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This is a silly question, but I can't resolve this:

$Y(1)$ is defined to be $\mathbb{H}/PSL_2(\mathbb{Z})$. So it seems that its universal cover should be $\mathbb{H}$.

On the other hand $Y(1)$ is isomorphic to $\mathbb{A}^1_{\mathbb{C}}$, and therefore is its own universal cover.

What am I missing?

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Why do you think the fact that $Y(1)$ is a quotient of the upper-half plane implies that its universal cover is the upper-half plane? –  Bruno Joyal Oct 29 '11 at 2:38
    
I can't think of points where it is branched, which would make it a covering map... –  Nicole Oct 29 '11 at 2:40
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Dear Nicole, There is branching at the points corresponding to $j=0$ and $j=1728$. Not coincidentally, these are the $j$-invariants corresponding to elliptic curves with extra automorphisms. Regards, –  Matt E Nov 25 '11 at 2:01

1 Answer 1

up vote 5 down vote accepted

To follow up on Bruno Joyal's comment, $PSL(2,\mathbb Z)$ does not act freely on $\mathbb H$. If you think of this group as being generated by $z\mapsto -1/z$ and $z\mapsto z+1$, then the complex number $i$ is fixed by the first map, so it has a nontrivial stabilizer.

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