Manifolds: local & global coordinate charts

[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)?!

 

[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 $\mathbb R$.

 

[micromass]

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)?!

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

 

[andrewkirk]

Noting Micromass's counterexample to Frank's hypothesis, it's salient to note that the following

the topology of a given coordinate chart is that of Euclidean space

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 $\mathbb R^2$.

 

[Frank Castle]

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 $\mathbb R$.

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?!)

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

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)?

 

[micromass]

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

 

[Frank Castle]

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

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:

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

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?!

 

[micromass]

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?

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

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?!

Yes.

 

[Frank Castle]

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



Yes.

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 $\mathbb{R}^{n}$?!

 

[micromass]

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 $\mathbb{R}^{n}$?!

Yes.

 

[Frank Castle]

Yes.

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?!)

 

[Orodruin]

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.

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

In the case of Minkowski space, it is globally homeomorphic to an open subset of RnRn\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.

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.

 

[Frank Castle]

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

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 traveling 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?!

 

[Orodruin]

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)?!

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

In the case of red shifted photon traveling from a distant galaxy

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.

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?!

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.

 

[Frank Castle]

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

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).

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.

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 traveling 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)

What in one coordinate system appears as a change in photon frequency might appear very differently in a different coordinate system.

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.)?!

 

[Orodruin]

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=0R=0R=0, then the geometry will be flat).

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.

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 traveling 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?

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.

In a gravitational field however a photon is definitely redshifted right? (Heuristically it loses energy to overcome the gravitational potential)

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 $\partial_t$.

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.)?!

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

 

[Frank Castle]

No, a Euclidean space is by definition the plane R2R2\mathbb R^2 or R3R3\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 R3R3\mathbb R^3. Furthermore, Minkowski space is flat and does not have a Euclidean metric.

Yes, sorry I hadn't really thought about that properly.

The things you mention, such as the (coordinate) velocity of distant photons are coordinate dependent.

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.

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.

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 realize 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.

 

[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 $\nabla_{\mu}T^{\mu\nu} =0$?!

No, a Euclidean space is by definition the plane R2R2\mathbb R^2 or R3R3\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 R3R3\mathbb R^3. Furthermore, Minkowski space is flat and does not have a Euclidean metric.

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?!

 

[Orodruin]

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?

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.

 

[Frank Castle]

No, it is true only if the manifold is simply connected.

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?!

 

[Orodruin]

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?!

They are simply connected and flat manifolds. For flat manifolds that are not simply connected, the parallel transport is the same for any two curves that are homotopic given fixed end points.

 

[Frank Castle]

They are simply connected and flat manifolds. For flat manifolds that are not simply connected, the parallel transport is the same for any two curves that are homotopic given fixed end points.

Ok, thanks for the details.

What are your thoughts on the other parts of my post #18 (the first paragraph)?

 

[Orodruin]

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.

This is a point already in SR. In GR it holds true in a local coordinate system as well. Even if two observers are at the same event where a light wave passes by, they will not measure the same frequency (unless they are at relative rest) due to the non-invariance of the energy under local Lorentz transformations.

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 p0=E′=∂x′0∂xμpμp0=E′=∂x′0∂xμpμp^{0}=E'=\frac{\partial x'^{0}}{\partial x^{\mu}}p^{\mu}?!

Do not be too quick in assigning an energy to the photon as $p^0$. This is only true if your coordinates are chosen appropriately. The energy for an observer is $E = g(V,P)$, where $V$ is the observer 4-velocity and $P$ the photon 4-momentum. This is not equal to the component $p^0$ unless the dual vector $g(V,\cdot)$ has $V_0 = 1$ as the only non-zero component.

Furthermore, unlike in SR, where the energy of a particle at point ppp as measured by an observer with velocity vμvμv^{\mu} at point qqq is given by E=−ημνvμ(q)pν(p

Even in SR, you should only take inner products of vectors at the same point. However, as we have discussed, the parallel transport to that point is unique in Minkowski space and in Minkowski coordinates the components of the vector are constant under this parallel transport.

This confuses me though since I thought energy-momentum conservation in GR was implied by ∇μTμν=0∇μTμν=0\nabla_{\mu}T^{\mu\nu} =0?!

This is energy-momentum conservation in a local frame. It does not imply global energy-momentum conservation. Globally the very definition of energy and momentum are doubtful and usually dependent on arbitrary assumptions.

 

[Frank Castle]

This is a point already in SR. In GR it holds true in a local coordinate system as well. Even if two observers are at the same event where a light wave passes by, they will

Is this because, for a local Lorentz transformation (between two local inertial frames), $E'=\Lambda^{0}_{\;\mu}p^{\mu}\neq E$ (unless the two frames are at rest with respect to one another)?!

Do not be too quick in assigning an energy to the photon as p0p0p^0.

I thought that the zeroth component of 4-momentum of a given particle was defined to be the energy (in units where $c=1$) though?

The energy for an observer is E=g(V,P)E=g(V,P)E = g(V,P),

Is the energy for an observer simply define as this out can this equation be derived?

