Is a Manifold with a Boundary Considered a True Manifold?

[PeterDonis]

Even the Wikipedia article you linked specifically describes that it's a generalisation of the field equations,

No, it doesn't. It says that for the particular case of a 4D Lorentzian spacetime with a nonzero cosmological constant and no other stress-energy present, i.e., de Sitter and anti-de Sitter spacetime, the solution of the Einstein Field Equations gives you an Einstein manifold. But there are plenty of solutions of the EFE that are not Einstein manifolds, so it is certainly wrong to say that "Einstein manifold" is a "generalization of the field equations".

Moreover, as I have pointed out multiple times now, the concept of "Einstein manifold" is not limited to 4D Lorentzian spacetimes. The concept is valid for manifolds of any dimension high enough to have a well-defined Ricci scalar (which means any dimension higher than two), and for Riemannian manifolds as well as pseudo-Riemannian manifolds.

You continue to maintain your position without having actually addressed either of those objections, which I have made repeatedly now.

 

[PeterDonis]

that is the vacuum equations plus a cosmological constant.

I have already addressed this point twice, first in the same post you quoted (#24), my second paragraph; and second in my response to @ergospherical just now.

 

[ergospherical]

Your objection is besides the point because it's nonetheless true that four-dimensional pseudo-Riemannian Einstein manifolds are vacuum solutions to the field equations with a cosmological constant. Just because that doesn't include every possible spacetime doesn't mean we can't study the spacetimes it does include!

 

[Dale]

The boundaries are differentiable manifolds (without a boundary) themselves.

Yes, that does come up in the various versions of Stokes theorem. I hadn’t thought that through fully.

 

[Dale]

It seems, at least at a first glance, that it would be hard to give physical meaning to the boundary.

I agree. But it is also hard to give a meaning to a geodesically incomplete manifold without a boundary. So I am not sure that the issue is the boundary. I think it is more the incompleteness.

One place where it might make a difference is at a singularity. Usually we remove the singularity from the manifold. But if we allow manifolds with boundaries then could we include the singularity as the boundary? I am not sure.

 

[PeterDonis]

Your objection is besides the point

No, it's not, because your claim was that the "Einstein equation" that defines an Einstein manifold, namely $R_{\mu \nu} = \lambda g_{\mu \nu}$, is a "generalization" of the Einstein field equation. That claim is wrong. I've already given one reason why--that there are plenty of solutions of the EFE which are not Einstein manifolds--but another even simpler reason why is that $R_{\mu \nu} = \lambda g_{\mu \nu}$ is obviously not a "generalization" of $G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}$, which is the Einstein Field Equation. In fact, in the context of the EFE, $R_{\mu \nu} = \lambda g_{\mu \nu}$ is a special case of the EFE, where $T_{\mu \nu} = 0$ and $\Lambda \neq 0$.

Moreover, the particular Einstein manifolds that are solutions to the EFE, namely de Sitter and anti-de Sitter spacetime, are manifolds without boundary, so they are beside the point in this thread, which is supposed to be discussing whether manifolds with boundary are relevant in GR.

Just because that doesn't include every possible spacetime doesn't mean we can't study the spacetimes it does include!

Of course we can study de Sitter and anti-de Sitter spacetime, but since, as above, those are manifolds without boundary, such study will not be using GR with manifolds with boundary, which is what this thread is supposed to be about.

 

[PeterDonis]

if we allow manifolds with boundaries then could we include the singularity as the boundary?

Not if we are using the standard EFE, since that will tell us that various invariants are infinite at the singularity, which means it cannot be treated as part of the manifold.

In some speculative quantum gravity models, I believe the effective field equation is modified near the singularity so that invariants remain finite, but in those models, IIRC, the reason for that is to be able to extend the manifold beyond what used to be the singularity, for example in models where black holes are supposed to spawn new baby universes. So in these models the singularity would not be a boundary to the manifold. I'm not aware of any models where the singularity is included as a boundary, i.e., no extension of the manifold beyond it, but has finite invariants there.

 

[PeterDonis]

it is also hard to give a meaning to a geodesically incomplete manifold without a boundary

Are there any examples of this?

