TITLE: Can the conformal structure on the projective light-cone detect hyperplane sections?
QUESTION [5 upvotes]: Let $(V,\langle\,\cdot\,,\,\cdot\,\rangle)$ be an $(n+1)$-dimensional real vector space, equipped with a nondegenerate symmetric bilinear form of indefinite signature, and denote by $\nu(v):=\langle v,v\rangle$ the corresponding quadratic form. 
As a homogeneous polynomial, $\nu$ cuts out the projective ``light cone'' (or null space) $\mathcal{N}:=\{[v]\in\mathbb{P}V\mid \nu(v)=0\}$, which is a smooth hypersurface in $\mathbb{P}V\equiv\mathbb{RP}^n$.

PRELIMINARY QUESTION: what is the ``remnant'' of the scalar product $\langle\,\cdot\,,\,\cdot\,\rangle$ on $\mathcal{N}$? I mean: does it induce a metric, a conformal structure, or something even weaker on $\mathcal{N}$? 
BIG EDIT: such a ``remnant" is a conformal structure $[g]$, according to Vít Tuček's answer below.

By a hyperplane section of $\mathcal{N}$ I simply mean the intersection $\mathcal{N}\cap\pi$ of $\mathcal{N}$ with a projective hyperplane $\pi\in \mathbb{RP}^{n\,\ast}$, and I'm interested in the possibility of characterising hyperplane sections by means of $[g]$.

MAIN QUESTION: Given a smooth hypersurface $S\subset \mathcal{N}$, is it possible to tell whether or not $S$ is an hyperplane section, just by using the conformal structure $[g]$ on $\mathcal{N}$?

Any reference/hint concerning the ``instrinsic geometry'' of this projective light cone will be warmly appreciated!
- - - COMMENTS & MOTIVATIONS - - -
This post is a major edit of a previous question, whose "no" answer was given in a very convincing way by Willie Wong (see below). Thanks to his geneorous explanation, I was able to rearrange my thoughts ane reformulate the question in a hopefully more robust way. The original question, which is a sort of "affine version" of the current one, is attached below, to keep sustaining Wong's answer and to provide some motivations as well. 
In my previous post, the light cone $\mathcal{N}$ was not projectivised, and I made the mistake of believing that $\langle\,\cdot\,,\,\cdot\,\rangle$ induced a metric $g$ on $\mathcal{N}$. Hence, I considered the Hessiian $\mathrm{hess}^V$ on $V$ (i.e., the one defined by means of the flat metric induced by $\langle\,\cdot\,,\,\cdot\,\rangle$) and the Hessian $\mathrm{hess}^\mathcal{N}$ on $\mathcal{N}$ (which in fact is ill-defined, as $g$ is degenerate). Finally, by ``hyperplane section'' I just meant the intersection of $\mathcal{N}$ with an affine hyperplane of $V$. Denoting smooth functions on $\mathcal{N}$ (resp., $V$) by $f$ (resp., $F$), I formulated the following question, which (if Wille Wong hadn't proved it wrong) would have answered the main question of the present post.

PREVIOUS QUESTION: is it true that  $\{f=0\}\subset \mathcal{N}$ is an hyperplane section if and only if $$\mathrm{hess}^\mathcal{N}f\textrm{ is proportional to }g\,\textrm{?} \quad\quad{(*)}$$

