Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
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
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
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
add comment

Élie Cartan proposes such interpretations in his fundamental paper Sur les variétés à connexion affine et la theorie de la relativité généralisée (Ann. Ec. Norm. 40 (1923), 325–412 and 41 (1924), 1–25). (These are reprinted in his collected words, Partie III.) It may be a bit hard to follow, so, especially if you are into physics, you might want to consult Misner, Thorne, and Wheeler's discussion of this in their famous book Gravitation before you dive into Cartan's paper.

share|improve this answer
MTW is a classic, but considering that it's 40 years old, and relativity (both experimental and theoretical) is a living field, IMO it's probably time to stop pointing people to it as the first "go to" book on GR. Carroll has a text that is up to date, and there is a version that's free online: arxiv.org/abs/gr-qc/?9712019 . It has a pretty complete discussion of torsion. Not that the interpretation of torsion has changed since 1973, or even since Cartan in 1923, but all other things being equal, it requires extra justification to point people to an obsolete \$136 book. –  Ben Crowell Jul 27 '12 at 18:34
@Ben: Is MTW really obsolete? (Atiyah-McDonald is over 40 years old, too, and I don't think anyone would call it 'obsolete'.) In any case, I wasn't recommending that Dox buy MTW, just consult it. I figure that, given MTW's 'classic' status, it should be readily available pretty much anywhere, many people are familiar with its language and notation, and, as you say, the standard interpretation of torsion hasn't changed in a long time. On the other hand, I'm happy to know of a free, up-to-date reference for these things, so thanks. (As you can probably tell, relativity is not my field.) –  Robert Bryant Jul 27 '12 at 18:46
MTW is unique, and will probably be unique forever, for a couple of reasons. One is its very cool and idiosyncratic style, and the other is that it forms a complete encyclopedia of techniques. An example of a place where it's completely obsolete is that it predates both the discovery of dark energy and the modern era of high-precision cosmology. It's from the era when some of the classic solar-system tests of GR were being done (e.g., tests of GR versus Brans-Dicke), and in terms of theory it's from the era when global methods were being rapidly developed (Hawking and Ellis was 1875). –  Ben Crowell Jul 27 '12 at 18:57
Oops, meant 1975, not 1875! –  Ben Crowell Jul 27 '12 at 18:58
MTW is a great book, of course one should focus in the mathematical framework and physical interpretation (which is always OK) rather than in the experimental facts (which vary radically with time). I've checked the exterior calculus section and it's very complete and the geometrical interpretation is quite nice! I'd also say that Sean Carroll's book presents the subject with ease, although the geometrical interpretation is not as impressive as in MTW. Thank you guys!!! Today has been a very productive day for my understanding of differential geometry. –  Dox Jul 27 '12 at 20:06
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.