A Manifolds: local & global coordinate charts

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1. Jul 8, 2016

Frank Castle

I'm fairly new to differential geometry (learning with a view to understanding general relativity at a deeper level) and hoping I can clear up some questions I have about coordinate charts on manifolds.

Is the reason why one can't construct global coordinate charts on manifolds in general because the topology of a given coordinate chart is that of Euclidean space (i.e. $\mathbb{R}^{n}$ with the standard topology), whereas, in general the global topology of the manifold will be much more complex?! If this is the case, can one only construct a global coordinate chart on a manifold if its global topology is the standard Euclidean topology (does Minkowski spacetime have the standard Euclidean topology with a Minkowski geometry)?!

2. Jul 8, 2016

andrewkirk

That's broadly it, but I wouldn't say it's about complexity. The simplest example I can think of is the one-dimensional manifold $S^1$, which is the perimeter of the unit circle in $\mathbb R^2$. It is not homeomorphic to any subset of $\mathbb R$ because it is path-connected but not simply connected, and there is no subset of $\mathbb R$ that is path-connected but not simply connected. So there can be no coordinate chart between the whole of $S^1$ and a subset of $\mathbb R$.

3. Jul 8, 2016

micromass

Staff Emeritus
No. There are nontrivial manifolds which do admit a global coordinate charts. For example, the cylinder.

4. Jul 8, 2016

andrewkirk

Noting Micromass's counterexample to Frank's hypothesis, it's salient to note that the following
is not quite correct. The domain of a coordinate chart is not required to have a topology that is homeomorphic to an entire Euclidean space, only to an open subset of a Euclidean space.

In the cylinder case, assuming the cylinder to be infinitely long so we don't have to worry about whether it's a manifold-with-boundary, the cylinder's topology is homoeomorphic to that of an open annulus or a pierced plane, both of which are proper subsets of $\mathbb R^2$.

5. Jul 9, 2016

Frank Castle

Ah ok. Is there a general reason though why one cannot cover a manifold with a single global coordinate chart, in general? I read in one set of notes that is because the structure of the manifold will, in general, be more complex than the structure of the coordinate chart (I'm not really sure what structure they are referring to?!)

Thanks for pointing this out.

Is the reason why one can only consider local properties in general relativity then because spacetime cannot be covered by a single coordinate chart in general and so one can only consider local quantities with respect to local coordinate charts (of course these quantities are coordinate independent, but their coordinate representative are local)?

6. Jul 9, 2016

micromass

Staff Emeritus
Why do you think we can't consider global properties in GR?

7. Jul 9, 2016

Frank Castle

I was referring to things such as velocities, metric tensor, energy-momentum tensor etc. I appreciate that one can have global quantities such as curvature as well, right?

With respect to your earlier comment:

Is it that the manifold has to be globally homeomorphic to an open subset of $\mathbb{R}^{n}$ in order for it to admit a global coordinate chart?!

8. Jul 9, 2016

micromass

Staff Emeritus
Right, we can talk about compactness of the spacetime for example.

Yes.

9. Jul 9, 2016

Frank Castle

Is there a general reason why one cannot cover a manifold with a single coordinate chart in general then? Is it simply that in general a manifold will not be globally homeomorphic to an open subset of $\mathbb{R}^{n}$?!

10. Jul 9, 2016

micromass

Staff Emeritus
Yes.

11. Jul 9, 2016

Frank Castle

Ok cool.

Also, my questions relating GR originally arose because I was trying to rationalise with myself why it doesn't make sense to consider things such as velocities of distant objects and energy on larger scales. My thought was that one can only compare velocities of objects meaningfully it they are compare at the same point, because vectors at different points are in different tangent spaces. To compare them we have to parallel transport one to the other, and thus is very much a part dependent procedure in general, hence velocities of distant objects are not well defined on a general manifold. In the case of Minkowski space, it is globally homeomorphic to an open subset of $\mathbb{R}^{n}$ and so one can construct global coordinate charts and tangent spaces at different points are naturally isomorphic (the path of parallel transports along its unique), and so one can meaningfully compare velocities of distant objects with one in your local neighbourhood.
A similar argument holds for comparing energies of distant objects, I'm guessing (as energy-momentum is only conserved locally)?! (The reason I ask here is to try and explain it is not meaningful to say that distant photons that have been redshifted have lost energy?!)