Even in SR, you should only take inner products of vectors at the same point.

Good point. I was a bit sloppy with that point - I implicitly assumed that the components of a vector at different points are identical in my statement.

the parallel transport to that point is unique in Minkowski space and in Minkowski coordinates the components of the vector are constant under this parallel transport

Incidentally, how does one prove both these statements?

 

[Orodruin]

s this because, for a local Lorentz transformation (between two local inertial frames), E′=Λ0μpμ≠EE′=Λμ0pμ≠EE'=\Lambda^{0}_{\;\mu}p^{\mu}\neq E (unless the two frames are at rest with respect to one another)?!

Essentially.

I thought that the zeroth component of 4-momentum of a given particle was defined to be the energy (in units where c=1c=1c=1) though?

In SR, this is only true in Minkowski coordinates. In GR, there may be such coordinates, but it will not be generally true.

Is the energy for an observer simply define as this out can this equation be derived?

It is a straight-forward generalisation of the corresponding SR expression. It is going to locally correspond to the same thing - the product of the wave vector with the observer 4-velocity.

Incidentally, how does one prove both these statements?

Construct the parallel transport around a loop as the contributions from several infinitesimal loops. This can be done since Minkowski space is simply connected. The change in the parallel transported vector around an infinitesimal loop is proportional to the curvature tensor, which for a flat space is zero everywhere.

 

[Frank Castle]

It is a straight-forward generalisation of the corresponding SR expression. It is going to locally correspond to the same thing - the product of the wave vector with the observer 4-velocity.

I'm ashamed to say that I've never seen the derivation of the expression in SR either. Would you be able to show me, and/or suggest a book to read up on it?!

Construct the parallel transport around a loop as the contributions from several infinitesimal loops. This can be done since Minkowski space is simply connected. The change in the parallel transported vector around an infinitesimal loop is proportional to the curvature tensor, which for a flat space is zero everywhere.

Is it essentially then, that $\delta v^{\mu}\propto R^{\mu}_{\;\nu}v^{\nu}$ (where $\delta v^{\mu}$ denotes the change in the components of a vector under parallel transport around an infinitesimal loop and $R^{\mu}_{\;\nu}$ is the Ricci curvature tensor) and for a flat space $R^{\mu}_{\;\nu}=0$, hence $\delta v^{\mu}=0$ and the components of a vector are invariant under parallel transport?!

 

[Orodruin]

I'm ashamed to say that I've never seen the derivation of the expression in SR either. Would you be able to show me, and/or suggest a book to read up on it?!

If you are considering a photon, it is not purely SR. You then also need to add QFT on top. If you want to do it hand-wavingly you can refer to the de Broglie frequency of the photon and study the corresponding 4-frequency, whose zeroth component is frequency that according to de Broglie theory is proportional to energy.

But if you want to fo it properly for a quantum particle, you need to study QFT. Referring to photons in SR without using QFT is often quite misleading. You may be better off computing the energy momentum tensor from classical electrodynamics to find the energy density and energy currents related to a particular wave.

Is it essentially then, that δvμ∝Rμνvνδvμ∝Rνμvν\delta v^{\mu}\propto R^{\mu}_{\;\nu}v^{\nu} (where δvμδvμ\delta v^{\mu} denotes the change in the components of a vector under parallel transport around an infinitesimal loop and RμνRνμR^{\mu}_{\;\nu} is the Ricci curvature tensor) and for a flat space Rμν=0Rνμ=0R^{\mu}_{\;\nu}=0, hence δvμ=0δvμ=0\delta v^{\mu}=0 and the components of a vector are invariant under parallel transport?!

No, this is true only in a two-dimensional manifold where the curvature tensor only has one independent component. In general, you will need the (rank 4) Riemann curvature tensor along with the directions that define the infinitesimal loops. But the concept is the same.

 

[Frank Castle]

If you are considering a photon, it is not purely SR. You then also need to add QFT on top. If you want to do it hand-wavingly you can refer to the de Broglie frequency of the photon and study the corresponding 4-frequency, whose zeroth component is frequency that according to de Broglie theory is proportional to energy.

But if you want to fo it properly for a quantum particle, you need to study QFT. Referring to photons in SR without using QFT is often quite misleading. You may be better off computing the energy momentum tensor from classical electrodynamics to find the energy density and energy currents related to a particular wave

Ah ok, I'll have to look into that then. Thanks for all the info.

No, this is true only in a two-dimensional manifold where the curvature tensor only has one independent component. In general, you will need the (rank 4) Riemann curvature tensor along with the directions that define the infinitesimal loops. But the concept is the same.

Would it be something like
$$\delta v^{\mu}= \delta x^{\alpha}\delta x^{\beta}R^{\mu}_{\;\alpha\nu\beta}v^{\beta}$$ where $\delta x^{\alpha}$ and $\delta x^{\beta}$ quantify infinitesimal displacements along the two different directions around an infinitesimal closed loop.

