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Fix a positive integer $n$. Let $f:\mathbf{H}\longrightarrow \mathbf{C}$ be a modular function with respect to the group $\Gamma(n)$. Is the derivative $$\frac{df}{d\tau}:\mathbf{H}\longrightarrow \mathbf{C}$$ also a modular function with respect to $\Gamma(n)$?

I think it's clear that $df/d\tau$ is meromorphic on $\mathbf{H}$ and that it is meromorphic at the cusp. I just don't know why it should be modular with respect to $\Gamma(n)$.

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1 Answer 1

Suppose $f(\tau)$ is modular function of weight $m$, i.e. for $\left( \begin{array}{cc} a & c \\ c & d \end{array} \right) \in \Gamma(n)$, $f\left( \frac{a \tau + b}{c \tau + d} \right) = \left(c \tau + d \right)^{m} f( \tau )$. Differentiating this equality:

$$ \begin{eqnarray} \frac{\mathrm{d}}{\mathrm{d} \tau}\left( f\left( \frac{a \tau + b}{c \tau + d} \right) \right) &=& \frac{\mathrm{d}}{\mathrm{d} \tau} \left( \left(c \tau + d \right)^{m} f( \tau ) \right) \\ f^\prime\left( \frac{a \tau + b}{c \tau + d} \right) \frac{\mathrm{d}}{\mathrm{d} \tau}\left( \frac{a \tau + b}{c \tau + d} \right) &=& \left(c \tau + d \right)^{m} f^\prime(\tau) + m c \left(c \tau + d \right)^{m-1} f(\tau)\\ f^\prime\left( \frac{a \tau + b}{c \tau + d} \right) \left( \frac{a d - b c}{(c \tau + d)^2} \right) &=& \left(c \tau + d \right)^{m} f^\prime(\tau) + m c \left(c \tau + d \right)^{m-1} f(\tau) \end{eqnarray} $$ Even though $ a d - b c = 1$, the resulting equation shows that the derivative is not a modular function of any weight, except $m=0$, in which case $f^\prime(\tau)$ is a modular function of weight $2$.

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Nice. What are some standard examples of modular functions of weight 0? I know that the j-invariant is of weight zero with respect to $\Gamma(1)$. Is the modular lambda function $\lambda$ of weight zero with respect to $\Gamma(2)$? More generally, is any hauptmodul for $\Gamma(n)$ of weight zero? –  Rayleigh Oct 9 '11 at 20:01
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@Rayleigh These are nice questions I would be interested in knowing an answer to as well. I would suggest you post them as a separate question on this site. –  Sasha Oct 9 '11 at 20:35

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