Consider \(\SL_2(\RR)\) acting on \(\RR^2\) by matrix multiplication. What is the stabiliser of a point? Does it depend which point? Do you know what sort of subgroup this is? What if \(\SL_2(\RR)\) acts by Möbius transformations instead?
Solution
For the linear action of \(\operatorname{SL}_2(\mathbb R)\) on \(\mathbb R^2\), the stabilizer of \(0\) is the whole group.
If \(v\ne0\), choose \(g\in\operatorname{SL}_2(\mathbb R)\) with \(g e_1=v\). Then \[ \operatorname{Stab}(v)=g\,\operatorname{Stab}(e_1)\,g^{-1}, \] and \[ \operatorname{Stab}(e_1) = \left\{ \begin{pmatrix}1&t\\0&1\end{pmatrix}:t\in\mathbb R \right\}. \] Thus all nonzero-point stabilizers are conjugate unipotent subgroups isomorphic to \((\mathbb R,+)\), while the stabilizer of \(0\) is all of \(\operatorname{SL}_2(\mathbb R)\).
For the Möbius action \[ z\longmapsto \frac{az+b}{cz+d}, \qquad \begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}_2(\mathbb R), \] there are two natural spaces to distinguish.
On the upper half-plane \(\mathbb H\), the action is transitive and \[ \operatorname{Stab}(i) = \left\{ \begin{pmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{pmatrix}:\theta\in\mathbb R \right\} \cong \operatorname{SO}(2). \] Hence every stabilizer of a point of \(\mathbb H\) is conjugate to \(\operatorname{SO}(2)\).
On the boundary \(\mathbb P^1(\mathbb R)=\mathbb R\cup\{\infty\}\), the action is also transitive, but the stabilizers are different. For example, \[ \operatorname{Stab}(\infty) = \left\{ \begin{pmatrix}a&b\\0&a^{-1}\end{pmatrix}:a\in\mathbb R^\times,\ b\in\mathbb R \right\}, \] a Borel subgroup; every boundary-point stabilizer is conjugate to it.