I Is a Manifold with a Boundary Considered a True Manifold?

[Dale]

<Moderator note: thread split from https://www.physicsforums.com/threads/speed-of-light.1012508/#post-6601734 >

The formalism does not exclude finite spacetimes which are manifolds with boundaries, no?

Is a manifold with a boundary still a manifold?

 

[ergospherical]

Well I mean for example, in dimension 2, the half-plane $\{ (x,y) \in \mathbf{R}^2 : x \geq 0 \}$, and anything isomorphic locally to this (using suitable coordinates), is a manifold with a boundary (e.g. a disc, or whatever.). The generalisation to higher dimensions follows the same idea.

 

[Dale]

Well I mean for example, in dimension 2, the half-plane $\{ (x,y) \in \mathbf{R}^2 : x \geq 0 \}$, and anything isomorphic locally to this (using suitable coordinates), is a manifold with a boundary (e.g. a disc, or whatever.). The generalisation to higher dimensions follows the same idea.

Interesting. I didn’t know that such spaces were considered manifolds. Is this a generalization of the concept?

My understanding is that a manifold is a topological space where around each point in the space you can map an open set in the manifold to an open set in R^n. So with a boundary the points at the boundary would lose that property. They couldn’t be mapped to an open set in R^n. So is there another definition or am I misunderstanding the application of that definition?

Thanks!

 

[ergospherical]

That's the difference between a manifold with and without a boundary, the former is locally isomorphic to the the region $\{ (x,y) \in \mathbf{R}^2 : x \geq 0 \}$ whilst the latter is locally isomorphic to $\mathbf{R}^2$. (But you can do GR on either.)

 

[Dale]

That's the difference between a manifold with and without a boundary, the former is locally isomorphic to the the region $\{ (x,y) \in \mathbf{R}^n : x \geq 0 \}$ whilst the latter is locally isomorphic to $\mathbf{R}^n$. (But you can do GR on either.)

Interesting, I definitely learned something new today! I had never heard of that concept before

 

[ergospherical]

Ah cool, glad to hear it! It's probably just the name that's unfamiliar because the idea itself crops up everywhere. Integration for example, if $M$ is a manifold with a boundary $\partial M$ then it is the theorem of Stokes which says $\int_M d\omega = \int_{\partial M} \omega$. For instance a segment $a \leq x \leq b$, $y=z=0$ of a straight line is a manifold with a boundary (in this case the boundary is just $\{ (a,0,0), (b,0,0) \}$), and $\int_M df = \int_a^b (df/dx) dx$ whilst on the other hand $\int_{\partial M} f = f(b) - f(a)$, i.e. the fundamental theorem of calculus.

 

[Dale]

It's probably just the name that's unfamiliar because the idea itself crops up everywhere

Yes, the idea is clear, I had just always understood the word “manifold” to explicitly exclude that idea.

 

[Dale]

No worries. I split the thread. It was clarifying a new concept for me. @ergospherical ’s response in the original thread was entirely B level until I asked for higher level clarification. Apologies for not splitting it immediately when I knew I was asking for higher level clarification

 

[PAllen]

The following clarifies various definitions of manifold, supporting the general practice that without qualification, it cannot have a boundary. Under examples, it gives generalizations.

http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/manifold

or the following, which is clearer:

https://faculty.math.illinois.edu/~kapovich/481-14/bry.pdf

 

[vanhees71]

But (oriehtable) (sub-)manifolds are very impotant in tensor calculus, particularly in the formulation of Stoke's general integral theorem mentioned above in #6.

 

[ergospherical]

Another GR-specific example arises during conformal compactification of an asymptotically-flat spacetime $(M,g)$, e.g. the classic example of transforming Minkowski to a subset of the Einstein static universe $\mathbf{R} \times S^3$. You introduce a new spacetime $(M', g')$ which contains the old spacetime $M$ as a subset (and with $g' = \Omega^2(x)g$), and then extend $M \cup \partial M$ to be a manifold with boundary inside $M'$. (The boundary represents "infinity" of the old spacetime).

 

[PeterDonis]

you can do GR on either

Can you? GR uses differential equations, and you can't do differential equations at the boundary of a manifold with boundary. You can only do differential equations at points which have open neighborhoods that are in the manifold. Boundary points don't.

