# Topology on flat space when a manifold is locally homeomorphic to it

• B

## Main Question or Discussion Point

[I urge the viewer to read the full post before trying to reply]

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.

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Slightly more elaboration on my thought process: 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.

Related Special and General Relativity News on Phys.org
Dale
Mentor
To define open balls we need to specify a metric
Hmm, I don’t think that this is generally correct. Open balls don’t need to be spherical, just connected. At least that is my understanding.

Edit: yes, it is not necessary to define a metric on a topological space to define an open ball. All that is necessary is that there must be a homeomorphism between the open ball in the topological space and an open ball in Rn

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• weirdoguy
DrGreg
Gold Member
this could be tricky to define since Minkowski metric isn't positive definite
I'm no expert in this area but I think that is the problem, and why, within each chart, we use the Euclidean-related metric not the Minkowski metric tensor. In the language of metric spaces, the metric tensor isn't a metric. (It should really be called the "pseudo-metric tensor", but no-one ever does.)

In what follows, I'm going to use the word "metric" in the "metric space" sense, so positive definite. I.e. I'm not referring to the "metric tensor".

Open balls don’t need to be spherical, just connected. At least that is my understanding.
I think the term "ball" does imply "spherical" (relative to the chosen metric), but the open sets of a topological space do not have to be balls, indeed there need not be any metric at all.

The definition of topological manifold given in the opening post specifies that each chart has a metric associated with it that is induced from the Euclidean metric on ##\mathbb{R}^4##. (In other words, a coordinate-dependent metric.) Where two charts overlap, you'll have two different metrics, so to establish that the definition makes sense, you'll need to verify that the two different metrics generate the same topology. I.e. the topology is coordinate-independent.

Caveat: I'm relying on my memory of metric spaces and topological spaces when I studied them 40 years ago at university. I've never had to use them since.

PeterDonis
Mentor
2019 Award
is assuming Euclidean metric necessary?
Assuming a Euclidean metric on ##\mathbb{R}^n## for this particular purpose is, yes. Note that the metric assumed on ##\mathbb{R}^n## in order to define the topology for purposes of constructing a coordinate chart on spacetime is not the same as the metric on spacetime.

I imagine this could be tricky to define since Minkowski metric isn't positive definite.
Not "tricky", impossible. As already noted, the Minkowski "metric" is not actually a metric, precisely because it is not positive definite. In other words, it is impossible to define a topology on spacetime using the Minkowski metric. So the topology on spacetime is defined by constructing coordinate charts whose topology is defined by the Euclidean metric on ##\mathbb{R}^4##, and then using the one-to-one correspondence between the coordinate charts and points in the spacetime to define the topology on the spacetime.

spacetime intervals (a measure of closeness of Minkowski spacetime points)
No, that's not what spacetime intervals are. They can't be, because the spacetime interval can be zero between distinct points in spacetime, and that property means the spacetime interval cannot define a topology. A topology can only be defined using a metric for which a zero "distance" is not possible between distinct points.

Where two charts overlap, you'll have two different metrics, so to establish that the definition makes sense, you'll need to verify that the two different metrics generate the same topology. I.e. the topology is coordinate-independent.
Yes.

• Shirish
robphy
Homework Helper
Gold Member
Here are some possibly useful old posts of mine with starting points into the literature on "spacetime topology":
UPDATE:
Here's a dissertation from a student in Fay Dowker's group (which works on https://en.wikipedia.org/wiki/Causal_sets ) that might be useful:
https://www.imperial.ac.uk/media/im...tions/2011/Guccione-Giovanni-Dissertation.pdf

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• • Shirish, Dale and PeterDonis
Dale
Mentor
I think the term "ball" does imply "spherical" (relative to the chosen metric), but the open sets of a topological space do not have to be balls, indeed there need not be any metric at all.
Yes, I should have been clear. I was talking about the open balls in the topological space where there isn’t even a metric that you could use to distinguish between a sphere and a cube, for example.

• DrGreg
Assuming a Euclidean metric on ##\mathbb{R}^n## for this particular purpose is, yes. Note that the metric assumed on ##\mathbb{R}^n## in order to define the topology for purposes of constructing a coordinate chart on spacetime is not the same as the metric on spacetime.

No, that's not what spacetime intervals are. They can't be, because the spacetime interval can be zero between distinct points in spacetime, and that property means the spacetime interval cannot define a topology. A topology can only be defined using a metric for which a zero "distance" is not possible between distinct points.
Thanks! So I was confusing two different notions. Here's what I understood: 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. The Minkowski metric is instead associated with a different structure (inner product space) on the spacetime manifold, whose purpose is to ensure that the required physical symmetry laws are obeyed (in this case the invariance of spacetime intervals).

