Levi-Civita connection and pseudoRiemannian metric

[TrickyDicky]

One of the properties of the unique Levi-Civita connection is that it preserves the metric tensor at each point's tangent space, allowing the definition of invariant intervals between points in the manifold. I'd be interested in clarifying: when the metric preserved by the L-C connection is a PseudoRiemannian metric, the PseudoRiemannian manifold is a:
-pseudometric space
-metric space
-none of the above

 

[micromass]

Every manifold (with or without pseudo-Riemannian metric or connections) is a metric space. So on every manifold you can define a metric such that the topology coincide with the metric topology.

This has nothing to do with the pseudo-Riemannian metric.

But maybe you have something else in mind?

 

[TrickyDicky]

Every manifold (with or without pseudo-Riemannian metric or connections) is a metric space. So on every manifold you can define a metric such that the topology coincide with the metric topology.

This has nothing to do with the pseudo-Riemannian metric.

But maybe you have something else in mind?

Ok, but the fact is that thru the unique connection in GR spacetimes a particular structure of distance functions between events is developed specially for null curves and geodesics, based on the light cone structure defined by the pseudoriemmanian metric in the space tangent at each event, very different from the distance functions in metric spaces and from which very important physical consequences are derived.

 

[WannabeNewton]

One of the properties of the unique Levi-Civita connection is that it preserves the metric tensor at each point's tangent space, allowing the definition of invariant intervals between points in the manifold.

Invariant intervals have nothing to do with the Levi-Civita connection. This is a trivial property of the metric tensor in and of itself. The Levi-civita connection preserves inner products along arbitrarily extended curves when the vectors in the inner product are parallel transported, which is much stronger. Note the inner product is not preserved along arbitrary curves if the transport is done differently e.g. Lie transport will not preserve the inner product, even if we substitute in the Levi-Civita connection, unless the curve is an integral curve of a Killing field.

I'd be interested in clarifying: when the metric preserved by the L-C connection is a PseudoRiemannian metric, the PseudoRiemannian manifold is a:
-pseudometric space
-metric space
-none of the above

All topological manifolds are metric spaces as has already been stated. This has nothing to do with your previous statements.

 

[TrickyDicky]

I know every manifold is metrizable, you are missing my point.
I'm trying to get an understabding on how Exactly the distance functions in spacetime are compatible with the metric space structure.
How for instance a lightlike interval, or an arbitrary null Curve in spacetime M fits in the requurements of the distance functions of metric spaces?

 

[TrickyDicky]

To be more specific, how is the distance in a lightlike geodesic that can be zero between 2 distinct points can be defined if the spacetime is a metric space in which the distance between 2 distinct points cannot be zero?

 

[TrickyDicky]

Invariant intervals have nothing to do with the Levi-Civita connection. This is a trivial property of the metric tensor in and of itself.

You need a connection preserving the metric tensor if you want to use properties of the metric(trivial or not) between points in a general spacetime, don't you?

I mean, the lightcone structure is only defined locally at a point's tangent space or is also defined globally in a curved lorentzian manifold ?(I leave out Minkowski spacetime since it is itself a vector space).

 

[micromass]

Do you mean to define a distance function

d(x,y) = \textrm{inf}\{\textrm{Length}(\gamma)~\vert~\gamma:[a,b]\rightarrow M~\text{smooth and}~\gamma(a) = x,~\gamma(b)=y\}

for $x,y\in M$? And whether $M$ becomes a (pseudo-)metric space under this $d$?

The definition and properties of $d$ still don't require a connection though.

 

[TrickyDicky]

Do you mean to define a distance function

d(x,y) = \textrm{inf}\{\textrm{Length}(\gamma)~\vert~\gamma:[a,b]\rightarrow M~\text{smooth and}~\gamma(a) = x,~\gamma(b)=y\}

for $x,y\in M$? And whether $M$ becomes a (pseudo-)metric space under this $d$?

The definition and properties of $d$ still don't require a connection though.

I'm asking if the pseudoRiemannian metric defines some distance function in a pseudoRiemannian manifold. I'm only mentioning the connection because it preserves the metric tensor.

