Is a Manifold with a Boundary Considered a True Manifold?

[PeterDonis]

I really don't see how this theorem is applicable.

The theorem is perfectly valid as far as it goes; it just doesn't go far enough to make the claims about inextendibility that @Nullstein is making.

For example, consider the exterior patch of Schwarzschild spacetime, i.e., the portion outside the horizon, and suppose we include the horizon itself as a boundary and view this patch as a manifold with boundary. It is straightforward to compute that every geodesic that intersects the boundary does so at a finite value of its affine parameter, and that all geometric invariants are finite at the horizon. So we have a case where, in the notation of the Wikipedia article, $x_\pm \neq \pm \infty$ and the "explosion in finite time" possibility does not apply. So the other possibility must be true. That is what the theorem says, and, as noted above, it is perfectly correct as far as it goes.

However, the Wikipedia article misdescribes this second possibility when it uses the phrase "leaves the domain of definition of $F$", because all of the possible functions $F$ that we could be using, which must be built from the metric and its derivatives, are perfectly well-defined on the boundary $\bar{\Omega}$ (the horizon) as well as on the open set $\Omega$. So the theorem is not telling us that the curve leaves the "domain of definition" of whatever differential equation it is an integral curve of (in this case, the geodesic equation). It is merely telling us that the curve leaves the "domain of definition" of the open interval that describes its affine parameter on the open set $\Omega$. Which tells us precisely nothing, in itself, about whether or not the curve is extendible past the boundary $\bar{\Omega}$. And in fact we know that all geodesics that intersect the horizon are extendible past the horizon, because the spacetime itself is.

 

[vanhees71]

But the theory of singularities in GR is different. You ask whether there exists a manifold without boundary within which the manifold with boundary is a submanifold, with the overall metric only needing to satisfy junction conditions at the boundary, such that the geodesics defined on each side of the boundary meet. In many cases, e.g the extension of exterior Schwarzschild manifold to Kruskal, not only smoothness but analyticity is possible. But as long as the first limited condition is met, we say the geodesic incompleteness is removable, and does not constitute a singularity. Further, the physical principle of equivalence requires that some such extension be made. So far as I know, if the geodesic is defined on the boundary of a manifold with boundary, then such an extension is possible. Nothing you have said so far seems relevant to this.

But aren't there these theorems saying that there are non-extensible singularities in the solutions of Einstein's field equations? E.g., isn't it true that the Schwarzschild manifold has a true singularity at the origin $r=0$, for which there is no extension? Of course, as @PeterDonis writes in #71, the singularities at the event horizon are coordinate singularities, which can by overcome just by choosing other coordinates like the Kruskal coordinates. So here the singularities are singularities of the coordinate chart defined by Schwarzschild coordinates, but you can simply use another chart to cover a "completed" Schwarzschild spacetime with, e.g., Kruskal coordinates.

 

[PeterDonis]

isn't it true that the Schwarzschild manifold has a true singularity at the origin $r = 0$, for which there is no extension?

Yes, but $r = 0$ itself is not part of the manifold, so the manifold itself is a manifold without boundary. $r = 0$ is called a singularity because geodesics approach a finite value of their affine parameter as a limit as they approach $r = 0$ as a limit.

 

[jbergman]

I need to look at the arguments more closely to refute Peter's claim, but my gut instinct is to side with Nullstein, here. As he stated, these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended. I believe if your Manifold is compact then you can always extend. But for the non-compact case that isn't the case.

 

[martinbn]

I need to look at the arguments more closely to refute Peter's claim, but my gut instinct is to side with Nullstein, here. As he stated, these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended. I believe if your Manifold is compact then you can always extend. But for the non-compact case that isn't the case.

No, if the maifold has a boundary and geodesics reach it in finite afine parameter, with all relevant quantities bounded, the manifold can be extended.

 

[jbergman]

No, if the maifold has a boundary and geodesics reach it in finite afine parameter, with all relevant quantities bounded, the manifold can be extended.

Did, i say otherwise?

 

[PeterDonis]

these are standard results in manifold and ODE theory, i.e., the existence of flows that can't be extended.

Nobody is saying that these standard results are wrong. We are simply explaining why those results don't help in assessing extensibility for the particular case of geodesics in a Lorentzian spacetime.

I believe if your Manifold is compact then you can always extend.

