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I'm watching Schuller's lectures on gravitation on youtube. It's mentioned that spacetime is modelled as a topological manifold (with a bunch of additional structure that's not relevant to this question).

A topological manifold is a set $M$ with a topology $\mathcal{O}_M$ such that each point in $M$ is covered by a chart $(U,x)$, where $U\in\mathcal{O}_M$ and $x:U\to x(U)\subset\mathbb{R}^n$ is a homeomorphism. To even talk about the map $x$ being homeomorphic, we need to be able to talk about open sets in, and hence a topology on, $\mathbb{R}^n$.

The instructor mentions (see here) that $\mathbb{R}^n$ is considered to have standard topology. Standard topology is defined on the basis of open balls around points in $\mathbb{R}^n$. To define open balls we need to specify a metric on $\mathbb{R}^n$, and the definition of open balls in lecture 1 of the series was given assuming a Euclidean metric on $\mathbb{R}^n$, i.e., $$B_r(p)=\{q\in\mathbb{R}^n\ |\ \|p-q\|_E<r\}$$ where $\|\cdot\|_E$ is the Euclidean norm.

So I wonder, is assuming Euclidean metric necessary? I've heard that curved spacetime is modeled as a manifold that locally looks like flat spacetime, which is modeled as Minkowski space as far as I know, which in turn has the Minkowski metric.

If that's the case, then charts on curved spacetime are locally homeomorphic to open sets in Minkowski space. Would we have to define the topology on $\mathbb{R}^4$ as a variant of the standard topology in which open balls are defined as per the Minkowski metric? i.e. $$B_r(p)=\{q\in\mathbb{R}^4\ |\ \|p-q\|_M<r\}$$ where $\|\cdot\|_M$ is the Minkowski norm corresponding to metric signature $\text{diag}(-1,1,1,1)$. I imagine this could be tricky to define since Minkowski metric isn't positive definite.


Slightly more elaboration on my thought process (thanks to mike stone for this): The topology is what decides the "closeness" of points in a set as far as I know. So essentially when we're approximating a small patch of curved spacetime by the flat Minkowski spacetime, if we're assuming standard topology characterized by the Euclidean metric, what we're saying is: the Euclidean metric decides the closeness of points in (locally approximated) Minkowski space.

This sounds contradictory because physical considerations scream at us that spacetime intervals (a measure of closeness of Minkowski spacetime points) are measured using the Minkowski metric.

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  • $\begingroup$ My conjecture is that there should be some way to obtain the standard topology using only the Minkowski metric, but I suspect it won't be a straightforward way. $\endgroup$ Jun 29, 2020 at 10:56
  • $\begingroup$ @MaximalIdeal: That's what I'm suspecting too, but Minkowski metric not being positive definite probably makes it really tricky (if at all possible) $\endgroup$ Jun 29, 2020 at 10:57

2 Answers 2

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  1. A pseudo-Riemannian manifold $(M,g)$ of signature $(r,s)$ is a differentiable manifold $M$ of dimension $n=r+s$ equipped with a metric tensor $g\in\Gamma({\rm Sym}^2T^{\ast}M)$ of signature $(r,s)$.

  2. A differentiable manifold $M$ is a topological manifold with a globally defined differential structure.

  3. A topological manifold $M$ of dimension $n$ is a locally Euclidean Hausdorff space, i.e. every point $p\in M$ has a neighbourhood which is homeomorphic to $\mathbb{R}^n$.

  4. Notice in particular that the underlying topological manifold $M$ is defined independently of the metric tensor $g$ (and its signature, causal structure, curvature, etc).

  5. Also one should not conflate a metric $d:M\times M\to [0,\infty[ $ in a metric space (within the framework of topological spaces and general topology) with a metric tensor $g$.

  6. If we try to use a metric tensor $g$ of indefinite signature to construct a metric $d$ from geodesic distance, it would for starters violate the Hausdorff property and possibly the non-negativity of $d$.

