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

[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.

 

[cianfa72]

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.

The claim about the metric unchanged applies in a neighborhood of each point along the integral curves of Killing vector field corresponding to the symmetry.

 

[PeterDonis]

The claim about the metric unchanged applies in a neighborhood of each point along the integral curves of Killing vector field corresponding to the symmetry.

No, it doesn't. It only applies on the integral curves themselves. The metric on two adjacent integral curves does not have to be the same.

Euclidean space is actually a bad example because it's flat everywhere. N-spheres with the standard metric are also bad examples because they have constant curvature. A better example would be, for example, Schwarzschild spacetime outside the horizon, where the metric is a function of $r$, so adjacent integral curves of the timelike KVF with slightly different values of $r$ have different metrics.

 

[cianfa72]

A better example would be, for example, Schwarzschild spacetime outside the horizon, where the metric is a function of $r$, so adjacent integral curves of the timelike KVF with slightly different values of $r$ have different metrics.

In this case the integral curves of the timelike KVF are the worldlines of hovering observers at fixed $(r,\theta,\phi)$. So the spacetime metric will be different for such two hovering observers at slightly different $r$ values (their $\theta, \phi$ coordinates are irrelevant).

 

[cianfa72]

Euclidean space is actually a bad example because it's flat everywhere.

So in Euclidean space any (smooth) curve joining two points is actually an integral curve of a KVF, I believe.

 

[PeterDonis]

In this case the integral curves of the timelike KVF are the worldlines of hovering observers at fixed $(r,\theta,\phi)$. So the spacetime metric will be different for such two hovering observers at slightly different $r$ values (their $\theta, \phi$ coordinates are irrelevant).

Yes, that's what I said.

 

[PeterDonis]

in Euclidean space any (smooth) curve joining two points is actually an integral curve of a KVF, I believe.

No. But you are illustrating why I said that Euclidean space is a bad example for understanding symmetries and KVFs.

Euclidean 3-space has a 3-parameter group of rotational KVFs and a 3-parameter group of translational KVFs. The rotational KVF integral curves are circles and the translational KVF integral curves are straight lines. The metric is of course constant along any integral curve of any of those KVFs; but the converse is not true, that any curve along which the metric is constant must be an integral curve of some KVF. But a manifold like Euclidean space, or indeed any manifold with constant curvature, can mislead one into thinking that the converse should be true, as it apparently has misled you.

 

[cianfa72]

The metric is of course constant along any integral curve of any of those KVFs; but the converse is not true, that any curve along which the metric is constant must be an integral curve of some KVF.

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

 

[jbergman]

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

I am new to KVF's but according to wikipedia the distinguishing feature is
$$\mathcal{L}_{X} g = 0$$
i.e., the Lie Derivative of the metric along the Killing Vector Field is 0.

An equivalent statement given in Lee's "Introduction to Riemannian Manifolds" is that the metric is invariant under the flow generated by the KVF $X$.
$$\theta_t^{*}g=g$$
So, the flow generated by a KVF gives us isometries which preserve the metric.

 

[DrGreg]

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

It's not just the spacetime that possesses the corresponding symmetry. The integral curves possesses the same symmetry.

For example, in Euclidean space, a family of parallel straight lines possesses translational symmetry in the common direction of the lines. A family of coaxial circles possesses rotational symmetry about the common axis.

For a rigorous definition, you have to look at the definition of a KVF.

There's also another way to look at it. If you find a coordinate system that has the property that, along every integral curve of the vector field, $(N-1)$ of the $N$ coordinates are constant, and all $N^2$ components of the metric tensor are independent of the $N$th coordinate, then your vector field is a KVF.

In the case of Schwarzschild coordinates, all metric components are independent of both $t$ and $\phi$ (but not $r$ and $\theta$), so that gives you two different families of integral curves corresponding to time translation symmetry (stationary spacetime) and rotational symmetry.

 

[jbergman]

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.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.

Thank you for this informative answer. I wasn't familiar with KVFs before this, but your point about distinguishing between coordinate transformations and invariant properties of the manifold was helpful.

That said, I might quibble a bit with your answer after looking into this. As I understand it, the KVFs only give us continuous symmetries. So something like a reflection in Euclidean space would not be included in them and that if we want to talk about all symmetries of a space then we would be talking about all isometries of that space.

 

[cianfa72]

I am new to KVF's but according to wikipedia the distinguishing feature is
$$\mathcal{L}_{X} g = 0$$
i.e., the Lie Derivative of the metric along the Killing Vector Field is 0.

So, in Euclidean space if we pick a generic curve (that is not the integral curve of any KVF) the Lie Derivative of the metric tensor $g$ along the curve's tangent vector field $X$ does not vanish, namely $\mathcal{L}_{X} g \neq 0$. Yet the metric tensor $g$ does not change along that curve.

Edit: since the curve's tangent vector is not defined off the curve, I believe we need a congruence filling the Euclidean space such that one of its curves is actually the given curve joining the points. Then we can evaluate the Lie derivative of the metric tensor $g$ along this vector field $X$.

 

[jbergman]

So, in Euclidean space if we pick a generic curve (that is not the integral curve of any KVF) the Lie Derivative of the metric tensor $g$ along the curve's tangent vector field $X$ does not vanish, namely $\mathcal{L}_{X} g \neq 0$. Yet the metric tensor $g$ does not change along that curve.

I don't think that's true. Think of a dilation, i.e. a vector field that points outward but gets larger the further out you go. The flow associated with that changes the distance between points and the metric. You have to flow all the points of the manifold the same flow parameter time.

 

[PAllen]

Just a quick point about the special case of a constant scalar curvature manifold (at every point, sectional curvature defined by a pair of vectors is the same for every pair, and also the same for every point). In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

 

[jbergman]

Just a quick point about the special case of a constant scalar curvature manifold (at every point, sectional curvature defined by a pair of vectors is the same for every pair, and also the same for every point). In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

Right, but you can construct vector fields that are not KVFs and the corresponding flow will not preserve the metric or have 0 Lie Derivative.

 

[cianfa72]

Just a quick point about the special case of a constant scalar curvature manifold. In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

So in this special case (i.e. constant scalar curvature manifold) the Lie Derivative of the metric tensor $g$ along every tangent vector at every point is always null (strictly speaking we need a vector field $X$ along which calculate the Lie derivative: any smooth vector field will do the job).