Timeline for Automorphisms of genus 6 surfaces
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 29, 2017 at 23:33 | comment | added | David E Speyer | Since $\mathbb{Z}/7$ can be generated $4$ elements of order $7$ with sum $0$, there is a degree $7$ cyclic cover $X \to \PP^1$ ramified over $4$ points and, by the Riemann-Hurwitz computation you describe, it will have genus $6$. This is the Riemann existence theorem: If you can described the covering map combinatorially, then it is always possible to choose the complex structure compatibly. | |
| Nov 29, 2017 at 22:05 | comment | added | roy smith | Sometimes the computation is easy. E.g. a curve of genus 5 cannot have an automorphism of genus 7 because the resulting quotient map cannot satisfy Hurwitz's formula. This rules out 84(g-1) in that case, but these genus 6 cases look harder. E.g,. an automorphism of order 7 of a genus 6 curve whose quotient has genus zero with 4 totally ramified branch points would seem to work numerically. I.e. 2g-2 = 10 = 7(-2) + 4(6). (Is that right?) So this doesn't even rule out 420. This stuff is all classical however since apparently Fricke knew these cases. I don't have his papers though. | |
| Nov 29, 2017 at 21:56 | comment | added | roy smith | Thanks Will! I didn't really do the computation, just piggybacked on what I assumed was an ordered sequence. | |
| Nov 29, 2017 at 20:53 | history | edited | j.c. | CC BY-SA 3.0 |
add arxiv link
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| Nov 29, 2017 at 19:12 | comment | added | David E Speyer | @roysmith Thank you! I was going to do this computation later today. Right, the computation is finite in theory; you just need to go through the groups of orders $420$, $240$, $200$ and $180$ and see if they can be generated as claimed. | |
| Nov 29, 2017 at 18:52 | comment | added | Will Sawin | @roysmith I believe you have missed $40(g-1)$ from the triple $(2,4,5)$, but not any others. | |
| Nov 29, 2017 at 18:34 | comment | added | roy smith | @abx. Maybe this is clear, but as I understand it, the maximum order of the group is either 84(g-1), or 48(g-1), or 36(g-1) or 30(g-1), ... according to whether the triple in David Speyer's answer can be (2,3,7), or (2,3,8), or (2,3,9) or (2,3,10),...Since these possibilities are 420, 240, 180, 150,...presumably you have to rule out the possibility of a curve of genus 6 having an automorphism (sub)group of order 4. ?? | |
| Nov 29, 2017 at 16:57 | comment | added | David E Speyer | Table 4 lists statistics for all automorphism groups of curves of genus $4 \leq g \leq 10$ with order $\geq 4(g-1)$. I don't understand what all the table entries mean, but the group order is the first element of the ordered pair in the second column. | |
| Nov 29, 2017 at 16:54 | comment | added | abx | Could you say more precisely where the bound 150 for genus 6 curves is obtained? | |
| Nov 29, 2017 at 16:19 | history | answered | Dan Petersen | CC BY-SA 3.0 |