4
$\begingroup$

My question arose after studying the article "John K. Beem: Conformal Changes and Geodesic Completeness". (http://projecteuclid.org/euclid.cmp/1103899983) One of the results there is:

Let $(M,g)$ be a causal spacetime which satisfies condition $N$. If $E$ is an open subset of $M$ with compact closure $\overline{E}$, then $(E,g)$ is stably causal.

To recall the necessary definitions:

For $(M,g)$ condition $N$ is satisfied, if for every compact subset $K \subset M$, there is no future inextendible nonspacelike curve which is totally future imprisoned in $K$.

$(E,g)$ is the metric $g$ restricted to the manifold $E\subset M$.

However I feel the proof of this is flawed, (The negation of stable causality is stated falsely in the proof of theorem 2, I think) yet the claim seems to be true. In fact even ignoring the condition $N$, I could not come up with a counterexample. Can anyone provide me with a counterexample? That is to answer the following question:

Can there be a causal spacetime $(M,g)$ and an open subset $E \subset M$ with compact closure $\overline{E}$, such that $(E,g)$ is not stably causal?

Thank you for your kind help.

$\endgroup$
1
  • $\begingroup$ Why do you think the negation of stable causality is stated incorrectly? $\endgroup$ Jun 5, 2015 at 13:15

1 Answer 1

3
$\begingroup$

Edit: In Beem's article on Page 180 before the introduction of condition (N) he mentions Carter's example (see also Hawking, Ellis The large scale structure of space-time, p. 195, Fig. 39). There you should get your counterexample by $E=(t_1,t_2)\times S^1 \times S^2$.

$\endgroup$
12
  • $\begingroup$ Yes, that answers the question. Thank you very much! $\endgroup$ Aug 16, 2014 at 15:23
  • $\begingroup$ Oh no! I just realized, that your example in fact does not seem to answer the question. Because the set E from your example indeed happens to be stably causal! (Because one can define a time function on E) Additionally you can see, that E is strongly causal, because there are no nearly closed non-spacelike curves completely contained in E. (They have to move around S^1 and therefor leave E) $\endgroup$ Aug 19, 2014 at 9:30
  • $\begingroup$ I meant $E$ to be a rectangle before projecting, i.e., $E$ should "loop around" too. I will think about it and post more details. $\endgroup$ Aug 19, 2014 at 9:42
  • $\begingroup$ Thank you for the clarification, @C_S, but unfortunately I still do not quite understand... My edition of the book doesn't have anything by the name of $\alpha$, but I assume you mean the lightlike curve whose endpoints are used for the definitions of $L_1$, $L_2$. That is the dotted line in the following picture: didaktik.mathematik.hu-berlin.de/files/… . I also tried to draw your suggestion of $E$, however I think I got you wrong here. Because the $E$ drawn in the picture is not compact. (for instance a sequence approaching $L_1$ has no limit) $\endgroup$ Aug 19, 2014 at 12:26
  • $\begingroup$ No, the curve drawn is called $\alpha$. I think it is not readable well in the picture and you apparently mistook if for an $x$. But you are right I forgot to take the removal of $L_1, L_2$ into account. $\endgroup$ Aug 19, 2014 at 12:36

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.