The reasoning below is what originally led me to suspect that hyperplane sections can be indeed characterised by means of formula $(*)$ above.
The hypersurface $\{F=0\}\subset V$ is an affine hyperplane if and only if $$\mathrm{hess}^VF=0\, ,\quad\quad{(**)}$$
so that it all comes down to show that $(*)$ is the ``restricted version'' of $(**)$.
Allow me to regard $\nu$ as a local coordinate function (normal to the light cone) and to decompose the Hessian of $F$ as follows:
$$
\mathrm{hess}^VF=\mathrm{hess}^{\mathcal{N}}(F|_{\mathcal{N}})+F_{\nu}\odot d\nu+F_{\nu\nu}\mathrm{hess}^V\nu\, .
$$
By restricting the last expression to the light cone, and observing that $\mathrm{hess}^V\nu=\langle\,\cdot\,,\,\cdot\,\rangle$, one gets $(*)$ for $f=F|_{\mathcal{N}}$.
Conversely, if $(*)$ is true for a scalar factor $\lambda$, then I can extend $f$ to a function $F$ on $V$, such that $(**)$ is valid: the idea is to set
$$
F:=\pi_{\mathcal{N}}^*(f)-\frac{1}{\lambda}\nu\, ,
$$
where $=\pi_{\mathcal{N}}$ is the projection of $V\stackrel{\textrm{loc.}}{\equiv}\mathcal{N}\times\mathbb{R}$ onto its first factor.
(The present question is linked to my previous one).

REPLY [5 votes]: For your new question, the answer is at least in part yes. I'll outline the proof here for the case your projective hyperplane is originally not a null hyperplane.  
First let us work in the ambient $\mathbb{R}^{n+1}$. Let $\Pi$ be a non-null hyperplane through the origin, this is our projective hyperplane. Let $\pi \in \Pi$ be a non-null point, and let $P$ be the orthogonal hyperplane through $\pi$, relative to the indefinite ambient metric. 
$P$ can be identified with $\mathbb{R}^n$ with $\pi$ sent to the origin, as a (plane) model of the projective space $\mathbb{R}P^n$ once you add the point at infinity. Since $P$ is non-null we have that the ambient metric induces a metric $\nu'$ on $P$ that is again pseudo-Riemannian; considering $P$ as a indefinite inner product space with $\pi$ at the origin you have that $\mathcal{N}$ is a hyperquadric in this picture. (It is the set of points $\nu'(x) = \pm c$ for some constant $c$.) In particular $\mathcal{N}$ is a pseudo-Riemannian manfiold. 
The projective hyperplane $\Pi$ now becomes a hyperplane through the origin in $P$. And it is well-known (see, e.g. O'Neill's Semi-Riemannian Geometry, pps 108 et seq) that $\Pi\cap \mathbb{N}$ is totally geodesic in $\mathbb{N}$. This is a characterisation in that a hypersurface in $\mathbb{N}$ is totally geodesic iff it is formed by intersecting with a plane through the origin. 

Next we use the mean-curvature change formula under conformal change of metrics. 
Let $(M,g)$ be a pseudo-Riemannian manifold, $(N,h)$ a non-null hypersurface. Let $\eta$ be a choice of unit normal along $N$. Let $f$ be a real function on $M$. 
Let $S$ be the second fundamental form of $N$ in $M$. 
Now let $\tilde{g} = e^{2f} g$ be a conformal metric and $\tilde{h} = e^{2f} h$ the induced conformal metric on $N$. And let $\tilde{S}$ be the second fundamental form of $N$ in $M$ relative to the conformal metric, we have that
$$ \tilde{S} = e^f S + \eta(e^f) h $$
where $\eta(e^f)$ is the derivative of $e^f$ in the direction of $\eta$. 
So we have the following statement:

Under a conformal change of metric $g \mapsto e^{2f} g$, the trace free part of the second fundamental form $A$ changes like $A \mapsto e^f A$. 

So it makes sense to speak of the conformal class of the trace free part of the second fundamental form. 

So from the above we have 

A necessary condition for $S \subset \mathcal{N}$ to be a hyperplane section is that the traceless part of the second fundamental form for the embedding vanishes. This condition is conformally invariant so can be checked using any realisation of the conformal class of $[g]$. 

Unfortunately I can't quite see whether this is a sufficient condition; my gut feeling is that probably it is not, but there probably is a slightly modified version of the statement for which the vanishing of the traceless part of the second fundamental form is both necessary and sufficient. 
At the very least you can now tell when $S$ is not a hyperplane section. :)