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