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, but I can't follow it. I've also had previous trouble 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 $T_p\overline{M} = T_pM \oplus T^\perp_pM$.
Update: I have proven that if $M^2\subseteq \Bbb S^4_1\cap \vec{L}_0^\perp$, then $M$ is spacelike, by brute-forcing my way through it. Applying $\Lambda \in {\rm O}_1 (4,\Bbb R)$ if necessary, we can assume that $\vec{L}_0 =(1,0,0,0,1)$. Then parametrize using $$\varphi(u,v)=(f(u,v),\cos g(u,v)\cos h(u,v), \cos g(u,v)\sin h(u,v),\sin g(u,v),f(u,v)), $$compute $(g_{ij})_{i,j=1}^2$ and apply Sylvester's criterion. But I'd prefer avoiding this approach in the general case.
