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How to see there are no nontrivial cusp forms for $\Gamma_0(4)$ of weight 2

I am searching for a proof of this that directly uses the provided set of generators for $M_2(\Gamma_0(4))$.

As mentioned in the other post, any cusp form in such space seems to be of the form $\lambda(3E_{2,2}-E_{2,4})$ for some $\lambda\in \mathbb{C}$.

Siegmeyer of Catarina
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    So all you need to do is show that $(3E_{2,2}-E_{2,4})\mid \begin{pmatrix}0&-1\1&0\end{pmatrix}$ does not go to zero at $\infty$. To show that, it should be enough to use the fact that $E_2(z)-\pi/Im(z)$ is $SL_2(\mathbb{Z})$-invariant to find an expansion at infinity of $E_2(N\cdot)\mid\begin{pmatrix}0&-1\1&0\end{pmatrix}=\leftE_2\mid \begin{pmatrix}0&-1\1&0\end{pmatrix} \right$. – Aphelli Jun 06 '23 at 07:31
  • @Aphelli What do you mean by $(3E_{2,2}-E_{2,4}) |$? I do not know that notation – Siegmeyer of Catarina Jun 06 '23 at 07:43
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    I meant the standard action of $SL_2(\mathbb{Z})$ on weight $k$ modular forms, $\leftf\mid \begin{pmatrix}a&b\c&d\end{pmatrix}\right=(cz+d)^{-k}f\left(\frac{az+b}{cz+d}\right)$. – Aphelli Jun 06 '23 at 10:34
  • @Aphelli I am sorry. I do not manage to obtain the expansion for $E_{2}|\begin{pmatrix} 0 & - 1 \1 & 0 \end{pmatrix}$. The function $\pi/Im(z)$ is not holomorphic, right? – Siegmeyer of Catarina Jun 06 '23 at 12:42
  • @Aphelli Also \begin{pmatrix} 0&-1\1&0\end{pmatrix} is not in $\Gamma_0(4)$ – Siegmeyer of Catarina Jun 06 '23 at 12:59
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    I think you didn’t understand the approach I suggested. It doesn’t matter that the matrix isn’t in $\Gamma_0(4)$ or that the modification to $E_2$ isn’t holomorphic. We just need to find an expansion at the other cusps for $E_{2,2}$ and $E_{2,4}$ so that we can tell whether $3E_{2,2}-E_{2,4}$ vanishes at all cusps. To find this expansion, we use that $E_2(\tau)-c/Im(\tau)$ is $SL_2(\mathbb{Z}$-invariant, for some constant $c$ that I’m not sure about any more (it’s shown in eg Diamond-Shurman). This yields an expansion of $E_2$ (and then we can tackle $E_2(2z),E_2(4z)$) at any other cusp. – Aphelli Jun 06 '23 at 15:09
  • @Aphelli Yes, with this I was able to solve it. Thank you! – Siegmeyer of Catarina Jun 06 '23 at 16:24

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