Differential Equations Satisfied by Modular Forms

In Verrill's paper preprint here, she has the following theorem which is from a paper of Stiller. It states that

Let $\Gamma$ be a discrete subgroup of $SL_{2}(\mathbb{R})$ commensurable with $SL_{2}(\mathbb{Z})$. For $f \in M_{k}(\Gamma)$ (the space of weight $k$ modular forms) and $t \in M_{0}(\Gamma)$ (the space of meromorphic weight 0 modular forms), if $f = \sum_{n \geq 0}b_{n}t^{n}$ near $t = 0$, then there is a linear order $k + 1$ differential equation satisfied by $g(x) = \sum_{n \geq 0} b_{n}x^{n}$, of the form $$P_{k + 1}(x)\frac{d^{k + 1}g}{dx^{k + 1}} + P_{k}(x)\frac{d^{k}g}{dx^{k}} + \cdots + P_{0}(x)g = 0 \tag{1}$$ where $P_{i}(x)$ are algebraic functions in $x$.

If we take $t$ to be a Hauptmodul for $\Gamma$, then $P_{i}(x)$ are rational functions. Hence by multiplying by a suitable polynomial, we can in fact assume that the $P_{i}(x)$'s are polynomials. My question is that how does one get explicit bounds on the degrees of these $P_{i}(x)$'s (specifically in the case when $k = 1$)?

(It seems Mr. 120487 has not been back here since the day he posted this. So I will take over the question... -- G. Edgar)

Mr. User's comment seems to say this is impossible. It's not.

The prototype for this may be $$f = \eta(\tau)^2 j(\tau)^{1/12} = 1+60q-4860q^2+660480q^3+\dots$$ with weight $1$, so we expect an order 2 differential equation. The relation we want is $$f = \;{}_2F_1\left(\frac{1}{12},\frac{5}{12};1;\frac{1728}{j}\right) \tag{2}$$ and the hypergeometric differential equation indeed has order 2.

I do not know how to predict the degree of the coefficients in the differential equation. It would be useful to know for computer calculations, though. So we can tell when it will do no good to run it again with higher degree...

Dror asks about writing $\Delta$as a function of $1/j$. This is practiacally already done in (2). Raise (2) to the power $12$, and use $\Delta = (2\pi)^{12} \eta^{24}$ to conclude $$\Delta = \frac{(2\pi)^{12}}{j}\;\;{}_2F_1\left(\frac{1}{12},\frac{5}{12};1;\frac{1728}{j}\right)^{12}$$

This may illustrate the question for us. Suppose we know $\Delta$ and $j$ to many terms of the Fourier series in $q$. (Integer coefficients!) Then can we find the differential equation? If we know a bound on the degree of the coefficients in the differential equation, then we can. Plug in the two $q$-expansions and solve a system of linear equations for the coefficients in the DE. Knowing the degree of the coefficients tells us how many coefficients of the $q$-series will suffice. Since $\Delta$ has weight $12$, we expect a differential equation of order $13$. If the coefficients have degree at most $5$, then we have to solve for around $(13+1)\cdot(5+1) = 84$ unknowns. Should be possible by computer. But if the coefficients have degree $25$? Maybe not.

• Considering only congruence groups, I think there are only a finite number of them with a hauptmodul, so also only a finite number of $k=1$ cases to check, and since hauptmodul is unique up to mobius transformations, really only a finite number of differential equations to check. As for large k, I would be interested to know an answer for $\Delta(1/j)$. – Dror Speiser Mar 22 '17 at 18:30
• @Edgar: nice! btw, I think the $(2\pi)^{12}$ factor can be removed. – Dror Speiser Mar 27 '17 at 7:47

If we write the differential equation in terms of $D_t = t\frac{\partial}{\partial_t}$ then the coefficients in the equation are polynomials in certain functions $p_1, p_2$, their derivatives and $t$. Calculation of these polynomials can be done by a polynomial time algorithm. For $k=1$ we get $D_t^2 f + p_1(t)D_tf + p_2f = 0$.
The functions $p_i$ are given explicitly as $$p_1(t) = \frac{D_q G_1 - 2G_1G_2/k}{G_1^2}, \quad p_2(t) = -\frac{D_q G_2 - G_2^2/k}{G_1^2},$$ where $D_t = q\frac{\partial}{\partial_q}$ and $$G_1 = \frac{D_q t}{t}, \quad G_2 = \frac{D_q f}{f}.$$
So the question boils down to bounding degrees in $p_1$ and $p_2$.