# Semi-riemannian hypersurfaces

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?

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.
• Take a critical point of the function $x_1+x_2-x_3$ restricted to the surface. – Gil Bor Jan 23 '15 at 15:36