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The Bring sextic, with 120 automorphisms, is the numerically most symmetric compact Riemann surface of genus 4. To cut it up into six pairs of pants, we need to cut along nine disjoint geodesic loops. How short can those loops be, and how symmetric can we make the decomposition?

I am studying the Bring sextic, by the way, because one can elegantly map all of the possible shapes of equilateral pentagons in the Euclidean plane to the points of the Bring sextic. (Calling it "the Bring sextic", by analogy with "the Klein quartic", obviates the clumsy need to choose between "the Bring surface" over $\mathbb{R}$ and "the Bring curve" over $\mathbb{C}$.)

For comparison, the Bolza quintic is the numerically most symmetric compact Riemann surface back in genus 2, with 48 automorphisms. The systole of the Bolza quintic is $2\operatorname{arccosh}(1+\sqrt{2})\approx 3.057$, and there are $12$ systolic loops (that is, loops of that length). There are triples of systolic loops that are disjoint, and such a triple cuts up the Bolza quintic into two isometric pairs of pants. In that decomposition, the three cuffs of one pair of pants are sewn to the three cuffs of the other, and all three twists, as a fraction of the systole, are $\operatorname{arccosh}((5+4\sqrt{2})/7)/(2\operatorname{arccosh}(1+\sqrt{2}))\approx 0.3213$.

Genus 3 is even prettier. The famous Klein quartic is the symmetry champion, with 168 automorphisms. It has $21$ systolic loops, each of length $8\operatorname{arccosh}(\frac{1}{2}+\cos(2\pi/7))\approx 3.936$. Some sextuples of them are disjoint, and such a sextuple cuts up the Klein quartic into four isometric pairs of pants. The 3-regular graph giving the sewing of the cuffs is $K_4$, the edge-graph of a tetrahedron, and all six twists are $1/8$.

With those cases as context, what happens in genus 4? The Bring sextic has $20$ systolic loops, each of length $2\operatorname{arccosh}((9+5\sqrt{5})/4)\approx 4.603$. Since our loops must be disjoint, however, we can take at most six of those $20$. Cutting along those six breaks the Bring sextic into three pieces, each with four loops of boundary. We need to cut each of those three pieces along another geodesic loop, to split it into two pairs of pants.

The Bring sextic has $30$ loops of length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$ (only slightly longer than systolic). For each of our three current pieces, there is one of those $30$ loops that splits it into two pairs of pants, leaving us with six isometric pairs of pants overall. The 3-regular graph giving the sewing of the cuffs in the resulting decomposition is $K_{3,3}$, a graph famous for its nonplanarity. The twists of the six systolic loops are $1/6$, while the twists of the three longer loops are $1/4$.

Is this the most symmetric way to cut up the Bring sextic? Or are there other decompositions that can compete with it for simplicity and symmetry?

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    $\begingroup$ Could you clarify your statement that “one can elegantly map all of the possible shapes of equilateral pentagons in the Euclidean plane to the points of the Bring sextic”? What is this map? (I just happened to be reading Matthias Weber's paper “Kepler's small stellated dodecahedron as a Riemann surface”, which is also about Bring's sextic, so I'm curious.) $\endgroup$
    – Gro-Tsen
    May 14, 2020 at 14:08
  • $\begingroup$ The polygon space for equilateral pentagons in the plane and the Bring sextic are both compact, orientable, smooth surfaces of genus 4, so there are lots of maps between them. I am drafting a 35-page paper about such a map, and I hope that people who are curious about this area will be interested in reading it. $\endgroup$ May 15, 2020 at 3:33
  • $\begingroup$ I am certainly curious to see this paper. But could you already explain what the automorphisms of order $2,4,5$ defining the triangle group $\Delta^0(2,4,5) \cong \mathfrak{S}_5$ acting on Bring's surface correspond to when acting on equilateral pentagons? (Or at least what are the fixed pentagon shapes under them?) Because the only action of $\mathfrak{S}_5$ on pentagons that I can think of is by permuting the vertices, and this doesn't preserve equilaterality, so I'm confused. $\endgroup$
    – Gro-Tsen
    May 15, 2020 at 14:05
  • $\begingroup$ Rather than permuting the vertices of the equilateral pentagon, permute the order in which its edge vectors get assembled, tip to tail. The 24 $\frac{\pi}{5}$ vertices of the Bring sextic then correspond to the 24 pentagons whose five edges are evenly distributed around the unit circle, while the 30 $\frac{\pi}{4}$ vertices correspond to the 30 pentagons that have two pairs of coincident edges. $\endgroup$ May 15, 2020 at 17:19
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    $\begingroup$ @Gro-Tsen, my paper about viewing the polygon space of equilateral pentagons as the Bring sextic is now available on arXiv: link. See, in particular, pages 10 and 11. $\endgroup$ May 21 at 14:18

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There is an alternative strategy for cutting up the Bring sextic into pairs of pants that might compete with the decomposition above in some ways.

Of the $30$ loops of the second-shortest length $2\operatorname{arccosh}((11+5\sqrt{5})/4)\approx 4.796$, there are sets of six that are disjoint. Cutting along such a set of six breaks the Bring sextic up into three pieces, each of which has four loops of boundary --- as happened also in the decomposition above.

The Bring sextic has $10$ loops of the third-shortest length, which is $2\operatorname{arccosh}((26+10\sqrt{5})/4)\approx 6.368$. For each of our current three pieces, there are two of those $10$ third-shortest loops, either of which can cut up that piece into two pairs of pants. With three binary choices, we get eight decompositions into isometric pairs of pants. The 3-regular graph that gives the sewing of the cuffs is $K_{3,3}$ in four of those eight decompositions, but is the edge graph of a triangular prism in the other four. The twists are somewhat simpler in all eight of those decompositions, however --- which might make them more attractive than the decomposition above for some purposes: The twists along the six shorter loops are $1/4$, as for the three loops of that same length in the decomposition above, but the twists along the three longer loops are $0$.

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