12. Jul 9, 2016

Orodruin

Staff Emeritus
Exactly. This has more to do with the geometry of the manifold than the possibility of covering it with a single chart.

No, the crucial thing is that Minkowski space is flat and therefore the parallel transport is path-independent. This depends on the affine connection and not on whether the manifold allows a global coordinate chart.

13. Jul 9, 2016

Frank Castle

So is the point that even if a given manifold can be converted by a single coordinate chart, if the geometry (encoded the metric tensor) is not flat (Euclidean) then parallel transport not path independent and so velocities of distant objects are not well defined (in the sense that it is not meaningful to compare locally velocities with distant velocities because the result will depend on the path)?!

In the case of red shifted photon travelling from a distant galaxy, is the point that the energy-momentum tensor is only locally conserved and so the energy lost by the photon due to the expansion of the universe is simply lost, it doesn't go anywhere?!

14. Jul 9, 2016

Orodruin

Staff Emeritus
Right, although generally "flat" is not the same as "Euclidean".

The photon itself is not redshifted. The observed frequency is a combination of the 4-momentum of the photon, which is parallel transported along a geodesic, and the 4-velocity of the observer.

In general, you cannot define something such as "the total energy of the Universe" in a meaningful way. In order to define something like that you must integrate the energy-momentum tensor over some arbitrary space-like surface. Again, the photon in itself is not red-shifted. The observed frequency depends on both the photon 4-momentum and the observer 4-velocity. What in one coordinate system appears as a change in photon frequency might appear very differently in a different coordinate system. The invariant measured quantity is $P\cdot V$, where $P$ is the photon 4-momentum and $V$ the observer 4-velocity at the observation event.

15. Jul 9, 2016

Frank Castle

I thought though that what makes a geometry flat is the fact that it's described by a Euclidean metric (although I understand that if one has "Ricci flatness", i.e. $R=0$, then the geometry will be flat).

So if the point that the observed redshift of the photon is a coordinate frame dependent observation and so there is no intrinsic meaning to it (since the frequency of the photon is an observer dependent quantity)?! Does this essentially all boil down to the that the velocity of distant photons not a well defined notion relative to us, the photons wavelength appears to be stretched due to the fact that the geometry between us and the photon is changing due to the expansion of the universe and hence the relative distance between us and the object from which the light is travelling from is constantly increasing, hence one end up with a Doppler shift effect, and no one says that energy is lost during a Doppler shift, it is simply due to the relative position of the observer to the propagating wave?
In a gravitational field however a photon is definitely redshifted right? (Heuristically it loses energy to overcome the gravitational potential)

Is this analogous to the reason why one can't have a notion of two objects being parallel (in the usual Euclidean sense), because what looks parallel in one coordinate system will not do so in another coordinate system (in the extreme, in the overlap between two neighbouring coordinate charts, two objects would appear both parallel and not parallel at the same time - obviously this is an ill defined concept.)?!

Last edited: Jul 9, 2016
16. Jul 9, 2016

Orodruin

Staff Emeritus
No, a Euclidean space is by definition the plane $\mathbb R^2$ or $\mathbb R^3$ or their higher dimensional generalisations. There are many other flat spaces, such as the typical cylinder or cone with the metric induced from the natural embedding in $\mathbb R^3$. Furthermore, Minkowski space is flat and does not have a Euclidean metric.

