Let $1 \leq \nu \leq n+1$ and $M^n \subseteq \Bbb S^{n+2}_\nu$ be a non-degenerate submanifold. Assume that $\renewcommand{\vec}[1]{{\bf #1}} \vec{L}_0 \in \Bbb R^{n+3}_\nu$ is lightlike and $M \subseteq \vec{L}_0^\perp \cap \Bbb S^{n+2}_\nu$. 

>Is is true that $M$'s Second Fundamental Form is always lightlike?

Since $\vec{L}_0$ is normal to $M$ at every point, we have that ${\rm coind}_{\Bbb R^{n+3}_\nu}(M) = 1$ or $2$. Since the orthogonal complement of $T_pM$ with respect to $\Bbb R^{n+3}_\nu$ is the orthogonal direct sum of the orthogonal complement with respect to $\Bbb S^{n+2}_\nu$ and the line spanned by $p$ (which is spacelike), we know that ${\rm coind}_{\Bbb S^{n+2}_\nu}(M)=1$ or $2$.

I'm having trouble checking that ${\rm coind}_{\Bbb S^{n+2}_\nu}(M) = 2$ cannot happen (this would forbid ${\rm II}$ being lightlike).

(Context: I've already proven that if $M^n \subseteq \Bbb S^{n+2}_\nu$ has lightlike Second Fundamental Form, then every point in $M$ admits a neighbourhood in $M$ contained in a intersection of the form $\vec{L}_0^\perp \cap \Bbb S^{n+2}_\nu$. I'm interested in the converse. Supposedly it is proven in [theorem 1 here](https://arxiv.org/pdf/1309.3875.pdf),
 but I can't follow it.
 I've also had [previous trouble](https://mathoverflow.net/questions/283417/possible-mistake-in-classification-of-marginally-trapped-submanifolds-of-bbb-r) with this paper, so I'm skeptic about it.)

---

If $(\overline{M}, \langle\cdot,\cdot\rangle)$ is pseudo-Riemannian, and $M\subseteq \overline{M}$ is non-degenerate, by ${\rm coind}_{\overline{M}}(M)$ I mean the index of the metric of $\overline{M}$ restricted to the normal spaces of $M$, following the decomposition $T_p\overline{M} = T_pM \oplus T^\perp_pM$.