sum of exterior angles of a closed broken line in space

Question

I am looking for a simple proof of the following fact: The sum of exterior angles of any closed broken line in space is at least $2 \pi$. I believe it equals $2 \pi$ if and only if the closed broken line equals a polygon.

Answer 1

Quoting Curves of Finite Total Curvature by J. Sullivan:

Lemma 2.1. (See[Mil50, Lemma 1.1] and [Bor47].)
Suppose $P$ is a polygon in $\mathbb E^d$. If $P'$ is obtained from $P$ by deleting one vertex $v_n$ then $\operatorname{TC}(P')\leq\operatorname{TC}(P)$. We have equality here if $v_{n-1}v_nv_{n+1}$ are collinear in that order, or if $v_{n-2}v_{n-1}v_nv_{n+1}v_{n+2}$ lie convexly in some two-plane, but never otherwise.

This total curvature $\operatorname{TC}$ is the sum of exterior angles you are asking about. You could reduce every closed broken line to a triangle by successive removal of vertices. Since the total curvature of a triangle is always $2\pi$, this gives the lower bound you assumed.

And with the other condition, you can argue that in the case of equality the last vertex you deleted must have been in the same plane as the triangle, and subsequently every vertex deleted before that, and hence the whole curve must have been a planar and convex polygon.