opposite sides identified. If we (illegally) take the edges of the square as C, then the quotient is certainly S^2. Now, the edges form an 8 rather than a theta, so deform it a little. Here's a simple way: let A = (1/4,1/4) and B = (3/4,3/4); draw lines NW,SW,SE from A and NW,NE,SE from B. If you draw copies of the square tiling the plane in the obvious way, these lines give you a sort of diagonal brick-wall pattern. Each brick is a fundamental domain, and quotienting by the edges of the brick gives you a sphere. This seems like just following the path of least resistance. Have I missed something that makes it not work? -- g
participants (3)
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Dan Asimov -
Gareth McCaughan -
James Propp