# Interpretation of Curvature and Torsion

Dear all,

When dealing with General Relativity one uses the Levi-Civita connection with is torsion-free. Thus the commutator of the covariant derivatives yields

$[\nabla_\mu,\nabla_\nu]V^\rho = R_{\mu\nu}{}^\rho{}_\lambda V^\lambda.$

Usually, it is said that curvature is the responsible of the change of the direction of the vector under parallel transport through the two different paths.

However, in general there are three different effects in the transportation:

• Change of direction of the vector.
• Non-closure of the path (say, if one moves 1meter along each direction).
• Rotation about its own axis.

whilst the general commutator is

$[\nabla_\mu,\nabla_\nu]V^\rho = R_{\mu\nu}{}^\rho{}_\lambda V^\lambda- T^\lambda{}_{\mu\nu}\nabla_\lambda V^\rho.$

Is it possible to give a meaning to the curvature and torsion in term of these intuitive geometry or is not possible in general?

Thank you!

-
I don't know the answer to this, but it seems to me best to focus on the 2-dimensional case first. –  Deane Yang Jul 27 '12 at 15:35
Yes Deane, I agree with you. Unfortunately I'm not sure if these objects have the expected interpretation or whether their higher dimensional analogs do. I think that the main point would be if their interpretation could be separated... at this level (2-dimensional manifolds) –  Dox Jul 27 '12 at 16:07

Here is my attempt to present the intuition behind torsion in an accessible way. Here is a similar, previous thread on MathOverflow.

In your question, you've described torsion in terms of its effect on parallel-transporting a vector along two different paths. The distinction between curvature and torsion may be more transparent if you think about scalars rather than vectors. Curvature effects vanish when you operate on a scalar, e.g., the mass of a hydrogen atom doesn't end up being different depending on which path you transport it along. But the covariant derivative does pick up an effect from the torsion when you compute the commutator of two derivatives acting on a scalar; the reason is that you're differentiating along two coordinate axes, and if there is torsion these axes themselves rotate as you move along.

Another nice way to distinguish between curvature and torsion is that nonvanishing torsion requires that the space have a detectable handedness to it, whereas curvature has no such handedness. E.g., in two dimensions, a bug living on a surface can never use measurements of curvature in the way we would use a magnetic compass to find north. In a real-world physical context, the experiment described at the end of 1 is looking for violations of the symmetry between left- and right-handedness.

-
Great web page Ben! Thank you very much. I already knew the post you mention, and that gave me some ideas, but it's a bit ahead of my knowledge. I got the point with the scalar field, very useful example. Cheers. –  Dox Jul 27 '12 at 18:33
My original answer had a paragraph discussing the two-dimensional case, but after further thought I lost confidence that it was correct, and I deleted the paragraph and made it into a question on math.se: math.stackexchange.com/questions/176543/… –  Ben Crowell Jul 29 '12 at 17:17