Sound alright or are there any points that I got wrong in the above?

PeterDonis
Mentor
2019 Award
the metric is (in the case of standard topology) used to define open sets
More precisely, the Euclidean metric on ##\mathbb{R}^4## is used to define open sets in ##\mathbb{R}^4##. Open sets in spacetime are then defined by using a coordinate chart, which is a mapping between points in ##\mathbb{R}^4## and points in spacetime.

Effectively then, to state whether or not the worldline of a particle is continuous, we're ignoring the Minkowski structure of flat spacetime.
Yes. Or, to put it another way, the topology we use on spacetime is not induced by the metric (more precisely, pseudo-metric) we use on spacetime.

• vanhees71 and Shirish
Yes. Or, to put it another way, the topology we use on spacetime is not induced by the metric (more precisely, pseudo-metric) we use on spacetime.
The Minkowski metric is instead associated with a different structure (inner product space) on the spacetime manifold, whose purpose is to ensure that the required physical symmetry laws are obeyed (in this case the invariance of spacetime intervals).
Any revision(s) required in the above quote? (My main takeaway is that there need not be a unique kind of metric on a space. Different metrics can be associated with different levels of structure imposed on a set.)

PeterDonis
Mentor
2019 Award
Any revision(s) required in the above quote?
The invariance of spacetime intervals is not a "physical symmetry law".

Note that flat Minkowski spacetime is not the only possible spacetime; there are also curved spacetimes, as treated in General Relativity, and many of those have no symmetries at all, and even the ones that do do not (with a small number of exceptions) have all of the symmetries of Minkowski spacetime. But all spacetimes still have invariant intervals between events (more precisely, invariant arc lengths along curves between events, since in a curved spacetime there is not a unique "interval" between two events in general, such things are path dependent).

• vanhees71
The invariance of spacetime intervals is not a "physical symmetry law".

Note that flat Minkowski spacetime is not the only possible spacetime; there are also curved spacetimes, as treated in General Relativity, and many of those have no symmetries at all, and even the ones that do do not (with a small number of exceptions) have all of the symmetries of Minkowski spacetime. But all spacetimes still have invariant intervals between events (more precisely, invariant arc lengths along curves between events, since in a curved spacetime there is not a unique "interval" between two events in general, such things are path dependent).
Thanks for the correction. I just checked myself as well and realized that I wrongly associated the conservation of some quantity with the presence of some symmetry. Turns out they're not the same because while presence of symmetry implies the conservation of some quantity, the converse isn't necessarily true.

PeterDonis
Mentor
2019 Award
while presence of symmetry implies the conservation of some quantity, the converse isn't necessarily true.
Can you give an example of a conserved quantity that does not have a corresponding symmetry? Note that "conserved" and "invariant" are not the same; arc lengths along curves are invariant (the same in all coordinate systems), but there is no meaningful sense in which they are conserved.

Can you give an example of a conserved quantity that does not have a corresponding symmetry? Note that "conserved" and "invariant" are not the same; arc lengths along curves are invariant (the same in all coordinate systems), but there is no meaningful sense in which they are conserved.
No idea, but I was resuming to watch this video and there's this part.

Oh, just saw your edit. And wow there are a lot of nuances that I need to get my head around. My understanding has been that special relativity involves symmetries associated with the Poincare group, and the transformations in that group have the property of preserving spacetime intervals.

So in my head, the requirement of invariance of spacetime intervals is the reason we specify the signature of the metric as the Minkowski one, then transformations orthogonal w.r.t. this metric were specified, and these transformations in turn comprise Poincare symmetry.

PeterDonis
Mentor
2019 Award
special relativity involves symmetries associated with the Poincare group, and the transformations in that group have the property of preserving spacetime intervals
The group here is the group of coordinate transformations. The fact that those transformations preserve spacetime intervals is simply because spacetime intervals are observable quantities, and coordinate transformations can't change observable quantities.

When we talk about symmetries as they are related to conservation laws by Noether's theorem, we are talking about a different sense of the term "symmetry". For example, time translation symmetry of physical laws is related to conservation of energy. Similarly, space translation symmetry of physical laws is related to conservation of momentum, and rotational symmetry of physical laws is related to conservation of angular momentum.

All of the transformations just talked about--time translations, space translations, and rotations--are in the Poincare group, so it does happen to be the case that those transformations are also "symmetries" in the sense of Noether's theorem in Minkowski spacetime, as well as being "symmetries" in the sense of preserving spacetime intervals. But, as noted, this is two different senses of the term "symmetry". Spacetime intervals are not a conserved quantity in the sense of Noether's theorem, as energy, momentum, and angular momentum are (as I noted in a previous post, spacetime intervals are invariant even in spacetimes that do not have the symmetries of Minkowski spacetime, and therefore do not have the corresponding conservation laws).

