Spacetime Metric Topology: Does g Induce O?

[nateHI]

Is it fair to say, when talking about spacetime with a given metric, it would be redundant to state that the associated set has the metric topology placed on it. In other words, let $M$ be a set, $O$ the metric topology, $\nabla$ a connection, $g$ a metric, and $T$ be the direction of time, then
$(M,O,\nabla,g,T)=(M,\nabla,g,T)$

Edit: I suppose what I'm asking is, does the metric $g$ induce the metric topology?

 

[George Jones]

What is "the metric topology" when $g$ is not positive-definite?

 

[robphy]

Possibly useful:
https://en.wikipedia.org/wiki/Spacetime_topology

More links from a recent thread:
https://www.physicsforums.com/threads/what-is-a-topology-intuitively.877771/#post-5513044

 

[nateHI]

What is "the metric topology" when $g$ is not positive-definite?

You're right, I suppose I should make that precise. The metric topology I'm referring to has as open sets
$B_r(m_0)=\{m_0\in M : g(m_0,m)<r\ \forall m\in M \text{ where } r>0\}$

 

[George Jones]

You're right, I suppose I should make that precise. The metric topology I'm referring to has as open sets
$B_r(m_0)=\{m_0\in M : g(m_0,m)<r\ \forall m\in M \text{ where } r>0\}$

I am sorry, but I still do not understand. The "metric" $g$ (i.e., not a metric in the sense of metric spaces) takes as input two tangent vectors at the same point of the manifold, not two points of the manifold.

 

[nateHI]

I'm referring to $g$ in two different contexts which is what I think made my question unclear. So, let $g$ be the metric from GR and let $d$ be the metric from math class. Does defining $g$ on a differentiable manifold automatically induce $d$?

As an aside, this is interesting to me since I believe it to be key in the notion of a topological field theory. Of course I'm just starting on the subject so I could be wrong.

 

[Orodruin]

The balls you refer to are not very good to use to build a topology when you have a pseudo Riemannian metric. You may want to use a topology based on the intersection of light cones instead.

 

[DrGreg]

As I understand it, the topology of $M$ is independent of $g$. It arises from the Euclidean topology of $\mathbb{R}^4$ mapped onto $M$ via the coordinate charts. You need the topology in order to define concepts such as "smooth" and "differentiable".

The metric tensor $g$ doesn't define a "metric" in the "metric space" sense because it isn't positive definite.

 

[nateHI]

OK, I think I get it now. Thanks everyone!

 

[vanhees71]

As I understand it, the topology of $M$ is independent of $g$. It arises from the Euclidean topology of $\mathbb{R}^4$ mapped onto $M$ via the coordinate charts. You need the topology in order to define concepts such as "smooth" and "differentiable".

The metric tensor $g$ doesn't define a "rmetric" in the "metric space" sense because it isn't positive definite.

That's why it's better to talk about pseudo-Riemannian manifolds and pseudo-metric to make this clear from the very beginning. Analysis is based on a true metric not a pseudo-metric.

 

[MeJennifer]

As far as I understand it a given metric in general relativity could be valid for more than one topology.