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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).

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    $\begingroup$ A few nit-picky comments: (a) the light cone is not a hypersurface: it is highly singular at the origin; away from the origin however you are right. (b) The metric $g$ induced on the hypersurface, if it is relative to the ambient inner product structure, is degenerate. Most people would not count this as pseudo- (or semi-) Riemannian. $\endgroup$ – Willie Wong Sep 17 '15 at 13:23
  • $\begingroup$ @WillieWong: you're absolutely right! I have reformulated the question in a ``projective'' way, and both the issues should have disappeared (though others may pop out). $\endgroup$ – Giovanni Moreno Sep 18 '15 at 10:10
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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. :)

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    $\begingroup$ In the paper of Burstall and Calderbank, this trace-free part of the second fundamental form is a big part of data that is induced by an immersion into a conformal manifold. One of their main results is a converse to this, i.e. that once such set of initial data satisfies conformal version of Gauss-Codazzi-Ricci equations, there is an immersion unique up to conformal transformation. $\endgroup$ – Vít Tuček Sep 18 '15 at 16:26
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    $\begingroup$ @VítTuček: neat! Now you've convinced me to read that paper. $\endgroup$ – Willie Wong Sep 18 '15 at 17:41
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    $\begingroup$ @VítTuček: yes, you're right. I've contacted David Calderbank on this concern and he confirms that the above necessary condition wrote down by Willie Wong is also a necessary one... neat indeed! $\endgroup$ – Giovanni Moreno Sep 23 '15 at 11:28
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The scalar product on $\langle\,\cdot\,,\,\cdot\,\rangle$ of signature $(p+1, q+1)$ on $V$ induces the conformal structure on the projectivized null-cone and in fact this projectivised null-cone is a (flat) model space for the conformal geometry of signature $(p,q)$ (in the sense of Cartan geometry). In fact, the orthogonal group of your scalar product acts by conformal mappings on this projectivised null-cone and a stabilizier of it's point is a prabolic subgroup.

If you look more closely at the case of signature $(1,n)$ you'll be able to see that the projectivized null-cone is diffeomorphic to a sphere and it's not hard to work out which elements of $SO(p+1, q+1)$ act by which conformal transformations. The details for general signatures are in Slovák's lecture notes that I have linked to in a previous answer of mine.

As for your main question, I do not know the answer on top of my head, but maybe you can check out Conformal submanifold geometry I-III by Burstall, Calderbank.

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  • $\begingroup$ Thanks for answering the preliminary question. Now I'll update the post and go through the Burstall-Calderbank long paper. $\endgroup$ – Giovanni Moreno Sep 18 '15 at 13:09
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The answer to the question posed is clearly no. There's some problem with defining what is the Hessian exactly, but supposing a definition has been made, here are first some trivial problems.

  1. Let $f$ be an arbitrary smooth function on $V$, and consider the function $f^3$; it is clearly a smooth function on $V$ and has the same zero section $\{f^3 = 0\} = \{f = 0\}$. But we have that $\mathrm{hess} f^3 = 0$ at $\{f^3 = 0\}$ due to the higher order degeneracy.

  2. Presumably you also want in the statement of the problem that you are only considering $\mathrm{hess}^N f |_{\{f = 0\}\cap N}$, since modifying $f$ outside of its level set will not change the character of the level set.

  3. Proportional is such a vague word... what if the constant of proportionality is zero? Let $f$ be a smooth function such that on some domain away from the origin, $f(x) = \langle x,x\rangle$. Then $\mathrm{hess}^N f = 0 \propto g$.

  4. At best your characterisation can be a local one. Take the trivial example of Minkowski space $\mathbb{R}^{1+2}$. Let $h:\mathbb{R} \to\mathbb{R}$ be a smooth step function, such that $h(x) = 1$ for all $x < -1/10$ and $h(x) = 2$ for all $x > 1/10$, and $h$ is increasing.

    Let $f(x) = x_1 - h(x_1)$.

    Along the intersection $\{f = 0\} \cap N$ locally the function $f$ is exactly $x_1$ minus a constant, and so appears to be a hyperplane section. But globally certainly it is not. Your local criterion can never detect such global structure.

To solve 1 and 3 you want at least to assume that $f$ is transverse to $\mathbb{N}$ along $\{f = 0\}$. To solve 2 and 4 you can localize your statement. So these are not fatal, but just imprecise.


A deeper problem with your question is:

What exactly do you mean by $\mathrm{hess}^N$? On an arbitrary degenerate (light-like) (sub-)manifold, you cannot necessarily define the Levi-Civita connection. (The connection coefficients computation requires the inverse metric, which now has components which are infinite; the Koszul formula cannot do the work, since the metric is degenerate; you cannot induce a connection from the ambient one since "normal projection to the tangent space" doesn't work for degenerate subspaces.)


In the Euclidean case your argument actually does work somewhat, and also in the indefinite case provided you replace the null cone by the level set $\{ \nu = c, c\neq 0\}$.

This follows from

  1. The general observation that the Hessian is expressed as $$ \mathrm{hess}_{X,Y} (f) = X(Y(f)) - (\nabla_X Y) f $$
  2. That on a non-degenerate submanifold the induced connection satisfies $$ \nabla_X^{(N)}Y = \nabla_X Y + II(X,Y) $$ where $II$ is the second fundamental form and
  3. For the level sets $\{\nu = c, c\neq 0\} = N$, the manifold $N$ is umbilic (the second fundamental form is proportional to the induced metric).

In short, if it weren't for the fact that the second fundamental form and the Hessian are tricky to define for degenerate hypersurfaces, your basic intuition is ok and probably would have worked.

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  • $\begingroup$ Very clear & exhaustive explanation! It allowed me to reformulate the original question in a more robust and hopefully convincing way. $\endgroup$ – Giovanni Moreno Sep 18 '15 at 10:06

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