2
$\begingroup$

Let us consider the pseudo-euclidean space $\mathbb{R}^3_1$; that is, the space $\mathbb{R}^3$ endowed with the metric

$$\langle x, y\rangle = x_1y_1+x_2y_2-x_3y_3.$$

I see in O'Neill's book that there does not exist any compact semi-riemannian hypersurface in this space. In fact, there does not exist any compact hypersurface in the general case $\mathbb{R}^n_\nu$ where $\nu$ is the index of the metric. Why does this kind of phenomenon occur in the pseudo-euclidean case?

$\endgroup$

1 Answer 1

6
$\begingroup$

The first thing to understand is what kind of quadratic form $q$ is induced on a vector plane $V\subset \mathbb{R}^3$ by the Lorentz product $\langle\cdot,\cdot\rangle$: if $V$ is vertical enough, then $q$ has signature $(1,1)$; if $V$ is horizontal enough, then $q$ has signature $(2,0)$, i.e. is Riemannian; but if $V$ is slanted with slope $1$, in between the other cases, i.e. if it is tangent to the isotropic cone, then $q$ has signature $(1,0)$ and is degenerate, not pseudo-Riemannian ("pseudo" is probably more common than "semi" here, nowadays).

Now, any compact surface of $\mathbb{R}^3$ must have at least one point (in fact even a curve) where its tangent plane has slope $1$, so that it cannot a pseudo-Riemannian surface.

$\endgroup$
7
  • $\begingroup$ It may seem obvious to some but is worth remarking that one of the results of pseudo-Riemannian geometry is that the signature of the metric must be constant over path connected components. $\endgroup$ Jan 23, 2015 at 14:04
  • $\begingroup$ @Benoit Kloeckner: Why the compact surface is tangent to the lightlike cone? $\endgroup$
    – Ergonvi
    Jan 23, 2015 at 15:08
  • $\begingroup$ Take a critical point of the function $x_1+x_2-x_3$ restricted to the surface. $\endgroup$
    – Gil Bor
    Jan 23, 2015 at 15:36
  • $\begingroup$ But that function hasn't critical points... $\endgroup$
    – Ergonvi
    Jan 23, 2015 at 22:38
  • $\begingroup$ @Ergonvi: it is a linear (hence smooth) function restricted over a compact smooth manifold. Of course it has critical points. $\endgroup$ Jan 24, 2015 at 1:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.