# Smooth proper action

Let $$G=SL(2)$$ the Lie group of real $$2\times 2$$ matrices with determinant equal to $$1$$ and let $$\mathbb{H}$$ be the Poincaré half-plane defined by $$\mathbb{H}=\{z\in\mathbb{C},\ \mathrm{Im}(z) > 0\}$$. The action of $$SL(2)$$ on $$\mathbb{H}$$ is given by $$\begin{pmatrix}a&b\\ c&d\end{pmatrix} \cdot z = \frac{az+b}{cz+d}.$$

Is the action of $$SL(2)$$ on $$\mathbb{H}$$ a proper action?

The isotropy group $$G_i$$ of the point $$i$$ under the action of $$SL(2)$$ on $$\mathbb{H}$$ is equal to the orthogonal group $$H=SO(2)$$. The action is transitive since the orbit of $$i$$ is the entire space $$\mathbb{H}$$. Therefore, all the isotropy groups are conjugate to $$SO(2)$$ and hence compacts.

Does the compactness of the isotropy groups imply that the action of $$G$$ on $$G/H$$ by left multiplication is proper?

If $$K$$ is a compact subset of $$G/H$$, is $$G_K=\{g\in G,\ gK\cap K\neq\varnothing\}$$ a compact subset of $$G$$?

• Technically the isotropy group is $O(2)$ which is why people usual consider the action of $PSL(2,\mathbb R)$ so that it is faithful
– LPZ
Commented May 9 at 8:23
• @LPZ We consider the action of $PSL(2,\mathbb{R})$ to get a $\textbf{free}$ action Commented May 9 at 18:26

Using the Iwasawa decomposition, any matrix in the special linear group $$SL(2,\mathbb{R})$$ can be uniquely represented as $$M=KAN$$, where $$k$$ belongs to the subgroup $$K=SO(2)$$, $$A\in\left\{\begin{pmatrix}\lambda&0\\0&1/\lambda\end{pmatrix},\ \lambda>0\right\}$$ and $$N\in\left\{\begin{pmatrix}1&a\\0&1\end{pmatrix},\ a\in\mathbb{R}\right\}$$.
An alternative definition of a proper action of a Lie group $$G$$ on a manifold $$M$$ is as follows: for any convergent sequences $$(x_n)$$ and $$(g_n\cdot x_n)$$ (in $$M$$), we can extract a convergent subsequence $$(g_{\varphi(n)})$$ in $$G$$.
Since $$SO(2)$$ is a compact group, we can assume that $$(z_n)$$ and $$(A_nN_n\cdot z_n)$$ are convergent sequences. Here, $$z_n=x_n+iy_n$$ where $$(x_n)$$ and $$(y_n)$$ are convergent sequences in $$\mathbb{R}$$ and $$]0,+\infty[$$ respectively. The convergence of $$A_nN_n\cdot z_n=\begin{pmatrix}\lambda_n&0\\0&1/\lambda_n\end{pmatrix}\begin{pmatrix}1&a_n\\0&1\end{pmatrix}\cdot z_n=\lambda_n^2(z_n+a_n)=\lambda_n^2(x_n+a_n)+i\lambda_n^2y_n$$ implies that $$(\lambda_n)$$ is a convergent sequence in $$(0,+\infty)$$, and therefore, $$a_n=(x_n+a_n)-x_n$$ is also a convergent sequence.