Re: [math-fun] A pretty surface with surprising symmetry
Michael Kleber wrote: ----- I have no intuition for where the genus of 73 comes from. Can you offer any way to intuit that, beyond just doing a calculation? Is that somehow 1 + 36 + 36, with a 36 for each 1-skeleton? ----- Not sure how to go beyond a calculation. But that calculation is this: The 1-skeleton of the 24-cell has 24 vertices and 96 edges, so its Euler characteristic is X = V - E = -72. Same for the dual 1-skeleton. Since a (tubular) neighborhood of the 1-skeleton in the 3-sphere S^3 can be deformed by shrinking down to the 1-skeleton itself, that also has Euler characteristic equal to -72. Now the tubular neighborhoods of the 1-skeleton and the dual 1-skeleton can be inflated until they just overlap in their common boundary, which is the surface in question, M_g. Now we can express the Euler characteristic of the 3-sphere S^3 (which, like for every odd-dimensional sphere, is 0) as the sum of the Euler characteristics of each tubular neighborhood, minus the Euler characteristic X(M_g) of their overlap, the surface M_g: 0 = X(S^3) = -72 + -72 - X(M_g) so X(M_g) = -144. Since the Euler characteristic and genus of an orientable (compact) surface M_g are always related by the equation X(M_g) = 2 - 2g it follows that g = 73. Sorry this isn't very intuitive. —Dan ----- I'm always struck by instances in math unexpected symmetry. ... ... Or consider the number 73. Not particularly conducive to symmetry, you might think. But recall the regular 4D polytope called the 24-cell: It's 24 octahedra arranged face-to-face to form the 3-dimensional sphere S^3. In fact it has 24 vertices, 96 edges, 96 2-dimensional faces, and 24 3-dimensional octahedra. This makes it self-dual. So if you draw the vertices and edges of this 24-cell on the 3-sphere S^3, you can also draw the *dual* vertices and edges as well — they will form a disjoint 1-dimensional figure in S^3. Now if you removed both of these "1-skeleta" from S^3, the portion of S^3 that remains is topologically equivalent to the product of a surface M_g of genus g with the real line R. And it turns out that the genus of g M_g is 73. Even nicer, it turns out that the surface of genus 73 that is halfway between the two dual 1-skeleta in S^3 is a *minimal surface*, meaning that if you drew a small simple closed curve C about any of its points, the surface minimizes the area among all surfaces in S^3 whose boundary is the curve C. Even nicer yet, this minimal surface M_73 has an isometry with itself for any isometry of the 24-cell. One of the octahedra can be taken to any of the 24 octahedra, and one of its triangles can go to any of the 8 triangular faces in any of 6 dihedral way, making for 24 x 24 x 2 = 1152 isometries. Nicest of all, the two halves of S^3 created when M_73 is removed can be interchanged, which carries M_73 to itself, giving it another 1152 isometries for a grand total of 2304 altogether. Who said 73 wasn't symmetrical. —Dan -----
My thought is that if the genus is 73, the Euler characteristic is -144. 73 may seem like a number that is not very conducive to fancy symmetry, but -144 seems a lot more promising. I'm guessing that the Euler characteristic is more important to allowable symmetries than the genus is. Other examples can be harvested from the thickened skeleta of the platonic solids treated as surfaces. I expect the genus plus one to be a nicely divisible number. The tetrahedron's skeleton has genus 3; 4 has better divisibility. The cube has genus 5 but 6 is better. The octahedron: 7 versus 8. The dodecahedron: 11 versus 12. The icosahedron: 19 versus 20. In other words, I don't think the symmetries are juggling a set whose size is the genus; I think the relevant size is the genus plus 1. On Sun, Nov 1, 2020 at 1:02 PM Dan Asimov <dasimov@earthlink.net> wrote:
Michael Kleber wrote: ----- I have no intuition for where the genus of 73 comes from. Can you offer any way to intuit that, beyond just doing a calculation? Is that somehow 1 + 36 + 36, with a 36 for each 1-skeleton? -----
Not sure how to go beyond a calculation. But that calculation is this:
The 1-skeleton of the 24-cell has 24 vertices and 96 edges, so its Euler characteristic is X = V - E = -72. Same for the dual 1-skeleton.
Since a (tubular) neighborhood of the 1-skeleton in the 3-sphere S^3 can be deformed by shrinking down to the 1-skeleton itself, that also has Euler characteristic equal to -72.
Now the tubular neighborhoods of the 1-skeleton and the dual 1-skeleton can be inflated until they just overlap in their common boundary, which is the surface in question, M_g.
Now we can express the Euler characteristic of the 3-sphere S^3 (which, like for every odd-dimensional sphere, is 0) as the sum of the Euler characteristics of each tubular neighborhood, minus the Euler characteristic X(M_g) of their overlap, the surface M_g:
0 = X(S^3) = -72 + -72 - X(M_g)
so X(M_g) = -144.
Since the Euler characteristic and genus of an orientable (compact) surface M_g are always related by the equation
X(M_g) = 2 - 2g
it follows that g = 73.
Sorry this isn't very intuitive.
—Dan
----- I'm always struck by instances in math unexpected symmetry. ... ...
Or consider the number 73. Not particularly conducive to symmetry, you might think. But recall the regular 4D polytope called the 24-cell: It's 24 octahedra arranged face-to-face to form the 3-dimensional sphere S^3. In fact it has 24 vertices, 96 edges, 96 2-dimensional faces, and 24 3-dimensional octahedra. This makes it self-dual.
So if you draw the vertices and edges of this 24-cell on the 3-sphere S^3, you can also draw the *dual* vertices and edges as well — they will form a disjoint 1-dimensional figure in S^3. Now if you removed both of these "1-skeleta" from S^3, the portion of S^3 that remains is topologically equivalent to the product of a surface M_g of genus g with the real line R. And it turns out that the genus of g M_g is 73.
Even nicer, it turns out that the surface of genus 73 that is halfway between the two dual 1-skeleta in S^3 is a *minimal surface*, meaning that if you drew a small simple closed curve C about any of its points, the surface minimizes the area among all surfaces in S^3 whose boundary is the curve C.
Even nicer yet, this minimal surface M_73 has an isometry with itself for any isometry of the 24-cell. One of the octahedra can be taken to any of the 24 octahedra, and one of its triangles can go to any of the 8 triangular faces in any of 6 dihedral way, making for 24 x 24 x 2 = 1152 isometries.
Nicest of all, the two halves of S^3 created when M_73 is removed can be interchanged, which carries M_73 to itself, giving it another 1152 isometries for a grand total of 2304 altogether.
Who said 73 wasn't symmetrical.
—Dan -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (2)
-
Allan Wechsler -
Dan Asimov