# What is the manifold topology of a spinning Cosmic String?

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.

## 2 Answers

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 $$M$$.

• Where does the topology introduced by this distance match that of the manifold topology on the causal hierarchy? Commented Apr 29 at 21:39
• But what sort of physical notion does this distance bring about? In case of interval topology it is connected with the notion of causality in spacetime? Commented Apr 29 at 21:57
• The topology induced by this metric will be the topology of $M$ and will be thus independent of the Lorentzian metric $g$. You asked if given a Lorentzian metric it was possible to recover the topology of $M$ in a similar fashion to what is done with a Riemannian metric. The answer is yes; you just need to change the definition of length of a path. Commented Apr 30 at 9:03
• And what about the differential structure? Can you recover it too by means of this metric and its induced topology? Does it fix the diff structure? I know that such thing is possible in case of strongly causal spacetimes pubs.aip.org/aip/jmp/article-abstract/17/2/174/224641/… Commented Apr 30 at 9:55
• And yet I have no clear idea about the physical meaning of such topology. Interval topology is physically the very result of time+space(Pseudo-Riemannian/Lorentzian) and causal structure instead of space alone(Riemannian geometry). Although I still admire the mathematical trick. Thanks Commented Apr 30 at 9:59

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.

• What do you mean by time offset? And are you sure this is the only topology consistent? I mean the topology that forgets about the time. (of course the manifold has to be time orientale). Commented Apr 28 at 13:33
• And I would be grateful if you introduce any reference where I can find more explanation regarding the manifold topology. Commented Apr 28 at 13:35
• The time offset does not affect the topology, which is Minkowski, so topologicaly ${\mathbb R}^4$. Commented Apr 28 at 13:47
• But wait a second please. I'm very much confused and maybe I'm too stupid. Not only I don't understand what you exactly mean by the "time offset" but also I don't understand how it can be topologically $\mathbb{R}^4$ while it has singularity. Pardon if I am not able enough. Commented Apr 28 at 13:52
• Doesn't it become a solid torus $\times \mathbb{R}$ perhaps due to the singularity? Commented Apr 28 at 14:02