disc, and mapped it to $\dot{\gamma}$ and to $-\dot{\gamma}$ correspondingly. Thus, we have a $\mathbb{Z}_{2}$ action on $\overline{C_{2}(S^{1})}$, whose action is to flip: $(x,y)\rightarrow (y,x)$. $\overline{\Phi_{H}}$ then descends to the quotient $$\label{sim}
\Phi_{H}:\overline{C_{2}(S^{1})}/\mathbb{Z}_{2}\rightarrow
S^{2}/\mathbb{Z}_{2}\simeq \mathbb{R}P^{2}$$ But $\overline{C_{2}(S^{1})}/\mathbb{Z}_{2}$ is also just $\mathbb{R}P^{2}$, so $\Phi_{H}$ is in fact a map from $\mathbb{R}P^{2}$ to itself.

$\mathbb{R}P^{2}$ is a non-orientable space, therefore only the degree of $\overline{\Phi_{\gamma}}$ is only defined mod 2. But the flipping action $(x,y)\rightarrow (y,x)$ is precisely the non-trivial path in $\mathbb{R}P^{2}$, hence the degree of $\overline{\Phi_{\gamma}}$ must be 1 (this follows from the topological assertion that the degree of a map is $\pi_{1}$ of that map).

This gives us a complete characterization of the framings for which $\phi$ lifts to $\sigma$- they are exactly those framings for which the mod 2 degree of the ‘extended Gauss mapping’ $\overline{\Phi_{\gamma}}$ is 1. This leads us to the rather startling conclusion that, given the blackboard framing, our ‘trivial’ knot turns out not to be the circle at all, but rather the boundary of the Moebius band.

A combinatorial description of the invariant
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By adding a correction term, we have shown that the Gauss self-linking integral can be made to be an invariant 