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I am trying to read the paper Marginally Trapped Submanifolds in Space Forms with Arbitrary Signature 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.

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    $\begingroup$ FWIW, I agree with your analysis in your question statement. $\endgroup$ – Willie Wong Oct 13 '17 at 20:27
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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.

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  • $\begingroup$ Glad to know I'm not crazy, thanks. If you actually meant $\xi =(0,\color{red}{2},0,0)$ then I got the same results as you pointed (which seems to hold in every point of that surface, not only for $w=z=0$). Apart from that, how did you came up with that example? I still don't have much intuition with lightlike coordinates. Thanks again :-) $\endgroup$ – Ivo Terek Oct 14 '17 at 2:20
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    $\begingroup$ I dislike normalizing $\nu\cdot \xi = 2$, because that means in computations, in addition to factors of $-1$ you now also have to keep factors of $2$, which just gives you more chance to make errors. // As to how I come up with the example: the statement is point-wise. So first identify the pointwise value of a second fundamental form that would violate it (which you already did most of the hard work for in your question statement), and then build a surface (by Taylor polynomial) using that the second fundamental form is essentially the Hessian. $\endgroup$ – Willie Wong Oct 15 '17 at 0:28

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