Chart for surface in $\Bbb L^4$ with positive relative nullity I am reading the paper Marginally trapped surfaces in Lorentzian space forms with positive relative nullity by Chen and Van der Veken.
The setup is roughly the following: we have that $M^2 \subseteq \Bbb L^4$ is a marginally trapped surface with positive relative nullity, and $E_1, E_2$ form an orthonormal frame field along (an open set of) $M$ such that $$\nabla_{E_2} E_1 = 0 \quad\mbox{and} \quad\nabla_{E_1}E_2= -\phi(E_2)E_1,$$where $\phi \in \Omega^1(M)$ is a closed $1$-form. If $\theta^1,\theta^2$ are the dual forms to $E_1$ and $E_2$, then ${\rm d}\theta^2 = 0$ (from structure equations), and there is a local smooth function $y$ such that ${\rm d}y=\theta^2$, ok. 
Then, in page 554, they claim that there is a coordinate system $(x,y)$ such the metric  takes the form $$\psi^2(x,y)\,{\rm d}x^2 + {\rm d}y^2.$$I am having trouble seeing why this is true. My first thought was to find a function $\psi$ such that $[\psi E_1,E_2] = 0$ and apply the Frobenius Theorem. But $$[\psi E_1,E_2] = -(\psi \phi(E_2) + {\rm d}\psi(E_2))E_1,$$and I don't see how to solve the differential equation in parenthesis. Maybe there is an easier way? Thanks. 
 A: Such coordinates are called semi-geodesic coordinates. They always exist locally on any Riemannian manifold. In the 2-d case, one way to construct them in a neighborhood of a point is to choose a smooth unit-speed curve $\sigma$ through the point, choose a unit normal vector field $N$ along the curve, and define $\Phi(x,y) = \exp_{\sigma(x)}(yN(x))$.
EDIT: Here's how to relate the coordinate vector fields in semi-geodesic coordinates to the given orthonormal frame $(E_1,E_2)$. If we choose $\sigma$ to be an integral curve of $E_1$ and $N=E_2$ along $\sigma$, the coordinate frame is $(\partial/\partial x,\ \partial/\partial y) = (\psi E_1,E_2)$ on a neighborhood of $\sigma$.
To see this, first we need to observe that $\nabla_{E_2}(E_2)\equiv 0$. This follows from the facts that $\langle E_2,E_2\rangle \equiv 1$ and $\langle E_2,E_1\rangle\equiv 0$ and the following computations:
\begin{align*}
0 &= \nabla_{E_2}\langle E_2,E_2\rangle = 2 \langle\nabla_{E_2}E_2,E_2\rangle;\\
0 &= \nabla_{E_2}\langle E_2,E_1\rangle = \langle \nabla_{E_2}E_2, E_1\rangle + \langle E_2,\nabla_{E_2}E_1\rangle = \langle \nabla_{E_2}E_2, E_1\rangle.
\end{align*}
Since $\nabla_{E_2}E_2$ is orthogonal to both $E_1$ and $E_2$, it is zero.
This implies that the integral curves of $E_2$ are unit-speed geodesics starting normal to $\sigma$. Since the same is true of the integral curves of $\partial/\partial y$, and both have initial velocity $N$ at points of $\sigma$, it follows that $E_2 \equiv \partial/\partial y$ where both are defined.
On the other hand, since $(1/\psi)\partial/\partial x$ is a unit vector field orthogonal to $E_2$, it must be equal to $\pm E_1.$ Along $\sigma$, our construction guarantees that $E_1 = \partial/\partial x$, so in fact we have the plus sign.
A: By previous lemmas in the paper, the surface has Second Fundamental Form always proportional to the same lightlike vector. Then its normal curvature $R^\perp$ is identically zero (by the Ricci equation), and $\phi$ is a closed form. By Poincaré's Lemma, there is a potential $\lambda$ such that $-\phi = {\rm d}\lambda$. Then $[e^\lambda E_1,E_2] = 0$ and the result follows from Frobenius as I had initially planned.
