defined as $\overline{\Phi_{\gamma}}(x,x^{\pm}):=\pm\dot{\gamma}$ on the boundary.

Checking invariance
-------------------

Let $H: S^{1}\times I\rightarrow S^{3}$ be a one parameter family of curves. For $t \in I$, let us define $$\label{ps:inw}
\eta_{t}(\gamma):= \frac{1}{4\pi}\int\limits_{
\overline{C_{2}(S^{1})}\times\{t\}} \overline{\Phi^{*}}\omega$$ Invariance of $\eta$ means that $\eta_{0}(\gamma)=\eta_{1}(\gamma)$ for all $\gamma$. But:

$$\label{ps:inv}
0=\int\limits_{\overline{C_{2}(S^{1})}\times I}
d\overline{\Phi_{H}^{*}}\omega=
\int\limits_{\partial(\overline{C_{2}(S^{1})}\times
I)}\overline{\Phi_{H}^{*}}\omega=
\eta_{1}-\eta_{0}+2\int\limits_{S^{1}\times I}
\overline{\Phi_{H}^{*}}\omega$$

The first equality holds because $d$ and $\overline{\Phi_{H}^{*}}$ commute, and since $\omega$ is a 2-form defined on a 2-manifold, $d\omega = 0$. The second equality is the Stokes theorem. The third equality is simply the fact that $\partial
\overline{C_{2}(S^{1})} = 2S^{1}$.

From (2.2), we learn that $\eta_{1}=\eta_{0}$ if and only if $\int\limits_{S^{1}\times I} \overline{\Phi_{H}^{*}}\omega =0$. But we have no reason to assume that this would generally be the case.

Introducing a correction term
-----------------------------

Let us then “cap off” the cylinder $S^{1}\times I$ by pasting two $D^{2}$’s to it, making it isomorphic to $S^{2}$, as shown in the illustration below.

![image](illu.eps){width="9cm"}

 

Let us define a “swaddling” map $\beta$ as a continuous map which wraps $S^{2}$ in $D^{2}$ such that on the boundary $\beta |_{\partial D^{2}}=\dot{\gamma}$. In our case we have two such maps, $\beta_{0}$ and $\beta_{1}$. $\overline{\Phi_{H}}$ maps the boundaries of $S^{1}\times I$ in “antipodally”— i.e. if $x\times 