 

[micromass]

I'm asking if the pseudoRiemannian metric defines some distance function in a pseudoRiemannian manifold. I'm only mentioning the connection because it preserves the metric tensor.

Can you tell us the distance function you're considering?

 

[DrGreg]

To be more specific, how is the distance in a lightlike geodesic that can be zero between 2 distinct points can be defined if the spacetime is a metric space in which the distance between 2 distinct points cannot be zero?

I think you are confusing "metric space" and "metric tensor". They are separate concepts which just happen to share the same name "metric". As I understand it, the metric space structure of a manifold is the topology inherited via coordinate charts from Euclidean \mathbb{R}^4. It's needed to define "continuity", "differentiability", "smoothness", etc, but isn't used for distance measurements via the metric tensor.

 

[micromass]

I think you are confusing "metric space" and "metric tensor". They are separate concepts which just happen to share the same name "metric". As I understand it, the metric space structure of a manifold is the topology inherited via coordinate charts from Euclidean \mathbb{R}^4.

You mean the right thing, but I want to be pedantic here. A manifold has (a priori) only a topological structure (and a smooth structure which is not interesting here). Not every topological space induces a metric space, but in a manifold that is the case. So we can show that there exists a distance function such that the topology generated by that distance function (meaning, the topology generated by the open balls $\{x\in M~\vert~d(x,a)<r\}$) is the original manifold topology.

I can already see this thread is going to be confusing because the word metric is used in two senses. So, I propose the following terminology:

1) A metric always means a metric tensor.
2) A distance function on a set $X$ is a function $d:X\times X\rightarrow \mathbb{R}$ such that $d(x,y)=0$ if and only if $x=y$, $d(x,y) = d(y,x)$ and $d(x,z)\leq d(x,y) + d(y,x)$. (Analogous definitions for pseudo-distance function)
3) A distance space is a pair $(X,d)$ where $X$ is a set and $d$ is a distance function on $X$. (Analogous definition for pseudo-distance space).

So I propose never to use the word "metric space" in this thread, or never to use the word "metric" when it actually means a distance function.

So the OP asks us whether a pseudo-Riemannian manifold is
a) A distance space
b) A pseudo-distance space
c) Neither

The question that I want to ask for the OP is which distance function $d$ he is considering concretely.

 

[TrickyDicky]

Can you tell us the distance function you're considering?

I said it ds^2=g_{\mu\nu} dx_\mu dx_\nu in a curved pseudoRiemannian manifold with g_{\mu\nu} a pseudoRiemannian metric tensor, allowing distances along curves to be determined through integration, in the specific case of a null curve.

I'm just trying to clear up the differences wrt the Riemannian case of the pseudoriemannian generalization and how it would affect if at all the metric space structure of the Riemannian case, because when M is any connected Riemannian manifold, then we can turn M into a metric space by defining the distance of two points as the infimum of the lengths of the paths (continuously differentiable curves) connecting them, but I wonder how can we define this for null curves in a curved pseudoRiemannian manifold.

 

[micromass]

I said it ds^2=g_{\mu\nu} dx_\mu dx_\nu in a curved pseudoRiemannian manifold with g_{\mu\nu} a pseudoRiemannian metric tensor, allowing distances along curves to be determined through integration, in the specific case of a null curve.

I'm just trying to clear up the differences wrt the Riemannian case of the pseudoriemannian generalization and how it would affect if at all the metric space structure of the Riemannian case, because when M is any connected Riemannian manifold, then we can turn M into a metric space by defining the distance of two points as the infimum of the lengths of the paths (continuously differentiable curves) connecting them, but I wonder how can we define this for null curves in a curved pseudoRiemannian manifold.

OK, so you don't have a specific distance function in mind. Your question is whether we can define a suitable (pseudo)-distance function from the metric tensor in the way that we can do this in Riemannian manifolds. Right?

 

[TrickyDicky]

OK, so you don't have a specific distance function in mind. Your question is whether we can define a suitable (pseudo)-distance function from the metric tensor in the way that we can do this in Riemannian manifolds. Right?

