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Recall that SL$_2(\mathbf R)$ acts on the complex upper half-plane $\mathbf H$. Let $\Gamma$ be a finite index subgroup of SL$_2(\mathbf Z)$. Then there is the quotient $Y_\Gamma = \Gamma \backslash \mathbf H$ which maps to the curve $Y(1)$ via the inclusion.

What is the degree of this map? Is it just the index of $\Gamma$ in SL$_2(\mathbf Z)$? If not, is the degree of this branched cover $Y_\Gamma \to Y(1)$ bounded by the index of $\Gamma$ in SL$_2(\mathbf Z)$?

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yes, it's $[SL_2(\mathbb{Z}):\Gamma]$. –  user27126 Apr 18 '13 at 22:35
    
That's what I thought. But is there a clear explanation? –  Maxim Apr 18 '13 at 23:03
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for most points the group acts freely, so you can count the number of points in preimage of a point directly. –  user27126 Apr 18 '13 at 23:08

1 Answer 1

By $Y(1)$, I'm assuming you mean the orbifold $SL_2(\mathbb{Z})\backslash \mathbf{H}$. In fact, the center of $SL_2(\mathbb{R})$ is $-I$, which acts trivially on $\mathbf{H}$ (so it's $PSL_2(\mathbb{R})$ that acts faithfully on $\mathbf{H}$). Given a subgroup $\Gamma < SL_2(\mathbb{Z})$, let $\pm\Gamma=\Gamma\cup -\Gamma$. The curve $Y_\Gamma$ will be the same as the curve $Y_{\pm \Gamma}$, and the degree of the cover $Y_\Gamma\to Y(1)$ will then be $[SL_2(\mathbb{Z}):\pm \Gamma]$. If $-I\in \Gamma$, then $\Gamma=\pm \Gamma$, and otherwise, $[\pm \Gamma:\Gamma]=2$, in which case the degree is $[SL_2(\mathbb{Z}):\Gamma]/2$.

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