 

[Dale]

Are there any examples of this?

Sure, like the portion of the Schwarzschild spacetime covered by the usual Schwarzschild coordinates.

various invariants are infinite at the singularity, which means it cannot be treated as part of the manifold

Why are those two connected? Why would an infinite invariant preclude something from being part of the manifold?

 

[PAllen]

Are there any examples of this?

All manifolds with singularities (meaning manifolds without boundary), e.g. all the BH manifolds. These are all manifolds in the default sense - without boundary, but are geodesically incomplete. (My earlier point was that any manifold with boundary, where the boundary was not at conformal infinity, is geodesically incomplete, and further, the existence of the boundary implies that at least some extension is possible, removing the inessential incompleteness.)

 

[PeterDonis]

the portion of the Schwarzschild spacetime covered by the usual Schwarzschild coordinates.

Ah, I see I should have been more specific in my question. I was looking for an example of a geodesically incomplete manifold without boundary (where "boundary" is interpreted such that the singularities in black hole spacetimes are boundaries) that cannot be extended. But now that I think about it, that question might not make sense--it would require that the manifold can be extended "to infinity" in every direction (heuristically speaking), but still have geodesics that can't be extended beyond some finite value of their affine parameter. That doesn't seem possible.

All manifolds with singularities (meaning manifolds without boundary)

Yes, I see that we are shifting the meaning of "boundary" here. See above.

 

[PeterDonis]

Why would an infinite invariant preclude something from being part of the manifold?

It would as far as the EFE is concerned because the EFE is not valid at points where any curvature invariant (i.e., an invariant derived from the metric) is infinite. That's why we can't treat singularities as part of the manifold in GR--in fact that was the chief stumbling block in defining singularities in the 1960s and early 1970s. The discovery that geodesic incompleteness could be used to define singularities in GR was important precisely because it gave a way of defining when a singularity was present based only on things that were part of the manifold. See, for example, the discussion in Wald, section 9.1.

If you discard the EFE, then yes, you could adjoin the boundary to the manifold, but then you wouldn't be doing GR any more.

 

[Dale]

If you discard the EFE, then yes, you could adjoin the boundary to the manifold, but then you wouldn't be doing GR any more.

Yes, that makes sense

 

[martinbn]

I have already addressed this point twice, first in the same post you quoted (#24), my second paragraph; and second in my response to @ergospherical just now.

Yes, and I understood you. My comment was about the part where you said that they are not solutions to the EFE, which they are. They are solutions to the vacuum equations. I also think that @ergospherical by a genaralization doesn't mean that they generalize the full EFE, but that they are a generalization of the vacuum equations to higher dimensions and other signatures.

 

[PeterDonis]

My comment was about the part where you said that they are not solutions to the EFE, which they are.

Two particular Einstein manifolds are solutions to the EFE, de Sitter and anti-de Sitter spacetime. But my point was that there are many Einstein manifolds that aren't, since an Einstein manifold does not even have to be a spacetime.

I also think that @ergospherical by a genaralization doesn't mean that they generalize the full EFE, but that they are a generalization of the vacuum equations to higher dimensions and other signatures.

The equation $R_{\mu \nu} = \lambda g_{\mu \nu}$ does not have to be derived from the vacuum EFE. That equation makes sense for manifolds where there isn't even a well-defined concept of "stress-energy tensor" or "vacuum".

 

[Nullstein]

But such cases typically have unnessessary geodesic incompleteness. So, for physical plausibility (removal of inessential singularities), you would extend the manifold.

While this is a reasonable physical assumption, it needn't be true in general. There are two ways in which an inextendible geodesic $\gamma:(-\infty,t_0)\rightarrow M$ (affinely parametrized) can fail to be complete (assuming a Lipschitz condition):

1. The solution of the geodesic equation blows up (in a coordinate system) as $t\rightarrow t_0$. In that case, one can't extend the spacetime.
2. The solution converges as $t\rightarrow t_0$.

In the second case, one may add the limit $\gamma(t_0) := \lim_{t\rightarrow t_0}\gamma(t)$ (yes, I'm being sloppy with the math here) to the spacetime manifold. The resulting set may have the structure of a manifold with boundary. In other words, geodesics may come to an end after finite proper time and end on the boundary of the manifold. I don't know any examples though. While such situations may be pathological, they are not strictly excluded.