 

[Orodruin]

Would it be something like

δvμ=δxαδxβRμανβvβδvμ=δxαδxβRανβμvβ​

\delta v^{\mu}= \delta x^{\alpha}\delta x^{\beta}R^{\mu}_{\;\alpha\nu\beta}v^{\beta} where δxαδxα\delta x^{\alpha} and δxβδxβ\delta x^{\beta} quantify infinitesimal displacements along the two different directions around an infinitesimal closed loop.

Something like that, but not exactly. Be careful with which indices you contract with what. With the definition $R(\partial_a,\partial_b)\partial_c = R_{cab}^d \partial_d$, the contraction should be $\delta v^a = \delta x^b \delta x^d R_{cbd}^a v^c$.

 

[Frank Castle]

Something like that, but not exactly. Be careful with which indices you contract with what. With the definition $R(\partial_a,\partial_b)\partial_c = R_{cab}^d \partial_d$, the contraction should be $\delta v^a = \delta x^b \delta x^d R_{cbd}^a v^c$.

Ah ok, what would it be exactly? (I know that parallel transport along a curve would be $\delta v^{\mu}=-\Gamma^{\mu}_{\;\nu\alpha}v^{\alpha}\delta^{\nu}$). Is the point essentially that, on a flat manifold, $R_{cbd}^a=0$, and so $\delta v^{\mu}=0$, i.e. the components of the vector remain unchanged under parallel transport in flat space?!
Yes, apologies about the indices, I was trying to think of it off the top of my head and couldn't exactly remember which indices contract with which

 

[Orodruin]

Ah ok, what would it be exactly?

What would what be? In order to derive the relation with the Riemann tensor, you need to do several parallel transports (4 to be exact) and remember to take into account that the Christoffel symbols will change (hence the appearance of the derivatives of the Christoffel symbols in the Riemann tensor).

Is the point essentially that, on a flat manifold, Racbd=0Rcbda=0 R_{cbd}^a=0, and so δvμ=0δvμ=0\delta v^{\mu}=0, i.e. the components of the vector remain unchanged under parallel transport in flat space?!

No, again we are talking about a parallel transport around a closed loop. If the components change during an arbitrary parallel transport depends on the coordinate system and whether the Christoffel symbols are zero in that particular system or not (remember that the Christoffel symbols being zero is not equivalent to zero curvature!). For example, the components of a vector in Euclidean space do change under parallel transport if you use spherical coordinates. The point is that when you parallel transport around a loop you end up with a vector in the same tangent space as you started in! (In other words, a closed loop defines a type (1,1) tensor at the start/end-point through parallel transport along it and so finding $\delta v = 0$ is a coordinate independent statement.)

 

[Frank Castle]

What would what be? In order to derive the relation with the Riemann tensor, you need to do several parallel transports (4 to be exact) and remember to take into account that the Christoffel symbols will change (hence the appearance of the derivatives of the Christoffel symbols in the Riemann tensor).

Ah ok, I'll have to work through it then.

No, again we are talking about a parallel transport around a closed loop. If the components change during an arbitrary parallel transport depends on the coordinate system and whether the Christoffel symbols are zero in that particular system or not (remember that the Christoffel symbols being zero is not equivalent to zero curvature!). For example, the components of a vector in Euclidean space do change under parallel transport if you use spherical coordinates. The point is that when you parallel transport around a loop you end up with a vector in the same tangent space as you started in! (In other words, a closed loop defines a type (1,1) tensor at the start/end-point through parallel transport along it and so finding δv=0δv=0\delta v = 0 is a coordinate independent statement.)

Good point, I hadn't fully thought that through.

Thanks for all of your help, I really appreciate it!

 

[Orodruin]

Ah ok, I'll have to work through it then.

One thing to note is that it is usually much (much!) easier to work it through if you start the parallel transport at the middle of the loop and perform the parallel transport in the two different directions. This should also be done in any introductory text on the calculus on manifolds.

 

[Frank Castle]

One thing to note is that it is usually much (much!) easier to work it through if you start the parallel transport at the middle of the loop and perform the parallel transport in the two different directions. This should also be done in any introductory text on the calculus on manifolds.

Thanks for the tips, are there any books that you would recommend in particular?

Also, one thing I'm slightly confused over now is, if one wishes to compare two vectors at different points in a flat space, then one can uniquely parallel transport one of the vectors to the other and compare them at the same point in a well defined manner. However, this is not around a closed loop and so the components of the parallel transported vector will change, in general (unless one uses Cartesian coordinates), so how can one meaningfully compare the two vectors (for example, suppose it is the same vector, but at two different points, with the same components at both points)? (Apologies, this may be a stupid question - it's a bit late at night and my brain has gone a bit to mush)

 

[Orodruin]

However, this is not around a closed loop and so the components of the parallel transported vector will change, in general (unless one uses Cartesian coordinates), so how can one meaningfully compare the two vectors (for example, suppose it is the same vector, but at two different points, with the same components at both points)?

If the vectors are in the same vector space, you can compare them using the normal addition/subtraction in that vector space. It does not matter what the components are in some arbitrary coordinate system. You simply do the parallel transport and compare the result with the other vector at the point.