$\text{SL}(2,\mathbb{R})$ is a double cover of $\text{SO}^+(2,1)$

Question

I am sticking with this:

Show that $\text{SL}(2,\mathbb{R})$ is a double cover of $\text{SO}^+(2,1)$, which is a component of the identity of a Lie group of $3\times 3$ matrices satisfying

$A^TJA=3\times 3$ diagonal matrix with values 1,1,-1 respectively.

Any thoughts?

Answer 1

The 2x2 matrixes with trace 0 and also with trace as the bilinear form has orthogonal basis $$\Bigg\{ \begin{pmatrix} 0&1\\ 1&0\end{pmatrix}, \begin{pmatrix} 0&1\\ -1&0\end{pmatrix}, \begin{pmatrix} 1&0\\ 0&-1\end{pmatrix} \Bigg\}.$$ The action of $SL_2\left(\mathbb{R}\right)$ on the space spanned by these matrices given by $g\cdot x=gxg^{-1}$ preserves the bilinear form since $gxg^{-1} gyg^{-1}= gxyg^{-1}$ and trace is invariant on conjugacy classes. The map to this copy of $SO(2,1)$ has kernel $\{ \pm 1\}$ so this is two copies, one to the connected component of the identity and one to the other.

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