9
$\begingroup$

In the book Quantum Gravity in 2+1 dimension by S. Carlip, in the second chapter (section 2.1), he comments that a compact 3-manifold with a flat time orientable Lorentzian metric and a purely spacelike boundary, necessarily has the topology $[0,1]\times \Sigma$, where $\Sigma$ is a closed surface that is homeomorphic to one of the boundary components.

Does this mean that all spacetime manifolds (flat) that we could allow in 2+1 dimension are necessarily of this topology (it seems to be a big restriction)?

Also, is it necessary for the boundary of a spacetime (for those that would have one) to be spacelike ? I have a rough argument (which I am not sure about) for this, that if we had a timelike boundary, in some coordinate system, we would have the boundary at some given value for the space co-ordinates which seems weird to me.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

Great question. I think these are the type of topologies that physicists like to deal with, certainly not true that this is all the spaces that can be written down .

For example if you demand global hyperbolicity then you will get something like what you describe.

In general, let $X$ be a smooth manifold, and $\{U_i\}_{i \in I}$ be an atlas, let $\mathcal T$ be the sheaf of symmetric positive rank-2 tensors over $X$. If you can provide

  • Initial conditions for Einstein equations on some subspace of $U_i$, and
  • sections $g_i \in \mathcal T(U_i)$ that solve the Einstein equations with these initial conditions such that
  • $g_i$ glue smoothly to a section $g \in \mathcal T(X)$

Then you're free to call the pair $(X, g)$ a spacetime.

Improve this answer
$\endgroup$
0
$\begingroup$

Does this mean that all spacetime manifolds (flat) that we could allow in 2+1 dimension are necessarily of this topology (it seems to be a big restriction)?

Carlip is talking about compact manifolds. If you eliminate the requirement of compactness, then his statement is no longer true, and doesn't even make sense, because it refers to a boundary.

As a concrete example, it's possible for a flat spacetime in 2+1 dimensions to have the topology of Euclidean three-space. It just wouldn't be compact.

To see the point of what Carlip is talking about, consider the example of GR in 2+1 dimensions on a 3-sphere, which is compact. On the sphere, you're not going to be able to define a time orientation.

Also, is it necessary for the boundary of a spacetime (for those that would have one) to be spacelike ?

Yes, I think there are counterexamples if the boundary is timelike. For example, consider a solid doughnut with the topology $B^2\times S^1$, where $B^2$ is a closed disc. Let the time orientation point around the doughnut, i.e., circling the doughnut hole. The boundary is timelike, and the topology does not have the form $[0,1]\times\Sigma$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Why should it be true for compact manifolds? For example the procedure outlined in my answer provides an abundance of compact manifolds not of that topology. $\endgroup$
    – zzz
    Jun 3, 2018 at 22:40
  • $\begingroup$ @bianchira: I haven't read Carlip's argument, so I don't know. However, your answer doesn't provide a counterexample to Carlip's claim, it only lists conditions that would have to be satisfied by such a counterexample. If you think Carlip is wrong, it would be interesting to see a counterexample. $\endgroup$
    – user4552
    Jun 3, 2018 at 23:25
  • $\begingroup$ Example: Take any comstamt curvature 3 manifold then wick rotate. $\endgroup$
    – zzz
    Jun 3, 2018 at 23:31
  • $\begingroup$ @bianchira: Carlip's claim is about flat manifolds. $\endgroup$
    – user4552
    Jun 3, 2018 at 23:37
  • 1
    $\begingroup$ @BenCrowell It is in fact possible to define a time orientation on the 3-sphere. Whilst it's not "topologically trivial", its tangent bundle is, so there's no obstruction to defining smooth non-zero vector fields, metrics, n-forms, etc. See arxiv.org/pdf/gr-qc/0609119.pdf. $\endgroup$
    – gj255
    Dec 25, 2018 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.