Given a smooth, unit-speed curve $\gamma \colon [0,L] \to \mathbb{R}^{3}$ and a unit vector $v$ in the normal plane $\gamma'(0)^{\perp}$, it is well-known that there is a unique normal parallel vector field $V$ along $\gamma$ such that $V(0)=v$. In other words, a vector field $V$ along $\gamma$ satisfying $\langle \gamma', V \rangle=0$ and $V' = f \gamma'$ for some function $f$.
It follows that, for any $v \in\gamma'(0)^{\perp}$, one can construct a smooth orthonormal frame $(\gamma', V,W)$ by extending $v$ and $w=v\times \gamma'(0)$ to normal parallel vector fields $(V,W)$.
My question is the following. If $\gamma$ is closed, under what conditions is the frame $(\gamma', V,W)$ closed as well?
Best Answer
Good question. Here's a hint. Look up total twist. (If you can't find it, look in the exercises of Section 3 of Chapter 1 of my differential geometry text, linked in my profile.)