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Suppose $f, g \in M_{k}(\Gamma)$ for some $\Gamma$ a congruence subgroup of $PSL_{2}(\mathbb{Z})$. If $k \geq 1$ then it is a finite computation to determine whether or not $f = g$. What if $k = 0$, that is, if $f$ and $g$ are both modular functions and I know their $q$-expansions, how can I tell when $f = g$?

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$M_0(\Gamma)$ is not the space of modular functions... – Qiaochu Yuan Jul 10 '11 at 4:27
up vote 6 down vote accepted

As Qiaochu has pointed out, the answer to this question depends on what exactly you mean by "modular function".

If you mean a function that satisfies the weight k functional equation for the action of $\Gamma$ and is holomorphic everywhere on the upper half-plane and the cusps, then this is what is generally called a modular form; and the only weight 0 modular forms are constants, so the question is trivial, and for weight $k \ge 0$ I gave an answer here.

If you mean only that your functions are meromorphic on $\mathcal{H}^* = \mathcal{H} \cup \{\text{cusps}\}$, then (whatever $k$ is) you can't answer the question just knowing the $q$-expansion, but you can if you know more about the functions involved. If you know some finite set of points of $\Gamma \backslash \mathcal{H}^*$ such that your functions have no poles away from these points, and you can bound the order of the poles at the bad points, then you're in good shape, since you can find a modular form $h$ of some weight $k$ such that $fh$ and $gh$ are modular forms, and that reduces you to the first case. (Alternatively, $f - g$ is a section of a line bundle on a compact Riemann surface, so if it vanishes at $\infty$ with multiplicity larger than the degree of the line bundle, it must be zero.)

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Ah I see now, sorry, I did mean meromorphic...Thanks for the response! – Shayla Jul 14 '11 at 1:07

The dimension of $\mathcal M_0(\Gamma)$ for a congruence subgroup is always $1$, so the only modular functions are the constant functions.

(I think; I'm a complete amateur at this stuff and something could be deathly wrong with what I say below)

In more detail, we know that a modular function $f$ is uniquely determined by its values on the quotient and Riemann surface $\Gamma\backslash H$. If $f$ has a $q$-expansion, this means that $f$ is holomorphic at the cusp, i.e. that it extends to a $\Gamma$-invariant holomorphic function on $H\cup\mathbb \Gamma(\infty)$. This means that it determines a holomorphic function on the compactificatied Riemann surface $\Gamma\backslash H^*$, which because the Riemann surface is compact has to be constant, and thus $f$ itself was constant.

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I think by "modular function" the OP means a function meromorphic, not holomorphic, at the cusps. – Qiaochu Yuan Jul 10 '11 at 4:28

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