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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.

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    $\begingroup$ $e^j \varphi$ where $j$ is the $j$-invariant and $\varphi$ is any form of weight $k$. $\endgroup$
    – Noam D. Elkies
    Commented Mar 28, 2021 at 16:55
  • $\begingroup$ @NoamD.Elkies Thanks! $\endgroup$
    – Shimrod
    Commented Mar 28, 2021 at 16:58

1 Answer 1

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[expanding on my comment to convert it to an answer]

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.

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  • $\begingroup$ 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. $\endgroup$
    – reuns
    Commented Apr 3, 2021 at 9:07
  • $\begingroup$ 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.) $\endgroup$
    – Noam D. Elkies
    Commented Apr 3, 2021 at 17:19
  • $\begingroup$ 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 $Hol(\Bbb{\Gamma\backslash H})=\sum_{l=0}^rh_l Hol(j)$. $\endgroup$
    – reuns
    Commented Apr 3, 2021 at 18:32

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