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I was going over the proof that all entire modular forms are expressible as polynomials of Eisenstein series $G_4$ and $G_6$. The proof is supported by inducting on the weight of modular forms. And an argument was used to prove that any entire modular form $f$ of weight $k$ can be written as $cG_k+\Delta h$ where $h$ is an entire modular function of weight $k-12$.

We pick $$c=\frac{f(i\infty)}{G_k(i\infty)}=\frac{c_f(0)}{c_{G_k}(0)}$$ where the $c(0)$ are the coefficients in the Fourier expansions of these functions. Thus, ${f-cG_k}$ is a cusp form. However, I cannot understand the following argument: we can write $f-cG_k=\Delta h$ where $h$ is an entire modular function of weight $k-12$. I tried to think about the ratio $\frac{f-cG_k}{\Delta h}$. If it's analytic and it's of weight $0$ then it is a constant function. The increase of weights can be explained by the fact that $\Delta$ has a zero of multiplicity $1$ at $i\infty$. However, we must find $h$ having zeros of the same multiplicities as $f$ at $i$, $\rho$ and inside the fundamental region $R_\Gamma$. How can we be sure that such $h$ exists?

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The function $h$ is necessarily $$h = \frac{f - cG_k}{\Delta}.$$ It's just a matter of checking that $h$ really is a modular form of weight $k - 12$. That $h$ satisfies the weight $k - 12$ transformation law is just a consequence of the fact that $f - cG_k$ is a weight $k$ form and $\Delta$ is a weight $12$ form. But you still need to check that $h$ is holomorphic on the upper half-plane and at $i\infty$. There is no possibility of a pole in the upper half-plane because $\Delta$ has no zeros in the upper half-plane. Since $f - cG_k$ is a cusp form, it has a zero at $i\infty$. On the other hand, $\Delta$ has an order $1$ zero at $i\infty$, therefore the quotient $h$ is holomorphic at $i\infty$.

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  • $\begingroup$ This really clears things up. $\endgroup$
    – justadzr
    Jul 16, 2020 at 4:31

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