Exactly.

 

[WannabeNewton]

You need a connection preserving the metric tensor if you want to use properties of the metric(trivial or not) between points in a general spacetime, don't you?

No. The metric is defined at all points of space-time and is completely independent of a connection. A connection (in this setting defined on sections of tangent bundles) is a way to transport vectors from one point to another along curves and is independent of a metric. They have two completely unrelated roles.

I mean, the lightcone structure is only defined locally at a point's tangent space or is also defined globally in a curved lorentzian manifold ?(I leave out Minkowski spacetime since it is itself a vector space).

It's defined globally based on how it varies from point to point as the metric varies from point to point. The metric is a field it isn't just defined at a single point. It defines a different light cone at each point and if the space-time is time orientable then the light cone variation will be smooth. The local light cone structure is trivial so physics wouldn't be interesting in curved space-times if all we had was local light cone structure.

 

[TrickyDicky]

Thanks for your clarifications WN.

No. The metric is defined at all points of space-time and is completely independent of a connection. A connection (in this setting defined on sections of tangent bundles) is a way to transport vectors from one point to another along curves and is independent of a metric. They have two completely unrelated roles.

I'm only concerned here with the unique L-C connection, that is related to the metric only in the sense that it is the only one compatible with it for torsionless manifolds. But you are right, in this thread the connection has basically been a distracting element. I shouldn't have even mentioned it. Sorry about that.

It's defined globally based on how it varies from point to point as the metric varies from point to point. The metric is a field it isn't just defined at a single point. It defines a different light cone at each point and if the space-time is time orientable then the light cone variation will be smooth. The local light cone structure is trivial so physics wouldn't be interesting in curved space-times if all we had was local light cone structure.

Right. And my question was how do the distances obtained globally with the pseudoriemannian metric, specifically in the case of null curves get along with the manifold being a distance(metric) space?

 

[micromass]

Given a piecewise smooth curve $\alpha:[a,b]\rightarrow M$. We can define the length of the curve as

\int_a^b \sqrt{|g_{\alpha(t)}(\dot{\alpha}(t),\dot{\alpha}(t)|}dt

See for example O'Neill "Semi-Riemannian Geometry". Thing is that this will violate the triangle inequality.

 

[TrickyDicky]

Given a piecewise smooth curve $\alpha:[a,b]\rightarrow M$. We can define the length of the curve as

\int_a^b \sqrt{|g_{\alpha(t)}(\dot{\alpha}(t),\dot{\alpha}(t)|}dt

See for example O'Neill "Semi-Riemannian Geometry". Thing is that this will violate the triangle inequality.

And in the case of the null curve the identity of indescirnibles is also violated.

 

[micromass]

And in the case of the null curve the identity of indescirnibles is also violated.

Right. But that is something I feel is a minor problem. The triangle inequality is much worse.

 

[George Jones]

Thing is that this will violate the triangle inequality.

This is a good thing.

 

[TrickyDicky]

This is a good thing.

It might, but it seems to clash with considering the pseudoRiemannian manifold a metric space. Wich I don't know if it is a good, bad or indifferent thing to be honest.
This is stated in Wikipedia:"If one drops the positive definiteness requirement of inner product spaces, then one obtains a pseudo-Riemannian metric tensor, which integrates to a pseudo-semimetric distance function".
A pseudo-semimetric space is one where distances can be defined that violate the triangle inequality d(x,z) ≤ d(x,y) + d(y,z) and the identity of indescirnibles d(x,y) = 0 iff x = y, the main problem here is that these pseudo-semimetric spaces are not even Hausdorff which I think it's not good. So it would be interesting to know how exactly a pseudoRiemannian manifold gets away not really being a pseudo-semimetric space but a metric space in the end.
The only way I see is by appealing to the fact GR concentrates on the local topology and geometry. But this approach only preserves all the physics that is related with curvature, since curvature is a local invariant, and leaves out a lot of physics usually considered predicted by GR.