• • etotheipi, vanhees71 and Shirish
Effectively then, to state whether or not the worldline of a particle is continuous, we're ignoring the Minkowski structure of flat spacetime. The Minkowski metric is instead associated with a different structure (inner product space)
I believe you got a little confused by knowing in advance that you're going to consider a pseudo-Riemannian manifold. If we're going to define things in the usual order, we introduce a topological manifold first, and we know nothing about any metric (in the sense of metric tensor) yet. Topological manifold is locally homeomorphic to the usual real n-dimensional space with its usual topology (induced by the usual Euclidean distance). Only after that we can start thinking about differential geometry and metric tensors. Topological manifold becomes a Riemannian or pseudo-Riemannian manifold when we endow it with additional structure - a bilinear map on the tangent vectors at each point. It doesn't have anything to do with the Euclidean distance function we used to define topology.

George Jones
Staff Emeritus
Gold Member
The only Hausdoff topology on ##\mathbb{R}^n## for which the vector space operations of ##\mathbb{R}^n## are continuous is the standard Euclidean topology for ##\mathbb{R}^n##.

• Shirish
The group here is the group of coordinate transformations. The fact that those transformations preserve spacetime intervals is simply because spacetime intervals are observable quantities, and coordinate transformations can't change observable quantities.

When we talk about symmetries as they are related to conservation laws by Noether's theorem, we are talking about a different sense of the term "symmetry". For example, time translation symmetry of physical laws is related to conservation of energy. Similarly, space translation symmetry of physical laws is related to conservation of momentum, and rotational symmetry of physical laws is related to conservation of angular momentum.

All of the transformations just talked about--time translations, space translations, and rotations--are in the Poincare group, so it does happen to be the case that those transformations are also "symmetries" in the sense of Noether's theorem in Minkowski spacetime, as well as being "symmetries" in the sense of preserving spacetime intervals. But, as noted, this is two different senses of the term "symmetry". Spacetime intervals are not a conserved quantity in the sense of Noether's theorem, as energy, momentum, and angular momentum are (as I noted in a previous post, spacetime intervals are invariant even in spacetimes that do not have the symmetries of Minkowski spacetime, and therefore do not have the corresponding conservation laws).
Very helpful, thank you! Regarding the first paragraph though - it seems a bit vague. I'm sure there's a more precise notion of "observable quantities" that you're trying to get at, but one that I'm not familiar with.

As in, 3-force or space separation are also observable quantities (something that an observer can record), but they're changed by Lorentz transformations. I'm guessing by "observable quantity" you mean a scalar (since it isn't affected by coordinate transformations)?

Finally, I thought the spacetime intervals in particular held a privileged position because their invariance is assumed as a postulate in some explanations of SR. And to make that postulate hold, we go through all the rigmarole of defining the Minkowski metric and coming up with transformations orthogonal w.r.t. it, etc.

But yeah, I know better than to throw words (like symmetry) about in an imprecise manner. Thanks again for the detailed explanation!

vanhees71
Gold Member
2019 Award
Tensors are also invariant objects and thus (can) describe observables. What changes are the tensor components when transforming from one basis to another, not the tensors themselves. That's how you get the transformation laws for tensor components by definition.

The problem is that physicists often call the tensor components simply tensor, but that's confusing as you can see in this discussion.

• Dale, Shirish and etotheipi
Ibix
As in, 3-force or space separation are also observable quantities (something that an observer can record),
Are they? How do you measure distance? One way is with a radar set. But what you record is not distance but your own proper time, which you interpret as a distance with the assumption that you are stationary and the speed of light is isotropic. Another way is with a ruler. But then you need to make two measurements - the position of one end of your distance and the position of the other end - and you need to make those measurements simultaneously, in general, so you need synchronised clocks along with your ruler. You record two position measurements and times and convert these into a distance.

Of course, if I analyse either of these systems in another frame of reference I'll find that they don't measure distance. They measure some mix of distance and duration. What you and I will agree on is that they measure the interval along a spacelike line perpendicular to your worldline.

Similar remarks apply to force measurement. I'll assess your 3-force measurer as not a 3-force measurer, but a some-mix-of-0th-and-ith-component-of-4-force measurer. I will agree that you are measuring the inner product of the 4-force with some spacelike vector perpendicular to your worldline.

And that's the general point. Any time you think you are measuring a component of a vector or tensor, you are actually measuring the contraction of the tensor with some vectors/one-forms - typically either your four-velocity or a spacelike vector perpendicular to it that you call ##\vec x##, ##\vec y##, or ##\vec z##. We're often sloppy and talk about "measuring the ##x##-component of the electric field", but what we actually mean is that we measured the inner product of the Faraday tensor with our four velocity and our ##x## basis vector. The former is shorthand.

