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I recently attended a talk where the person stated a uniqueness result for static vacuum spacetimes whereby he came to a conclusion about a type of spacetime (a 4-manifold) by studying 3-manifolds which are embedded as hypersurfaces in the 4-manifold (similar to the analysis by Schoen and Yau for the positive mass theorem).

However, he made the assumption that the 3-manifold in the spacetime always has a vanishing second fundamental form (similar to Part I of the Schoen-Yau proof). I believe in the literature that this a special case known as the time-symmetric case, but when I asked if his argument could then be generalized to the case where this is not assumed (perhaps using a PDE), he stated that it could not as the spacetime being static implies that it contains a 3-manifold with zero second fundamental form.

I would like to confirm if that is true. Surely a manifold can be 'time-symmetric' in some sense without being 'static'. Time-symmetric is just talking about symmetry under reversal of time, whereas static means it does not change at all: it cannot even rotate as with stationary state metrics like the Kerr metric.

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    $\begingroup$ Your last paragraph doesn't contradict any of the rest; static implies time symmetric but the converse is not true. $\endgroup$ – Javier Dec 2 '19 at 22:09
  • $\begingroup$ Yes, it seems like Ben in his post has given two examples which show that time-symmetric does not imply static, but static does imply time symmetric, so the geometric assumption is necessary. $\endgroup$ – Tom Dec 3 '19 at 1:41
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A stationary spacetime is one that has a timelike Killing vector. There is also a notion of an asymptotically stationary spacetime, which is what some authors mean by "stationary."Although a stationary spacetime does not have a uniquely pre- ferred time, it does prefer some time coordinates over others. In a stationary spacetime, it is always possible to find a “nice” t such that the metric can be expressed without any t-dependence in its components. A static spacetime is one that is not only stationary but also has the property that coordinates exist in which it is diagonal. (Coordinates will also exist in which it is not diagonal.)

GR does not have a notion of time-reversal that applies in all cases. Basically the structure of GR does not allow the concept of discrete symmetries to be applied.

Surely a manifold can be 'time-symmetric' in some sense without being 'static'.

Yes, this is certainly true. For example, the maximal extension of the Schwarzschild spacetime has a preferred notion of time reversal, under which the black hole and white hole regions are interchanged, and the time coordinate of a static observer in one of the exterior regions is reversed. However, this spacetime is not static, because the interior regions are not static.

Similarly, you can have an FLRW spacetime in which there is a preferred time (the time of an observer at rest with respect to the Hubble flow), and a time-reversal symmetry (big crunch or big bounce cosmologies), but it's not static.

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  • $\begingroup$ Thanks for your answer. I am not an expert on GR but my feeling from the talk was that the person was sweeping something under the carpet by assuming in his proof of uniqueness for the 4-manifold that the second fundamental form of the embedded hypersurface in question was zero, and then saying that he had to assume this because the spacetime is static, whereas in fact I do not think this needs to be assumed and so the result is not as general as it could be. $\endgroup$ – Tom Dec 3 '19 at 0:48
  • $\begingroup$ And actually your examples explicitly show that neither implies the other, so his claim that he could only prove the time-symmetric version of his result because the spacetime is static is either misguided or misleading. $\endgroup$ – Tom Dec 3 '19 at 0:50
  • $\begingroup$ @Tom: I'm not completely understanding the context of the talk, so there could be something we're missing or misinterpreting here. All I can vouch for is that I think what I wrote is correct for GR. A lot of the language you're using in the question is the kind of old-fashioned language people would use in old-school differential geometry, where you talk a lot about 2-surfaces embedded in 3-space, and use a lot of coordinate-dependent language. I'm not sure how much of this translates into GR, since I'm not fluent in that language. $\endgroup$ – user4552 Dec 3 '19 at 0:55
  • $\begingroup$ I'm not sure either. I could attach the pre-print if that helps as the content of the talk was identical to what is in the pre-print. I would see myself as a geometer rather than a GR expert, but I simply noticed that in the paper the main theorem is proved for uniqueness of a particular static vacuum spacetime, but he assumes that the spacetime contains a compact 3D spacelike totally geodesic slice bounded by Killing horizons. $\endgroup$ – Tom Dec 3 '19 at 1:04
  • $\begingroup$ There are various assumptions in there (the spacelike hypersurface is compact, the spacelike hypersurface is totally geodesic and has zero second fundamental form). That's fine, but the second assumption is a huge geometric restriction. Perhaps in the context it is OK, I don't know that much about GR. $\endgroup$ – Tom Dec 3 '19 at 1:08

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