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?


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.

  • $\begingroup$ it should be $Area(\sigma)$ for the surface $\sigma$ with boundary $\gamma$ above? (its a particular case of Ambrose-Singer formula for the parallel transport) $\endgroup$ – valeri Jun 26 '14 at 15:51

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