# Confusion wiht Curvature Tensor

facenian
I Have a problem understanding that vanishing of the curvature tensor implies that paralel transport is independent of path. With the converse of this assertion I have no problem.
The text I'm reading(Lovelock and Rund) explains the converse but treats the direct assertion as trivial. Can someone shed some light on this?

Mentz114
I can see what you mean. The covariant derivative depends on the path ( a vector V) through the connections, and the connections can be non-zero when the Riemann tensor is zero.

Mentor
2021 Award
Yes, but if the curvature tensor vanishes everywhere then it is always possible to transform to a global coordinate system where the connections are all zero.

Mentz114
Yes, but if the curvature tensor vanishes everywhere then it is always possible to transform to a global coordinate system where the connections are all zero.

That sounds highly plausible.

What about the spherical surface where the transformation to rectilinear coords is undefined at the 'poles' ?

Or maybe that's not a suitable transformation.

Mentor
2021 Award
What about the spherical surface where the transformation to rectilinear coords is undefined at the 'poles' ?
The curvature tensor doesn't vanish anywhere on a sphere.

Mentz114
The curvature tensor doesn't vanish anywhere on a sphere.

I had to check this, by writing flat spacetime in spherical polar coords with a constant r coordinate and calculating Rabcd. You're right, the Riemann tensor is not zero. I was surprised.

Mentor
2021 Award
Yeah, it can be surprising to realize that an embedded space may be curved even if the embedding space is flat. I had a guy a while back that assumed that space was flat in a rotating reference frame simply because the spacetime was flat. Couldn't get it through to him even posting a link to the math fully worked out.

Staff Emeritus
Gold Member
I Have a problem understanding that vanishing of the curvature tensor implies that paralel transport is independent of path. With the converse of this assertion I have no problem.
The text I'm reading(Lovelock and Rund) explains the converse but treats the direct assertion as trivial. Can someone shed some light on this?

The answer to your question might depend a lot on what you take to be the definition of the curvature tensor. The definition I would give, boiled down into nonmathematical terms, is that the curvature tensor tells you the extent to which parallel transport is path dependent, for certain types of infinitesimal paths starting from a given point. With that definition, the only issue I can see is that path-independence for infinitesimal paths needs to be extended to path-independence for finite paths.

Are we assuming no torsion?

facenian
The answer to your question might depend a lot on what you take to be the definition of the curvature tensor. The definition I would give, boiled down into nonmathematical terms, is that the curvature tensor tells you the extent to which parallel transport is path dependent, for certain types of infinitesimal paths starting from a given point. With that definition, the only issue I can see is that path-independence for infinitesimal paths needs to be extended to path-independence for finite paths.

Are we assuming no torsion?

We are not assuming torsion=0. The definition of curvature tensor we are assuming is that which gives the expresion for the diference between the second covariant derivatives o a tensor field(torsion does not appear in this expression when the field is of type (1,0) ). The text I metioned also explains that the same tensor must vanish for paralel transport be path independent. My problem is the sufficient part.

Mentz114
facenian said:
The definition of curvature tensor we are assuming is that which gives the expresion for the diference between the second covariant derivatives o a tensor field(torsion does not appear in this expression when the field is of type (1,0) ).

Is this the one ?

$${\lambda^a}_{;bc}-{\lambda^a}_{;cb}={R^a}_{dbc}\lambda^d$$

This equation can be interpreted

1. If in the the left-hand side we replace the covariant derivative with a partial drivative, then Radbc = 0

2. If on the RHS we have Radbc = 0, then the commutator is zero.

facenian
Is this the one ?

$${\lambda^a}_{;bc}-{\lambda^a}_{;cb}={R^a}_{dbc}\lambda^d$$

Yes, it is

Is this the one ?
This equation can be interpreted

1. If in the the left-hand side we replace the covariant derivative with a partial drivative, then Radbc = 0

2. If on the RHS we have Radbc = 0, then the commutator is zero.

1. I think that if you replace the covariant derivates by partial derivates then identity is no longer valid

2. Yes, nulity of the curvature tensor implies comutativity of covariant derivates, my question is: why the nulity of tensor curvature is sufficient condition for parallel transport being path independent?

