I am trying to read the paper [*Marginally Trapped Submanifolds in Space Forms with Arbitrary Signature*](https://arxiv.org/pdf/1309.3875.pdf) by Henri Anciaux, but I think that there is a mistake in Lemma $1$, in page $5$:



> 
> The second fundamental form $h$ is collinear to $\nu$ if and only if the mean curvature vector $H$ is collinear to $\nu$ and $\nu$ has rank at most $1$.
> 
> 
> 


Here $\Sigma$ is a $n$-dimensional submanifold of $\Bbb R^{n+2}\_{p+1}$ such that metric induced in the normal spaces to $\Sigma$ is Lorentzian, and $(\nu,\xi)$ is a null frame normal to $\Sigma$ satisfying $\langle \nu, \xi\rangle = 2$, and $(e\_1,\ldots,e\_n)$ is an orthonormal local frame tangent to $\Sigma$. The author puts $h^1\_{ij} = \langle h(e\_i,e\_j), \nu\rangle = -\langle d\nu(e\_i),e\_j\rangle$, so far, so good. Then he states that $$d\nu(e\_i) = -\sum\_{j=1}^n h^1\_{ij}e\_j + \frac{1}{2}\langle d\nu(e\_i),\xi\rangle \nu.$$


I disagree: by orthonormal expansion, we should have $$d\nu(e\_i) = -\sum\_{j=1}^n \color{red}{\epsilon\_j}h^1\_{ij}e\_j + \frac{1}{2}\langle d\nu(e\_i),\xi\rangle \nu,$$where $\epsilon\_j$ denotes the indicator of $e\_j$ ($1$ if $e\_j$ is spacelike, $-1$ if timelike).
His idea was to use that $h^1\_{ij}$ is symmetric with rank $1$ to write $h^1\_{ij} = c\lambda\_i\lambda\_j$ for some certain constants, and proceed with the argument. As it is written, his reasoning seems to hold only if $\Sigma$ is itself spacelike.


This detail seems to put the rest of the proof in jeopardy: we'll have that the matrix $(\epsilon\_ih^1\_{ij})$ has rank $1$ instead of $(h^1\_{ij})$, but it is not anymore symmetric, so the argument fails. We'll have that $$c\sum\_{i=1}^n\epsilon\_i \lambda\_i^2 = 0$$instead, and if $c \neq 0$ we can't conclude that $(\lambda\_1,\ldots,\lambda\_n) = (0,\ldots,0)$.


I don't know if this is fixable. Please help.



Possible mistake in classification of marginally trapped submanifolds of $\Bbb R^{n+2}_{p+1}$
The proof of the forward implication is perhaps more complicated than necessary, but it is true. 


The proof of the reverse implication is wrong. In fact here's a counterexample. Consider the space $\mathbb{R}^{4}\_2$ with the quadruple null coordinates $(u,v,w,z)$ where the metric is $du\otimes dv + dv \otimes du + dw \otimes dz + dz \otimes dw)$. 


Consider the surface 
$$ (0, w^2, w, z) $$
Denote by $\nu$ the null vector field $(1, 0, 0, -2w)$ and $\xi$ the null vector field $(0,1,0,0)$. You can check that the second fundamental form at $w = z = 0$ is 


$$II\_p = \xi \otimes dw \otimes dw $$


The mean curvature vanishes, but $d\nu|\_p = -2 \partial\_z \otimes d w$ is rank 1. 


