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Let $\Gamma$ be a finite index subgroup of $\Gamma(1)=SL_2(\mathbf{Z})$ and $f$ a modular function for $\Gamma$. By this I mean a meromorphic function defined on the upper half-plane $f: \mathfrak{h} \to \mathbf{C}$, satisfying a growth condition at infinity, which is invariant by $\Gamma$: for every $\gamma \in \Gamma$, $f(\gamma(\tau))=f(\tau)$.

My question is: what can be said about the action of the full modular group $\Gamma(1)$ on $f$? For $\gamma \in \Gamma(1)$, can we always express the function $\tau \mapsto f(\gamma(\tau))$ in terms of $f$?

For example, take the modular lambda function $\lambda$. It is an everywhere holomorphic modular function (or better, modular form of weight zero) for $\Gamma(2) = \{\gamma \in \Gamma(1) \mid \gamma \equiv id \mod 2\}$. It can be proven that \begin{align*} \lambda(-1/\tau) = 1-\lambda(\tau) \end{align*} and \begin{align*} \lambda(\tau +1 ) = \frac{\lambda(\tau)}{1-\lambda(\tau)} \end{align*} so that we actually know how $\lambda$ transforms by $\Gamma(1)$, and not just $\Gamma(2)$. I'm asking if this is general.

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No, this cannot always be done. For instance, the function $$f(z) = \left(\Delta(5z) / \Delta(z)\right)^{1/4}$$ is a modular function of level $\Gamma_0(5)$, but if you act on it by an element of $\Gamma(1)$ that doesn't normalise $\Gamma_0(5)$, such as $z \mapsto -1/z$, the result won't be $\Gamma_0(5)$-invariant, so it certainly can't be expressed as a rational function in $f$.

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  • $\begingroup$ [Using the notations from the question.] I don't undersand. If $g(\tau) = f(S\tau)$ where $S$ is the transformation $\tau \mapsto -1/\tau$ (or any other element of $\Gamma(1)$ if you like), then isn't it trivially true that $g$ is invariant by $\Gamma$ if $f$ is ? $\endgroup$ – user1971 Jun 11 '15 at 12:43
  • $\begingroup$ No, it is not true (trivially or otherwise). $\endgroup$ – David Loeffler Jun 11 '15 at 20:53
  • $\begingroup$ I see, I'm sorry. I missplaced the position of $\tau$ in the formula (and this makes the whole difference). $\endgroup$ – user1971 Jun 12 '15 at 8:47

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