A common proof of the lack of foliation of the Gödel universe, apparently mostly copy pasted from Hawking and Ellis, goes thusly :

  • A closed timelike curve must cross a spacelike hypersurface without boundary an odd number of times
  • A continuous deformation of the curve can only change the number of crossings an even number of times, meaning that the curve cannot be deformed to 0
  • As the manifold is topologically trivial, this is a contradiction

This proof only mentions two references, Gödel's original paper (which doesn't use that method but Frobinius's theorem) and Kundt's 1956 paper, "Tragheitsbahnen in einem von Godel angegebenen kosmologischen Modell", which does not seem to be available online. No other paper seems to redo that proof, so I am unable to find any details.

What motivates the assumption that a closed timelike curve must cross a spacelike slice an odd number of times? And what theorem says that a continuous deformation crossing a surface changes the number of crossings an even number?

  • $\begingroup$ That the number of crossings is invariant $\mod 2$ under homotopy is something that is discussed in e.g. Guillemin & Pollack, if you're talking about what I think you're talking. $\endgroup$ – Danu Mar 11 '16 at 20:53
  • $\begingroup$ It should be noted that Kundt's paper does not give the proof and Gödel's original paper proves a weaker version of the stated result. $\endgroup$ – Ryan Unger Mar 20 '16 at 2:28

I don't see how to do this using mod 2 intersection, but I believe it can be done over $\mathbb{Z}$ in the following way.

Since $\mathbb{R}^4$ is simply-connected, it's time-orientable, i.e. we can continuously choose at every point the positive half of the light cone. Similarly, it is space-orientable: any 3-dimensional space-like submanifold is orientable.

Now take a leaf of the assumed foliation, call it $L$, and a closed time-like curve $\gamma$. The previous paragraph implies that $\gamma$ intersects $L$ "positively" everywhere. The total number of intersection points taken with sign is a homotopy invariant, if $L$ is a proper submanifold. It's a standard fact if $L$ is compact and without boundary, but it also works if $L$ is just proper (then the normal exponential map has the good properties).

Since $\mathbb{R}^4$ is simply-connected, it follows that $L$ can not intersect $\gamma$ at all (all intersections are positive and therefore can not cancel each other). A contradiction, if we can find a proper $L$ intersecting $\gamma$.

I don't know how to claim properness (it might follow from the existence of Lorentzian metric), but it is essential for this kind of argument. See Reeb foliation, there you have a closed curve intersecting a non-proper leaf once - the intersection number is not a homotopy invariant.

Edit: the previous attempt at trying to make "mod 2" work was incorrect.

  • $\begingroup$ Is $\gamma$ required to be a geodesic? Since the Gödel solution admits no closed timelike geodesic $\endgroup$ – Slereah Nov 7 '17 at 11:50
  • $\begingroup$ @Slereah I was under the wrong impression that it did. But I haven't used that fact. Thank you, I will correct my answer. $\endgroup$ – mathquest Nov 7 '17 at 11:56
  • $\begingroup$ @Slereah found this tab in my browser and felt like I had correct my answer! I'm not sure if it is useful now, but here it is anyway. $\endgroup$ – mathquest Jun 27 at 4:16

What motivates the assumption that a closed timelike curve must cross a spacelike slice an odd number of times?

Its not an assumption. And it isn't true of all manifolds. Consider $\mathbb S^2\times\mathbb R^2$ as a subset of $\mathbb R^6,$ or just $$\{(a,b,A,B,y,z)\in\mathbb R^6:a^2+b^2=A^2+B^2=1\}$$ with the metric $$d\tau^2=da^2+db^2-dA^2-dB^2-dy^2-dz^2.$$

Then there is a spacelike surface $a=1.$

And there is a closed timelike curve that crosses it twice, namely the curve $$\theta\mapsto (\cos (2\theta), \sin(2\theta),\cos \theta, \sin \theta,0,0).$$

So why is that different than Gödel's solution? Gödel's solution is homeomorphic to $\mathbb R^4$ so you can imagine a curve in $\mathbb R^4$ that starts and stops in the same place. Even better you can put it into the one point compactification and identify $\mathbb R^4$ as that portion of $\mathbb S^4$ without the north pole. Then the CTC is a curve in $\mathbb R^4$ or a curve in $\mathbb S^4$ that avoids the north pole.

