Simple GR Question

  • I
  • Thread starter nateHI
  • Start date
  • #1
146
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:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,519
1,307
What is "the metric topology" when ##g## is not positive-definite?
 
  • #4
146
4
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\}##
 
  • #5
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,519
1,307
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.
 
  • #6
146
4
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.
 
  • #7
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
2021 Award
17,670
7,632
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 QuantumQuest
  • #8
DrGreg
Science Advisor
Gold Member
2,360
1,272
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.
 
  • #9
146
4
OK, I think I get it now. Thanks everyone!
 
  • #10
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,501
10,254
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
2,000
5
As far as I understand it a given metric in general relativity could be valid for more than one topology.
 

Related Threads on Simple GR Question

  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
825
Replies
8
Views
3K
  • Last Post
Replies
16
Views
4K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
12
Views
3K
Replies
1
Views
716
  • Last Post
Replies
15
Views
3K
Top