 

[PAllen]

While this is a reasonable physical assumption, it needn't be true in general. There are two ways in which an inextendible geodesic $\gamma:(-\infty,t_0)\rightarrow M$ (affinely parametrized) can fail to be complete (assuming a Lipschitz condition):

1. The solution of the geodesic equation blows up (in a coordinate system) as $t\rightarrow t_0$. In that case, one can't extend the spacetime.
2. The solution converges as $t\rightarrow t_0$.

In the second case, one may add the limit $\gamma(t_0) := \lim_{t\rightarrow t_0}\gamma(t)$ (yes, I'm being sloppy with the math here) to the spacetime manifold. The resulting set may have the structure of a manifold with boundary. In other words, geodesics may come to an end after finite proper time and end on the boundary of the manifold. I don't know any examples though. While such situations may be pathological, they are not strictly excluded

What I was referring to was a manifold with boundary not at conformal infinity, equipped with a pseudo-Reimannian metric, defined on the boundary. In this case, every geodesic reaching the boundary is incomplete, but there is no physical reason for this. That the geodesic is defined on the boundary rules out any "true" singularity. Thus the manifold with metric can be extended. And this would be necessary for physical plausibility.

 

[Nullstein]

What I was referring to was a manifold with boundary not at conformal infinity, equipped with a pseudo-Reimannian metric, defined on the boundary. In this case, every geodesic reaching the boundary is incomplete, but there is no physical reason for this. That the geodesic is defined on the boundary rules out any "true" singularity. Thus the manifold with metric can be extended. And this would be necessary for physical plausibility.

But it's not guaranteed that the geodesic will be extendible. The geodesic may end on the boundary, yet the solution to the geodesic equation may not be extendible beyond the boundary. A compact interval $[t_1,t_2]$ may already be the maximal interval of existence of the solution.

 

[PAllen]

But it's not guaranteed that the geodesic will be extendible. The geodesic may end on the boundary, yet the solution to the geodesic equation may not be extendible beyond the boundary. A compact interval $[t_1,t_2]$ may already be the maximal interval of existence of the solution.

How is this possible? For example, if a continuous function of reals to reals is defined on a closed interval, it is always possible to continuously extend it. This is trivially false, in general, for an open interval. To my knowledge, the same would be true for geodesics in GR, excluding boundaries at conformal infinity. Can you provide any example of what you claim, or a clear argument or link to proof that it’s possible.

 

[Nullstein]

How is this possible? For example, if a continuous function of reals to reals is defined on a closed interval, it is always possible to continuously extend it. This is trivially false, in general, for an open interval. To my knowledge, the same would be true for geodesics in GR, excluding boundaries at conformal infinity. Can you provide any example of what you claim, or a clear argument or link to proof that it’s possible.

You can always extend a curve, but the extension may no longer solve a particular differential equation, in this case the geodesic equation. There are existence and uniqueness theorems that guarantee the existence of solutions of differential equations. In particular, the Picard-Lindelöf theorem is relevant here and it guarantees the existence of a local solution of an ODE. Once you have a local solution, you can try to extend it further, but this may not always be possible. In particular, the solution may blow up after finite time or it may just not be extendible beyond a finite time, because no extension of the solution beyond a particular finite time $t_0$ may be able to solve the geodesic equation. The first case corresponds to a singularity and the second case corresponds to a boundary. The geodesic ends without developing a singularity. It ends, because no continuous extension satisfies the geodesic equation.

 

[PAllen]

You can always extend a curve, but the extension may no longer solve a particular differential equation, in this case the geodesic equation. There are existence and uniqueness theorems that guarantee the existence of solutions of differential equations. In particular, the Picard-Lindelöf theorem is relevant here and it guarantees the existence of a local solution of an ODE. Once you have a local solution, you can try to extend it further, but this may not always be possible. In particular, the solution may blow up after finite time or it may just not be extendible beyond a finite time, because no extension of the solution beyond a particular finite time $t_0$ may be able to solve the geodesic equation. The first case corresponds to a singularity and the second case corresponds to a boundary. The geodesic ends without developing a singularity. It ends, because no continuous extension satisfies the geodesic equation.

