Edge of Polyhedron has no Curvature

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Theorem

The edge of a polyhedron has zero 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:

$R = \dfrac {\delta \theta} {\delta a}$

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 given by:

\(\ds R\)

\(=\)

\(\ds \frac {\delta \theta_1 + \delta \theta_2} {\delta a}\)

\(\ds \)

\(=\)

\(\ds \frac {\delta \theta_1 - \delta \theta_1} {\delta a}\)

\(\ds \)

\(=\)

\(\ds \frac 0 {\delta a}\)

\(\ds \)

\(=\)

\(\ds 0\)

The result follows.

$\blacksquare$