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.