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Suppose one is given the spacetime $(M,g)$ where $M$ is a fixed differentiable manifold and $g$ is a Lorentzian metric whose local expression is:

$$g= -(dt + a \, d \phi)^2 + d\rho^2 + \kappa^2 \rho^2 \, d\phi^2 + dz^2 \ \text{where} \ \ a > 0$$

This is the spacetime of an infinite straight Spinning Cosmic String.

Suppose that the spacetime $(N,h)$ is locally isometric to the spacetime $(M,g)$.

Since there are distinct definitions of local isometry in the mathematical literature it's better to clarify one's definition explicitly:

Spcetime $(M,g)$ is locally isometric to $(N , h)$ if and only if each point $p \in M$ has an open neighborhood $U_p$ and an open neighborhood $V \subset N$ and a diffeomorphism $f: U_p \rightarrow V$ such that: $$g \big |_{U_P}=f^*h \big|_{U_p}$$

The question is:

what can in general be extracted anout the differentiable manifold $N$ by a particular choice of $M$ inclusive of its differential structure?

This applies both to the differential structure over the manifold $N$ and its topology.

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    $\begingroup$ I suspect because this question seems like a more general version of your other question. Again, you haven't specified anything about the domain of the explicitly-given metric. Can we take $M$ to be literally $\mathbb{R}^4 \setminus \mathbb{R}^2$, or any open subset of this? A concrete question like "can we say anything about a maximal [adjectives] extension containing $\mathbb{R}^4 \setminus \mathbb{R}^2$ with this metric?" might go down better. The definition of local isometry is really not the problem here. $\endgroup$
    – David Roberts ♦
    Commented May 2 at 12:21
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    $\begingroup$ This question is actually quite different, and I hope that readers will notice the difference. For example, if instead of the metric given, imagine that we take the de Sitter metric. So $N$ is required to be locally isometric to de Sitter space, and that is an interesting class of Lorentz manifolds, especially if we ask that $N$ not be isometric to a proper open set in any Lorentz manifold. To me, this is an interesting question. $\endgroup$
    – Ben McKay
    Commented May 2 at 12:41
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    $\begingroup$ @DavidRoberts This question is improved with respect to the previous one, and I don't think it deserves the downvotes. OP is just requiring a certain structure of the metric locally (in a chart), so $M$ can a priori be anything. It seems to me that the question asks what can be deduced about $N$ (topologically, smoothly). $\endgroup$
    – Overflowian
    Commented May 2 at 13:03
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    $\begingroup$ @BenMcKay: the question in your comment is certainly interesting, but I don't see any mention in the OP about any maximality condition. Without it I really don't see how anything meaningful can be said topologically. $\endgroup$
    – Willie Wong
    Commented May 2 at 14:14
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    $\begingroup$ @BastamTajik: If you don’t understand very basic things like why a tensor can’t be locally zero but globally nonzero, then I think you need to go learn basic manifold theory before you try to use MO. I’m not trying to be rude, but from this and other things you have said it sounds like you don’t even have the prerequisites to understand the technical meanings of the terms in your questions or in the answers you would get here. That is probably the source of all the conflicts you are having. $\endgroup$
    – Andy Putman
    Commented May 2 at 21:53

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I think in your question, as currently formulated, the whole rotating cosmic string is a red herring.

If I interpret your notation correctly, $a$ and $\kappa$ are constants. And hence locally you can define $\psi = \kappa \phi$ and $s = t + a \phi$ to get that the metric is locally the same as $- ds^2 + d\rho^2 + \rho^2 d\psi^2 + dz^2$ which is just the Minkowski metric in disguise.

Therefore, for a manifold to be locally isometric to your rotating cosmic string is the same as asking for the manifold to be flat.

Absent other geometric conditions, I don't think there's anything meaningful that you can say about the topology of such a manifold. There's lots of "cut-and-glue" type operation you can do on subsets of Minkowski space to build such manifolds. In addition to the the operation that makes the rotating cosmic strong, you can also do something like "cut a closed three-dimensional disk from $\mathbb{R}^{1,3}$, double the manifold, and glue the two copies together by crossing the 'interiors'." You can use this to also glue on copies of the four torus and other stuff.

I think the key to the rotating cosmic string solution is precisely in its global structure that generates the closed time-like curves, but this is not something detectable at the local isometry level and so probably this means you are not asking the question you think you are asking.

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    $\begingroup$ Are these manifolds also known as flat Lorentzian $5$-manifolds? People certainly study complete flat Lorentzian manifolds. Without completeness it is harder to get off the ground even though some things can still be said topologically, e.g. one always has the developing map, so the universal cover of a flat Lorentzian manifold is stably parallelizable because it immerses into a Euclidean space of the same dimension. $\endgroup$
    – Igor Belegradek
    Commented May 2 at 15:27
  • $\begingroup$ @IgorBelegradek probably $4$-manifolds and not five. // Given the recent question of the OP on homotopy group, I am not sure whether looking at universal cover is useful. Remember that the starting point of this question is the spinning cosmic string, which is neither complete nor simply connected. // Feel free however to provide some references on results about complete flat Lorentzian manifolds; at least I would be interested to read more. $\endgroup$
    – Willie Wong
    Commented May 2 at 15:52
  • $\begingroup$ @IgorBelegradek: ah, I see I had a typo which confused the dimension for you. I fixed it. Thanks. $\endgroup$
    – Willie Wong
    Commented May 2 at 15:53
  • $\begingroup$ I am not an expert, but e.g. "Complete Flat Affine And Lorentzian Manifolds" by Charette, Drumm, Goldman, researchgate.net/publication/…, and Goldman and Kamishima, "The fundamental group of a compact flat Lorentz space form is virtually polycyclic" J. Differential Geom (1984) or web.ma.utexas.edu/users/jdanciger/pdfs/RNC-slides.pdf. $\endgroup$
    – Igor Belegradek
    Commented May 2 at 16:42
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    $\begingroup$ @BastamTajik: I don't think you understood my answer. The condition that your manifold is flat allows fairly wild topologies. Certainly Minkowski space and the cosmic string spacetime are locally isometric but do not admit global diffeomorphisms. The point of my answer is that there's enough freedom topologically that it is hard to say anything universal about all of them. $\endgroup$
    – Willie Wong
    Commented May 5 at 9:20

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