7
$\begingroup$

It is not clear for me why a positive definite metric is necessary to define a topology as noted in some textbooks like the one by Carroll.

Does this imply that in cosmology, say through FLRW metric, we can only discuss the topology or global geometry of space, or spatial hypersurface, instead of spacetime?

Also related to this question is that we know there "exists" a coordinate system in which the pseudo-Riemannian metric in GR becomes, locally, a Lorentzian one, thus having canonical signature - + + +.

In FLRW metric we assume an isotropic and homogeneous cosmos based on observation in the, well, "observable" universe.

But how about breaking this assumption and imagine that a global Riemannian metric or coordinates system "exists" for spacetime and only demand that it locally becomes Lorentzian?

Which part of my understanding is correct and which one incorrect?

$\endgroup$
1
  • 1
    $\begingroup$ The claim is probably more accurately that pseudo-metrics cannot induce anything useful by considering balls (the standard way of inducing the usual topology on $\mathbb{R}^n$ from the usual metric). See for example here. $\endgroup$
    – user10851
    Jan 11, 2016 at 1:21

1 Answer 1

15
$\begingroup$

It is simply false, at least written as it stands.

The point is that the relation between the topology and the metric is more complicated than in the Riemannian case, where the geodesical balls form a basis of the topology$^1$.

As a matter of fact, a (connected) Lorentzian smooth metric $g$ over the time-oriented smooth manifold $M$ does define a topology, the same already present on $M$ if the spacetime is strongly causal.

Fix a point $p\in M$ and consider all (smooth) timelike future-directed curves through $p$ and denote by $L(\gamma)$ the Lorentzian length of $\gamma= \gamma(\xi)$, $\xi \in [a,b]$. $$L(\gamma) = \int_a^b \sqrt{|g(\dot{\gamma},\dot{\gamma})|} d \xi$$ If $q \in M$ define the so called Lorentzian distance of $q$ from $p$ as $$\tau(q,p) := \sup \{L(\gamma) \:|\: \mbox{$\gamma$ timelike future-directed from $p$ to $q$}\}$$ If no timelike future-directed from $p$ to $q$ exists, $\tau(q,p) :=0$.

Next define $I^+(p) := \{q \in M \:|\: \tau(q,p)>0\}$ and $I^-(p) := \{q \in M \:|\: \tau(p,q)>0\}$.

It is possible to prove that the family of sets $I(p,q):= I^+(p) \cap I^-(q)$ (with the suitable chronological order of the arguments) is a basis of the topology of $M$ if the spacetime is strongly causal [Kronheimer and Penrose (1967)]. (Strongly causal means that every open neighborhood $U$ of every event $p\in M$ includes another open neighborhood $V$ of $p$ such that $J^+(r) \cap J^-(s) \subset V$ if $r,s \in V$, Minkwski spacetime and all globally hyperbolic spacetimes like Kruskal's one are strongly causal.)

This is a well known result of semi-Riemannian geometry (Theorem 4.9 in Global Lorentzian Geometry second edition 1996 by J.K. Beem, P.E. Ehrlich, K.L. Easley)

ADDENDUM. To answer to a comment to my answer, in view of the quoted result, the topology induced by sets $I(p,q)$ is metrizable since it coincides with the natural topology of the manifold $M$ viewed as a smooth manifold regardless any (semi)-Riemannian structure thereon, which is metrizable. In particular, if $M$ is Minkowski spacetime a distance producing the said topology can be constructed explicitly: $$d((t,\vec{x}), (t',\vec{x}')) = ||\vec{x}-\vec{x}'||+c|t-t'|$$ The balls of this distance are evidently the sets $I((t,\vec{x}),(t',\vec{x}'))$. This distance evidently depends on the choice or the Minkowkian reference frame.


(1) In a connected Riemannian manifold $M$ whose metric is denoted by $g$, $d(p,q) = \inf \{ L(\gamma) \:|\: \gamma \mbox{ smooth curve joining $p$ and $q$}\}$ where $$L(\gamma) := \int_a^b \sqrt{g(\dot{\gamma},\dot{\gamma})} d \xi$$ is a distance making $M$ a metrical space. All the open balls $B_\delta(p):=\{ q\in M \:|\: d(p,q)<\delta\}$, varying $p\in M$ and $\delta \in (0,+\infty)$, form a basis of the topology already present in $M$.

$\endgroup$
5
  • $\begingroup$ Comments are not for extended discussion; this (now obsolete) conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Apr 11, 2017 at 15:10
  • $\begingroup$ Beautiful answer. My question is then: if such topology in general(not only for the strongly causal spacetimes) metric? $\endgroup$ Aug 16, 2023 at 7:39
  • $\begingroup$ @Bastam Tajik As I pointed out in my footnote, every (connected) smooth manifold $M$ admits a Riemannian metric $h$. ($h$ is evidently different from any Lorentzian metric $g$ already defined on $M$.) This Riemannian metric defines a metric distance $d_h(p,q)= \inf \{L_h(\gamma) \:|\: \gamma \: \mbox{joins} \:p \:\mbox{and} q\}$. This metric distance always induces the same topology already present on the manifold $M$ in its own right. There are however infinitely many Riemannian metrics $h$ and associated distances $d_h$ compatible with the topology of $M$. $\endgroup$ Aug 16, 2023 at 9:20
  • $\begingroup$ @ValterMoretti I meant if the topology induced by the Lorentzian metric is metric. $\endgroup$ Aug 16, 2023 at 10:31
  • $\begingroup$ For strongly causal spacetimes it is, otherwise I do not know, sorry. $\endgroup$ Aug 16, 2023 at 12:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.