Where did I do mistakes in the following:
So the real special linear group $SL(3,\ \mathbb R)$ acts transitively on the sphere $S^2$ and the isotropy of the point $(1,0,0)\in S^2$ is $H=\begin{pmatrix} 1 & \mathbb R & \mathbb R\\ 0 & * & * \\ 0 & * & * \end{pmatrix}$. Hence, $H$ is homeomorphic to $\mathbb R^2\times SL(2,\ \mathbb R)$ and so $\pi_1(H)=\mathbb Z$.
Now the exact sequence of the principal bundle $SL(3,\ \mathbb R)\to SL(3,\ \mathbb R)/H=S^2$ is
$$0\to \mathbb Z\to \pi_1(H)\to \mathbb Z_2\to 0$$
which gives us that $\pi_1(H)=\mathbb Z\times \mathbb Z_2$??
Best Answer
$SL(3,\Bbb R)$ does not act on the sphere $S^2$. The groups $O(3)$ and $SO(3)$ do, of course.