Consider $\Bbb L^3 = (\Bbb R^3, {\rm d}s^2)$, where: $${\rm d}s^2 = {\rm d}x^2 + {\rm d}y^2 - {\rm d}z^2.$$
We have both the hyperbolic space: $$\Bbb H^2(-1) = \{(x,y,z) \in \Bbb L^3 \mid x^2+y^2-z^2 =- 1\},$$ which has constant Gaussian Curvature $K = -1$, and the De Sitter's space: $$\Bbb S^2_1(1) = \{(x,y,z) \in \Bbb L^3 \mid x^2+y^2-z^2 = 1\},$$ which has Gaussian Curvature $K=1$.
My problem is: I'm trying to actually compute these curvatures, and I'm getting $K=-1$ for both of them. I'm using connection forms just like in $\Bbb R^3$ and bi-dimensional Riemannian manifolds, but I think stuff is going wrong because the De Sitter's space is a timelike surface, while the hyperbolic space is spacelike..
I have not seen a rigorous treatment of this in semi-Riemannian surfaces, so I would like to know exactly where my calculations are failing. Additional explanations would be very welcome, too. Enough talk, let's go to the action:
Rotating a hyperbola, we parametrize the De Sitter's space by $${\bf x}(u,v) = ( \cosh u \cos v, \cosh u \sin v, \sinh u),$$ and so: $$\begin{align} {\bf x}_u(u,v) &= (\sinh u \cos v, \sinh u \sin v, \cosh u) \\ {\bf x}_v(u,v) &= (-\cosh u \sin v, \cosh u \cos v, 0) \end{align}.$$
Dropping the point $(u,v)$ from now on, we have: $$E = -1 \quad F = 0 \quad G = \cosh^2u.$$
So, we can take a frame $$E_1 = {\bf x}_u \quad E_2 = \frac{{\bf x}_v}{\cosh u},$$ with dual forms $$\theta_1 = {\rm d}u \quad \theta_2 = \cosh u \ {\rm d}v.$$
So $ {\rm d}\theta_1 = 0 $ and $ {\rm d}\theta_2 = \sinh u \ {\rm d}u \wedge {\rm d}v.$ Now I need the connection form $\omega_{12}$. Write: $$\omega_{12} = \alpha \ {\rm d}u + \beta \ {\rm d}v.$$
Since ${\rm d}\theta_1 = \omega_{12} \wedge \theta_2$, we get $\alpha = 0$. And from $ {\rm d}\theta_2 = - \omega_{12} \wedge \theta_1$, we obtain: $$\begin{align} \sinh u \ {\rm d}u \wedge {\rm d}v &= -(\beta \ {\rm d}v) \wedge {\rm d}u \\ \sinh u \ {\rm d}u \wedge {\rm d}v &= \beta \ {\rm d}u \wedge {\rm d}v \end{align},$$ so $\beta = \sinh u$. Now $\omega_{12} = \sinh u \ {\rm d}v$ gives us $ {\rm d}\omega_{12} = \cosh u \ {\rm d}u \wedge {\rm d}v$. Rewriting: $$ \cosh u \ {\rm d}u \wedge {\rm d}v = \cosh u \ \theta_1 \wedge \left(\frac{\theta_2}{\cosh u}\right) = -(-1) \ \theta_1 \wedge \theta_2, $$ and ${\rm d}\omega_{12} = -K \ \theta_1 \wedge \theta_2$ wields $K=-1$.
What is going wrong? The same calculation gives the right answer for the hyperbolic space. Where does the causal character comes in? Please help.
Edit: I have redone the calculations using $E_1 = \frac{{\bf x}_u}{\rm i}$ instead of my previous $E_1$, and I got the right answer. However, using complex numbers felt like cheating, so I'll rephrase my question: is there any way that I can avoid this approach? From what I've studied about $\Bbb L^3$ so far, complex numbers weren't used, so I reckon that must be another way to tackle this.
Ivo, the point is that with the Lorentz metric the structure equations change. I'm going to give you the appropriate moving frames computation without using an explicit parametrization.
Consider the isometry group of the metric, $O(2,1)$. Let $e_0,e_1,e_2$ be a moving frame on your surface $M$, so that $e_0$ projects to $x\in M$ and $e_1,e_2$ project to the tangent plane. The structure equations for $O(2,1)$ are different from those for $O(3)$.
If $M$ is the hyperboloid $x^2+y^2-z^2=-1$, we have $\langle e_0,e_0\rangle = -1$, and, letting $\omega_{AB} = \langle de_A,e_B\rangle$, (here $1\le i,j\le 2$, $0\le A,B,C\le 2$) we obtain $$\omega_{0i} = \omega_{i0},\quad \omega_{ij} = -\omega_{ji},\quad d\omega_{AB} = \sum\omega_{AC}\wedge\omega_{CB}.$$ Here $\omega_{01}$ and $\omega_{02}$ are the standard orthonormal coframe on $M$. In particular, the restriction of the Lorentz metric to $M$ is positive definite and we have $$\Omega_{12}=d\omega_{12}-\omega_{12}\wedge\omega_{21} = \omega_{10}\wedge\omega_{02} = \omega_{01}\wedge\omega_{02},$$ so $K=-1$.
If $M$ is the deSitter space, we have $\langle e_0,e_0\rangle = +1$, and now the restriction of the Lorentz metric is indefinite. In particular, choosing $e_1,e_2$ with $\langle e_1,e_1\rangle = +1$ and $\langle e_2,e_2\rangle = -1$, we have $\omega_{01} = -\omega_{10}$ and $\omega_{02}=\omega_{20}$. So the comparable computation of the curvature gives $$\Omega_{12} = \omega_{10}\wedge\omega_{02} = -\omega_{01}\wedge\omega_{02},$$ so $K=1$.
The key thing for you to contemplate here is the difference between the Lie algebras $\mathfrak o(3)$ and $\mathfrak o(2,1)$. Work that out explicitly, and you'll see what's going on with the Maurer-Cartan forms.
SLIGHT EDIT: To conform with your notation, you should think about $e_0$ as the position vector, so $\omega_{0i} = \theta_i$ give the usual orthonormal coframe. As I suggested in the comments, it's classical to think of deriving the structure equations by differentiating the position vector and frame fields as vector-valued functions. By definition, $de_A$ is an $\Bbb R^3$-valued $1$-form, whose $e_B$ coefficient is $\omega_{AB}$.
Now let me try to elaborate on where the sign change affects the structure equations. We have $$0 = d\langle e_A,e_B \rangle = \langle \sum \omega_{AC}e_C,e_B \rangle + \langle e_A,\sum\omega_{BC}e_C\rangle = \langle e_B,e_B \rangle\omega_{AB} + \langle e_A,e_A \rangle \omega_{BA}.$$ So, in the case of the hyperboloid, with $A=0$ and $B=i$, we get $\omega_{0i} - \omega_{i0} = 0$. In the case of the deSitter space, with $A=0$ and $B=1$ and $2$ respectively, we get $\omega_{01} + \omega_{10} = 0$, but $\omega_{02}-\omega_{20} = 0$.
The capital $\Omega_{ij}$ is the notation used for the (matrix of) curvature $2$-forms. For surfaces, it's just $\Omega_{12} = -K\theta_1\wedge\theta_2$. Generally, we have $$\Omega_{ij} = d\omega_{ij} - \sum\omega_{ik}\wedge\omega_{kj}.$$ Of course, there are different sign conventions in all the books, depending on whether our group acts on the left or on the right on the bundle of orthonormal frames. I thought you had already alluded to different sign conventions in structure equations in an earlier comment.