I am wondering what conditions a Lorentzian manifold $(M,g)$ must satisfy to ensure the existence of a global proper-time foliation (i.e. a decomposition of $M$ into spacelike Cauchy hypersurfaces and a global time function whose tangent vector field $\vec{t}$ satisfies $g(\vec{t},\vec{t})=-1$ and $g(\vec{t},\vec{x})=0$ for all $\vec{x}$ tangent to one of the spacelike hypersurfaces. Is global hyperbolicity enough? If anyone could recommend a good reference, that would be much appreciated!


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    $\begingroup$ What do you mean by "tangent vector field" $\vec{t}$ of a "time function"? Do you just mean the metric dual vector to the differential? $\endgroup$ Aug 13, 2015 at 15:02
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    $\begingroup$ I agree with Willie's remarks below. In general, global hyperbolicity is neither necessary nor sufficient for the existence of such a proper time function. Other relevant key-words are "Gaussian normal coordinates", "geodesic slicing" or "synchronous slicing". Unfortunately, doing a quick search to see if there are any papers that discuss the existence of a "global" structure of this type unfortunately didn't give any obvious hits. $\endgroup$ Aug 13, 2015 at 17:12
  • $\begingroup$ @IgorKhavkine thank you for the search terms! I'll have a go with those and see what I can turn up. $\endgroup$
    – Idempotent
    Aug 14, 2015 at 1:14
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    $\begingroup$ You may want to look at Geroch, J Math Phys 11 (1970) 437. The result is described in Hawking and Ellis, p. 212. Global hyperbolicity only guarantees that spacetime is homeomorphic to some structure like the one you have in mind. You want such much stronger condition than global hyperbolicity. Staticity would work. $\endgroup$
    – user21349
    Sep 3, 2015 at 21:05
  • $\begingroup$ @BenCrowell Thank you! I am going to look up these references - very helpful comment!! $\endgroup$
    – Idempotent
    Sep 6, 2015 at 16:34

2 Answers 2


Global hyperbolicity only gives you a Cauchy time function, whose gradient is past directed timelike. See Smoothness of Time Functions and the Metric Splitting of Globally Hyperbolic Spacetimes - Antonio N. Bernal, Miguel Sanchez - Commun. Math. Phys. 257, 43–50 (2005)

  • $\begingroup$ Thank you so much! In glancing through the reference, I don't see any mention of the time function being a proper time, but I guess once one has the global time function and its level surfaces as spacelike Cauchy hypersurfaces, one can construct a global timelike unit normal vector field to the hypersurfaces by projecting the tangent vectors to the existing global time function onto the hypersurfaces, subtracting that off, and normalizing? Then I guess it shouldn't be a problem to define the global proper time function in terms of integral curves to the timelike unit normal field? $\endgroup$
    – Idempotent
    Aug 13, 2015 at 13:07
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    $\begingroup$ Yes, sorry, I didn't say anything about proper time. Note that $\vec{t}$ is normal to the Cauchy hypersurfaces, so I think normalizing and looking at the flow should work out. I am just not quite sure about the details yet. $\endgroup$ Aug 13, 2015 at 13:54
  • $\begingroup$ Aha, yes, I see now from the form of the splitting of the metric as $-\beta dT^2 + \bar{g}$ that indeed $\vec{t}$ is already normal to the hypersurfaces - thanks!! Then since $\vec{t}$ is smooth, I agree, it should be OK to normalize it and look at the flow as you say. Will think more about details. Thanks again!! $\endgroup$
    – Idempotent
    Aug 13, 2015 at 13:57
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    $\begingroup$ Flowing is not enough. After normalizing the flow may not be complete. Furthermore the incompleteness may occur at different "times". For an extremely simple example, look at the diamond $|x| + |t| < 1$ in 1+1 Minkowski space which is globally hyperbolic. Pick any GH foliation such that along $\{x = 0\}$ the normal vector lines up with the $t$-axis. A propertime foliation built from this GH foliation will have total proper time length of $2$. But for every $\epsilon$ there exists a point $p$ in the diamond such that any timelike curve of length $>\epsilon$ through $p$ cannot be contained... $\endgroup$ Aug 13, 2015 at 14:57
  • $\begingroup$ ... in the diamond. $\endgroup$ Aug 13, 2015 at 14:59

Global hyperbolicity is certainly not enough.

Consider the causal diamond $$ D = \{(t,x) \in \mathbb{R}^{1+1} | |t|+|x| < 1\} $$ in 1+1 dimensional Minkowski space, which is certainly globally hyperbolic. I claim that this set does not admit the so-called "global proper-time foliation".

  1. Observe that with a proper-time foliation, the total elapsed proper-time must be at most 2, this being the length of the longest time-like geodesic you can fit in the causal diamond $D$.

  2. Supposing such a foliation exists with the global time function denoted $\tau$ (so that the level sets of $\tau$ are Cauchy, and $|\mathrm{d}\tau|_g = 1$), and is at least $C^1$ smooth, we can start with an arbitrary point $p\in D$ and follow the flow of $p$ by $\mathrm{d}\tau^\sharp$. By the first step we know that the total length of the flow-line is finite, and by definition this length is non-zero. Call it $\epsilon_p$.

  3. Claim: there exists a point $q\in D$ such that any time-like curve $\gamma$ through $q$ must have the total length of $\gamma\cap D$ be strictly less than $\epsilon_p$.

    Proof: Consider the point $q = (t = 0, x = 1 - \delta)$ for some $\delta$ to be specified later. Since time-like geodesics maximize length among time-like curves, the maximum length that can be attained by a curve $\gamma$ contained in $D$ through $q$ is bounded above by the lengths of the two segments joining $q$ to $(t = 1, x = 0)$ and $(t = -1,x = 0)$. By taking $\delta$ arbitrarily small we see that this maximum length can be made arbitrarily close to zero.

  4. Steps 2 and 3 above contradict, since by assumption the flow of $q$ by $\mathrm{d}\tau^\sharp$ should have the same length as the flow of $p$.

The key is step 3: the same argument carries through whenever your global hyperbolic spacetime $D$ satisfies the condition that "for every $\epsilon$ there exists a point $q\in D$ such that no timelike curves through $q$ in $D$ has length greater than $\epsilon$."

  • $\begingroup$ @ Willie Wow, very illuminating, thank you! I'll have to take some time to think this through. I really appreciate the concrete example. $\endgroup$
    – Idempotent
    Aug 14, 2015 at 1:12

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