Is the Lorentz metric compatible with the topology of flat spacetime?

[dslowik]

Could someone clarify, and/or point me to some reference on:
Lorentz metric is not really a metric in the sense of metric spaces of a topology course since it admits negative values. If I use it to define the usual open sphere about a point, that sphere includes the entire light-cone through that point. so space time is not even Hausdorff?

 

[robphy]

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

 

[dslowik]

Is the topology of (flat) ST generated by a metric? i.e. is spacetime a metric space? If so, what is the metric? (Lorentz 'metric' goes negative -not a metric! is my confusion)

 

[yenchin]

Does this help?

https://www.physicsforums.com/showthread.php?t=199427

 

[PAllen]

Is the topology of (flat) ST generated by a metric? i.e. is spacetime a metric space? If so, what is the metric? (Lorentz 'metric' goes negative -not a metric! is my confusion)

Are you confusing topology and geometry? A topological manifold doesn't have a metric at all. As for being a metric space, technically, neither SR flat spactime nor GR spacetime are Riemanian metric spaces. Instead, they are semi-Riemannian or pseudo-Riemannian; specifically, this allows a metric signature other than ++++.

 

[dslowik]

Right, a topological space may not have a metric.
A metric is additional structure on a set.
A set with a metric gives rise to a metric topology on that set. (I mean usual non-neg, symmetric, non-deg metric).

I am reading John Lee's Topological Spaces. There he defines a top manifold as:
1) topological space locally homeomorphic to R^n, 2) Hausdorff & 3) second countable.
So, assuming ST is a topological space (by this def), can't we use the local homeomorphism to R^n to define (locally) a metric on ST via pullback of the Euclidean metric on R^n? And likewise any topological manifold is a (local)metric space (but not general topological spaces).

It seems that ST is a topological manifold with a locally Euclidean metric. This describes its topological structure as a metric space. We than add further structure to this metric/topological space by adding the non-Riemanian Lorentz metric. Thus we are using one metric and corresponding open balls to describe the topology, and another metric to describe the 'physical' distance between points. the physical distance between some points is 0, which is a very different topology than the locally Euclidean one; but the Lorentz metric can't be used to describe open balls for a topology?

as Chris Hillman said:
"Lorentzian metrics get their topology from the (locally euclidean) topological manifold structure, not from the bundled indefinite bilinear form."
But it seems odd to me that we impose a locally euclidean topology, then use a quite different metric to describe physical separation of points..