TITLE: Curvature and Failure to return to starting point
QUESTION [5 upvotes]: Assume I have a geodesic polygon $P$ in a Riemannian manifold $M$ that is given by the image of a piecewise geodesic closed curve $\gamma(t)$ (parametrized by arclength), with vertices $x_i = \gamma(t_i)$, $i=0, \dots, N$. Let 
$$X_i = \lim_{t \searrow t_{i-1}}\dot{\gamma}(t) \in T_{x_{i-1}}M, ~~~~~~~~~~~~~~~~i = 1, \dots, N$$
be the velocity vectors of $\gamma$ at the beginning of the respective geodesic segments.
Now I can use the parallel transport the $X_i$ backwards along $\gamma$ to obtain $N$ vectors in $T_{x_0} M$ (which I also call $X_i$ by abuse of notation). Now I can look at the piecewise polygon curve given by this data, the end point of which is
$$ v = \sum_{i=1}^N (t_i - t_{i-1}) X_i.$$
If $M$ is flat, then $v = 0$, as the geodesic polygon was closed. Otherwise, $v=0$ may be not equal to zero, and is somehow given in terms of the curvature of $M$. For example
$$ |v|^2 = \int_0^{t_N} \!\!\!\int_0^{t_N} \bigl\langle \dot{\gamma}(t), [\gamma\|_s^t]\dot{\gamma}(s)\bigr\rangle ds dt,$$
where $[\gamma\|_s^t]$ denotes parallel transport along $\gamma$.
Question(s): What are other expressions for $|v|$? Can I express this value somehow in terms of the curvature integral over surfaces whose boundary is $P$? Are there other things that come to mind regarding this situation?

REPLY [2 votes]: Your question reminds me of the following result: let $X$ and $Y$ be tangent vectors in $T_p M$ such that $|X \wedge Y| = 1$ and let $\gamma$ be the exponential of a piecewise smooth closed curve in $T_p M$ based at the origin which lies in the plane spanned by $X$ and $Y$.  Then:
$$P_\gamma Z - Z = R(X,Y)Z \text{Area}(\gamma) + o(\text{Area}(\gamma))$$
This result has generalizations in which $\gamma$ is the boundary of a smooth surface in $T_p M$.  There are formulas for the area of a surface with polygonal boundary obtained via Stokes' theorem, and I bet these formulas combined with the result above would yield something interesting.