 

[George Jones]

We have been around and around this circle before. For example, consider this important comment from a thread just over a year ago:

So even without a smooth structure or a metric tensor, we already have that our manifold is metrizable. Again: the metric of the metric space might not be physical or might not have anything to do with a metric tensor!

 

[TrickyDicky]

We have been around and around this circle before. For example, consider this important comment from a thread just over a year ago:

"So even without a smooth structure or a metric tensor, we already have that our manifold is metrizable. Again: the metric of the metric space might not be physical or might not have anything to do with a metric tensor!"

We all agree the topological manifold is metrizable. The issue is how we deal with the distance functions we define from a pseudoRiemannian metric tensor.
You seem to be suggesting that pseudoRiemannian manifolds can be both metric and pseudo-semimetric space at the same time, I don't know if that is posible at all. But I think the fact the triangle inequality is violated obstructs the possibility of it being a metric space.

 

[micromass]

We all agree the topological manifold is metrizable. The issue is how we deal with the distance functions we define from a pseudoRiemannian metric tensor.
You seem to be suggesting that pseudoRiemannian manifolds can be both metric and pseudo-semimetric space at the same time, I don't know if that is posible at all. But I think the fact the triangle inequality is violated obstructs the possibility of it being a metric space.

So the PseudoRiemannian manifold is a metric space sure.
Now you have defined some distance function $d$ which makes the manifold into a pseudo-semi-metric space.

They key part is however that this distance function $d$ has nothing to do with the original topology on the manifold. It's not even sure that it is physically very important.

So the situation is much alike this. Consider the set $X=\{0,1\}$. Define the metric $d(x,y) = 1$ if $x\neq y$ and $d(x,y) = 0$ elsewhere.
However, we can also define the pseudometric $e(x,y) = 0$ for all $x,y$.
Are you surprised that the set $X$ can be made both in a metric and a pseudometric space? Of course not, since there is no connection between the $d$ and the $e$. Nothing say that they should be related.

The same thing happens with our manifold. Our pseudo-semi metric is just some function on the manifold. It is not clear to me why you expect this to be somehow related to the original topology of the manifold?

 

[TrickyDicky]

So the PseudoRiemannian manifold is a metric space sure.
Now you have defined some distance function $d$ which makes the manifold into a pseudo-semi-metric space.

They key part is however that this distance function $d$ has nothing to do with the original topology on the manifold. It's not even sure that it is physically very important.

So the situation is much alike this. Consider the set $X=\{0,1\}$. Define the metric $d(x,y) = 1$ if $x\neq y$ and $d(x,y) = 0$ elsewhere.
However, we can also define the pseudometric $e(x,y) = 0$ for all $x,y$.
Are you surprised that the set $X$ can be made both in a metric and a pseudometric space? Of course not, since there is no connection between the $d$ and the $e$. Nothing say that they should be related.

The same thing happens with our manifold. Our pseudo-semi metric is just some function on the manifold.

Ok, as I was saying I was not even aware the possibility existed that being metric or pseudometric spaces weren't mutually exclusive. It looks a bit arbitrary but that's just my subjective impression.

It is not clear to me why you expect this to be somehow related to the original topology of the manifold?

Well, not exactly, the Wikipedia reference talks about pseudometric spaces not being Hausdorff, that's all. Of course being Hausdorf for manifolds is an arbitrary requirement made by definition to avoid pathological cases, and I guess in retrospect that is done because all metric spaces are Hausdorff.

 

[TrickyDicky]

Also from Wikipedia:"The difference between pseudometrics and metrics is entirely topological."
I understand that when in doubt the original topology of the manifold, that is the usual topology, prevails.
What I don't understand is how so much physics is derived from the topology of the pseudo-semimetrics of pseudoRiemannian manifolds when it is mathematically unfounded if the manifold is a proper metric space.

 

[micromass]

Also from Wikipedia:"The difference between pseudometrics and metrics is entirely topological."
I understand that when in doubt the original topology of the manifold, that is the usual topology, prevails.
What I don't understand is how so much physics is derived from the topology of the pseudo-semimetrics of pseudoRiemannian manifolds when it is mathematically unfounded if the manifold is a proper metric space.