 

[PeterDonis]

The boundary represents "infinity" of the old spacetime

But you don't actually solve the Einstein Field Equation for the original spacetime at the boundary. The only thing the boundary is used for is to define distinct "points at infinity" for certain useful purposes. If you actually wanted to solve the Einstein Field Equation on the new spacetime, you would not be able to do it just on the manifold with boundary; you would have to include the larger containing spacetime (for example, you would have to solve the Einstein static universe for the case you mention, where Minkowski spacetime is mapped to a subset of the Einstein static universe). And of course this solution would give a different geometry than that of the original spacetime (that's why you have to have a conformal factor in the metric).

In short, none of this involves treating the actual, physical spacetime as a manifold with boundary. I'm not aware of any example where you can do that in GR. Nor does it involve constructing a self-contained manifold with boundary for the conformally compactified spacetime, with no extension outside the boundary: the boundary is just arbitrarily chosen to make the conformal compactification work as desired and does not represent an actual geometric boundary.

 

[Dale]

you can't do differential equations at the boundary of a manifold with boundary

Interesting. I have not worked with these manifolds with boundaries. Is it really not possible to do differential equations on such a manifold? Or is it just necessary to specify the boundary conditions at such a boundary? (As opposed to a normal manifold where you can specify the boundary conditions anywhere)

 

[ergospherical]

Can you? GR uses differential equations, and you can't do differential equations at the boundary of a manifold with boundary. You can only do differential equations at points which have open neighborhoods that are in the manifold. Boundary points don't.

Where did you get that idea from? In undergrad, everybody learns how to solve the heat equation on a rectangle, Schrödinger's equation in a cube, Laplace's equation in a sphere, etc., subject to specified Dirichlet/Neumann/Robin conditions on the boundary of the domain...

 

[PeterDonis]

Or is it just necessary to specify the boundary conditions at such a boundary?

Yes, you could. But see below.

In undergrad, everybody learns how to solve the heat equation on a rectangle, Schrödinger's equation in a cube, Laplace's equation in a sphere, etc., subject to specified Dirichlet/Neumann/Robin conditions on the boundary of the domain...

Yes, that's true. I should have qualified my original statement since I was talking specifically about GR. In GR, I am not aware of any example where we solve the EFE in this way and consider the resulting solution the entire spacetime. We do construct solutions that contain different regions with different geometries, separated by boundaries at which particular junction conditions hold. But the underlying manifold itself is never a manifold with boundary. Even in the case of conformal compactification, as I said, the boundary is not part of the original, physical spacetime (e.g., Minkowski spacetime in your example), and is arbitrarily chosen in terms of the containing spacetime (e.g., the Einstein static universe in your example), which is always a manifold without boundary.

 

[ergospherical]

That seems too strong a claim. I managed to find lots of papers studying the properties Einstein manifolds with boundaries, e.g.
https://arxiv.org/pdf/math/0612647.pdf

also
https://www.math.stonybrook.edu/~anderson/isomext2.pdf
https://link.springer.com/content/pdf/10.1007/s00229-021-01340-4.pdf

 

[PeterDonis]

I managed to find lots of papers studying the properties Einstein manifolds with boundaries

Thanks for the references, I'll take a look.

 

[PAllen]

This is a physically reasonable assumption but I don't believe you can state it with certainty. The formalism does not exclude finite spacetimes which are manifolds with boundaries, no?

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

 

[PeterDonis]

Einstein manifolds

Note that "Einstein manifolds" in all of these papers does not mean "manifolds obtained by solving the Einstein Field Equation". It means "manifolds for which the Ricci tensor is a multiple of the metric". This is a recognized term [1], but it can be confusing. None of these papers are talking about solving the Einstein Field Equation on manifolds with boundary; they are simply investigating general mathematical properties of Einstein manifolds with boundary.

[1] https://en.wikipedia.org/wiki/Einstein_manifold

 

[romsofia]

Can you?

Would you consider the addition of the Gibbons-Hawking-York term a solution to doing GR on manifolds with boundaries?

 

[PeterDonis]

Would you consider the addition of the Gibbons-Hawking-York term a solution to doing GR on manifolds with boundaries?

