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  Is the metric-induced topology relevant at all in a (psuedo) Riemannian manifold?

+ 5 like - 0 dislike

A (pseudo) Riemannian manifold is a tuple: $$(M,g)$$ where $M$ is a smooth manifold (in particular, a topological space with an atlas) and $g$ is a (pseudo) Riemannian metric tensor.

It is apparent that by this definition, to define a pseudo Riemannian manifold, one need to first fix the topology of $M$ with a collection of open sets, and then give a metric tensor.

How is it then that often in physics (mostly GR), people are able to say things about the topological structure of their spacetime (a Lorentzian manifold) from looking at the metric?

For example, "schwartzchild spacetime has a singularity at r= 0" seems to be a topological statement.

A guess: is there some convention where when we equip the manifold with a metric, we also add to its open sets the metric topology induced by g?

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user bechira

asked Dec 5, 2014 in Mathematics by bechira (80 points) [ revision history ]
recategorized Apr 25, 2015 by Dilaton
Minor comment on question (v1) to stem any confusion readers may have: $M$ is required to be more than just a topological space, it's required to be a smooth manifold.

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user joshphysics
Right, I was trying to be clear but I see how it can be confusing, I'll change it.

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user bechira
a GR singularity is not necessarily topological: possibly, it's 'just' a metric degeneracy; @CristiStoica probably has something to say about that...

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user Christoph
also note that if $g$ is a solution of the Einstein field equations, it needs to be differentiable at least twice, and there's a pre-existing topology due to the diferential structure of the manifold, independent of any metric tensor; there are other kinds of topologies that might be of interest to physicist (Wikipedia lists two of them)

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user Christoph
possible duplicate of Global Properties of Spacetime Manifolds

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user JamalS
Thanks @JamalS looks relevant

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user bechira
Also related question linked to @JamalS's question, Global Properties of Spacetime Manifolds

This post imported from StackExchange Physics at 2015-04-25 21:56 (UTC), posted by SE-user bechira

2 Answers

+ 3 like - 0 dislike

In a mathematical sense, (differential) topology is imposed on the space-time, it is the "background" of the theory and the metric cannot change the topology of the background manifold. I.e. every metric should come with a "manual of topology" which specifies things such as coordinate ranges and identified coordinate values. However, this is not always the case, coordinates such as $\phi$ are taken as the $(0,2 \pi]$ identification without saying and in many cases there are multiple topological interpretations of a metric even after imposing the conditions sketched below. (btw. see what happens when you do not do the $(0,2 \pi]$ thing.)
Nevertheless, the properties of the metric "feed back" into the manifold topology through the person studying the metric who chooses a different background topology based on physical motivation or convenience. 

A few examples. Consider the Schwarzschild metric: $${\rm d}s^2 = -(1-2M/r) {\rm d}t^2 + \frac{1}{1-2M/r} {\rm d}r^2 + r^2 {\rm d}\Omega^2$$

By physical argumentation we find out that there is something fishy about the horizon, it has zero volume, particles which fly through pass through $t=\infty$ but at finite proper time etc. The coordinate transformation known as Kruskalization (see Kruskal-Szekeres coordinates) which makes the horizon regular actually changes the differential structure of the manifold. (alternatively see Eddington-Finkelstein coordinates)

Another canonical example would be the Kerr metric in Boyer-Lindquist coordinates representing a spinning black hole. There one finds a singularity at the disc $r=0$ with a field jump in the interior parts of the disc $r=0, \theta \neq \pi/2$ and a genuine metric singularity at the edge of the disc $r=0, \theta=\pi/2$. The field jump on the interior, when confronted with Einstein equations, would mean negative matter density. On the other hand, one can choose to avoid the negative matter density by introducing different topology, and by saying that by going through the disc one enters a new $r<0$ region. I.e. going through the "top" of the disc $r=0, \theta<\pi/2$ one does not enter the "bottom" of the disc $r=0, \theta>\pi/2$ but a completely new region and vice versa.

In the case of Schwarzschild introducing an $r<0$ region would be contrived; all geodesics end in $r=0$ and there would be no causal communication with $r<0$ whatsoever. However, the nature of the $r=0$ singularity in Kerr allows one to change the topology in a physically natural sense. It is then possible to give physically motivated conditions on the singularities which are "terminal" and which are not allowed at all.

Every mathematician knows from theorems such as the hairy-ball theorem that restrictions on singularities of differential systems restrict the allowed topology. It is also in this sense that the properties of the metric such as asymptotic flatness along with such conditions can restrict the topology of the manifold. The exact-solver of Einstein equations then usually starts with one coordinate patch and tries to solve the equations without encountering the "wrong" singularities - this, on the other hand, already determines the topology and I believe it is exactly in this sense in which you have encountered the interactions of the metric/topology.

answered Apr 26, 2015 by Void (1,645 points) [ no revision ]
+ 0 like - 0 dislike

There is a line on wikipedia about this  at http://en.wikipedia.org/wiki/Spacetime_topology

This topology coincides with the manifold topology if and only if the manifold is strongly causal but in general it is coarser

answered Apr 30, 2015 by anonymous [ revision history ]
edited Apr 30, 2015 by dimension10

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