In complex analysis, it is possible to decompose any given entire function into a product of linear factors using the zeros, using the Weierstrass factorization theorem. For example, $\sin(z)$ is factorized into an infinite product as follows:
$$\sin(z) = z \prod_{k = 1}^{\infty} \left( 1 - \frac{z^2}{\left( \pi k \right)^2} \right)$$
Note that to provide convergence, some linear factors may be combined to form factors of higher order than the linear factor but simpler than the original function.
Also, in the case of rational functions, factorization can be performed using all zeros and poles as well. For example, $\cot(z)$ can be factorized in a similar way to $\sin(z)$. This factorization of complex functions has the advantage of revealing the fundamental structure of complex functions through zeros and poles, and also sometimes providing very useful and mysterious relations.
Then I've thought about modular forms.
Since the modular form has a natural boundary, the domain is limited to the open upper half-plane $\mathbb{H}$. For that reason, we can't apply the Weierstrass factorization theorem to a modular form, even if the modular form is holomorphic in $\mathbb{H}$. And even though the modular form may not have any zeros or poles in $\mathbb{H}$, since these functions can exhibit behavior such as vanishing or divergence at the cusps $\mathbb{Q} \cup \left\{ i \infty \right\}$, they can't be of the form $\exp(f(z))$ where $f(z)$ is an entire function.
An important example of a modular form that has an infinite product that is similar to a factorization is the modular discriminant (and the Dedekind eta function) . Modular discriminant is defined as follows. $$\Delta(z) = e^{2 \pi i z} \prod_{k = 1}^{\infty} \left( 1 - e^{2 \pi i k z} \right)^{24}$$ This is an infinite product resulting from the fact that $\Delta(z)$ has no zeros or poles in $\mathbb{H}$, and 'becomes zero at $\mathbb{Q} \cup \left\{ i \infty \right\}$'. It should be noted that in this infinite product, one 'zero' (actually one cusp) does not make only one factor $0$, but an infinite number of factors $0$, thus to be precise, this expression is not a factorization of $\Delta(z)$. (Moreover, the structure of $\Delta(z)$ at $z \in \mathbb{Q} \cup \left\{ i \infty \right\}$ is more of an essential singularity rather than a multiple zero.)
Besides $\Delta(z)$, there are many modular forms whose factorization, or even just an infinite product expression is not well-known. The most basic modular forms are those of level $1$ and that are holomorphic in $\mathbb{H}$ and $i \infty$. Since these modular forms can be expressed as polynomials of $E_4(z)$ and $E_6(z)$, to know the factorization of them, we need to know the factorization of functions $E_4(z)$, $E_6(z)$, and arbitrary linear combinations of $E_4(z)^3$ and $E_6(z)^2$, i.e. of the form $a E_4(z)^3 + b E_6(z)^2$ ($a$, $b$: constant) . (The last one is the same as factorizing $j(z) - c$ ($c$: constant) .)
These modular forms have an infinite number of zeros in $\mathbb{H}$. I would like to express the modular form through the infinite product that reflect the structure of the zeros of the function using only simple basic factors. For example, first using the periodicity of modular forms, we can think of a basic factor of the form $1 - e^{2 \pi i (z - \rho)}$ for a zero of the modular form $\rho$ where $0 \leq re(\rho) < 1$. However, simply multiplying all of these factors for all $\rho$ does not give the correct infinite product. For instance, when factoring $E_6(z)$, the set of all zeros $\rho$ with $0 \leq re(\rho) < 1$ is $\left\{ \frac{n + i}{m} \, \middle| \, m \in \mathbb{N}, n \in \mathbb{Z}, 0 \le n < m, m \vert \left( n^2 + 1 \right) \right\}$, so by multiplying all factors $1 - e^{2 \pi i (z - \rho)}$ we can construct an infinite product as follows:
$$\prod_{\substack{m \in \mathbb{N}, n \in \mathbb{Z} \\ 0 \le n < m \\ m \vert \left( n^2 + 1 \right)}} \left( 1 - e^{2 \pi i \left( z - \frac{n + i}{m} \right)} \right)$$
But this is not the same as $E_6(z)$. This equation has the same zeros as $E_6(z)$ in $\mathbb{H}$ and converges to $1$ as $z \to i \infty$, but when $z$ is near the real axis, this infinite product behaves differently from $E_6(z)$. As such, some basic attempts don't give the factorization of modular forms. (Sometimes the infinite product does not converge, and sometimes it converges but the convergence value is different from the correct one.)
Therefore, I wonder how, given a modular form with some zeros and poles in $\mathbb{H}$ and cusps $\mathbb{Q} \cup \left\{ i \infty \right\}$, it can be expressed as an infinite product that reflects the structure of zeros and poles of it. Are such factorizations known for $E_4(z)$, $E_6(z)$, and for all linear combinations of $E_4(z)^3$ and $E_6(z)^2$? I strongly believe that such a factorization should exist. The factorization of modular forms is a fundamental and natural question, and I guess many people would have thought of it independently before, but I haven't found any articles or papers discussing it.
Additional questions:
· What about the more general modular forms? The factorization of modular forms of level $N$? The factorization of quasimodular forms? (In particular, the structure of the zeros of the basic quasimodular form $E_2(z)$ is not perfectly regular and has an interesting shape.) Is there anything known about them too?
· In general, when the natural domain of a complex function is smaller than $\mathbb{C}$ for reasons such as the natural boundary, or when the complex function has a Riemann surface other than $\mathbb{C}$ (such as a quotient space of $\mathbb{C}$ or a covering space of $\mathbb{C}$) as its natural domain, it is difficult to apply the Weierstrass factorization theorem to the functions. Is there such a thing as a generalization of the Weierstrass factorization theorem that we can use to find the factorization of these complex functions? Or even in the case of these complex functions, I wonder if there is at least known results about factorization of them through the structure of the zeros and poles.
Thank you for reading this question. Any answers will be greatly appreciated.