This is obviously false as you state it: any n-sphere for n >= 1 is a counterexample.

 

[jbergman]

Nobody is saying that these standard results are wrong. We are simply explaining why those results don't help in assessing extensibility for the particular case of geodesics in a Lorentzian spacetime.This is obviously false as you state it: any n-sphere for n >= 1 is a counterexample.

It's a theorem in Lee's Smooth Manifolds book. Can you more precisely state your counterexample? I think you are probably getting hung up on my choice of the wording. The technical definition is that one can extend a flow such that it is parametrized by a parameter such that the parameters domain is $[-\infty, \infty]$.

 

[martinbn]

Did, i say otherwise?

Yes, that is what Nullstein said and you agree with him.

 

[jbergman]

Yes, that is what Nullstein said and you agree with him.

I'll have to go back and look at his claims more closely. I agree with your statement here.

 

[PeterDonis]

Can you more precisely state your counterexample?

You said any manifold that is compact can always be extended. Any n-sphere for n >=1 is a compact manifold but cannot be extended.

The technical definition is that one can extend a flow such that it is parametrized by a parameter such that the parameters domain is $[-\infty, \infty]$.

Suppose I have a geodesic on a 2-sphere, a great circle. For concreteness, say it's the equator. It is parameterized by longitude $\varphi$, with parameters $0 \le \varphi \lt 2 \pi$ denoting distinct points. I can of course extend this parameterization to cover $- \infty < \varphi < \infty$, but only if I am ok with having multiple values of $\varphi$ denoting the same points. I cannot extend the geodesic itself.

 

[jbergman]

You said any manifold that is compact can always be extended. Any n-sphere for n >=1 is a compact manifold but cannot be extended.Suppose I have a geodesic on a 2-sphere, a great circle. For concreteness, say it's the equator. It is parameterized by longitude $\varphi$, with parameters $0 \le \varphi \lt 2 \pi$ denoting distinct points. I can of course extend this parameterization to cover $- \infty < \varphi < \infty$, but only if I am ok with having multiple values of $\varphi$ denoting the same points. I cannot extend the geodesic itself.

Of course. I am talking about the domain. What you are describing is a periodic orbit in the flow domain. You can also have stationary points that never move but have a solution parametrized for all of $$\mathbb{R}$$.

When talking about extending a flow, I'm talking about it's domain, not whether the resulting path is injective.

 

[PeterDonis]

When talking about extending a flow, I'm talking about it's domain, not whether the resulting path is injective.

Then what you are saying is irrelevant to the discussion in this thread, which is about whether and under what conditions a manifold with boundary can be extended. You seem to be saying that theorems about whether a parameterization can be extended do not say anything about whether the underlying manifold can be extended, which I agree with and I think @PAllen would agree with. But @Nullstein has been claiming the opposite.

 

[jbergman]

Then what you are saying is irrelevant to the discussion in this thread, which is about whether and under what conditions a manifold with boundary can be extended. You seem to be saying that theorems about whether a parameterization can be extended do not say anything about whether the underlying manifold can be extended, which I agree with and I think @PAllen would agree with. But @Nullstein has been claiming the opposite.

Well, there are only three kinds of flows, stationary, periodic and, aperiodic. For the periodic and constant cases you can always extend your domain. For the aperiodic case, whether or not you can extend your domain does say something about the extension of the image.

The case I'm thinking of that might cause trouble would be if you can't extend to the boundary of the manifold your flow. For example, if the flow on the boundary is either stationary or only lives in the boundary this might be the case.

 

[PeterDonis]

For the periodic and constant cases you can always extend your domain.

Yes, but without extending the image (to use your terminology), so these cases are irrelevant to the discussion in this thread.

For the aperiodic case, whether or not you can extend your domain does say something about the extension of the image.

Ok, then what, specifically, does it say?

The case I'm thinking of that might cause trouble would be if you can't extend to the boundary of the manifold your flow.

This is also irrelevant to the discussion in this thread, because we are talking about cases where we know this potential problem does not occur. The theorem @Nullstein referenced explicitly has this as one of its assumptions.