Mentz114
1. I think that if you replace the covariant derivates by partial derivates then identity is no longer valid

If parallel transport is path-independent the covariant derivatives become partial derivatives, don't they ?

I'm sorry, I'm not a mathematician and your problem is too subtle for me.

Mentor
2021 Award
Me too, I have only worked with torsion free connections.

Gold Member
I Have a problem understanding that vanishing of the curvature tensor implies that paralel transport is independent of path. With the converse of this assertion I have no problem.
The text I'm reading(Lovelock and Rund) explains the converse but treats the direct assertion as trivial. Can someone shed some light on this?

In general it is not true that parallel translation is independent of path on a flat manifold.
What is true for flat (zero curvature tensor) Riemannian manifolds is that the there are only finitely many possible linear transformations of the tangent space at any point that are obtained from parallel translation around closed loops.

A good example is the flat Klein bottle. Here parallel translation around some paths maps the tangent space into itself by reflection. So some vectors return to their negative.

For flat tori, Euclidean spaces and their Cartesian products it is true that parallel translation is independent of path. This means that locally in small regions parallel transalation is indepndent of the path.

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Gold Member
Perhaps a proof for Euclidean space is to show that the exponential map is an isometry from standard flat Euclidean space onto the manifold.

facenian
If parallel transport is path-independent the covariant derivatives become partial derivatives, don't they ?
I don't think so because the conection components do not necesarily vanish

facenian
In general it is not true that parallel translation is independent of path on a flat manifold.
So the answer is that vanishing of the curvature tensor is not suffient condiciont for parallel transport to be path independent.
Thank you

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goedelite
In his little monograph, "The General Theory, P.A.M. Dirac wrote on p. 22 (top): "If the space is flat, we may choose a system of coordinates that is rectilinear, and then the g (matrix elements) are constant. The tensor R (curvature tensor) then vanishes." This is because its elements contain 1st and 2nd derivatives of the g's. Alternatively, your question premises that the "parallel transport" is path independent. To me, that means that the vector does not change infinitesimally under parallel transport. This is true since any closed path line-integral of the vector in the space must be zero no matter how small the loop. Thus, dA = 0 = Gamma x A x dx . Consequently, Gamma (the Riemann-Christoffel tensor) must be zero. The 1st derviatives of all the g's are then zero. The curvature tensor vanishes.

goedelite
The converse is treated in detail in the same reference as cited in my prior message. Dirac shows one can always find, when R (curvature tensor) is zero, a transformation to a coordinate system in which the derivatives, with respect to the old coordinates, of g (the metric tensor) are zero. Dirac leaves it at that. One may transform to derivatives of g with respect to the new coordinates by means of the chain rule. The transformation equations are homogeneous because the derivatives (above) are zero. Thus, the only solution is the trivial one in which all the derivatives of g with respect to the coordinates of the new system are also zero.

facenian
Helo goedelite,
The question was not whether the vanishing of the tensor curvature is necessary and sufficient condition for flatness but whether it is necesary and sufficient condition for parallel transport to be path independent.
The answer given by lavinia is "no"

goedelite
facenian: My knowledge of topology is, regrettably, too weak for my to appreciate lavinia's reply more than on the surface (superficially!). Thanks for your correction.

Phrak
Helo goedelite,
The question was not whether the vanishing of the tensor curvature is necessary and sufficient condition for flatness but whether it is necesary and sufficient condition for parallel transport to be path independent.
The answer given by lavinia is "no"

Then apparently this is true:

The vanishing of the curvature tensor a necessary and sufficient condition for parallel transport to be path independent on an orientable manifold.​