# Manifold with a boundary

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• Dale
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#### Dale

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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 and vanhees71
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!

• PeroK
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.)

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• • PeroK, vanhees71 and 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.

• Demystifier and vanhees71
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.

• ergospherical
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

• ergospherical and Dale
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).

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.

• vanhees71 and Dale
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.

• vanhees71 and 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)

• vanhees71
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...

• jbergman, dextercioby, Astronuc and 1 other person
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.

• Dale
I managed to find lots of papers studying the properties Einstein manifolds with boundaries
Thanks for the references, I'll take a look.

• Dale
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.

• ergospherical
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 , 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.

 https://en.wikipedia.org/wiki/Einstein_manifold

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

• haushofer and ergospherical
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 , 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.

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

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).

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).

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.

• Dale
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.

• 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
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 and ergospherical
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.

• jbergman, Dale and ergospherical
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.

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.

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

• 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.
Then you have PDE on manifolds without boundaries.

• vanhees71
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.

• vanhees71
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.

• dextercioby and vanhees71
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).

• vanhees71