Here's the first half of a different solution: Take a geometrically different torus whose fundamental domain is a rhombus with angles of 60 and 120 degrees. View it as having a regular hexagon as its fundamental domain. (More details: Create a regular hexagons with vertices alternately colored black and white; identify opposite edges of the hexagon with each other, using the coloring of the vertices to assign consistent orientations of the edges. The universal cover of this torus is the plane, tiled by translates of the hexagon.) If we rotate the hexagon by 120 degrees around its center, we get a three-fold rotation of the torus, whose fixed points are the black vertex (there are three of them but they're all identified with one another by the gluing), the white vertex, and the center. BUT: this is a different torus! Topologically it's the same, but geometrically it's different. So I ask: is there a way to turn this into an answer to Dan's question by conjugating my 120-rotation using a homomeomorphism between the "square torus" and the "hexagonal torus"? Jim On Monday, February 3, 2014, Adam P. Goucher <apgoucher@gmx.com> wrote:
Torus puzzle:
Find a homeomorphism h: T^2 -> T^2 from the torus T^2 to itself that's periodic of least period 3, and that has exactly 3 fixed points.
--Dan
Very beautiful.
Let T^2 = R/Z x R/Z. Then the homeomorphism corresponding to the matrix:
[0 1] [-1 -1]
is period-3 and has fixed points (0,0), (1/3,1/3) and (2/3,2/3).
This corresponds to the order-3 element of the modular group SL(2,Z), which in turn corresponds to an order-3 rotation in the (2,3,infinity) tiling of the hyperbolic plane (which is how I discovered this homeomorphism).
Sincerely,
Adam P. Goucher
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