  7. For a Lorentzian manifold $(M,g)$, diamond sets of the form $$ I^+(p)\cap I^-(q) , \qquad p,q\in M, $$ and their finite intersections generate all open sets $\{G\subseteq M \mid G\in\tau\}$ for the underlying locally Euclidean topology $\tau$. Here $I^{\pm}(p)$ is the chronological future/past of the point $p\in M$, respectively.

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  • $\begingroup$ I agree that a topological manifold, in general, doesn't necessarily need a metric to be defined. But I'm specifically talking about the standard topology on $\mathbb{R}^n$ here, in which open sets are defined as open balls. An open ball, by definition, needs the underlying $\mathbb{R}^n$ to be equipped with a Euclidean metric. $\endgroup$ Jun 29, 2020 at 10:51
  • $\begingroup$ Your fifth point seems to tackle the question. Could you elaborate on that, if possible? The metric is (in the case of standard topology) used to define open sets, which in turn are used to specify continuity of maps. So a curve $\gamma:\mathbb{R}\supset I\to\mathbb{R}^4$, which physically corresponds to the worldline of a particle, can be called continuous / non-continuous based on standard topology, which assumes a Euclidean metric. Effectively then, to state whether or not the worldline of a particle is continuous, we're ignoring the Minkowski structure of flat spacetime. Is that correct? $\endgroup$ Jun 29, 2020 at 11:49
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$
    – Qmechanic
    Jun 29, 2020 at 11:53
  • $\begingroup$ Awesome! My main takeaway so far is that I was wrong to assume that there can only be one metric on a set (spacetime in this case). There can be different metrics corresponding to different structures on it. And each structure serves a different purpose. Topology (structure for which Euclidean is used) serves the purpose of us being able to talk about continuity of curves. Inner product space (structure for which Minkowski is used) serves the purpose of us enforcing that spacetime interval is invariant. That's the rough idea I have. Looking forward to your update! $\endgroup$ Jun 30, 2020 at 8:18
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Jun 30, 2020 at 10:49
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The topology on the mathematical model of spacetime used in general relativity is the standard topology of ${\mathbb R}^4$ that is induced from the usual Euclidean metric on ${\mathbb R}^4$. It is not a topology induced from the Minkowski metric.

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  • $\begingroup$ Could you elaborate on why we don't use topology induced from Minkowski metric? Using the Minkowski metric for one purpose and Euclidean for another seems inconsistent to a beginner like me (even though I'm sure there must be a good reason for this which is obvious to an expert). So whenever you get time, I'd be grateful for an (as detailed as possible) explanation on how we can get away with using different metrics for different purposes. $\endgroup$ Jun 29, 2020 at 9:59
  • $\begingroup$ If we try to use the Minkowski metric to induce a topology, we would have to identify (i,e regard as the same) all points that are light-like separated. I imagine that the resulting topology would not be of much use. The mathematics is not there to describe reality. It is a mathematical model of reality You can't for example learn anything about spacetime by proving some theorem about the real numbers, but using ${\mathbb R}$ as a model turns out to be useful for many purposes. $\endgroup$
    – mike stone
    Jun 29, 2020 at 10:09
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    $\begingroup$ Sorry if I'm being a pain (I'm sure I am), but the topology is what decides the "closeness" of points in a set as far as I know. So essentially when we're approximating a small patch of curved spacetime by the flat Minkowski spacetime, if we're assuming standard topology characterized by the Euclidean metric, what we're saying is: the Euclidean metric decides the closeness of points in Minkowski space. This sounds contradictory because physical considerations scream at us that spacetime intervals are measured using the Minkowski metric. I'll edit the question to include this too. Thanks! $\endgroup$ Jun 29, 2020 at 10:43
  • $\begingroup$ The Minkowski metric decides how light and other waves propagate; the spacetime point-set topology decides what functions are regarded as continuous. Wave equations can have discontinuous solutions. $\endgroup$
    – mike stone
    Jun 29, 2020 at 12:16
  • $\begingroup$ What do you mean when you say the "Minkowski metric decides how light and other waves propagate"? (It's not immediately obvious to me how that's so) $\endgroup$ Jun 29, 2020 at 12:22

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