Lens identity: the Gauss integral of a loop that sweeps a window forth along one curve and back along another is the difference of the two Gauss integrals

Let $N\in S^3$ and let $A_1,A_0,\Lambda,\gamma:\mathbb R\to S^3\subset\mathbb R^4$ be $C^2$ loops (with $A_1,A_0$ of period $1$) that avoid $N$, where each of $A_1,A_0,\Lambda$ is disjoint from $\gamma$. Let $d\le\tfrac12$ and $s(\sigma)=s_0-d\cos2\pi\sigma$, which sweeps $[s_0-d,s_0+d]$ forward for $\sigma\in[0,\tfrac12]$ and backward for $\sigma\in[\tfrac12,1]$. Suppose:

• $A_1=A_0$ near every point of $[s_0+d,\,s_0-d+1]$ (the curves differ only inside the window);
• $\Lambda=A_1\circ s$ near every point of $[0,\tfrac12]$;
• $\Lambda=A_0\circ s$ near every point of $[\tfrac12,1]$.

Then $\mathrm{Gauss}_N(\Lambda,\gamma)=\mathrm{Gauss}_N(A_1,\gamma)-\mathrm{Gauss}_N(A_0,\gamma)$ (gaussLinkingIntegral, stereographic projection from $N$).

Proof (substitution only, no homotopy). After stereographic projection the outer integrand is $J_A(x)\,=\,\int_0^1 f(A(x),A'(x),\gamma(t),\gamma'(t))\,dt$, which is linear in the velocity $A'(x)$. On $[0,\tfrac12]$ we have $\Lambda'(\sigma)=s'(\sigma)A_1'(s(\sigma))$, so substituting $x=s(\sigma)$ gives $\int_0^{1/2}J_\Lambda=\int_{s_0-d}^{s_0+d}J_{A_1}$. Likewise $\int_{1/2}^1J_\Lambda=-\int_{s_0-d}^{s_0+d}J_{A_0}$. By periodicity, $\int_0^1J_{A_i}=\int_{s_0-d}^{s_0+d}J_{A_i}+\int_{s_0+d}^{s_0-d+1}J_{A_i}$, and the last integrals agree for $i=0,1$.