1
$\begingroup$

Problem: Let $M = \mathbb{R}^2 \setminus \left\{ (0,0 \right\}$ be the pseudo-Riemannian manifold with metric $$ ds^2 = \frac{2 du dv}{ u^2 + v^2}. $$ Let $\mu(u, v) = (2u, 2v)$. This is an isometry (trivially). I wish to show that the group $ \Gamma = \left\{ \mu^n \right\}$ generated by $\mu$ acts properly discontinuous on $M$. Then $T = M/\Gamma$ is a Lorentz surface, called the Clifton-Pohl Torus.

So by definition I need to prove that: every $p \in M$ has a neighborhood $U$ such that $\phi(U) \cap U = \emptyset$ for all $\phi \in \Gamma$ with $\phi \neq Id$. (do Carmo, Riemannian Geometry, p. 165).

The way I wished to show this is as follows. If $p = (u_0, v_0) \in M$, then let $\epsilon = d((u_0, v_0), (2u_0, 2v_0))$ where $d$ denotes the distance between $(u_0, v_0)$ and $(2u_0, 2 v_0)$ for the metric $ds^2$ on $M$. Then if I take as a neighborhood $U = B_{\epsilon/2} (p)$ , i.e. the 'ball' centered at $p$ with radial distance $\epsilon/2$, then any action of a $\phi \in \Gamma$ will move all of the points of $U$ out of $U$, i.e. $\phi(U) \cap U = \emptyset$.

The problem that I have is that I'm not sure if $\epsilon > 0$, since in a semi-Riemannian manifold the distance between two distinct points can be zero.

Moreover, does one define distance in Lorentzian geometry the same way as in Riemannian geometry? In Riemannian geometry the Riemannian distance $d(p,q)$ between two points $p$ and $q$ is defined as the greatest lower bound of $\left\{ L(\alpha): \alpha \in \Omega(p, q) \right\}$, where $\Omega(p,q)$ is the set of all piecewise smooth curve segments in $M$ from $p$ to $q$ and $L(\alpha)$ denotes the arc length of $\alpha$.

Does this definition extend to Lorentzian geometry?

$\endgroup$
3
  • $\begingroup$ What you wrote on the 2nd line is a Riemannian metric, not Lorentzian one. Did you mean to use $u^2-v^2$ or $uv$ in the denominator? In any case, do not use the Lorentzian metric to define your fundamental domain, use the Euclidean annulus $\{(u,v): 1\le u^2+v^2\le 4\}$. The thing is that Lorentzian metrics are not good for metrizing topological spaces. $\endgroup$ – Moishe Kohan Apr 13 '19 at 13:07
  • $\begingroup$ No I didn't mean $u^2 - v^2$; I meant exactly what I wrote. See page 193 of the book of O'Neill : Semi-Riemannian geometry. $\endgroup$ – Kamil Apr 15 '19 at 7:31
  • $\begingroup$ Oh, you are right of course. The rest of my comment stands as is. $\endgroup$ – Moishe Kohan Apr 15 '19 at 23:25
0
$\begingroup$

The distance used in the definition of a properly discontinuous group action should be compatible with the topology on of the manifold. The Lorentzian distance is the wrong thing to use here. Since the group acts on $\mathbb{R}\setminus\{(0,0)\}$ you should use the euclidean distance, $\rho$.

Then given any point $(u,v)$ we can take $U=\{(x,y): \rho((u,v),(x,y))<1\}$. In your answer you seem to know what to do at this point to check the definition.

The issue in your understanding is that the Lorentzian distance is not actually a distance. It does not satisfy the definition of a distance. It is called a distance, but it is not a distance. Some authors call it the time separation function to avoid this awkwardness.

Unlike a Riemannian manifold there is no canonical choice of Lorentzian distance. There are lots of distances on the manifold that are compatible with the topology of the manifold but there is no "given" one. You just need to pick one and work with it.

In the particular case of the Clifton-Pohl torus life is easy as the construction is explicit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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