What is the manifold topology of a spinning Cosmic String?

Question

Given the following metric which is that of a rotating Cosmic String:

$$g=-c^2 dt^2 + d\rho^2 + (\kappa^2 \rho^2 - a^2) d\phi^2 - 2ac d \phi dt + dz^2.$$

can one determine the manifold topology underlying this metric?

Mathematically one might pose lots of opposition since they fix the topology before assigning a metric to the topological manifold and in case of Lorentzian geometry the topology induced by the Lorentzian distance does not match the initial manifold topology one started with, and so there's no guarantee that the manifold topology can be recovered from the local metric form (when the spacetime is not strongly casual, the manifold topology is strictly finer than the Lorentzian distance induced topology just like in the case of spinning Cosmic String).

But physically I think knowing enough symmetries, the local metric form (and so local Riemann curvature form and so the topology by theorems like Gauss-Bonnet, Chern-Gauss-Bonnet or more general Index theorems and a complete classification of topological surfaces at one's disposal) can narrow down all the consistent non-homeomorphic topologies if not determining it uniquely [in our case] (NOTE: I know by now that these possible topologies can not be finer or coarser in comparison to each other, but I don't know if they can be distinct).

I'd be thankful if anyone sheds light on this problem.

Answer 1

I'm not sure this is the answer you are looking for.

Given a Lorentzian manifold $(M,g)$ suppose that it's time orientable. Then there exists a never-vanishing time-like vector field $v$.

The vector field $v$ and $g$ split the tangent space as $TM = \mathbb R v \oplus v^\perp$. Now define the Riemannian metric $\tilde g$ as $\tilde g = -g|_{\mathbb R v} \oplus g|_{v^\perp}$.

In this way we have contructed a Riemannian metric in an essentially unique way out of $(M,g)$ which induces a distance compatible with the topology of the manifold.

Putting this in another way: instead of considering the usual definition of distance, there is another notion of distance, that you can build up from your Lorentzian-metric, which gives a metric compatible with $M$.

Answer 2

I think it is just ordinary Minkowski with a singular spacetime surface (one direction being timelike) at which a spatial wedge is cut out and the sides of the wedge glued back together with a time offset.

You can see the time offset by writing the metric as $$ ds^2 = - (dt+a d\phi)^2 + d\rho^2 + \kappa^2 \rho^2 d\phi^2 +dz^2 $$ so as you circle the string along the curve $dt + a d\phi=0$ until $\phi = 2\pi$ there is zero spacetime interval between $(t, \phi=0)$ and $(t = -2\pi a, \phi=2\pi)$, so those points are identified.