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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?

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- Thread starter facenian
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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?

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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.

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

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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 R

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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?

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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.

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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 ?

[tex]

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

[/tex]

This equation can be interpreted

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

2. If on the RHS we have R

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Is this the one ?

[tex]

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

[/tex]

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 R^{a}_{dbc}= 0

2. If on the RHS we have R^{a}_{dbc}= 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?

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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.

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

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lavinia

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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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lavinia

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

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

Thank you

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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"

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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.

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