I'm not convinced at all that the topology of the pseudo-semimetric space is very applicable in physics at all!

And what do you mean that it's unfounded that the manifold is a proper metric space?

 

[WannabeNewton]

What I don't understand is how so much physics is derived from the topology of the pseudo-semimetrics of pseudoRiemannian manifolds when it is mathematically unfounded if the manifold is a proper metric space.

They have absolutely no relation to what we deal with in GR. The topologies we deal with are the manifold topology inherited from the smooth structure or Hawking's path topology. Other standard topologies exist but none of them have anything to do with the metric induced topology you're talking about.

 

[TrickyDicky]

And what do you mean that it's unfounded that the manifold is a proper metric space?

No, I meant the putative physics derived from the pseudometric geometric topology would be unfounded if the pseudoRiemannian manifold was a truly metric space.

But WN says all the GR physics is based on the standard manifold topology. I would have to look for examples where I think this would not be the case so that I can correct my judgment.

 

[TrickyDicky]

But actually nobody has attempted so far giving an explanation about how null geodesics with their distance ds=0 are compatible with the standard topology with d(x,y) = 0 iff x = y. If they need not be compatible it makes seem unnecessary to impose a metric space structure on the manifold, it's like anything goes.

 

[TrickyDicky]

They have absolutely no relation to what we deal with in GR. The topologies we deal with are the manifold topology inherited from the smooth structure or Hawking's path topology. Other standard topologies exist but none of them have anything to do with the metric induced topology you're talking about.

Are not null hypersurfaces for instance objects derived from the pseudoriemannian metric pseudometric properties and their induced topology?

 

[TrickyDicky]

This section from Kevin Brown's Reflections on relativity book(hopefully a valid source, these pages have been profusely commented in PF ) on mathpages http://mathpages.com/rr/s9-01/9-01.htm argues as evident that the physics of Minkowski spacetime (the simplest pseudoRiemannian manifold of Lorentzian signature, having vanishing curvature) comes from its particular pseudosemimetric induced topology, which he says it's not even a topological space topology.

The article also states :"It is, of course, possible to assign the Euclidean topology to Minkowski spacetime, but only by ignoring the non-transitive null structure implied by the Lorentz-invariant distance function. "
If one ignores this structure it seems to me that when we go to the case with curvature much of the physics related with this structure like event horizons and their associated black holes or conformal null infinity are left baseless.

 

[TrickyDicky]

Also extracted from mathpages in relation with a posible way of ignoring the topology induced by the Lorentzian pseudosemimetric that implies considering the spatial and temporal parts of spacetime as having each an absolute meaning by themselves:"In the case of a spacetime theory, we need to consider whether the temporal and spatial components of intervals have absolute significance, or whether it is only the absolute intervals themselves that are significant."
I see two problems. First of all this seems contrary to the spirit of the relativity theory, and besides in the general case it is only possible for static spacetimes, and we know our universe is not static even if it can be approximated partially by static models. What do you guys think?

 

[micromass]

But actually nobody has attempted so far giving an explanation about how null geodesics with their distance ds=0 are compatible with the standard topology with d(x,y) = 0 iff x = y. If they need not be compatible it makes seem unnecessary to impose a metric space structure on the manifold, it's like anything goes.

They're not compatible. In fact, while it is possible to impose a metric space structure on the manifold, the distance function will be very much unphysical. It will be completely useless in physics. So yes, all what matters is the topology and not the metric space structure.

 

[TrickyDicky]

They're not compatible.

That's my impression too.

In fact, while it is possible to impose a metric space structure on the manifold, the distance function will be very much unphysical. It will be completely useless in physics.

I tend to agree, but as I said it is routinely used in relativist literature (I gave the event horizon example) ever since the early works of Hawking, Penrose, Geroch...Of course Hawking just recently changed his mind about black holes so go figure...

So yes, all what matters is the topology and not the metric space structure.

Did you read the mathpages article? If so, do you find it accurate?

