$SL(2,\mathbb{R})$ as a Lorentz Group $O(1,2)$

Question

Define $$X = \bigg\{ \begin{pmatrix} x_0+x_1 & x_2 \\ x_2 & x_0-x_1 \end{pmatrix} | x_i \in \mathbb{R} \bigg\}.$$

Given $g\in SL(2,\mathbb{R})$, consider $s(g)$ which is a transformation $x \rightarrow gxg^T$ where $x\in X$.

We can view $s$ as a map from $SL(2,\mathbb{R})$ to its image, and I want to show that the $s$ is a map onto a connected component $G_+(2)$of indefinite orthogonal group $O(1,2)$ that contains the identity.

i.e. I want to show that $s:SL(2,\mathbb{R}) \rightarrow G_+(2)$.

And here's my attempt. Define $k=diag(1,-1,-1)$. My thought was let's first show that $s$ is a map onto $O(1,2)$ and then show that its a map onto its connected component. To start that, I tried to show $((gxg^T)^T)k(gxg^T)=k$ but I got stuck right away. Question: is there a (better) way to approach this?

Appendix questions:

1. What does it mean connected for the component of $O(1,2)$?

2. I showed that $s$ is a homomorphism, but how can I show that it is a continuous homomorphism? (Is the answer one of 'the codomain is continuous' or 'for every open subset in the codomain there exists a open preimage'?)

Answer 1

1. Consider the vector space $$ X~:=~\{ x\in {\rm Mat}_{2 \times 2}(\mathbb{R}) \mid x^T=x \}.\tag{1}$$ If we use the determinant as a quadratic form, then the vector space $X\cong \mathbb{R}^{1,2}$ becomes isomorphic to the 1+2D Minkowski space.

2. OP's map $$ SL(2,\mathbb{R})\quad\stackrel{s}{\longrightarrow}\quad O(X)~\cong~O(1,2;\mathbb{R}) \tag{2}$$ is given by $$ s(g)x~:=~gxg^T, \qquad g~\in~SL(2,\mathbb{R}), \qquad x~\in~X. \tag{3} $$

3. Since $s$ is a continuous map from a path-connected set $SL(2,\mathbb{R})$, the image $$G_+(2)~:=~ {\rm Im}(s)\tag{4}$$ is also path-connected, and in fact the connected component of the Lorentz group $O(1,2;\mathbb{R})$ that contains the identity.

4. For more details, see my related Phys.SE answer here.