I'm trying to solve this exercise:
Let $\mathbb{H}^3=\{(x,y,z)\in \mathbb{R}^3: z>0\}$ with a hyperbolic metric $g_{ij} = \dfrac{\delta_{ij}}{z^2}$. Use the mobile referential method to find the curvatures (intrinsic, mean and extrinsic) of a open half-plan whose boundary it's in the plan $\{z=0\}$ and whose normal (pointing up) forms an angle $\theta$ with a vertical direction (pointing up).
Well, I used the referential: $$\begin{cases} t_1= z(1,0,0) \\ t_2 = z(0,-\cos(\theta), \sin (\theta)) \\ t_3 = z(0, \sin (\theta), \cos (\theta)) \end{cases}$$
And I get:
$$ w_1=\dfrac{dx}{z}, \quad w_2= -\dfrac{\cos(\theta)}{z}dy+\dfrac{\sin(\theta)}{z}dz,$$ $$ w_3= \dfrac{\sin(\theta)}{z}dy+\dfrac{\cos(\theta)}{z}dz, \quad w_{12} = \dfrac{\sin(\theta)}{z}dx$$
But when I do $dw_{12} = -k w_1 \wedge w_2$ it doesn't work
PS: $w_1 \wedge w_2$ denotes the exterior product.