Here $\mathbb{L}^3$ is $\mathbb{R}^3$ with the metric $\langle u, v\rangle_\mathbb{L} = -u_1v_1 + u_2v_2 + u_3v_3$, where $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$ and $\mathbb{H}^1 \times \mathbb{R} = \{(x, y, z) \in \mathbb{L}^3 \ \vert \ y^2 - x^2 = 1, x>0 \} \subset \mathbb{L}^3$.
We know that for a curve $\alpha: I \to S \subset \mathbb{R}^3$ with the usual metric (where $S$ is a surface) parameterized by arclength, it's geodesic curvature is given by (where $N$ is the normal to the surface and $T = \alpha'$): $$\kappa_g = \left\langle \alpha'', N \times T\right\rangle$$
But is the same valid for curves on $\mathbb{H}^1 \times \mathbb{R}$? Do I just compute the dot procut with the metric from $\mathbb{L}^3$ and proceed as usual?
Also, assuming the answer is yes, how would I go about computing the cross product in $\mathbb{L}^3$ (update: I've just figured out it's pretty easy to define an analogous of the cross product)?