I've been assigned the following exercise:
Prove that $S^2$ is a homogeneous space w.r.t. the adjoint action of $SU(2)$.
We're obviously referring to $Ad:G\to\operatorname{Aut}(\mathfrak{g}), Ad_g:X\mapsto gXg^{-1}$. I believe that what has been asked is to consider the isomorphism of Lie algebras $$\mathfrak{su}(2)=\left\{\begin{pmatrix}it & -\overline{z}\\z &-it\end{pmatrix} \ \middle\vert\ t\in\Bbb{R},\ z\in\Bbb{C}\right\} \cong\Bbb{R}^3$$ (with the vector product as Lie bracket), and interpreting $\{X\in\mathfrak{su}(2)\mid \det X=1\}$ as $S^2$.
I feel like the difficult part is using specifically the adjoint action of $SU(2)$.
What I tried: we're basically asked to prove that $Ad$ acts transitively on $\mathfrak{su}(2)$, i.e. for all $X,Y\in\mathfrak{su}(2)$ there is some $A\in SU(2)$ s.t. $AXA^\dagger=Y$ (here $A^{-1}=A^\dagger$ denotes the conjugate transpose matrix). I tried playing around with decompositions such as $$SU(2)\ni\begin{pmatrix}\alpha&-\overline{\beta}\\ \beta&\overline{\alpha}\end{pmatrix}=\begin{pmatrix}e^{i\theta}&0\\0&e^{-i\theta}\end{pmatrix}\begin{pmatrix}t&-\overline{\beta}e^{-i\theta}\\ \beta e^{i\theta}&t\end{pmatrix},\quad \alpha=te^{i\theta}.$$ Unfortunately, I keep getting stuck. I even tried choosing one particular $Y$ and finding explicitly an $A$ in function of $X$, but I've had no luck with this method either so far. For example: $$\begin{pmatrix}\alpha&-\overline{\beta}\\ \beta&\overline{\alpha}\end{pmatrix}\begin{pmatrix}it & -\overline{z}\\z &-it\end{pmatrix}\begin{pmatrix}\alpha&-\overline{\beta}\\ \beta&\overline{\alpha}\end{pmatrix} = \begin{pmatrix}Z&-\overline{W}\\ W&\overline{Z}\end{pmatrix},$$ $$ Z=(|\alpha|^2-|\beta|^2)it+\alpha\beta\overline{z}-\overline{\alpha\beta}z$$ $$ W=\overline{\alpha}^2z+\beta^2\overline{z}+(\overline{\alpha}\beta) it $$
I wonder if there's a simpler (or more elegant) way to solve this problem, or if I'm just awful at this.
Thanks in advance!