 

[WannabeNewton]

Are not null hypersurfaces for instance objects derived from the pseudoriemannian metric pseudometric properties and their induced topology?

It's derived from the pseudo-riemannian metric tensor properties. It has nothing to do with a pseudo-metric. The metric tensor only induces a topology in each tangent space. The topology we work with for space-time (usually) is the manifold topology of the smooth atlas. We never use metrics in GR, period. They're pointless in the GR context.

 

[TrickyDicky]

It's derived from the pseudo-riemannian metric tensor properties.

Agreed.

It has nothing to do with a pseudo-metric. The metric tensor only induces a topology in each tangent space.

Null hypersurfaces topology is not confined to the tangent space, they are submanifolds of M.

The topology we work with for space-time (usually) is the manifold topology of the smooth atlas.

Yes, usually, for instance when applied to anything related to curvature, but usually is not always.

We never use metrics in GR, period. They're pointless in the GR context.

That's a strong statement that doesn't seem right without qualifications. In this context you are saying "we never use distances in GR, they are pointless", are you sure?

 

[WannabeNewton]

Null hypersurfaces topology is not confined to the tangent space, they are submanifolds of M.

Where did I say otherwise? You talked about the topology induced by the metric tensor. This has nothing to do with null hypersurfaces of the subspace topologies of null hypersurfaces. The metric tensor only induces topologies in the tangent spaces.

That's a strong statement that doesn't seem right without qualifications.

Give one example of a metric being used in GR. Metrics are useful in QM not in GR.

In this context you are saying "we never use distances in GR, they are pointless", are you sure?

You're confusing different notions of distance. George already pointed this out.

 

[TrickyDicky]

Where did I say otherwise? You talked about the topology induced by the metric tensor. This has nothing to do with null hypersurfaces of the subspace topologies of null hypersurfaces. The metric tensor only induces topologies in the tangent spaces.

Forget the topology for a minute, I meant the obvious fact that without pseudoRiemannian metric tensor there is no null hypersurface at all, are you disagreeing with this?

Give one example of a metric being used in GR. Metrics are useful in QM not in GR.

Metric tensors can be thouth of as infinitesimal metrics, you just have to integrate them, as is done often in GR to get distances that accommodate or not the requirements of metric spaces. As pointed out in the linked article one can ignore those integrated distances, but that should be consistent with the physical predictions of GR one uses.

You're confusing different notions of distance. George already pointed this out.

Please, enumerate those different notions and tell me how exactly I confuse them so I can get corrected.

 

[micromass]

BTW, there are no metrics in QM, the spaces used thre are seminormed, not normed.

The main used spaces in QM seem to be $\mathbb{R}^n$, $\ell^2$ and $L^2(M)$ for some $M\subseteq \mathbb{R}^n$. How are these not normed.

 

[WannabeNewton]

I'm sorry but you're lacking a lot of analysis knowledge, especially if you think there are no norms or metrics in QM. You need to learn analysis first. This discussion is pointless.

 

[TrickyDicky]

The main used spaces in QM seem to be $\mathbb{R}^n$, $\ell^2$ and $L^2(M)$ for some $M\subseteq \mathbb{R}^n$. How are these not normed.

I'm sorry but you're lacking a lot of analysis knowledge, especially if you think there are no norms or metrics in QM. You need to learn analysis first. This discussion is pointless.

You are right , I wrote that sentence in a rush, I meant that in functional anlysis pseudometrics also arise naturally, thru seminormed spaces. I'll erase it.

Anyway WN this is just a tangential goof that has nothing to do with what we are discussing. Your using it as excuse not to continue discussing is rather odd.

 

[TrickyDicky]

The metric tensor only induces topologies in the tangent spaces.

The structures and objects that I comment below are not confined to the tangent space, they belong to the manifold.

Give one example of a metric being used in GR.

I'll give you examples where distances between events(whatever you want to call those distance functions:metrics, pseudometrics...) based on the pseudoriemannian metric tensor are used: Timelike and null geodesics, and all the objects in GR that are based on any of those geodesics, for instance in the case of null geodesics the notion of null infinity, used to define event horizons or the notion of conformal infinity. In the case of timelike geodesics, the very concept of proper time.