I still don't buy it. You can extend manifold and metric such that the geodesic is extensible. I don't find this believable without an example or an existence proof. I have seen no hint that such a thing is possible in any GR literature I have seen. (Also, the most common definition of singularity in GR is the presence of incomplete inextensible geodesics, so either case would be a singularity by the standard definition).

 

[Nullstein]

I still don't buy it. You can extend manifold and metric such that the geodesic is extensible. I don't find this believable without an example or an existence proof. I have seen no hint that such a thing is possible in any GR literature I have seen.

It's just simple ODE theory, e.g. see this Wikipedia link:

In the case that $x_\pm \neq \pm\infty$, there are exactly two possibilities
- explosion in finite time: $\limsup_{x \to x_\pm} \|y(x)\| \to \infty$
- leaves domain of definition: $\lim_{x \to x_\pm} y(x)\ \in \partial \bar{\Omega}$

 

[PeterDonis]

It's just simple ODE theory, e.g. see this Wikipedia link:

The second alternative requires that the curve leaves the domain in which $F$ is defined. But for the geodesic equation, $F$ is a function composed entirely of the metric and its derivatives, and any such function is defined everywhere the metric and its derivatives are defined, i.e., everywhere in the spacetime. So it is impossible for this case to be realized for the geodesic equation.

 

[Nullstein]

The second alternative requires that the curve leaves the domain in which $F$ is defined. But for the geodesic equation, $F$ is a function composed entirely of the metric and its derivatives, and any such function is defined everywhere the metric and its derivatives are defined, i.e., everywhere in the spacetime. So it is impossible for this case to be realized for the geodesic equation.

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

 

[PeterDonis]

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

You are not responding to the actual issue I raised. Go back and read my post again. Responding by simply repeating your position is not sufficient.

 

[PAllen]

No, we're exactly in the situation of a manifold with boundary. $\bar\Omega$ is the manifold with boundary, $\Omega$ is its interior and $\partial\bar\Omega$ is the boundary. ODE theory requires $F$ to be defined on an open set, in this case $\Omega$, not $\bar\Omega$. You solve the ODE there and then study its behavior on $\bar\Omega$ by taking limits.

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.

 

[Nullstein]

You are not responding to the actual issue I raised. Go back and read my post again. Responding by simply repeating your position is not sufficient.

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

 

[PeterDonis]

The post is addressing what you said.

No, it isn't. See below.

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

The spacetime is $\Omega$, not $\bar{\Omega}$. I explained why, for the particular case of the geodesic equation, $F$ is guaranteed to be defined on $\Omega$. You have not responded to what I said about that at all.

 

[PAllen]

The post is addressing what you said. It doesn't matter whether $F$ is defined on $\partial\bar\Omega$. "Leaving the domain of definition" in the second case doesn't refer to leaving $\bar\Omega$, but leaving $\Omega$.

But that is then not relevant to the question essential geodesic incompleteness as used in GR, nor is relevant to the question of extending a manifold with boundary to satisfy the POE (which is the post of mine you originally responded to). If the geodesics don't reach the boundary, then physically, the boundary is not part of manifold for GR purposes. If the geodesics do reach the boundary, then this is simply to be considered a submanifold of some larger manifold without boundary.

 

[Nullstein]

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

No, you don't necessarily ask this. You can also just have a manifold with boundary, which isn't embedded into a larger manifold without boundary. It may not be possible to extend the manifold beyond the boundary.

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 the second case I discussed, it is not possible for two geodesics to meet on this boundary, because then the full curve would constitute extensions of the original solutions, which is in contradiction to the assumption that the original solutions were already maximal.

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.

So can you provide a reference that proves that such an extension is always possible? So far, you have only made a claim while I have provided literature.

If the geodesics don't reach the boundary, then physically, the boundary is not part of manifold for GR purposes. If the geodesics do reach the boundary, then this is simply to be considered a submanifold of some larger manifold without boundary.

