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.

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

1 Answer 1


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 $E=(t_1,t_2)\times S^1 \times S^2$.

  • $\begingroup$ 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

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