 

[PeterDonis]

Every ODE $\ddot x(t) + f(x(t))=0$ can be transformed into a geodesic equation by introducing a new function $t(s)$ (reparametrizing time) and writing $\ddot x(s)+\dot t(s) \dot t(s) f(x(s))=0$ and $\ddot t(s) = 0$ (without loss of generality, we choose inital conditions that include $\dot t(0) = 1$). Then we just have $\Gamma^x_{tt}=f(x)$ and all other Christoffel symbols are $0$. So if the second case can never occur for a geodesic equation, is can also automatically never occur for the initial ODE in the first place and then the Wikipedia article would be lying.

Once again you are failing to address what we are actually saying. The issue is not the form of the equation but the domain of definition of what you are here calling $f$, and the Wikipedia article calls $F$. The second case is described in the Wikipedia article as the curve "leaving the domain of definition" of $F$. But in the cases we have been discussing, of spacetime regions that are manifolds with boundary, $F$ is defined on the boundary, so the curve reaching the boundary at a finite value of its curve parameter is not "leaving the domain of definition" of $F$.

Here's another way of putting it. The theorem says that there exists a maximal extension of a curve given certain assumptions, and that if that maximal extension stops at a finite value of its parameter, either case 1 or case 2 must apply. But it does not tell you how to prove that some particular parameterization you have that covers the curve in some region is the maximal extension of the curve. In particular, the theorem does not say that any parameterization that reaches the boundary of some region at a finite value of the parameter and does not "blow up" there, so it can't be an instance of case 1 of the theorem, must therefore be an instance of case 2 of the theorem. But that is what you are claiming.

 

[jbergman]

Yes, but without extending the image (to use your terminology), so these cases are irrelevant to the discussion in this thread.Ok, then what, specifically, does it say?

Basically there is a limit point which the curve cannot reach. Typically it's speed would also get slower and slower as it approaches that point.

 

[PeterDonis]

there is a limit point which the curve cannot reach. Typically it's speed would also get slower and slower as it approaches that point.

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

 

[jbergman]

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

I'll need more time to more carefully look at the particular case you mention. I'll try and look at this evening.

 

[jbergman]

How does this relate to whether the image can be extended or not?

Please bear in mind that this is the relativity forum, and we are talking about spacetime, not general abstract math. For example, look at my post #71 and the curve I described there, a geodesic that approaches the horizon. In Schwarzschild coordinates, your verbal description, quoted above, appears to apply to this curve: the Schwarzschild coordinate $t$ goes to infinity as the horizon is approached, and the "speed" of the curve, $dr / dt$, goes to zero. But this curve can be extended to and beyond the horizon; the appearance of "not being able to reach the horizon" is an artifact of a particular choice of coordinates. (And we have threads all the time in this subforum where we have to explain this to newbies who misunderstand it.) So, again, what exactly does whatever math you are relying on say about what you call the aperiodic case and what it says about whether or not the image can be extended?

This question turned out to be harder than I expected as I wasn't familiar with coordinate singularities. There is a nice discussion about this at https://math.stackexchange.com/questions/852678/known-methods-to-detect-coordinate-singularities

I particularly like Jack Lee's answer on this question, https://math.stackexchange.com/a/853519

The TLDR I am taking away from this discussion is that coordinate singularities are hard to detect, but if you can and can remove them then what I said above, I believe is correct.

 

[PeterDonis]

The TLDR I am taking away from this discussion is that coordinate singularities are hard to detect, but if you can and can remove them then what I said above, I believe is correct.

What you said above where?

Note that, for the particular example I gave, of the exterior patch of Schwarzschild spacetime in Schwarzschild coordinates, we know what the maximal analytic extension is, and it is a manifold without boundary.

The question is whether there are any solutions of the Einstein Field Equation which are (a) manifolds with boundary, and (b) not extendible.

If we assume that we can always discover and remove coordinate singularities, how does what you said wherever you said it bear on this question?

 

[PeterDonis]

I particularly like Jack Lee's answer on this question, https://math.stackexchange.com/a/853519

His answer is irrelevant to this discussion because he is talking about a case where we have an open manifold, which will not include any boundary. The claim at issue in this thread concerns a manifold with boundary.

 

[jbergman]

What you said above where?

Note that, for the particular example I gave, of the exterior patch of Schwarzschild spacetime in Schwarzschild coordinates, we know what the maximal analytic extension is, and it is a manifold without boundary.

The question is whether there are any solutions of the Einstein Field Equation which are (a) manifolds with boundary, and (b) not extendible.

If we assume that we can always discover and remove coordinate singularities, how does what you said wherever you said it bear on this question?