• PeterDonis and Shirish
Are they? How do you measure distance? One way is with a radar set. But what you record is not distance but your own proper time, which you interpret as a distance with the assumption that you are stationary and the speed of light is isotropic. Another way is with a ruler. But then you need to make two measurements - the position of one end of your distance and the position of the other end - and you need to make those measurements simultaneously, in general, so you need synchronised clocks along with your ruler. You record two position measurements and times and convert these into a distance.

Of course, if I analyse either of these systems in another frame of reference I'll find that they don't measure distance. They measure some mix of distance and duration. What you and I will agree on is that they measure the interval along a spacelike line perpendicular to your worldline.

Similar remarks apply to force measurement. I'll assess your 3-force measurer as not a 3-force measurer, but a some-mix-of-0th-and-ith-component-of-4-force measurer. I will agree that you are measuring the inner product of the 4-force with some spacelike vector perpendicular to your worldline.

And that's the general point. Any time you think you are measuring a component of a vector or tensor, you are actually measuring the contraction of the tensor with some vectors/one-forms - typically either your four-velocity or a spacelike vector perpendicular to it that you call ##\vec x##, ##\vec y##, or ##\vec z##. We're often sloppy and talk about "measuring the ##x##-component of the electric field", but what we actually mean is that we measured the inner product of the Faraday tensor with our four velocity and our ##x## basis vector. The former is shorthand.
Could you clarify just a bit more on how your comments apply to the spacetime interval? As in, what makes spacetime interval in particular an observable quantity

martinbn
Just to add, that the Euclidiean metric and norm are in a way not necessary. All norms on ##\mathbb R^n## are equivalent and give the same topology. Obviousely not all of them come from a dot product. Also you could start with ##\mathbb R## and its topology and give ##\mathbb R^n## the product topology.

• lavinia, vanhees71 and Dale
A.T.
As in, what makes spacetime interval in particular an observable quantity
Time-like spacetime intervals correspond to proper time.

• vanhees71
Time-like spacetime intervals correspond to proper time.
I'm very sorry if I sound rude, but I'm not at a level where I can understand one-line answers. I should've prefaced my question by saying that I'm not an expert at the subject.

What I'm saying is, you'll need to spell it out (for a beginner like me) how the quoted line implies that spacetime interval is an observable quantity vis-a-vis, say, spacetime separation, spacial separation, 3-force, etc. (On the plus side, that'll also benefit any future readers viewing this thread.)

Ibix
What I'm saying is, you'll need to spell it out (for a beginner like me) how the quoted line implies that spacetime interval is an observable quantity vis-a-vis, say, spacetime separation, spacial separation, 3-force, etc. (On the plus side, that'll also benefit any
Let's say you make a measurement. You then analyse your experiment and use theory to predict the reading on your instrument. I do the same, but using a different reference frame. Any workable theory must predict the same result according to both analyses: the instrument reads only one thing and a theory that makes different predictions from different frames is wrong. What is the only thing that is the same in all frames in a tensor theory? A scalar - the result of a calculation with no free indices.

Thus anything which you can measure must correspond to a Lorentz scalar - or else the theory is broken. Proper time, the integral of ##\sqrt{g_{\mu\nu}U^\mu U^\nu}## along your worldline, is one such. So is the distance measurement we talked about. But the difference with your distance measurement is that it involves a spacelike basis vector - and the basis vectors aren't anything physical, and I use different ones from you. Hence if we inform each other of what basis vectors we are using then we can both calculate the other's measure of distance. But if we just talk naively about "distance" then you naturally use your basis vector in your calculation and I naturally use mine and we get different results.

• • Shirish and vanhees71
Let's say you make a measurement. You then analyse your experiment and use theory to predict the reading on your instrument. I do the same, but using a different reference frame. Any workable theory must predict the same result according to both analyses: the instrument reads only one thing and a theory that makes different predictions from different frames is wrong. What is the only thing that is the same in all frames in a tensor theory? A scalar - the result of a calculation with no free indices.

Thus anything which you can measure must correspond to a Lorentz scalar - or else the theory is broken. Proper time, the integral of ##\sqrt{g_{\mu\nu}U^\mu U^\nu}## along your worldline, is one such. So is the distance measurement we talked about. But the difference with your distance measurement is that it involves a spacelike basis vector - and the basis vectors aren't anything physical, and I use different ones from you. Hence if we inform each other of what basis vectors we are using then we can both calculate the other's measure of distance. But if we just talk naively about "distance" then you naturally use your basis vector in your calculation and I naturally use mine and we get different results.
Perfectly clear! Thanks as usual for the awesome explanation

• Ibix