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Hi,

I am interested in the following fact:

Suppose that $f(z)$ is a modular function over $\text{SL}_2(\mathbb{Z})$ such that it has the $q$-expansion

$f(z) = q^{-m} + \displaystyle \sum_{n=1}^\infty a_m(n) q^n$

It is claimed in Ono, K, "The partition function and Hecke operators", Advances in Mathematics 228 (2011), 527-534 that this function (which he denotes $j_m(z)$) is unique.

I know that modularity imposes strict restrictions on what modular functions must look like, but I have a problem seeing this claim. Can anyone offer a short proof or some ideas as to why this is the case?

Thanks in advance.

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The difference of two such functions will be a cusp form of weight 0 on the full modular group -- i.e., 0 (I think). –  alpoge Mar 19 '12 at 19:21
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@alpoge: That's nice. The computational approach is to use the theorem that $f(z)$ is a polynomial in $j$. Start with $j^m$, and for each $k=m-1,m-2,m-3,\ldots,0$ use the vanishing of the $q^{-k}$ coefficient to determine the coefficient of $j^k$. At the end you've found the polynomial and proven it unique. –  Noam D. Elkies Mar 19 '12 at 20:16
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1 Answer

up vote 5 down vote accepted

I would say that the basic fact underlying this is that the only regular functions on a compact Riemann surface are constant functions, but I'm guessing that this isn't the missing detail in your case.

A modular function for $\mathrm{SL}_2(\mathbb{Z})$ is (basically by definition in your context) a holomorphic function on the quotient of the upper half-plane by its usual action of $\mathrm{SL}_2(\mathbb{Z})$ by fractional linear transformations. At the end of the day, this amounts to a holomorphic function on the Riemann sphere minus the point $\infty$ (if you haven't done so already, reading about the modular function known as the $j$-invariant will help to explain this). The $q$-expansion of such a function is its local expansion about $\infty$. In particular, if two such functions have $q$-expansions of the type you specify, then their difference is regular everywhere and vanishes at $\infty$, and hence must be the constant zero.

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