I wasn't trying to answer the full question yet. I don't know if there is a solution to Einstein Field Equations that is a manifold with boundary. I was just trying to address the question of extendability of the metric.

What I've concluded so far is the same as most others in the thread. That is if we can extend to a boundary we can always extend it further.

I was trying to ascertain if you could have a weird case where geodesics would remain on the boundary or in the interior but never cross between these regions. It seems unlikely, but personally, I still want to think about that case.

 

[WWGD]

Just curious here, hope not to derail the post: what are the differences from Euclidean 4-space and Space-time, being a product of 3-space and (1-dimensional) time?

 

[PeterDonis]

Just curious here, hope not to derail the post: what are the differences from Euclidean 4-space and Space-time, being a product of 3-space and (1-dimensional) time?

Euclidean 4-space has a flat Riemannian metric, with signature (+, +, +, +). Minkowski spacetime has a flat Lorentzian metric (strictly speaking a pseudo-metric since it is not positive definite) with signature (-, +, +, +) (using the spacelike signature convention) or (+, -, -, -) (using the timelike signature convention). Both of them have the same underlying manifold, $\mathbb{R}^4$.

 

[sysprog]

What I've concluded so far is the same as most others in the thread. That is if we can extend to a boundary we can always extend it further.

Are you contending that a majority of those who have posted in this thread have in it concluded that in the case of every 'manifold with a boundary', extensibility of the manifold to its boundary entails that the manifold or the boundary thereto is ispo facto further extensible? I don't see where anyone else in this thread has exposited such a conclusion.

 

[PAllen]

Are you contending that a majority of those who have posted in this thread have in it concluded that in the case of every 'manifold with a boundary', extensibility of the manifold to its boundary entails that the manifold or the boundary thereto is ispo facto further extensible? I don't see where anyone else in this thread has exposited such a conclusion.

What most on this thread claim is that a pseudo-Riemannian manifold with boundary (excluding boundaries at conformal infinity), such that the metric is well defined everywhere including the boundary, and curvature scalars don't blow up on approach to the boundary, then:

- there is inessential geodesic incompleteness at the boundary, which, in GR terms, is an inessential singularity. It is physically implausible to accept this as a complete spacetime.
- under these circumstances, it is always possible to extend the manifold and metric (typically, in an infinite number of different ways), to become an manifold without boundary, such that any remaining geodesic incompleteness is irremovable (no further exension that would remove the icnompleteness is possible).

 

[jbergman]

Euclidean 4-space has a flat Riemannian metric, with signature (+, +, +, +). Minkowski spacetime has a flat Lorentzian metric (strictly speaking a pseudo-metric since it is not positive definite) with signature (-, +, +, +) (using the spacelike signature convention) or (+, -, -, -) (using the timelike signature convention). Both of them have the same underlying manifold, $\mathbb{R}^4$.

It might be also worth mentioning that in Minkowski spacetime we are typically interested only in the transformations that preserve the direction of time (along with orientation) which is different from a generic Euclidean 4-space.

 

[WWGD]

It might be also worth mentioning that in Minkowski spacetime we are typically interested only in the transformations that preserve the direction of time (along with orientation) which is different from a generic Euclidean 4-space.

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group; i.e., are these transformations even invertible? Sorry for my ignorance on this topic?

 

[PAllen]

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group; i.e., are these transformations even invertible? Sorry for my ignorance on this topic?

General coordinate transforms form a group. There is really no restriction on them, because pullbacks associated with the transform ensure all tensors (including the metric tensor) represent the same geometric object. That is, in the framework of GR, there is no such thing as an excluded coordinate transform. While I mention GR, the same is true in SR. Absolutely coordinate transform may is allowed, but the pullbacks ensure the spacetime has zero curvature and all invariants are preserved.

I like to give the following example for SR:

Consider the metric: $ds^2 = du dv + du dw + du da + dv dw + dv da + dw da$. This is just standard Minkowski spacetime with timelike signature in 4-lightlike coordinates. All of SR physics can be done in these coordinate with no problem

 

[PAllen]

It might be also worth mentioning that in Minkowski spacetime we are typically interested only in the transformations that preserve the direction of time (along with orientation) which is different from a generic Euclidean 4-space.

