Re: [math-fun] Torus puzzle (spoiler)
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
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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I hadn't thought of Adam's example, though I was indeed thinking of the same ((0 1),(-1 -1)) matrix just a couple of weeks ago. Well, actually it was ((0 -1), (1 -1)). I was thinking exactly of Jim's example when I asked the question. Good question, how to conjugate one map to get the other. It's possible to see that they're topologically conjugate. I suspect the conjugacy function can be found from their both have the same real Jordan normal form. (Or maybe one needs to use the matrix I recently noticed that by translating the hexagon in the plane, we can think of the same picture more symmetrically: Divide the regular hexagon into six equilateral triangles the usual way. Consider 3 centers of alternate triangles: * * * * * * * o * * * * * * * * * * * * * * * * * * * * o * * o * * * * * * * * * * * * * * On the torus, the Voronoi region of each of these points (the set of points that is closer to that point than to either of the other two) is a hexagon with side 1/sqrt(3) times as large as the original one. Now rotate each of these smaller hexagons by 1/3 rev. That's another description of the same map on the hexagonal torus, with the three alternate triangle centers being the set of fixed points. --Dan On 2014-02-03, at 11:36 AM, James Propp wrote:
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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Could we just use a skew transformation that converts a square into a rhombus, viewed as a bijection between fundamental domains? Jim On Monday, February 3, 2014, Dan Asimov <dasimov@earthlink.net> wrote:
I hadn't thought of Adam's example, though I was indeed thinking of the same ((0 1),(-1 -1)) matrix just a couple of weeks ago. Well, actually it was ((0 -1), (1 -1)).
I was thinking exactly of Jim's example when I asked the question.
Good question, how to conjugate one map to get the other. It's possible to see that they're topologically conjugate. I suspect the conjugacy function can be found from their both have the same real Jordan normal form. (Or maybe one needs to use the matrix
I recently noticed that by translating the hexagon in the plane, we can think of the same picture more symmetrically: Divide the regular hexagon into six equilateral triangles the usual way. Consider 3 centers of alternate triangles:
* * * * * * * o * * * * * * * * * * * * * * * * * * * * o * * o * * * * * * * * * * * * * *
On the torus, the Voronoi region of each of these points (the set of points that is closer to that point than to either of the other two) is a hexagon with side 1/sqrt(3) times as large as the original one. Now rotate each of these smaller hexagons by 1/3 rev.
That's another description of the same map on the hexagonal torus, with the three alternate triangle centers being the set of fixed points.
--Dan
On 2014-02-03, at 11:36 AM, James Propp wrote:
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<javascript:;>> 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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[Having just woken from nap}: Yes, that sounds very likely! --Dan On 2014-02-03, at 12:51 PM, James Propp wrote:
Could we just use a skew transformation that converts a square into a rhombus, viewed as a bijection between fundamental domains?
Jim
On Monday, February 3, 2014, Dan Asimov <dasimov@earthlink.net> wrote:
I hadn't thought of Adam's example, though I was indeed thinking of the same ((0 1),(-1 -1)) matrix just a couple of weeks ago. Well, actually it was ((0 -1), (1 -1)).
I was thinking exactly of Jim's example when I asked the question.
Good question, how to conjugate one map to get the other. It's possible to see that they're topologically conjugate. I suspect the conjugacy function can be found from their both have the same real Jordan normal form. (Or maybe one needs to use the matrix
I recently noticed that by translating the hexagon in the plane, we can think of the same picture more symmetrically: Divide the regular hexagon into six equilateral triangles the usual way. Consider 3 centers of alternate triangles:
* * * * * * * o * * * * * * * * * * * * * * * * * * * * o * * o * * * * * * * * * * * * * *
On the torus, the Voronoi region of each of these points (the set of points that is closer to that point than to either of the other two) is a hexagon with side 1/sqrt(3) times as large as the original one. Now rotate each of these smaller hexagons by 1/3 rev.
That's another description of the same map on the hexagonal torus, with the three alternate triangle centers being the set of fixed points.
--Dan
On 2014-02-03, at 11:36 AM, James Propp wrote:
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<javascript:;>> 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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participants (3)
-
Adam P. Goucher -
Dan Asimov -
James Propp