# Extended Upper Half plane and Modular Curves

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?

• Hint: What are the fixed points of the parabolic elements of the modular group? – Moishe Kohan Dec 1 '16 at 9:29
• 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. – reuns Dec 4 '16 at 10:55

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.
• 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. – BharatRam Mar 20 '17 at 8:39