I 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.
My solution a few lines below Gareth's solution is correct, of course, and essentially the same as mine, which looks a little more symmetrical then the brick arrangement. Namely, consider the tiling of the plane by regular hexagons. The group G of translations preserving this tiling must be a lattice in the plane. Hence the quotient of the plane by the lattice must be a torus, topologically. Call the torus T. Let one of the hexagons be called H. Since every point of the plane is equivalent to some point of H, we can obtain the same quotient torus T by just identifying the points of H that are equivalent by the group G of translations. This means: Take the (closed) hexagon H and identify opposite edges in the simplest way -- identify each point of each edge with its closest poiwwnt on the opposite edge. This identification turns H into the torus T. It's easy to check that the boundary of H becomes a theta curve C on T. It's also clear that identifying the boundary of H to a point gives a topological sphere S^2 as the quotient space. Which means that taking the theta curve C on the torus T, and identifying it to a point, will also give the same topological sphere S^2 as *its* quotient space. Puzzle solved. * * * ((( Interestingly, it's easy to check (from orientability and Euler characteristic) that taking any regular polygon with an even number 2n of sides, and identifying its opposite sides (in the simplest way), has as quotient space the surface having its genus = floor(n/2). I.e., 2n-gon where 2n = 4 6 8 10 12 14 16 18 20 22 24 26 28 30 . . . -------------------------------------------------------------------------------- genus 1 1 2 2 3 3 4 4 6 6 8 8 10 10 . . . To get a highly symmetric *metric* surface of constant curvature, this already works for the torus using a square or regular hexagon in the plane. But for surfaces of higher genus, we need to start with the regular 2n-gon of constant negative curvature (say = -1) whose vertex angles are each pi/n if n is even, and 2pi/n if n is odd (with geodesic edges). These are unique up to isometry. ))) --Dan ________________________________________________________________________________________ It goes without saying that .
My comments a few lines below. I thought it might be worth mentioning that Dan's hexagonal representation of the torus appears "naturally" in the context of chessboard complexes. The chessboard complex corresponding to a chessboard of size m x n is a simplicial complex with one vertex for each square of the chessboard. A set of squares forms a face of the complex if and only if no two squares in the set appear in the same row or column. (Equivalently, rooks positioned on the given squares are pairwise non-attacking.) For example, if we label the columns with letters and the rows with integers, we have that the set {a1, b3, c2} forms a face, whereas {a1, b1, c2} and {a1, a3, c2} do not. In the case of a 4x3 chessboard with the four columns indexed by the letters A, B, C and D and the three rows indexed by the colors red, blue and yellow, we may represent the complex geometrically in the following manner. http://www.math.kth.se/~jakobj/pics/M43.jpg (Each small triangle in the picture, filled or not, belongs to the complex.) As the picture indicates, one may view this complex as a triangulation of a hexagonal region with opposite sides identified in the manner described by Dan:
Gareth's solution is correct, of course, and essentially the same as mine, which looks a little more symmetrical then the brick arrangement.
Namely, consider the tiling of the plane by regular hexagons. The group G of translations preserving this tiling must be a lattice in the plane. Hence the quotient of the plane by the lattice must be a torus, topologically.
Call the torus T.
Let one of the hexagons be called H. Since every point of the plane is equivalent to some point of H, we can obtain the same quotient torus T by just identifying the points of H that are equivalent by the group G of translations.
This means: Take the (closed) hexagon H and identify opposite edges in the simplest way -- identify each point of each edge with its closest poiwwnt on the opposite edge. This identification turns H into the torus T.
Some references for the chessboard complex can be found here: http://mathoverflow.net/questions/36791/is-the-4x5-chessboard-complex-a-link... Jakob
participants (2)
-
Dan Asimov -
Jakob Jonsson