Spacetime Metric Topology: Does g Induce O?

  • Context: Undergrad 
  • Thread starter Thread starter nateHI
  • Start date Start date
  • Tags Tags
    Gr
Click For Summary

Discussion Overview

The discussion revolves around the relationship between a metric tensor ##g## and the induced metric topology on a spacetime manifold ##M##. Participants explore whether the metric ##g## inherently induces a topology, particularly in the context of general relativity and pseudo-Riemannian metrics. The scope includes theoretical considerations and conceptual clarifications regarding topology in relation to metrics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants question whether it is redundant to state that a set has the metric topology when discussing a spacetime with a given metric.
  • There is a query about the nature of "the metric topology" when the metric ##g## is not positive-definite.
  • One participant suggests that the open sets of the metric topology can be defined using balls based on the metric ##g##.
  • Another participant clarifies that the metric ##g## takes two tangent vectors as input rather than two points of the manifold, which complicates the discussion.
  • Some participants differentiate between the metric ##g## from general relativity and a traditional metric from mathematics, questioning if the former induces the latter on a differentiable manifold.
  • There is a suggestion that using light cones may provide a better basis for topology in the context of pseudo-Riemannian metrics.
  • One participant asserts that the topology of ##M## is independent of the metric ##g##, arising instead from the Euclidean topology of ##\mathbb{R}^4## through coordinate charts.
  • Another participant emphasizes the importance of distinguishing between true metrics and pseudo-metrics in the context of analysis.
  • It is noted that a given metric in general relativity could correspond to multiple topologies.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the metric ##g## and the induced topology, with no consensus reached on whether ##g## inherently induces a metric topology. The discussion remains unresolved regarding the implications of non-positive-definite metrics.

Contextual Notes

Some limitations include the dependence on definitions of metrics and topologies, as well as the unresolved nature of mathematical steps regarding the relationship between metrics and topologies.

nateHI
Messages
145
Reaction score
4
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?
 
Last edited:
Physics news on Phys.org
What is "the metric topology" when ##g## is not positive-definite?
 
George Jones said:
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\}##
 
nateHI said:
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.
 
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.
 
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.
 
  • Like
Likes   Reactions: QuantumQuest
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.
 
OK, I think I get it now. Thanks everyone!
 
  • #10
DrGreg said:
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.
 
  • #11
As far as I understand it a given metric in general relativity could be valid for more than one topology.
 

Similar threads

  • · Replies 39 ·
2
Replies
39
Views
3K
  • · Replies 25 ·
Replies
25
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K