# On pseudo rational modular forms of weight 2 and level N

So consider the $\mathbb{Q}$-vector space $V$ of functions which satisfy the following conditions

(1) $f:\mathbb{H}\rightarrow\mathbb{C}$ is holomorphic. Here $\mathbb{H}$ stands for the upper half plane.

(2) $f(z+1)=f(z)$

(3) The Fourier series of $f$ at infinity has the form $\sum_{n\geq 1} a_nq^n$ where $q=e^{2\pi iz}$ where the $a_n$'s are rational numbers

(4) $\frac{1}{Nz^2}f(\frac{-1}{Nz})=\pm f(z)$

Examples of non-zero elements of $V$ are given for example by $\mathbb{Q}$-linear combinations of modular forms associated to rational elliptic curves of conductor $N$ that have the same sign in their functional equation. Let us denote this sub vector space by $W$. Unless I made a mistake in my calculation, an example of an element of $V$ which is not in $W$ could be  $$\sum_{d|N} d a_d E_2(dz)$$  with $\sum d a_d=0$ and $a_d=a_{N/d}$. If $N$ is sufficiently composite then we may find such $a_d$'s. Here $E_2$ is the weight $2$ Eisenstein series suitably normalized.

Q: How big is $V$ and is it possible to describe it in some interesting way?

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Not in general — not even after correcting the factor $1/(N^2z^2)$ to $1/Nz^2$. There are several problems here. First, not every elliptic curve of conductor $N$ works: you must impose a sign condition. Second, the modular forms of level $N$ have a rational basis but usually most of them come from abelian varieties of dimension $>1$. Finally, once $N$ is at all large the transformations $z \mapsto z+1$ and $z \mapsto -1/Nz$ generate only a tiny part of the relevant modular group, so there should be plenty of other "pseudo modular" cusp forms. – Noam D. Elkies Sep 6 '11 at 23:38
Yes I just edited. The transformation $z\mapsto \frac{-1}{Nz}$ is not in the modular group unless $N=1$ but I guess you could work in a bigger group which contains the involution of level $N$. – Hugo Chapdelaine Sep 6 '11 at 23:43
So @Noam do you think you could use Poincare's trick and average out over the group generated by $z\mapsto z+1$ and $z\mapsto \frac{-1}{Nz}$. Of course, one has to be careful about convergence issues. – Hugo Chapdelaine Sep 6 '11 at 23:45
@Hugo C.'s edit: indeed the Fricke involution $z \mapsto -1/Nz$ is not in $\Gamma_0(N)$ but is in the normalizer of $\Gamma_0(N)$ in ${\rm SL}_2({\bf R})$. However, once $N>4$ Fricke and $z \mapsto z+1$ do not generate the full normalizer.  Also I don't think $E_2(z) - NE_2(Nz)$ works: that's a modular form of weight $2$ for all of $\Gamma_0(N)$ [proportional to the logarithmic differential of the modular function $\Delta(z) / \Delta(Nz)$], but it does not vanish at $q=0$. – Noam D. Elkies Sep 6 '11 at 23:47
@Hugo C.'s question: It's not clear to me whether the Poincaré trick would work well enough to produce a modular form with rational coefficients. But you can probably find a subgroup $G$ contained in $\Gamma_0(N)$ with finite index and use holomorphic differentials on the Riemann surface ${\cal H}^* / G$. – Noam D. Elkies Sep 6 '11 at 23:49

To summarize some remarks in the comments. The function $f \cdot d \tau$ will be a differential on $Y_N:=\mathbf{H}/\Gamma$, where $\Gamma_N$ is the group generated by the two matrices $$\left( \begin{matrix} 1 & 1 \\\ 0 & 1 \end{matrix} \right) \ , \ \left( \begin{matrix} 0 & -1 \\\ N & 0 \end{matrix} \right)$$ If $N > 4$, then $\Gamma_N$ will have infinite index in $\mathrm{SL}_2(\mathbf{Z})$. In particular, if $X_N$ is the complex curve obtained by filling in the puncture at $\infty$ (to account for the condition that the $q$-expansion has positive coefficients), then $X_N$ (for $N > 4$) will be an open curve, and so $H^1(X_N,\Omega^1)$ will be infinite dimensional (and not particularly interesting). It's a theorem of Hecke that $\Gamma_N$ is the normalizer of $\Gamma_0(N)$ for $N \le 4$. In this way, Hecke was able to show that an $L$-series $L(f,s) = \sum a_n n^{-s}$ which satisfied a functional equation of a certain kind (with "conductor" $N \le 4$) was exactly the Mellin transform of a modular form. Being modular is thus (for $N > 4$) a stronger condition than simply asking that the $L$-series satisfies the appropriate functional equation. Historically this was interesting, because one could conjecture that the $L$-series of elliptic curves satisfied a functional equation of a certain kind, which is (a priori) a weaker conjecture than asking that elliptic curves are modular. Weil, however, showed that if the $L$-series attached to an elliptic curve $E/\mathbf{Q}$ satisfied the right functional equation, and the same was true of all the quadratic twists of $E$, then $E$ was actually modular (the so called converse theorem, which has been vastly generalized).
No, it will be much much bigger. A good model to think of is the space of holomorphic differentials on the disc $|z| < 1$, which is isomorphic to the space of power series with radius of convergence $\ge 1$ (in particular, it has uncountable dimension). – Michael Sep 7 '11 at 3:15
I agree that the space will likely be too huge to say much about, but it's not as simple as Michael's answer suggests, because Hugo C. asked for rational $q$-expansions, and it's not immediately obvious there'll be lots of non-modular forms for "$\Gamma_N$" with rational coefficients, let alone an uncountable-dimensional space of such forms. – Noam D. Elkies Sep 7 '11 at 4:09
Noam makes a good point - I actually didn't notice that you asked for $\mathbf{Q}$-coefficients. However, as soon as you have a single holomorphic function $z$ on $X_N$ with rational coefficients, then, after normalizing $z$ to have norm $\le 1$, and letting $f$ be any classical weight $2$ form with rational coefficients on $X_0(N)^{+}$, then $f \cdot B(z,1)$ is a subspace of the holomorphic weight $2$ form with rational coefficients, where $B(z,1)$ denotes the power series in $\mathbf{Q}[[z]]$ with radius of convergence at least $1$> – Michael Sep 7 '11 at 5:35
Right, right so this space will be incredibly huge! So too construct your function $z$ you might try to take the quotient of two linearly independent "rational "Poincare series of the same weight. Something like $\sum_{\gamma\in \Gamma_N} H(\gamma(z))$ for two suitable $H$'s should work. – Hugo Chapdelaine Sep 7 '11 at 11:33