But that just isn't possible in general if we're in the second case. It is possible for a geodesic to reach a boundary and not possesses an extension. That's what the theorem in Wikipedia says.

 

[Nullstein]

The spacetime is $\Omega$, not $\bar{\Omega}$. I explained why, for the particular case of the geodesic equation, $F$ is guaranteed to be defined on $\Omega$. You have not responded to what I said about that at all.

No, if we're talking about a spacetime manifold with boundary, then the spacetime is $\bar\Omega$. $\Omega$ doesn't have a boundary, so it's not the spacetime manifold we're talking about.

 

[Nullstein]

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.

 

[PAllen]

I really don't see how this theorem is applicable. If we start by assuming a manifold with boundary equipped with a metric, that means metric is defined on the boundary. That is, in the context of the theorem, F is defined on a closed interval rather than an open interval. Thus, I reject the applicability of this theorem to the case at hand. As I understand it, @PeterDonis has also tried to make this same point, without getting a satisfactory answer.

 

[Nullstein]

I really don't see how this theorem is applicable. If we start by assuming a manifold with boundary equipped with a metric, that means metric is defined on the boundary. That is, in the context of the theorem, F is defined on a closed interval rather than an open interval. Thus, I reject the applicability of this theorem to the case at hand. As I understand it, @PeterDonis has also tried to make this same point, without getting a satisfactory answer.

It is not necessary for the metric not to be defined on a boundary for the theorem to apply. Whether the metric is defined on the boundary or not is not relevant for the applicability of the theorem.

Suppose we chose initial conditions $x(0)=x_0\in\Omega$. Then the Picard-Lindelöf theorem guarantees the existence of a solution to the geodesic equation on a maximal open interval: $x:(t_-,t_+)\rightarrow\Omega$. It may then be the case that $\lim_{t\rightarrow t_+} x(t) = x_+$ exists and that $x_+\in\partial\bar\Omega$. In this case, the solution ends on the boundary and can't be extended beyond it. It's not relevant whether the metric is defined on $\partial\bar\Omega$ for this situation to be possible.

 

[PeterDonis]

No, if we're talking about a spacetime manifold with boundary, then the spacetime is $\bar\Omega$. $\Omega$ doesn't have a boundary, so it's not the spacetime manifold we're talking about.

You evidently failed to comprehend the Wikipedia article you yourself linked to.

The article explicitly says (right after it gives the two cases you quoted):

$\Omega$ is the open set in which F is defined, and $\bar{\Omega}$ is its boundary.

For a spacetime, $F$, as I said, is a function built entirely from the metric and its derivatives, and $F$ must be defined on whatever the entire spacetime is. That must at least include $\Omega$. If the spacetime has a boundary $\bar{\Omega}$ that is part of the spacetime--i.e., if the spacetime is a manifold with boundary, which is the case you are making claims about--then $F$ must also be defined on $\bar{\Omega}$. If it isn't, then we are talking about the first possibility ("explosion in finite time"), not the second. And if it is, then the second possibility cannot happen, because evaluating the geodesic equation on $\bar{\Omega}$ is not "leaving the domain of definition" of $F$.

It may then be the case that $\lim_{t\rightarrow t_+} x(t) = x_+$ exists and that $x_+\in\partial\bar\Omega$.

Yes, this part is fine, and in fact it will always be true in the case of a spacetime where the boundary is included in the manifold, so the metric is defined on the boundary.

In this case, the solution ends on the boundary and can't be extended beyond it.

This does not follow; the premises you have stated are not sufficient to establish this conclusion. In fact, for a spacetime, I believe the opposite conclusion will hold: if the affine parameter along a geodesic has a finite value at the boundary, and all geometric invariants are finite (so the first case, "explosion in finite time", does not apply), then there must exist a further extension of the spacetime beyond the boundary. (I think this may be proved in Hawking & Ellis.)

It's not relevant whether the metric is defined on $\partial\bar\Omega$ for this situation to be possible.

Yes, it is, because the metric being defined on the boundary, and all geometric invariants being finite there (which will always be true if the metric is defined on the boundary), is what grounds the conclusion that there must be an extension of the spacetime beyond the boundary.