The observed red-shift of light depends on: emitter motion, the geometry of the space-time, and the motion of the receiver. Without those three, you cannot compare the emitted frequency with the observed frequency. It boils down to the geometry of space-time, nothing else. The things you mention, such as the (coordinate) velocity of distant photons are coordinate dependent. It is unclear what you mean by "eometry between us and the photon is changing due to the expansion of the universe". In a particular coordinate system (comoving coordinates), it is well defined what you mean by stretching the wavelength of the radiation because there is an implicit assumption of comoving observers and emitters. This description is however coordinate dependent and will not be true in any coordinate system.

No. Again, this depends on your choice of coordinates. The typical description here is in terms of stationary observers for which the photon frequency would be well defined. The "energy" is a constant of motion along the light-like geodesic which takes a particular form for the stationary observers, i.e., related to the time-like Killing field $\partial_t$.

What do you mean by "look parallel"? A vector field is either parallel or not, this is not a coordinate dependent statement. It only depends on the vector field and the affine connection.

Edit: I suggest reading the paper that is linked in this thread: On the relativity of red shift

Last edited: Jul 9, 2016
17. Jul 9, 2016

Frank Castle

By this I meant that because parallel transport is path dependent in non Euclidean geometry, the velocities of distant objects are not meaningful relative to our frame of reference.

By this I meant that two curves (lines) that appear parallel (using the Euclidean notion of parallel) to one another in one coordinate frame with not necessarily appear parallel to one another in another coordinate frame. In hindsight I realise I was speaking complete rubbish (Apologies for this!), since parallelism is a geometrical notion, and like you say, is consistently introduced in non Euclidean geometry through the structure of an affine connection.

Thanks for the link by the way, I'll take a look now.

18. Jul 9, 2016

Frank Castle

Is the point then that energy is a relative (i.e. frame dependent) quantity and so it is not meaningful to say that a photon has lost energy due to redshifting. The energy that one measures a photon to have its relative to a given reference frame (it is the 4-momentum that is frame independent) and the energies measured in two different frames of reference (for the same photon) are related by $p^{0}=E'=\frac{\partial x'^{0}}{\partial x^{\mu}}p^{\mu}$?! Furthermore, unlike in SR, where the energy of a particle at point $p$ as measured by an observer with velocity $v^{\mu}$ at point $q$ is given by $E=-\eta_{\mu\nu}v^{\mu}(q)p^{\nu}(p)$ (where $p^{\mu}$ is the 4-momentum of the particle. By the way, is it possible to derive this equation of its it simply a definition?!), the energy of a particle at one point in spacetime relative to an observer at another point is not well defined since one cannot compare vectors residing in two different tangent spaces - one has to parallel translate one vector to the other and this procedure is path dependent in GR.

I read a post by Sean Carroll http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/ in which he argues that energy is not conserved in GR in the sense that when the space through which particles move changes, the energy of those particles is not conserved. This confuses me though since I thought energy-momentum conservation in GR was implied by $\nabla_{\mu}T^{\mu\nu} =0$?!

On a side note, it the geometry defined on a manifold is flat, is it always true parallel transport of vectors is path independent, such that velocity vectors at arbitrary points can be meaningfully compared? In Minkowski spacetime think this is true right? And in this case one can compare ones velocity with object at an arbitrary spacetime point because parallel transport is path independent and thus quantifying the velocity of a distant object relative to yourself is well defined?!

Last edited: Jul 9, 2016
19. Jul 9, 2016

Orodruin

Staff Emeritus
No, it is guaranteed only if the manifold is simply connected. Then the parallel transport around any loop gives back the original vector. An example of a flat smooth manifold where this is not the case is a cone with the apex removed.

Edit: For closed curves with non-trivial homotopy (i.e., that go around the apex) it will not be true, although it will still be true for loops not going around the apex.

Edit 2: Edited for preciseness. There are clearly some manifolds that are not simply connected where parallel transport around any loop also gives back the same vector, but it is true for all flat and simply connected manifolds.

Last edited: Jul 12, 2016
20. Jul 9, 2016

Frank Castle

Ah ok. So, the reason why parallel transport is path independent in Euclidean space and Minkowski spacetime is because they are both simply connected manifolds then?!