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I know that $M_2(\Gamma_0(4))$ is generated by

$E_{2,2}(z)=E_2(z)-2E_2(2z)$ and

$E_{2,4}(z)=E_2(z)-4E_2(4z)$,

where $E_2$ is the Eisenstein's series.

This space is supposed to only have $0$ as cusp form (i.e. $S_2(\Gamma_0(4))=\{0\}$), but I am struggling to find a proof of this.

I believe any cusp form must be of the form $\lambda(3E_{2,2}-E_{2,4})$ for some $\lambda\in \mathbb{C}$, but I do not manage to obtain nothing from that.

I am searching for a proof that uses the fact that $E_{2,2}$ and $E_{2, 4}$ generate $M_2(\Gamma_0(4))$.

Siegmeyer of Catarina
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1 Answers1

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Quite analogously to the $SL_2(\mathbb{Z})$ case, where the argument principle gives the relation $\nu_i/2+\nu_\omega/3+\sum \nu_z+\nu_\infty=\mathrm{weight}/12$, using the standard explicit fundamental domain, for the conjugate $\Gamma_\theta$ of $\Gamma_o(4)$, there is $\nu_i/2+\nu_\infty+\sum \nu_z = \mathrm{weight}/4$. Thus, analogous to the corresponding discussion for $SL_2(\mathbb Z)$, for suitably low weights there can be no vanishing at infinity.

For example, my notes https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2013-14/10_thetas_equi.pdf treat this (well-known) situation.

Yes, this is an application of Riemann-Roch in low genus, so gives explicit formulas easily.

paul garrett
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  • Do you know of a proof that uses this set of generators? – Siegmeyer of Catarina Jun 06 '23 at 06:53
  • If we really know (by some device) that the weight two holo modular forms of that level are linear combinations of those two Eisenstein series... if we also check that the two-tuples of their values at (the two) cusps are linearly independent, then the only linear combination vanishing at both cusps is $0$. – paul garrett Jun 06 '23 at 15:58