A periodic immersion onto the circle is a $k$-fold cover with $k\neq0$, and the Gauss integral of its image loop is $k$ times that of the standard circle

Let $u:\mathbb{R}\to S^1$ be a $C^2$ map of period $1$ with $u'\neq0$ everywhere. Then $u$ is onto $S^1$, and $u(s)=(\cos2\pi\theta(s),\sin2\pi\theta(s))$ for a lift $\theta$ with $\theta(s+1)=\theta(s)+k$, $k\in\mathbb{Z}\setminus\{0\}$ (if $k=0$, $\theta$ would be periodic, hence have a critical point, where $u'=0$).

Let $E$ be $C^2$ near $S^1$ with $|E|=1$ on $S^1$, let $\gamma:\mathbb{R}\to S^3$ be a $C^2$ loop of period $1$ missing $E(u(\mathbb{R}))=E(S^1)$, and let $N\in S^3$ miss both. Then

with gaussLinkingIntegral as $\mathrm{Gauss}_N$.

Suggested proof. The straight-line homotopy $\theta_\tau=(1-\tau)\theta+\tau ks$ keeps the loops $E\circ c\circ\theta_\tau$ inside $E(S^1)$, so homotopy invariance (gaussIntegral_eq_of_homotopy) gives $\mathrm{Gauss}_N(E\circ u,\gamma)=\mathrm{Gauss}_N(E\circ c(k\,\cdot),\gamma)$. For $k>0$ use the symmetry of the Gauss integrand and gaussLinkingIntegral_comp_mul_nat; for $k<0$ additionally reverse the loop, which changes the sign.