Edge of Polyhedron has no Curvature

Theorem
The edges of polyhedra have no curvature.

Proof


Let $X$ and $Y$ be two separate faces of a polyhedron separated by the edge $l$.

Let $P$ be a point on $X$ and let $Q$ be a point on $Y$.

The curvature inside an infinitesimal region $\delta a$ is given by the net angular displacement $\delta\theta$ a vector $v$ experiences as it is parallel transported along a closed path around $\delta a$.

The curvature is then given by $R=\frac{\delta \theta}{\delta a}$.

We must then prove that the vector $v$ experiences no net angular displacement as it is parallel transported from $P$ to $Q$ and back to $P$.

The two open curves $r$ and $s$ make a closed curve.

As the vector is parallel transported along the open curve $r$, it crosses the edge between the two faces $X$ and $Y$. In doing so, it gains a finite angular displacement $\delta\theta_1$.

Then, when the vector is transported back along the open curve $s$, it gains another angular displacement $\delta\theta_2$. Notice that because it is not being transported the other way (from $Y$ to $X$), the new angular displacement will be $\delta\theta_2=-\delta\theta_1$.

The curvature inside the region $\delta a$ is therefore $R=\frac{\delta\theta_1+\delta\theta_2}{\delta a}=\frac{\delta\theta_1-\delta\theta_1}{\delta a}=\frac{0}{\delta a}=0$.

The result follows.