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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?

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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 '10 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 '10 at 11:34
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 '10 at 11:45
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 '10 at 11:50
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 '10 at 12:36

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