# Showing that the action of the Lie group $SL(2,\mathbb{R})$ on the upper half plane $\mathbb{H}$ is smooth

Let us consider the group action of the special linear group $SL(2,\mathbb{R})$ on the upper half plane $\mathbb{H} := \{z \in \mathbb{C} : \mathrm{Im}(z) > 0\}$ defined by $$\begin{pmatrix} a & b\\ c & d\end{pmatrix} \cdot z := \frac{az +b}{cz + d}$$ where $ad - bc = 1$. Now this is a smooth action and I would like to show this formally, i.e. that the defining map $\theta : SL(2,\mathbb{R}) \times \mathbb{H} \to \mathbb{H}$ is smooth. Has anyone a hint for me? I mean one can write any such fractional transformation as a composition of inversion, dilatations, etc. but I would like to do it formally with respect to the Lie group structure of the special linear group and especially with respect to differential topology.

Fix $(a,b,c,d)\in SL(2,\mathbb{R})$. w.l.o.g. say $a\not=0$. Then since $\partial_d(ad-bc)=a\not=0$ we have by the implicit function theorem a $C^\infty$ map $\mathbb{R}^3\supset U\ni(x,y,z)\stackrel{f}{\to} \mathbb{R}$ centered at $(a,b,c)$, such that $(x,y,z,f(x,y,z))\in SL(2,\mathbb{R})$. In other words, $f$ is a chart for $SL(2,\mathbb{R})$ around $(a,b,c,d)$. Taking some open $V$ around $q\in \mathbb{H}$ we see that the action takes on the form: $$(x,y,z)\times s\mapsto \frac{xs+y}{zs+f(x,y,z)}$$ and this is clearly a smooth map (is a composition of smooth maps: complex multiplication, addition, division away from zero, $f$).
• In that case it suffices to say $((a,b,c,d),z) \mapsto \frac{az+b}{cz+d}$ is smooth around $z_0 \in \mathbb{C}, (A,B,C,D) \in \mathbb{R}^4,\Im(z_0) > 0, AD-BC \ne 0$ – reuns Aug 17 '17 at 11:28
• For the assumptions of the implicit function theorem, the considered set should be open. Is that really the case for $SL(2,\mathbb{R})$? I mean it is just the preimage of $1$ under the continuous determinant function, so it should be closed. – TheGeekGreek Aug 17 '17 at 11:34
• @reuns Well yea, but since the OP asks for a "formal proof", I figured it'd be better to stick to the definition of smoothness: take a chart on the manifold, and verify that in local cooridnates the map is smooth. You "take a chart" for the manifold $GL(2,\mathbb{R})$ instead of $SL(2,\mathbb{R})$, which, when explicitly asked for a formal proof, seems to defeat the point – user2520938 Aug 17 '17 at 11:35
• @TheGeekGreek You apply the implicit function theorem to the function $(a,b,c,d)\mapsto ad-bc$, which you consider as a function on $\mathbb{R}^4$. $\mathbb{R}^4$ is clearly an open set. – user2520938 Aug 17 '17 at 11:36
• @TheGeekGreek Note that this aproach using the implicit function theorem is the definition of the smooth structure on $SL(2,\mathbb{R})$, and hence the map $(a,b,c)\mapsto(a,b,c,f(a,b,c))$ is indeed a chart (again, by the very definition of the smooth structure on $SL(2,\mathbb{R})$) – user2520938 Aug 17 '17 at 11:49