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Let $k\geq 0$, $N\geq 2$ be integers. I suspect that there is an action of $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$ on the finite-dimensional $\mathbb{Q}$-vector space of holomorphic modular forms of level $\Gamma(N)$ and weight $k$ such that the invariant subspace is the space of holomorphic modular forms of level $\Gamma(1)$ and weight $k$ (considered as level $\Gamma(N)$ modular forms).

I think so because in the moduli interpretation of the modular curve $X(N)$ the group $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$ acts on the set of pairs of generators of the $N$-torsion group on a generalized elliptic curve (but for $N=2$ the curve is stacky so I am not sure).

First question: does such an action indeed exist and where can I read more about it? Second question: can I explicitly write down what this action does to the Fourier coefficients of modular forms?

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    $\begingroup$ Tiny quibble: you mention a $\mathbf{Q}$-vector space. If you're thinking of the space of $\Gamma(N)$ modular forms with Fourier coefficients in $\mathbb{Q}$, then this space is not preserved by the $SL_2$ action; you need to work over $\mathbb{Q}(\zeta_N)$. This also prevents there being a "nice" formula for the action on q-expansions. $\endgroup$
    – David Loeffler
    Aug 20, 2019 at 6:31
  • $\begingroup$ @DavidLoeffler I don't understand one thing. For a prime level $p\geq 3$, does there exist a geometrically connected smooth projective curve $X$ over $\mathbb{Q}$ whose base change to $\mathbb{C}$ is the (connected) modular curve $X(p)$ and such that the $\mathrm{PSL}_2(\mathbb{F}_p)$ automorphisms of $X(p)$ are base changed from automorphisms of $X$? $\endgroup$
    – user693936
    Aug 20, 2019 at 7:45
  • $\begingroup$ I don't know if such an $X$ exists. But if you add the requirement that the field of rational functions on $X / \mathbb{Q}$ should be the meromorphic modular functions of level $\Gamma(p)$ with rational Fourier coefficients, then the answer is definitely "no". $\endgroup$
    – David Loeffler
    Aug 20, 2019 at 10:29

2 Answers 2

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There is an action of $\text{SL}_2(\mathbb{Z}/N\mathbb{Z})$. Let $f$ be a modular form of weight $k$, invariant for $\Gamma(N)$. Suppose $\gamma \in \text{SL}_2(\mathbb{Z}/N\mathbb{Z})$ and lift $\gamma$ to a matrix $\widetilde{\gamma} \in \text{SL}_2(\mathbb{Z})$ (this is the hard part of the strong approximation theorem).

Then $f|_k \widetilde{\gamma}$ is clearly independent of the choice of lift (another choice differs on the left by something in $\Gamma(N)$), and it's invariant for the action of $\Gamma(N)$ because $\Gamma(N)$ is normal in $\text{SL}_2(\mathbb{Z})$. I'm not sure if there's a nice formula for the Fourier coefficients -- probably there is!

Your argument with the moduli interpretation suggests this extends to an action of $\mathbb{GL}_2(\mathbb{Z}/N\mathbb{Z})$, but this is misleading because the fine moduli space for this curve is geometrically disconnected and its complex points are isomorphic to $\phi(N)$ copies of $\mathcal{H}/\Gamma(N)$. This is where you get an action of $GL_2$. See more at The modular curve X(N).

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  • $\begingroup$ (Just a remark that the first two paragraphs have very little to do with modular forms. All that's going on is that if $G$ is a group acting on a set $S$ and $H$ is a normal subgroup of $G$, then $G/H$ acts on the invariants $S^H$.) $\endgroup$
    – hunter
    Aug 20, 2019 at 3:58
  • $\begingroup$ if a modular form of level $\Gamma(N)$ is invariant under the action of $\mathrm{SL}_2(\mathbb{Z}/N\mathbb{Z})$ is it actually of level $\Gamma(1)$ (in the sense that its Fourier expansion is that of a level $\Gamma(1)$ modular form)? $\endgroup$
    – user693936
    Aug 20, 2019 at 7:13
  • $\begingroup$ also what does the notation $f|_k \gamma$ mean? $\endgroup$
    – user693936
    Aug 20, 2019 at 7:29
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    $\begingroup$ @hello $f|_k(\gamma)(z) = f(\gamma z) j(\gamma, z)^{-k}$ where $j(\gamma, z) = (\gamma_{21}z + \gamma_{22})$. A modular form of weight $k$ is precisely a fixed point of this action. $\endgroup$
    – hunter
    Aug 20, 2019 at 11:48
  • $\begingroup$ (I edited the notation in the answer just now to make it clear that it's the choice of lift that's acting by this formula). $\endgroup$
    – hunter
    Aug 20, 2019 at 11:49
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The Riemann surface $X(N)$ can be obtained from $Y(N)$ plus its few cusps $$Y(N)= \{ \Bbb{C}^*(\Bbb{Z}+z\Bbb{Z}), \Bbb{C}^*(p+\Bbb{Z}+z\Bbb{Z}) ,\Bbb{C}^*(q+\Bbb{Z}+z\Bbb{Z})\}$$ where $p, q \in \Bbb{C}/(\Bbb{Z}+z\Bbb{Z})$ are two points that generate the $N$ torsion of the complex torus $\Bbb{C}/(\Bbb{Z}+z\Bbb{Z})$ and $\Bbb{C}^*(\Bbb{Z}+z\Bbb{Z})$ is the set of all lattices (=complex tori) isomorphic to $\Bbb{Z}+z\Bbb{Z}$.

If $ \pmatrix{a & b \\ c & d} \in GL_2(\Bbb{Z}/N\Bbb{Z})$ you might want to try the map $$( z,p,q) \mapsto ( z, ap+bq, cp+dq)$$ which does permute the points of $Y(N)$.

If this map is holomorphic then it is continuous, and since locally $z,p,q$ are complex numbers, the map is given by some $a,b,c,d \in \Bbb{Z}$, which means it extends holomorphically to the map $$\Bbb{Z}+z\Bbb{Z} \mapsto (a+bz)\Bbb{Z}+(c+dz) \Bbb{Z} $$ and that map is well-defined on $\Bbb{C}^* (\Bbb{Z}+z\Bbb{Z})$ iff $ \pmatrix{a & b \\ c & d} \in GL_2(\Bbb{Z})= \pm SL_2(\Bbb{Z})$.

Whence the group of holomorphic automorphisms of $Y(N)$ is $$\pm SL_2(\Bbb{Z}) / \ker(\pm SL_2(\Bbb{Z}) \to GL_2(\Bbb{Z}/N\Bbb{Z})) = \pm SL_2(\Bbb{Z}/N\Bbb{Z})$$

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