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In his notebooks Ramanujan mentions something called a "complete series" which is some power series $\sum_{n = 0}^{\infty}a_{n}q^{n}$ in terms of $q = e^{-y}$ with $y = \pi K'/K$ and $z = 2K/\pi$ such that the expression $$S = \frac{1}{z^{p}}\sum_{n = 0}^{\infty}a_{n}q^{n}$$ is an algebraic function of $k = \vartheta_{2}^{2}(q)/\vartheta_{3}^{2}(q)$

The exact quote of Ramanujan is given in Entry 1, Chap 21, Ramanujan's Notebooks vol 3 by Bruce C. Berndt. Here Ramanujan just says that the value $S$ defined above can be expressed in terms of radicals.

Then Ramanujan gives examples of Eisenstein Series $E_{4}$ and $E_{6}$ defined as $$E_{4}(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}, E_{6}(q) = 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}q^{n}}{1 - q^{n}}$$ Indeed with factors $1/z^{4}$ and $1/z^{6}$ we can see that $E_{4}/z^{4}$ and $E_{6}/z^{6}$ are in fact polynomials in $k = \vartheta_{2}^{2}(q)/\vartheta_{3}^{2}(q)$.

I believe that Ramanujan did have his own sense of a modular form and he called it a "complete series". Going by his ideas let's formulate the following definition.

Let $0 < q < 1$ and let $k = \vartheta_{2}^{2}(q)/\vartheta_{3}^{2}(q)$ and $z = \vartheta_{3}^{2}(q)$. A function $f(q)$ defined by a power series $f(q) = \sum_{n = 0}^{\infty}a_{n}q^{n}$ is said to be modular form of weight $p > 0$ if the expression $f(q)/z^{p}$ is an algebraic function of $k$.

By using the power series we just want to make sure that $f(q)$ is analytic in $|q| < 1$. I want to know whether such a formulation is acceptable (at least to some extent)? How far does this match with the modern definition of a "modular form"?

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I would say that the definition you formulate (which as you indicate is very much in the spirit of what Ramanujan said) is close to the modern definition of a modular form. In fact, it is a bit more inclusive.

More precisely, given a meromorphic modular form $f$ of weight $p$ for a finite index subgroup $\Gamma$, for some finite index subgroup $\Gamma$ containing the group of transformations for $k$, then $f/z^{p}$ is a meromorphic modular function for $\Gamma$. Let $K$ be the field of modular functions for $\Gamma$. This field is a transcendence degree $1$ extension of $\mathbb{C}$ and so $K/\mathbb{C}(k)$ is a finite, algebraic extension with Galois group that can be expressed in terms of $\Gamma$. In particular, $f/z^{p}$ is an algebraic function of $k$. (You can find more details about this in "A first course in modular forms" by Diamond and Shurman.)

However, it is not necessarily true that every algebraic function of $f/z^{p}$ is a modular function. The reason for this is that the Galois group of $\mathbb{C}(k)$ is too large for all algebraic functions to come from modular functions. Every finite index, torsion-free subgroup of ${\rm PSL}_{2}(\mathbb{Z})$ is a free group. Since every finite group is arises as the Galois group of some extension of $\mathbb{C}(k)$, but not every finite group is a quotient of a free group on $r$ generators, there will be algebraic functions that don't come from modular forms for any subgroup (congruence or otherwise).

One final comment on the definition: normally when someone refers to a modular form, they normally mean a holomorphic modular form. This requires that $f(q)$ is holomorphic for $|q| < 1$, satisfies the transformation law for a subgroup $\Gamma$, and is also holomorphic at the cusps of $\Gamma$. The conditions you specify above enforce the first two, but not the third. As a simple example, $k^{2}$ satisfies all of your conditions, but it is not a holomorphic modular form of weight zero, because it has a pole at one of the cusps of $\Gamma$. (This third condition is equivalent to polynomial growth of the coefficients $a_{n}$.)

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  • $\begingroup$ Thanks Jeremy Rouse for a very comprehensive answer. Exactly what I needed and +1 for that. From the answer it is clear that Ramanujan did have very good idea of the importance of such functions as "complete series/modular forms" and even more he was able to express these functions directly as algebraic functions of $k$. $\endgroup$ – Paramanand Singh Apr 24 '14 at 6:22

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