The orientability and causal structure of a spacetime (e.g. Minkowski) are not affected by any coordinate transform at all because of how tensors, including the metric tensor, transform. Even if you say $t'=-t$ as a coordinate transform, you simply have that the unit future pointing timelike vector in primed coordinates is (-1,0,0,0). And if you have the transform $x'=t$, all you've done is to have a timelike coordinate called x'.

 

[PeterDonis]

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group

Yes, these transformations form a group: the group of proper orthochronous Lorentz transformations. They are the component of the full Lorentz group that is connected to the identity, which is why they form a group in their own right. The other three components of the Lorentz group are obtained from the proper orthochronous subgroup by applying P (parity inversion), T (time reversal), and PT (both), respectively; each of these last three are disconnected components and none of them form a group in themselves because none of them contain the identity.

i.e., are these transformations even invertible?

Invertibility is a necessary but not sufficient condition for being a group. The transformations in the three disconnected (from the identity) components of the Lorentz group are all invertible even though none of those components forms a group in itself. Being a group also requires an identity element and closure (meaning the composition of any two transformations in the set is also in the set), and these components of the Lorentz group lack those properties as well, considered as sets in themselves. (Note that these transformations are also part of the full Lorentz group, which is a group.)

 

[vanhees71]

Is there a name for these transformations that preserve the directiin, orientation of space, time in Minkowski Space? Is it even a group; i.e., are these transformations even invertible? Sorry for my ignorance on this topic?

That's the proper orthochronous Lorentz group, $\mathrm{SO}(1,3)^{\uparrow}$. It's isomorphic to the group of real $4 \times 4$ matrices $({\Lambda^{\mu}}_{\nu}$ that satisfy
$$\eta_{\rho \sigma} {\Lambda^{\rho}}_{\mu} {\Lambda^{\sigma}}_{\nu}=\eta_{\mu \nu},$$
which defines the Lorentz group $\mathrm{O}(1,3)$ as a subgroup of $\text{GL}(4)$.

Then there is the proper Lorentz group, where in addition you have
$$\mathrm{det} \hat{\Lambda}=+1,$$
which defines the subgroup $\mathrm{SO}(1,3)$, and finally if in addition also
$${\Lambda^0}_{0} \geq 1,$$
then you have the proper orthochronous Lorentz group $\mathrm{SO}(1,3)^{\uparrow}$, which forms a Lie group, and that's the only symmetry group any special-relativistic theory must obey in order to compatible with the Minkowski spacetime structure.

In nature it's indeed the only symmetry group related to Minkowski spacetime since the weak interaction breaks symmetry under parity (space reflections), which is an orthochronous but not proper Lorentz transformation, under time-reflection, which is neither orthochronous nor proper, under PT, which is proper but not orthochronous.

 

[jbergman]

The orientability and causal structure of a spacetime (e.g. Minkowski) are not affected by any coordinate transform at all because of how tensors, including the metric tensor, transform. Even if you say $t'=-t$ as a coordinate transform, you simply have that the unit future pointing timelike vector in primed coordinates is (-1,0,0,0). And if you have the transform $x'=t$, all you've done is to have a timelike coordinate called x'.

Your answer while correct left me confused. What is the difference between a coordinate transformation and a symmetry of a space?

For Euclidean space, rotations, reflections and translations are symmetries. In other words, applying a rotation to all points results in the same space and our metric is such that the distance between any two points is the same. Practically this means that the form of the metric is unchanged for these types of transformations. For more general coordinate transformations the form of the metric must change.

For space time, the symmetries are the transformations in the proper orthochronous Lorentz group.

How do you think about the symmetry groups of a space versus the coordinate transformations?

 

[PeterDonis]

What is the difference between a coordinate transformation and a symmetry of a space?

A coordinate chart is a labeling of points in the space by n-tuples of real numbers, with certain requirements about continuity, etc. A coordinate transformation is just a change of labeling.

A symmetry of a space corresponds to a Killing vector field, which is an invariant property that is independent of any choice of coordinates.

Practically this means that the form of the metric is unchanged for these types of transformations.

No. It means that the metric is unchanged along integral curves of the Killing vector field that corresponds to the symmetry. For example, in Euclidean 3-space, there is a 3-parameter group of Killing vector fields corresponding to rotational symmetry. Integral curves of any particular one of these Killing vector fields correspond to circles centered on a particular axis of rotation. Moving along the flow of such integral curves is not a coordinate transformation.