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Perhaps an introductory question, but, recently a came across the notion of Modular Curve and I read about its compactification. Now, every Modular Curve is by definition a quotient of the form $\mathbb{H}/ \Gamma$, for $\mathbb{H}$ the upper-half plane and $\Gamma$ a suitable (congruence) subgroup of $SL_{2}(\mathbb{Z}$). Since this becomes a Hausdorff space thourgh the quotient topology given by the natural projection $\pi: \mathbb{H} \rightarrow \mathbb{H} / \Gamma$, we can ask about its compactification. We can define that compactification to be the quotient of the topological space $\mathbb{H}^{*}= \mathbb{H} \cup \mathbb{P}^{1} (\mathbb{Q})$ called the extended upper half plane, by $\Gamma$, with the "natural" action of the latter on $\mathbb{P}^{1} (\mathbb{Q})$ (induced by the action on each coordinate of points of $\mathbb{Q}^{2}$ via Möbius Transformation). Now, my question is, why do we choose $\mathbb{Q}$ and not another field instead, such as $\mathbb{R}$ for instance?

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    $\begingroup$ Hint: What are the fixed points of the parabolic elements of the modular group? $\endgroup$ – Moishe Kohan Dec 1 '16 at 9:29
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    $\begingroup$ Let $Y(\Gamma) = H/\Gamma$ and $X(\Gamma)$ its compactification (i.e. $Y(\Gamma)$ plus a few points). Then $X(\Gamma) = H^* / \Gamma$ with $H^* = H \cup \mathbb{Q} \cup \{\infty\}$ because $H^* / \Gamma$ is a compact Riemann surface and $Y(\Gamma)$ is dense in it. $\endgroup$ – reuns Dec 4 '16 at 10:55
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Here is one answer. Take $\Gamma = \text{SL}_2(\mathbb{Z})$. Then $\mathbb{H}/\Gamma$ is identified with $\mathbb{C}$ (via the $j$-function). If you take a path in $\mathbb{H}/\Gamma$ going to the missing point (infinity), you can lift it to a path in $\mathbb{H}$ that also goes to infinity. This suggests we should add infinity to $\mathbb{H}$ to get a compact quotient. But $\mathbb{H} \cup \{\infty\}$ is not stable by $\Gamma$, so we should actually add the entire $\Gamma$ orbit of infinity to $\mathbb{H}$, and this orbit is exactly $\mathbb{P}^1(\mathbb{Q})$.

A second, less satisfying answer: there is no way to use a bigger field (like $\mathbb{R}$) in such a way that the quotient will be nice—I think you will always have some non-Hausdorff or noncompact behavior.

If you are trying to learn more about this sort of thing, I highly recommend Shimura's book "Introduction to the arithmetic theory of automorphic functions" here.

Shimura, Goro. Introduction to the arithmetic theory of automorphic functions. Kanô Memorial Lectures, No. 1. Publications of the Mathematical Society of Japan, No. 11. Iwanami Shoten, Publishers, Tokyo; Princeton University Press, Princeton, N.J., 1971. xiv+267 pp.

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  • $\begingroup$ Isn't your answer a bit circular? We know that the only cusp of the modular curve is $\infty$ whose preimage orbit in $\mathbb{C}$ is $\mathbb{Q} \cup \{\infty\}$. I am unable to see why we do not need to even consider other elements of $\mathbb{R}$. Please do clarify if possible. $\endgroup$ – BharatRam Mar 20 '17 at 8:39

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