You're confusing different notions of distance.

If you refer to the fact that formally spacetimes are defined as metric spaces with the standard topology, I'm aware of that. I'm just being consequent with it and being strict this definition implies ignoring any distance function incompatible with the definition of metric spaces one could apply to the manifold based on the pseudoriemannian tensor. But I'm highlighting the fact that if one ignores those distance notions one shouldn't be able to use them in physics. The above examples apparently show that is not the case, and I'm just reflecting it.

 

[TrickyDicky]

Quoting from mathpages again: "we can simply take as our basis sets all the finite intersections of Minkowski neighborhoods. Since the contents of an e-neighborhood of a given point are invariant under Lorentz transformations, it follows that the contents of the intersection of the e-neighborhoods of two given points are also invariant. Thus we can define each basis set by specifying a finite collection of events with a specific value of e for each one, and the resulting set of points is invariant under Lorentz transformations.[...] In the case of a spacetime theory, we need to consider whether the temporal and spatial components of intervals have absolute significance, or whether it is only the absolute intervals themselves that are significant."

It would seem this is basically how the issue is solved in general for validly using the physics derived from structures generated by a pseudoRiemannian metric instead of a Riemannian metric, right? considering temporal and spatial components separately?

Is this related to what is called metric identification as explained in Wikipedia http://en.wikipedia.org/wiki/Metric_identification ?

 

[robphy]

Possibly interesting reading:
http://en.wikipedia.org/wiki/Spacetime_topology

 

[TrickyDicky]

Possibly interesting reading:
http://en.wikipedia.org/wiki/Spacetime_topology

Yes, the topology of spacetime must coincide with the manifold topology R4.

I had a pertinent (I think) doubt about how the fact that pseudoriemannian metric tensors can in principle be integrated to distance functions with topology different from the natural manifold topology (that is by definition a metric space) was handled in relativity.
Apparently this question is enough to irritate certain people. And no, I'm not confusing different notions of distance. Anyway it is probably not so hard, but it wasn't obvious to me when I asked. The basic way I think I could answer my own question is the following (wich by the way makes use of the L-C connection, so my initial intuition maybe wasn't so mistaken):
I was concerned with integrated lengths in the case of timelike and mostly null geodesics. I was also stuck at spacetime intervals derived from this integrations, considering the concept as something absolute in itself rather than its spatial and temporal components as it is usually stressed in relativity pedagogy.

But the fact is that the geodesic equation in its canonical form in GR and Riemannian geometry, that is using the unique Levi-Civita affine connection, demands the use of affine parameters, and this simple fact is important here because it only allows geodesics to be expressed in terms of either temporal or spatial lengths that are never zero for distinct temporal or spatial points. And wrt the triangle inequality, one just has to stick to the equality part(test particles). So the manifold topology is preserved.

Ironically it seems to salvage the spacetime topology one has to think in terms of space and time separately.

But it makes one wonder if say explanations based on the reverse triangle inequality for the twin paradox are then valid explanations.

 

[PeterDonis]

pseudoriemannian metric tensors can in principle be integrated to distance functions with topology different from the natural manifold topology

This isn't quite correct as you state it. The correct statement is that the distance function associated with a pseudoriemannian metric does not induce a valid topology on the underlying manifold, because that distance function assigns zero distance to distinct points. Only a riemannian metric can induce a valid topology on the underlying manifold.

But the fact is that the geodesic equation in its canonical form in GR and Riemannian geometry, that is using the unique Levi-Civita affine connection, demands the use of affine parameters, and this simple fact is important here because it only allows geodesics to be expressed in terms of either temporal or spatial lengths that are never zero for distinct temporal or spatial points.

Huh? The geodesic equation applies perfectly well to null geodesics, which have zero length. Null geodesics still have affine parameters; the valid affine parameters for null geodesics just don't include the "distance" along them in terms of the pseudoriemannian metric.

And wrt the triangle inequality, one just has to stick to the equality part(test particles).

