Missing data on genus-2 curves

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Linden Disney

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Jun 20, 2023, 8:07:25 AM6/20/23
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In the list of genus-2 curves with a quotient to a genus-1 curve (http://www.lmfdb.org/HigherGenus/C/Aut/?genus=2&g0=1&search_type=List), there is only one curve given corresponding to the equation y^2 = (x^2 - 1)(x^2 - a)(x^2 - b) (http://www.lmfdb.org/HigherGenus/C/Aut/2.2-1.1.2-2.1), where the quotient by the automorphism x -> -x is genus 1. There should be additional curves, for example the curve given by y^2 = x^6 - ax^3 + 1 which has the involution (x,y) -> (1/x, y/x^2) which also quotients to an elliptic curve. This can be verified in Sage. 

By my understanding of http://www.lmfdb.org/HigherGenus/C/Aut/Completeness, this quotient should be included in the database. Is this an error with my understanding, or are there gaps in the higher genus curve data.


Jennifer Paulhus

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Jun 20, 2023, 12:18:16 PM6/20/23
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Hi Linden,

I haven't worked out the details, but I would assume that the curves in the family y^2=x^6-ax^3+1 are a 1 dimensional subfamily of the dimension 2 family you wrote above: y^2 = (x^2 - 1)(x^2 - a)(x^2 - b).  Do you have a reason to believe they are not isomorphic to some curves in this family? (I also didn't check the details about the curve y^2=x^6-ax^3+1 but I trust your results in Sage, with tiny typo that I think you mean (x,y)->(1/x,y/x^3) instead of y/x^2).

The reason I am confident the data is correct (at least for genus 2) is that by the Riemann-Hurwitz formula there are only 3 possible signatures for actions in genus 2 with a quotient genus of 1:  [1;2,2],  [1;3], and [1;2].  The latter 2 can be ruled out because they only work as signatures for nonabelian group actions, but the group orders of the group acting with these signatures would be 3 and 4, respectively (again by Riemann-Hurwitz) which, of course, are orders containing only abelian groups.

And signature [1;2,2] represents a 2 dimensional family (3*quotient genus + # branch points-3) so we've accounted for all the curves.

It may not be relevant, but  note that in all these cases, C2 is never the full automorphism group, there is always another C2 action  so the full automorphism group of all these curves is at least C2xC2 (although that acts with quotient genus 0).

Hope that helps,
Jen Paulhus

Linden Disney

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Jun 21, 2023, 10:15:25 AM6/21/23
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Hi Jen,

Thanks for the quick reply, you've clarified where I was misinterpreting the data. You are right that the family y^2 = x^6 - ax^3 + 1 is isomorphic to a subset of curves of the form y^2 = (x^2 - 1)(x^2 - a)(x^2 - b), but I had expected a separate entry in the database corresponding to a different "family containing this refined passport", i.e. a different entry if there was a family with a different full automorphism group. 

Kind regards,

Linden

Jennifer Paulhus

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Jun 24, 2023, 12:36:01 PM6/24/23
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Hi Linden,

Ah, I see.  Thank you for explaining your confusion. It's helpful to know where I can be more clear in the description of these pages. I say a bit more below about  the pages.

Each family page represents a branched locus in the moduli space of curves of a fixed genus. The loci are defined by the group and signature pair acting on those curves.  So in this example, the family is a dimension 2 locus inside \mathcal{M}_2 consisting of all curves of genus 2 which have a C2 action with signature [1;2,2].  (The refined passport pages distinguish between the lists of conjugacy classes in the group that the monodromy comes from.)

Sitting inside any locus, there might be curves that have even larger  automorphism group acting on them.  Like you note, there is a one dimensional family of curves in this locus that has a D_6 action on it.  But when you consider the whole action of D_6 on those curves, you have a quotient genus of 0 now.  Those curves do have their own page, representing to locus of curves in \mathcal{M}_2 which have an action of D_6 with signature [0;2,2,2,3]. It's here:

Similarly, a curve like y^2=x^6+1 will have an even larger automorphism group and that will form a dimension 0 locus inside of the original locus and it will also have its own page.

I have started a project to connect these actions up, so that if someone were to click on the page for the dimension 2 locus, they could see what loci were inside of it and vice versa (in effect "mapping out" part of the singular locus of moduli spaces of curves).  It's early days on the project, but hopefully once implemented, that new feature will be helpful to researchers.

Thanks again for your comment and do let me know if you run across any other issues,
Jen

Linden Disney

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Jul 4, 2023, 12:57:10 PM7/4/23
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Hi Jen, following up on this, I don't know if you're able to help explain the following misunderstanding:

In genus 3 there should are non-hyperelliptic curves with automorphism group C2 of the form z^4 + z^2 * f2(x, y) + f4(x, y) = 0 in projective coordinates [x: y: z] where f2, f4 are homogeneous polynomials of degree 2 and 4 respectively and the C2 is generated by [x: y: z] -> [x: y: -z]. The genus of the quotient is 1 so the corresponding signature should be (1; 2, 2, 2, 2) (about these facts I am confident see https://doi.org/10.1016/S0019-3577(99)80015-9 or using Sage). This 4-dimensional family is identified in the LMFDB at http://www.lmfdb.org/HigherGenus/C/Aut/3.2-1.1.2-2-2-2,  where there is only one passport given http://www.lmfdb.org/HigherGenus/C/Aut/3.2-1.1.2-2-2-2.1 which corresponds to a 1-dimensional family of hyperelliptic curves http://www.lmfdb.org/HigherGenus/C/Aut/3.16-11.0.2-2-2-4 with full automorphism group C2 x D4. I understand that as there is trivially only one order-2 conjugacy class in C2 there can only be one possible generating vector for this signature. 

What I am confused about is how to interpret the "full automorphism" data associated to the passport 3.2-1.1.2-2-2-2.1. Clearly there are curves which have a smaller full automorphism group which fall into the corresponding branched locus, so where do these appear in the database? 

Many thanks in advance for your help again. 

Best, Linden

Jennifer Paulhus

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Jul 4, 2023, 2:15:38 PM7/4/23
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I responded to Linden directly.
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