For purposes of the GHY term, the "boundary" could be at infinity, e.g., an asymptotically flat spacetime. (For example, as noted in the Wikipedia article [1], you need to include this term to get the correct ADM energy, which makes sense since in the vacuum case, e.g., Schwarzschild spacetime, the Ricci scalar vanishes everywhere, so without the GHY term the ADM energy would vanish and we would predict that all black holes have zero externally measured mass.) In such cases, you do not need the GHY term to actually solve the EFE. For example, you don't need it to obtain the Schwarzschild solution for the case of vacuum and spherical symmetry. You only need it to make sense of things like the nonzero ADM energy when the EFE is derived by varying the Einstein-Hilbert action.

The other applications described in the Wikipedia article appear to involve "boundaries" to the past or future (or both) of a spacetime region of interest, which is not the entire spacetime. I am not aware of any applications where the GHY boundary term is evaluated on the boundary of a compact manifold with boundary (i.e., the boundary is not at infinity) which is the entire spacetime.

[1] https://en.wikipedia.org/wiki/Gibbons–Hawking–York_boundary_term

 

[ergospherical]

Note that "Einstein manifolds" in all of these papers does not mean "manifolds obtained by solving the Einstein Field Equation". It means "manifolds for which the Ricci tensor is a multiple of the metric".

That condition is a generalisation of Einstein's equations, and it is referred to as simply Einstein's equations in much of the literature. Vacuum solutions of Einstein's equations are Einstein manifolds, for $R_{\mu \nu} \propto g_{\mu \nu}$ with proportionality constant $\Lambda$ (in dimension $n=4$).

None of these papers are talking about solving the Einstein Field Equation on manifolds with boundary; they are simply investigating general mathematical properties of Einstein manifolds with boundary.

No. The first paper specifically states that if $\partial M$ is connected* then the solutions to Einstein's equations on manifolds with boundaries, i.e. the metrics, form an infinite dimensional Banach manifold.

*(the condition $\pi_1(M, \partial M)$ also seems to be a topological requirement for some Einstein metrics).

 

[PeterDonis]

That condition is a generalisation of Einstein's equations

No, it isn't. It's a general condition on manifolds with metric. It doesn't even require a manifold to be a solution of the Einstein field equations. It can be true, for example, of Riemannian manifolds, i.e., manifolds with positive definite metrics.

Vacuum solutions of Einstein's equations are Einstein manifolds, for $R_{\mu \nu} \propto g_{\mu \nu}$ with proportionality constant $\Lambda$ (in dimension $n=4$).

If you include de Sitter and anti-de Sitter spacetimes as "vacuum" solutions, yes, this is true for those particular "vacuum" solutions (but not all relativity physicists would agree with such terminology). But it is not true for, e.g., Schwarzschild spacetime or Kerr spacetime, except in the vacuous sense that the Ricci tensor vanishes for those spacetimes, so it is "proportional" to the metric with proportionality constant zero. But that is not the intended meaning of "Einstein manifold".

The first paper specifically states that if is connected* then the solutions to Einstein's equations on manifolds with boundaries, i.e. the metrics, form an infinite dimensional Banach manifold.

No, it doesn't. By "Einstein metrics" the paper does not mean "solutions to the Einstein Field Equations". It means "metrics for which the Ricci tensor is a multiple of the metric". That definition is given in Equation 1.1 of the paper and the text immediately above it.

As noted above, such metrics do not have to be solutions of the Einstein Field Equations--they do not even have to be spacetimes (since they can be manifolds with positive definite metrics).

 

[ergospherical]

I totally disagree, so it appears we're an an impasse. Even the Wikipedia article you linked specifically describes that it's a generalisation of the field equations,

In differential geometry and mathematical physics, an Einstein manifold is a Riemannian or pseudo-Riemannian differentiable manifold whose Ricci tensor is proportional to the metric. They are named after Albert Einstein because this condition is equivalent to saying that the metric is a solution of the vacuum Einstein field equations (with cosmological constant), although both the dimension and the signature of the metric can be arbitrary, thus not being restricted to the four-dimensional Lorentzian manifolds usually studied in general relativity. Einstein manifolds in four Euclidean dimensions are studied as gravitational instantons.

 

[martinbn]

No, it doesn't. By "Einstein metrics" the paper does not mean "solutions to the Einstein Field Equations". It means "metrics for which the Ricci tensor is a multiple of the metric". That definition is given in Equation 1.1 of the paper and the text immediately above it.