I don't understand what you're talking about here.

But it makes one wonder if say explanations based on the reverse triangle inequality for the twin paradox are then valid explanations.

Of course they are. Why wouldn't they be? What does the fact that the pseudoriemannian metric doesn't induce a valid topology have to do with whether the reverse triangle inequality works as an explanation for the twin paradox?

 

[TrickyDicky]

This isn't quite correct as you state it. The correct statement is that the distance function associated with a pseudoriemannian metric does not induce a valid topology on the underlying manifold, because that distance function assigns zero distance to distinct points. Only a riemannian metric can induce a valid topology on the underlying manifold.

What you write is exactly what I wanted to mean, so if the way you word it is clearer or more correct, I make it mine.

Huh? The geodesic equation applies perfectly well to null geodesics, which have zero length. Null geodesics still have affine parameters; the valid affine parameters for null geodesics just don't include the "distance" along them in terms of the pseudoriemannian metric.

Again, it seems my poor command on the english language must be playing here. What you say here is in perfect agreement with what I'm saying. My point was that it is precisely the fact that the geodesic equation demands that null geodesics are affinely parametrized what makes them perfectly compatible with the manifold topology. This is related to some paragraphs in the mathpages article I linked where towards the end it addresses how to avoid the issues that could arise from the distance function associated with a pseudoriemannian metric. In the end as WN said(although he could have explained it in a less dismissive way (if only for the community spirit) (pseudo)metrics are not used in GR. As simple as that.

I don't understand what you're talking about here.

Yes, I was a bit cryptic here. I meant that for test particles the triangle inequality which is ≤, in a non-euclidean setting the inequality is not strict so d(x,z) can be equal to d(x,y)+d(y,z). Because there are no non-gravitational forces at play.

Of course they are. Why wouldn't they be? What does the fact that the pseudoriemannian metric doesn't induce a valid topology have to do with whether the reverse triangle inequality works as an explanation for the twin paradox?

Well, as I was saying I just meant that the reverse triangle inequality is a property of distance functions that don't induce a valid topology on the manifold, so being rigorous since in the twin paradox are involved accelerations(that is non-gravitational forces) one must use the strict reverse inequality and it feels like an explanation not very appropriate if one is stressing the fact that distances with that property are topologically incompatible with the manifold as a metric space. I don't know maybe my reasoning is too involved.

 

[PeterDonis]

What you write is exactly what I wanted to mean

Ok.

My point was that it is precisely the fact that the geodesic equation demands that null geodesics are affinely parametrized what makes them perfectly compatible with the manifold topology.

Ok.

I meant that for test particles the triangle inequality which is ≤, in a non-euclidean setting the inequality is not strict so d(x,z) can be equal to d(x,y)+d(y,z). Because there are no non-gravitational forces at play.

I still don't understand. Can you give a specific example? That is, can you give three points in Minkowski spacetime (specified by coordinates given in some inertial frame) such that d(x, z) = d(x, y) + d(y, z)? (With d being the Minkowski distance function, of course.)

it feels like an explanation not very appropriate if one is stressing the fact that distances with that property are topologically incompatible with the manifold as a metric space.

First of all, I don't see what this concern has to do with whether or not the reverse triangle inequality must be strict, or whether the equality can hold. (I am unable to think of an example where the equality does hold, which is why I asked you for one above.)

Second, I don't see what the concern is. The pseudo-Riemannian distance function is just a function on the manifold--more precisely, it's a symmetric 2nd-rank tensor field on the manifold, which gets contracted with tangent vector fields and integrated along curves to obtain physical distances, i.e., distances that are used to make physical predictions. There's no requirement that every function or tensor field on a manifold has to induce a valid topology on the manifold; nor is there a requirement that topological "distances" (i.e., distances computed using the metric that corresponds to whatever topology on the manifold we are using) must match "distances" that are used to make physical predictions. The fact that the term "distance" can be used for both concepts does not mean there has to be any relationship between them. It looks to me like you are being confused by terminology issues that have nothing to do with the physics involved.