For genus 3, we have the Klein Quartic based on {7,3} which wins in symmetry counting, but we also have Schmutz’s M(3) which is based on {12,3}. If you are inside the surface, what you see is the covering hyperbolic plane: in the Klein quartic you see 14 (closest) copies of yourself arranged symmetrically around you, in M(3) you see only 12 of them, but the layout in M(3) is more symmetric: in KQ the distance is 4.080620 if you are standing in the center of the heptagon and 3.935946 if you are standing optimally (and that number is the systole); in M(3) the distance is 4.317779 in the center of dodecagon and 3.983305 optimally.

For genus 2 we have Bolza’s Surface which is based on {8,3} and also has the greatest systole, for genus 4 we have Bring’s surface based on {5,4} with 5*8*(4-1) automorphisms and Schmutz’s M(4) based on {12,3} with 12*2*(4-1) automorphisms. M(4) again wins in terms of systole.

Wikipedia lists the maximum number of isometries for each genus from 2 to 4: https://en.wikipedia.org/wiki/Riemann_surface#Isometries_of_Riemann_surfaces I have learned about M(3) and M(4) from pages on Klein’s Quartic and Bring’s Curve pages respectively.

I have just implemented MacBeath’s Surface, Bring’s Surface, M(3), M(4), so I know what they look like :) I already had Klein Quartic and a surface which turned out to be one of the first Hurwitz triplet (one with the smallest systole, unfortunately).

(By look like, I mean the universal cover with orbits marked… I think this is much more informative than a “donut with g holes” display)

]]>The answer for life, universe and everything. Since long people believed that Douglas Adams had just made that up, but that was not the case. The guy was truly a genius, and that is what I am going to write here. […]

]]>A finite group G is called factorable if G = HK where H and K are both proper subgroups of G. If G is factorable, H and K may be assumed, without loss of generality, to be maximal subgroups of G. But the maximal subgroups of are isomorphic to , , and dihedral groups of order 12 or 14. The only possibility that might work out, order-wise, is and a dihedral group of order 14 — but these contain an element of order 2 in common, so factorability fails.

Any transitive permutation group on a prime-power number of points (say it’s a power of the prime p) which is not cyclic is factorable, since H may be chosen to be a point stabilizer and K may be chosen to be a Sylow p-subgroup of G. This covers the nonabelian simple groups (transitive on 5 points), (transitive on 7 or 8 points), (transitive on 9 points), and (transitive on 11 points). For , both H and K can be chosen to be isomorphic to (so then any automorphism of taking H to K is outer, as an automorphism of ).

Factorability is interesting for those of us who want to find newer (and maybe easier) proofs of statements — such as a proof of the conjugacy of complements in Schur-Zassenhaus which avoids the appeal to the Odd Order Theorem. I think a minimal (in terms of the order of the complement) counterexample has unfactorable complements, though I need to be more careful before asserting that.

]]>My attempt to link the two appearances of the number 17 hit a roadblock even before noticing *this* problem. There’s an interesting numerical way of showing that there are 17 wallpaper groups: it’s Conway’s Magic Theorem. However, I can’t figure out how to link this to the 17 different ways to find n whole numbers whose reciprocals add up to one less than n/2.

By the way, the triples of regular polygons meeting snugly at a point that *do* give tilings of the plane:

1/6 + 1/6 + 1/6 = 1/2

1/4 + 1/6 + 1/12 = 1/2

1/4 + 1/8 + 1/8 = 1/2

1/3 + 1/12 + 1/12 = 1/2

have a simple numerical feature in common which distinguishes them from the triples that don’t:

1/5 + 1/5 + 1/10 = 1/2

1/4 + 1/5 +1/20 = 1/2

1/3 + 1/10 + 1/15 = 1/2

1/3 + 1/9 + 1/18 = 1/2

1/3 + 1/8 + 1/24 = 1/2

1/3 + 1/7 + 1/42 = 1/2

**Puzzle:** what is this feature and why?

Anyone who gets stuck can read the Wikipedia article…

]]>Glue together the tetrahedral ends to close the surface. Although each heptagon is identically situated in the figure, the 7 sides of a heptagon do not see identical landscapes. So its not Platonic, nevertheless interesting.

Gerard

]]>• Archimedean tilings and Egyptian fractions.

• week182: ADE classifications, Egyptian fractions and tilings of the plane, sphere and hyperbolic plane.

• 42: the special role of the number 42.

The last one has a big appendix at the end, not included in this blog article.

]]>I think these are the ‘translations’ you’re talking about. For any point in any connected symmetric space, the symmetry group of that space is generated by the rotations around that point (that is, the symmetries fixing that point), together with the transvections for that point.

]]>So for example, for , , using the second through sixth terms of Sylvester’s sequence you could fit a triangle, a heptagon, a 43-gon, an 1807-gon and a 3,263,442-gon (subtracting 1 from the last term) together to form a local double cover of the flat plane.

I wonder if you could get some kind of interesting hyperbolic Riemann surface by omitting one of the polygons, as you can with .

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