As noted above, such metrics do not have to be solutions of the Einstein Field Equations--they do not even have to be spacetimes (since they can be manifolds with positive definite metrics).

Yes, but that is the vacuum equations plus a cosmological constant.

 

[martinbn]

That seems too strong a claim. I managed to find lots of papers studying the properties Einstein manifolds with boundaries, e.g.
https://arxiv.org/pdf/math/0612647.pdf

also
https://www.math.stonybrook.edu/~anderson/isomext2.pdf
https://link.springer.com/content/pdf/10.1007/s00229-021-01340-4.pdf

But your original point was about GR for spacetime with a boundary. It seems, at least at a first glance, that it would be hard to give physical meaning to the boundary.

On the other hand you shouldn't be confused by the situation where one studies only a portion of the spacetime and includes boundary conditions. Say the extirior/interior of a star plus boundary conditions at the boundary of the star.

 

[vanhees71]

Interesting. I have not worked with these manifolds with boundaries. Is it really not possible to do differential equations on such a manifold? Or is it just necessary to specify the boundary conditions at such a boundary? (As opposed to a normal manifold where you can specify the boundary conditions anywhere)

Of course you can do differential equations "on boundaries". We do this all the time in physics. E.g., you can solve the wave equation for a spherical shell, which is the boundary of the corresponding "ball". The boundaries are differentiable manifolds (without a boundary) themselves.

 

[martinbn]

Of course you can do differential equations "on boundaries". We do this all the time in physics. E.g., you can solve the wave equation for a spherical shell, which is the boundary of the corresponding "ball". The boundaries are differentiable manifolds (without a boundary) themselves.

This seems to be a terminology issue. What is meant by "to do differential equations"? In your example the solution satisfies the equation in the interior, not on the boundary, and the boundary conditions on the boundary.

 

[vanhees71]

No, my example describes the motion of the spherical shell in terms of a partial differential equation defined on the boundary. The boundary of a differentiable manifold with a boundary is a differentiable manifold without a boundary, and thus you can formulate field theories with partial differential equations of motion on them.

Of course, whether you can make physical sense of a pseudo-Riemannian spacetime with a boundary is a different question, particularly what this boundary means, is a different question.

 

[ergospherical]

But your original point was about GR for spacetime with a boundary. It seems, at least at a first glance, that it would be hard to give physical meaning to the boundary.

Yeah, I agree. I think it's somewhat an interesting thing to think about though, at least from a mathematical perspective, because there do exist physical theories where spacetime is modeled by a manifold-with-boundary, e.g. Horava-Witten supergravity with spacetime as an 11-dimensional manifold-with-boundary!

https://arxiv.org/pdf/hep-th/9603142.pdf

 

[martinbn]

No, my example describes the motion of the spherical shell in terms of a partial differential equation defined on the boundary. The boundary of a differentiable manifold with a boundary is a differentiable manifold without a boundary, and thus you can formulate field theories with partial differential equations of motion on them.

Then you have PDE on manifolds without boundaries.

 

[martinbn]

Yeah, I agree. I think it's somewhat an interesting thing to think about though, at least from a mathematical perspective, because there do exist physical theories where spacetime is modeled by a manifold-with-boundary, e.g. Horava-Witten supergravity with spacetime as an 11-dimensional manifold-with-boundary!

https://arxiv.org/pdf/hep-th/9603142.pdf

What exactly does he do?

It may not be on topic, but there are ways to consider the singularities as boundaries of the usual spacetimes.

 

[martinbn]

Going back to classical GR, if the spacetime has a boundary, then what happens to the equivalence principle? The spacetime will not look like Minkowski locally around an event on the boundary.

 

[ergospherical]

What exactly does he do?

It may not be on topic, but there are ways to consider the singularities as boundaries of the usual spacetimes.

I found some more thorough explanations in these lecture notes: https://arxiv.org/abs/hep-th/0201032
The idea seems that instead of working on a $11\mathrm{d}$ manifold $M$ with a $Z_2$ symmetry (i.e. in this case reflection symmetry of $S^1$), they work with the quotient $M/Z_2$ manifold-with-boundary because it's apparently more intuitive (but harder computationally...) and in this new picture the boundaries have their own dynamics (in the form of a Yang-Mills field theory).

 

[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".