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You have a closed curve on a Euclidean manifold $c: S^1 \to M$ and a tangent vector $v \in T_{x_0} M$ which you parallel translate around $c$ using the Levi-Civita connection. The monodromy representative of this curve should be a rotation on $T_{x_0}M$ (which is a plane in our case). Could you have gotten this angle by integrating the curvature over the area enclosed by the curve? What is the name of that result?

flag	asked Dec 21 2010 at 10:48 John Mangual 6,150●4●40

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I suppose you are asking about 2-dim case. I think it is Gauss-Bonnet theorem for 2-dim manifolds with boundaries (check it on Wiki). In two dimensional case one can define the curvature form such that its integral over a piece D is equal to the angle of rotation of a vector transporting along ∂D. See V.I. Arnold's Mathematical Methods of Classical Mechanics. Anyway Gauss-Bonnet is a keyword here. – Petya Dec 21 2010 at 11:26

That is likely. Also thanks for your guess on how to turn it into Stokes. I wonder what mechanical interpretation it has. – John Mangual Dec 21 2010 at 11:34

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It is not my guess, I know it from the cited Arnold's book. As I remember the key observation is that angle of rotation is additive with respect to a splitting of piece into parts. Thus one can wait that it is an integral of 2-form (it remains to check the details on an infinitely small parallelogram). Big part of "Mathematical methods" is devoted to Mathematics not only Mechanics! – Petya Dec 21 2010 at 11:45

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One can speculate - Geodesic lines on Riemmanian manifold are trajectories of the free motion of particles, so in that sense the curvature (and all Riemanian geometry) belongs to Mechanics. – Petya Dec 21 2010 at 11:50

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I read the russian edition - in it it was the first additional chapter entitled "Riemannian curvature". Googlebooks gives page 301 Appendix 1, "Riemannian curvature". – Petya Dec 21 2010 at 12:36

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“Curvature” variant of Stoke’s Theorem

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