3
$\begingroup$

Let $p\geq 3$ be a prime number. The modular curve $X(p)$ can be considered as a connected smooth projective curve over complex numbers. There is a subgroup $\mathrm{PSL}_2(\mathbb{F}_p)$ inside its group of automorphisms (Serre has proved that for $p\geq 7$ it is the full group of automorphisms).

For which $p$ does there exist a geometrically connected smooth projective curve $X$ over $\mathbb{Q}$ such that $X(p)$ is the base change of $X$ and such that all $\mathrm{PSL}_2(\mathbb{F}_p)$-automorphisms of $X(p)$ are base changed from automorphisms of $X$?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Let the function field $F_n= \Bbb{Q}(j,\{ j(\frac{a.+b}{c.+d}, ad-bc| n\})$ then $Aut(F_n/\Bbb{Q}(j)) = SL_2(\Bbb{Z}/n\Bbb{Z})/\pm I$ and it becomes automorphisms of $C/\Bbb{Q}$ any smooth projective curve with function field $F_n$ $\endgroup$
    – reuns
    Commented Aug 20, 2019 at 12:19
  • $\begingroup$ @reuns can one give this curve a moduli interpretation? $\endgroup$
    – user144105
    Commented Aug 20, 2019 at 12:22
  • 1
    $\begingroup$ @hello I believe this is equivalent to the moduli space of elliptic curves with an isomorphism between $( \mathbb Z/n)^2$ and $E[n]$ well-defined up to scalar multiplication, i.e. a 1-dimensional subspace of $\operatorname{Hom} ( ( \mathbb Z/n)^2,E[n])$ whose generic elements are isomorphisms. $\endgroup$
    – Will Sawin
    Commented Aug 20, 2019 at 12:53
  • $\begingroup$ This is not the same curve as $X(n)$ (except if $\phi(n)=n$). $\endgroup$
    – Will Sawin
    Commented Aug 20, 2019 at 12:55
  • $\begingroup$ In conversation with Johan de Jong we noticed that the curve with this moduli interpretation is not geometrically connected so would not be a good choice. $\endgroup$
    – Will Sawin
    Commented Aug 20, 2019 at 18:43

1 Answer 1

Reset to default
3
$\begingroup$

A nontrivial unipotent element of $PSL_2(\mathbb F_p)$ fixes $(p-1)/2$ cuspidal points of this modular curve. Its eigenvalue on the tangent space of those points is a $p$th root of unity.

If the curve and the action are defined over $\mathbb Q$, then this set of roots of unity must be Galois-invariant. However, there is no set of $(p-1)/2$ Galois-invariant roots of unity, as the Galois action on them is transitive.

In fact, I think one can check that the roots of unity obtained are a Galois orbit under the absolute Galois group of the quadratic extension of $\mathbb Q$ with conductor $p$, suggesting that the curve and the action may be defined over that field, as is true at least for $p=3$ and $p=5$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ why is there at least one eigenvalue not equal to $1$? $\endgroup$
    – user144105
    Commented Aug 20, 2019 at 19:33
  • $\begingroup$ @hello If you work in the Tate curve model, the elliptic curve can be viewed as $\mathbb G_m / q^p$ where $q$ is a local paramater, and the $p$-torsion is generated by $q$ and the $p$th roots of unity. Then the unipotent element fixing the point fixes the $p$th roots of unity subgroup and moving the other generator, and thus it multiplies $q$ by some fixed $p$th roots of unity. $\endgroup$
    – Will Sawin
    Commented Aug 20, 2019 at 20:25

You must log in to answer this question.