Now when you have a 3d surface in Gödel's manifold there is a corresponding 3d surface in $\mathbb R^4$ or a surface in $\mathbb S^4$ (and the surface can go to the north pole if otherwise it looks like it has a boundary since later you will only consider deformations that avoid the north pole so it doesn't matter if the surface is there or not).

So you have a different manifold with a closed curved and a 3d surface. But the new manifold is homeomorphic to $\mathbb R^4$ and that is why a curve that goes through the 3d surface has to cross it an odd number of times. And even then its also because since the original curve was everywhere timelike so it can't be tangent to the original surface (as the surface was everywhere spacelike) and so the old and new curves must pierce through the old and new surfaces respectively.

So we do go that if it a curve goes through a surface and is never tangent (so locally goes through it) and you are in $\mathbb R^4$ globally and the 3d surface doesn't have a boundary then the curve crosses the surface an odd number of times. Well, it must pierce through an odd number of times if they pierce through at least once.

But what theorem says this? Now it's about $\mathbb R^4$ so it might seem like a pure math question. But math theorems are organized by which techniques are used to prove them, and so you can have theorems that assume your manifold is smooth (and $\mathbb R^4$ is smooth). And while the original 3d surface in Gödel's universe has a differentiable surface (it was spacelike so had tangents) but a mere homeomorphism doesn't make the corresponding 3d surface in $\mathbb R^4$ smooth. So if you picked up a random book on Morse theory topology it won't be one of the easier theorems to have an arbitrary 3d surface. Even getting a finite number of crossings is a thing you need to prove, let alone that it is odd if nonzero.

But if you started out assuming your spacetime was smooth (some people categorically do this, and physicslly its wrong to do so, assuming smoothness sometimes forces a manifold to develop closed time like curves when otherwise Einstein's Equation didn't require it. And to me assuming you have time travel when it isn't required is close to the absolute worst thing you can do. Just like using the space I gave above instead of one with a linear time would be 100% and completely unacceptable as a mere assumption). Then starting with an assumption of the original manifold and surface being smooth could give the result right there. So its not clear to me what assumptions you want to make.

Rightly you shouldn't assume Gödel's universe is smooth, you should prove it if you think it. But then that's yet another theorem you need, one that isn't a pure mathematics theorem.

  • 1
    $\begingroup$ Unfortunately, I don't see where the argument for an odd number of crossings is. $\endgroup$ – Ryan Unger Nov 5 '15 at 20:30
  • $\begingroup$ Could you please highlight the part where you answer the question? I still don't see it. $\endgroup$ – Ryan Unger Nov 6 '15 at 1:36
  • $\begingroup$ @0celo7 The structure of the proof should be taken as this: from 2 and 3 one sees that in $\mathbb R^4$ a closed curve can only cross a hypersurface transversally an even number of times (this is intuitively very plausible and shouldn't be hard to prove). On the other hand, a timelike curve has to cross a spacelike hypersurface an odd number of times (I suspect this is not hard to prove either, but I don't know how. I would be interested to know). Since in the Gödel universe there are closed timelike curves, there cannot be spacelike hypersurfaces. $\endgroup$ – doetoe Nov 6 '15 at 9:52
  • $\begingroup$ @doetoe I understand this. Slereah is asking for a proof for the odd number of crossings. This is not present in this post, as far as I can tell. $\endgroup$ – Ryan Unger Nov 6 '15 at 13:39
  • $\begingroup$ @0celo7 I don't think it is present in the post either $\endgroup$ – doetoe Nov 6 '15 at 17:17

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