Direct proof that there is no Modular form of weight $2$ for $SL_2(\mathbb{Z})$.

Question

I want to show that there are no Modular forms of weight $2$ for $SL_2(\mathbb{Z})$ without using the dimension formular or the $\frac{k}{12}$- formular. I was given some hints, too. However, there are two things missing.

The idea is to look at a antiderivative of $f \in M_2(SL_2(\mathbb{Z}))$ which is given by $$F(\tau)=a_0\tau+\frac{1}{2\pi i} \sum_{n=1}^\infty \frac{a_n}{n}q^n$$ where $f=\sum_{n=0}^\infty a_nq^n$. First I want to show that $F(\gamma\tau)=F(\tau)+C(\gamma)$ where $\gamma\tau$ denotes a mobius transformation and $C(\gamma)$ does not depend on $\tau$.

My approach: $$F'(\gamma \tau)=f(\gamma \tau)*(c\tau+d)^{-2} \ \ (chain \ rule)$$ Since $f(\gamma\tau)=f(\tau)*(c\tau+d)^2$ it follows that $$F'(\gamma \tau)=f(\tau)$$ And after integrating both sides I got $$F(\gamma \tau)=F(\tau)+ C(\gamma).$$ The next step is to show that $C(\gamma_1\gamma_2)=C(\gamma_1)+C(\gamma_2)$ (for $\gamma_1,\gamma_2 \in SL_2(\mathbb{Z}$)) but I have no idea how to show that.

Additionally, I am not sure how to show that $F$ is holomorphic. I see that $F(\tau+1)=F(\tau)$, so $F$ is bounded along the real axis. However, what happens if $\lim y \rightarrow \infty$, especially since $a_0 \tau$ might be unbounded? Thanks for your help

Answer 1

Throughout "form" will mean a modular form for the whole modular group holomorphic everywhere including at cusps.

Let us assume that there is no non-zero form of negative weight.

Let $f$ be a weight two form, and let $a=f(i\infty)$. Then $(f^2-a^2E_4)/\Delta$ is a form of weight $-8$ and so is zero. Therefore $f^2=a^2E_4$. Similarly $f^3=a^3E_6$. So $1728 a^6\Delta=a^6(E_4^3-E_6^2)=f^6-f^6=0$ and so $a=0$. Then $f/\Delta$ has negative weight and so $f=0$.

To show there are no nonzero negative weight forms, note that if there were then there would be a nonzero weight zero form with $g(i\infty)=0$. But every weight zero form is constant (this can be proved using the maximum modulus theorem).

Answer 2

I think we can try a more algebraic answer in the case $\Gamma = SL_2(\mathbb{Z})$, $f \in M_2(\Gamma)$.

$f \in M_2(\Gamma)$ means $f$ is analytic on $\mathcal{H} = \{ z \in \mathbb{C}, \Im(z) > 0\}$ and $\forall \gamma \in \Gamma$,$\gamma(z) = \frac{az+b}{cz+d}$, $f(\gamma (z)) = (cz+d)^2 f(z)$. Since $ad-bc=1$ then $d \gamma (z) = \frac{dz}{(cz+d)^2}$ so that $f(z)dz = f(\gamma( z))d\gamma (z)$ and hence $$F(z) = \int_i^z f(s)ds$$

is an holomorphic map $\mathcal{H} \to \mathbb{C}$ satisfying $$F(\gamma (i))=\int_i^{\gamma(i)}f(s)ds=\int_i^{\gamma(i)}f(\gamma_2(s))d\gamma_2(s)=\int_{\gamma_2(i)}^{\gamma_2(\gamma(i))}f(s)ds =F(\gamma_2(\gamma (i)))-F(\gamma_2 (i))$$ Thus $\phi : \gamma \to F(\gamma (i))$ is an homomorphism $\Gamma \to \mathbb{C}$, since $\mathbb{C}$ is an abelian group, $\phi$ factors through the abelianization $\Gamma^{ab}$.

Here they show $[\Gamma,\Gamma]$ is a subgroup of index $12$ so that the abelianization $\Gamma^{ab}= \Gamma/[\Gamma,\Gamma]$ is a finite group.

Therefore $\forall \gamma \in \Gamma, F(\gamma (i)) \in \phi(\Gamma^{ab})$ which is a finite set of complex values and $F$ is now an holomorphic map $\mathcal{H}^*/[\Gamma,\Gamma] \to \mathbb{C}$.

Since $\mathcal{H}^*/[\Gamma,\Gamma]$ is compact we can take $z \in \mathcal{H}^*/[\Gamma,\Gamma]$ where $|F(z)|$ is maximum, and hence $F$ is constant by the maximum modulus principle, so that $f = 0$.

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If the above argument is correct, it would be nice to explain how the commutator is different for $\Gamma$ a congruence subgroup (allowing in those case $f \in S_2(\Gamma)$ to be an holomorphic map $\mathcal{H}^* / \Gamma \to \mathbb{C}/\Lambda \cong E$ an elliptic curve)