# Automorphisms of the modular curve defined over $\mathbb{Q}$

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 $$X$$?

• 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 $F_n$ – reuns Aug 20 '19 at 12:19
• @reuns can one give this curve a moduli interpretation? – user144105 Aug 20 '19 at 12:22
• @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. – Will Sawin Aug 20 '19 at 12:53
• This is not the same curve as $X(n)$ (except if $\phi(n)=n$). – Will Sawin Aug 20 '19 at 12:55
• 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. – Will Sawin Aug 20 '19 at 18:43

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 $$p=5$$.
• why is there at least one eigenvalue not equal to $1$? – user144105 Aug 20 '19 at 19:33
• @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. – Will Sawin Aug 20 '19 at 20:25