3
$\begingroup$

Theorem Let $\beta\colon [0,1] \to M$ be a null geodesic. If $\beta(t_0)$ is conjugate to $\beta(0)$ along $\beta$ for some $t_0\in (0,1)$, then there is a timelike curve from $\beta(0)$ to $\beta(1)$.

This classical theorem can be found in

  • Hawking and Ellis, The large scale structure of spacetime, Prop. 4.5.12,
  • O'Neill, Semi-Riemannian geometry, Prop 10.48,
  • Beem, Ehrlich and Easley, Global Lorentzian geometry, Theor. 10.72
  • Kriele, Spacetime, Lemma 4.6.15

and the proofs are more or less refinements over Hawking and Ellis'. I am following the notation by Beem et al. I do not understand a step in the proof.

They construct a variation $\alpha\colon [0,1]\times (-\epsilon,\epsilon)\to M$, $(t,s)\mapsto \alpha(t,s)$, $\alpha(\cdot,0)=\beta$, $\alpha(0,\cdot)=\beta(0)$ and $\alpha(t_2,\cdot)=\beta(t_2)$ for some $t_0<t_2\le 1$. Denoting $T=\alpha_* \partial/\partial t$, $V=\alpha_*\partial/\partial s$, in all these references it is claimed that if

  • $\frac{d }{d s} g(T,T)\vert_{s=0}=0$, $t\in [0,t_2]$
  • $\frac{d^2 }{d s^2} g(T,T)\vert_{s=0}\le 0$, $t\in [0,t_2]$ with strict inequality in $(0,t_2)$,

then for sufficiently small $s\ne 0$ the varied curves are timelike.

I assume that they are using here Taylor's theorem together with a continuity and compactness argument. However, it seems to me that this argument does not work as it stand, at most one can conclude that for every $\delta>0$ there is some $b$, $0<b<\epsilon$, such that the varied curves $\alpha_s$, $s\in (0, b]$ are timelike on the time interval $[\delta,t_2-\delta]$. Indeed the third derivative could be diverging for $t\to 0,t_2$. Observe that trying to bound the second derivative by a negative constant $-c$, $c>0$, as done e.g. by Beem et al. does not solve this problem as long as this bound holds only over $\beta$ (that is for $s=0$) (by the way the second derivative would not be continuous at $t=0,t_2$ since there $\frac{d^2 }{d s^2} g(T,T)=0$).

Am I missing something? Incidentally, do you have completely different proofs, perhaps more topological?

Improve this question
$\endgroup$
2
  • $\begingroup$ Let me see if I understand correctly. You are worried about the curve $\alpha(\cdot,b)$ being timelike on the open interval $(0,t_2)$ but asymptotically becoming null as $t$ approaches the extremes $0$ and $t_2$? $\endgroup$
    – Igor Khavkine
    Commented Dec 30, 2014 at 17:01
  • $\begingroup$ No. For every $t$ the varied curve at $(t,s)$ is timelike for sufficiently small $s$ namely for $s<z(t)$ where $z$ might depend on $t$. I am worried that $z(t)\to 0$ for $t\to 0,t_2$. Thus the subset on the s-t space where $g(T,T)\le 0$ might not contain a rectangle in the s-t space. The variation is casual only if there is such rectangle. $\endgroup$
    – Ettore Minguzzi
    Commented Dec 30, 2014 at 19:14

3 Answers 3

Reset to default
3
$\begingroup$

It seems that your main argument against the correctness of the proof in Beem, Ehrlich and Easley, Global Lorentzian geometry, Theor. 10.72, is that $\frac{d^2}{ds^2}g(T,T)|_{s=0}=0$ at $t=0$ and $t=1$. Anyway, I don't see why $\frac{d^2}{ds^2}g(T,T)|_{s=0}=0$ should be zero there. In my opinion the proof by Beem, Ehrlich and Easley is correct (and also the one in Hawking and Ellis). Please observe that finding a negative bound on the second derivative along $\beta$ in $(0,1)$ is enough, by continuity, to have negative second derivative on a small rectangle $ [0,1]\times (-\epsilon,\epsilon)$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Let me illustrate the problem with an example. Consider the function $f(t,s):=g(T,T)$. Suppose that $$f(t,s)=-[t(t_2-t)]^2s^2+[t(t_2-t)] s^3.$$ It satisfies: $f(t,0)=0$; $\frac{d f}{d s}(t,0)=0$; $\frac{d^2 f}{d s^2}(t,0)=-2 [t(t_2-t)]^2$. Thus $\frac{d^2 f}{d s^2}\le 0$ on $[0,t_2]$ with strict inequality in $(0,t_2)$. You might notice that the second derivative of $f$ is not negative on any rectangle containing s=0. Furthermore, $f(t,s)>0$ for $s>t(t-t_2)$ thus no $\alpha_s$ would be causal. Still all the reference assumptions are complied. $\endgroup$
    – Ettore Minguzzi
    Commented Mar 8, 2015 at 17:41
  • $\begingroup$ Concerning Beem et al.'s proof you are probably right. I am now realizing a naive mistake I was making, I assumed that since $V=\alpha* \partial_s$ vanishes at $t=0$ it had to be (my mistake, the first equality does not hold) $\frac{d^2 f}{d s^2}(0,0)=V(\frac{d f}{d s})=0$, thus I did not see how one could apply a compacteness argument if $f$ was not $C^2$. Now, is $f$ $C^2$ also for $t=0$? I would say so because of the properties of the exponential map. I will think about it and correct my answer. Thank you! $\endgroup$
    – Ettore Minguzzi
    Commented Mar 8, 2015 at 19:10
3
$\begingroup$

A few years ago I gave a course on causality theory based on O'Neill's exposition. At the time I also noticed a problem with the proof of his Prop. 10.48. Back then I worked out an extended version of that proof which I believe fixes the problem you address. That proof can be found here (handwritten): http://www.mat.univie.ac.at/~mike/oneill_10.48.pdf

Improve this answer
$\endgroup$
2
$\begingroup$

The first question is whether there is really a gap in the classical proofs. The answer is 'no' in the version by Beem et al (thanks to Erasmo who spotted my mistake) while the other versions are somewhat incomplete. Greg Galloway told me that he also noticed some gaps in O'Neill's version. This fact motivated him to prove a related statement (the causality lemma)

G. J. Galloway. A finite infinity version of topological censorship. Class.Quantum Grav., 13:1471–1478., 1996.

It is fairly easy to check that the causality lemma implies the usual chronality statement mentioned in my question.

In arXiv:1502.02313 the interested reader can find another proof of the chronality results and of the causality lemma. I worked it in the context of Finsler spacetimes but the specialization to the Lorentzian case is straightforward. The classical proofs try to construct a curve variation, and the difficult part is to control the causal character of the varied curves. The new strategy is completely different and based on the comparison of two null hypersurfaces away from the focusing/conjugate points.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .