First of all, I'm considering $\Bbb L^3$ with the convention: $$\langle (x_1,y_1,z_1),(x_2,y_2,z_2)\rangle = x_1x_2+y_1y_2 - z_1z_2$$ Let $\Lambda = (\lambda_{ij})$ be an isometry of $\Bbb L^3$. I want to prove that $\Lambda$ preserves time-orientation if, and only if $\lambda_{33} > 0$. I don't know exactly how to go about it, but I made some calculations, and got the following:
If we write $\Lambda = \begin{pmatrix} \ell_1 & \ell_2 & \ell_3 \end{pmatrix}$, with $\ell_i = (\lambda_{1i}, \lambda_{2i}, \lambda_{3i})$ being the columns of $\Lambda$, and using the fact that $\Lambda^tG\Lambda = G$, where $G = \mathrm{diag}[1,1,-1]$, I got that $$\langle \ell_3, \ell_3 \rangle = \lambda_{13}^2 + \lambda_{23}^2 - \lambda_{33}^2 = -1$$ that is, $\ell_3$ is a timelike vector. And so, $\ell_3$ would be future-directed or past-directed according to the sign of $\lambda_{33}$. Having this in mind, the result becomes quite intuitive.
Can someone help me organize my ideas, and give me a little push? I'm using this text to study. If requested, I can add some of the main definitions here, or my calculations, for commodity. Thanks in advance.
Had some progress. If $\Lambda$ preserves time orientation, then $\lambda_{33} > 0$ because since $(0,0,1)$ is future-directed, then $\Lambda(0,0,1) = (\lambda_{13}, \lambda_{23}, \lambda_{33})$ must be future-directed too.
The set of timelike vectors, $\mathcal T$, has two connected components. Every isometry of the space maps $\mathcal T$ onto itself. Since $\Lambda$ is an invertible linear map (hence homeomorphism), it can either fix each component of $\mathcal T$ or exchange them.