Sufficient conditions for a curve on the sphere to be the Gauß map of a closed path

Question

I was wondering which curves on the $n-1$ sphere arise as the Gauss maps of closed paths in $\Bbb R^n$. Necessary conditions are obviously that the path on the sphere is the image of some smooth $S^1\to S^{n-1}$ and that there is no half sphere which contains the complete path (otherwise, if the image of the Gauss map would lie entirely in the, say, upper hemisphere, the particle in $\Bbb R^n$ would only move upwards and never downwards).

Are these to conditions already sufficient for the existence of a smooth path $S^1\to \Bbb R^n$ with the prescribed Gauss map?

Answer 1

$\def\conv{\mathop{\rm conv}}$Your condition is almost sufficient.

Let $g\colon S^1\to S^{n-1}$ be the given path on the sphere. The `no half-sphere' condition tells that $0$ lies in the convex hull of $g(S^1)$ (if a half-shpere is open) of that $0$ lies in the interior of $g(S^1)$ (if a half-sphere is closed). The latter condition is not necessary for a path lying in a hyperplane (but then we may pass to a subspace!). The former one is not sufficient: if $0$ lies on the boundary of $\conv(g(S^1))$, but the interior of this convex hull is nonempty, then a required path does not exist.

So, let us assume that $\mathop{\rm lin}(g(S^1))$ is the whole space; then the necessary condition is that $0$ lies in the interior of $\conv(g(S^1))$. Let us show that it is also sufficient. To perform this, we need to find a positive smooth density function $p$ on $S^1$ (say, parametrized by angle) such that $$ \int_{0}^{2\pi}p(x)g(x)\,dx=0; $$ then $S^1\ni t\mapsto \int_0^t p(x)g(x)\,dx$ is a required path. But the set of values of such integrals is convex and meets any neighborhood of any point on $g(S^1)$; thus it covers the whole interior of $\conv(g(S^1))$.