# Modular form not meromorphic at $\infty$

Is there a function $$f$$ with the following properties

1. $$f$$ meromorphic at the upper half plane $$\mathfrak h$$,
2. $$f$$ is of weight $$k$$ under a congruence subgroup of $$\operatorname{SL}_2(\mathbb Z)$$,
3. $$f$$ has an essential singularity at $$\infty$$.

Modular forms or functions are defined to have good behaviour at $$\infty$$. But I have not seen an example showing that this is a necessary part of the definition.

• $e^j \varphi$ where $j$ is the $j$-invariant and $\varphi$ is any form of weight $k$. – Noam D. Elkies Mar 28 at 16:55
• @NoamD.Elkies Thanks! – Shimrod Mar 28 at 16:58

An example is $$e^j \varphi$$ where $$j$$ is the $$j$$-invariant and $$\varphi$$ is any nonzero form of weight $$k$$.
In general a holomorphic (or meromorphic) function on the upper half-plane that is invariant under $${\rm SL}_2({\bf Z})$$ is the same as an entire holomorphic (or meromorphic) functon of $$j$$. So for $$k=0$$ all we need is a entire function (or a meromorphic function on $$\bf C$$) that is not a polynomial (or rational) function. The exponential is a simple example; there are plenty of others, including functions such as $$\tan j$$ which has infinitely many poles in each fundamental domain.
• Also I think that for $k=12 a+b,4\le b\le 14$ the weight $k$ level $1$ modular forms with any kind of singularity at $i\infty$ are of the form $\Delta(z)^a G_b(z)f(j(z))$ with $f$ entire. – reuns Apr 3 at 9:07
• Almost: $b$ must be even, it doesn't quite work for $b=12$ because the Eisenstein series $G_{12}$ has a zero in each fundamental region. So say $b \in \{0, 4, 6, 8, 10, 14\}$ and declare that $G_0 = 1$. (Also, the OP allowed modularity under a congruence subgroup of ${\rm SL}_2({\bf Z})$, and those get harder to describe once the associated modular curve has positive genus.) – Noam D. Elkies Apr 3 at 17:19
• Yes obviously I messed up with $b=0,12$. For the level $n$ and weight $0$ case: the coordinate ring of $\Bbb{\Gamma\backslash H}$ is a free $\Bbb{C}[j]$ module $\sum_{l=0}^r h_l \Bbb{C}[j]$. Is $\sum_{l=0}^r h_l Hol(j)$ integrally closed? If so: $Hol(\Bbb{\Gamma\backslash H})$ is integral over $Hol(j)$ and we'd get $Hol(\Bbb{\Gamma\backslash H})=\sum_{l=0}^rh_l Hol(j)$. – reuns Apr 3 at 18:32