Smooth proper action

Question

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 $G$?

Answer 1

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.