therefore we have $de_{1}\cdot
e_{2}\wedge de_{1}\cdot e_{3}$.

But $de_{i}\cdot e_{i}=0$, and by differentiating this equality we find that $de_{i}\cdot e_{j}=-de_{j}\cdot e_{i}$ and so this equals $$\label{de:wej} (de_{3}\cdot
e_{1})\wedge(de_{2}\cdot e_{1})$$

Now we remember that the $e^{i}$’s are an orthonormal to one another, and therefore they satisfy the equality $$de_{3}\cdot de_{2}=\sum_{i=1}^{3}(de_{3}\cdot
e_{i})\wedge(de_{2}\cdot e_{i}) \label{wedg}$$

But according to (3.1) this is exactly $e_{1}^{*}\omega$, and so we have found that $$e_{1}^{*}\omega=de_{3}\cdot de_{2}=de_{3}\cdot de_{2}+
d^{2}e_{3}\cdot e_{2}=d(de_{3}\wedge e_{2}) \label{wege}$$

### Relating $\tau_{\phi}$ to the total torsion of a space curve

In $\gamma$ has nowhere vanishing curvature, we can use Lemma (3.2) to show that this correction term as well is equal to the total torsion of the curve $\gamma$. Here $e_{1}$ is the tangent, $e_{2}$ the normal, and $e_{3}$ the binormal, so $$de_{3}\cdot e_{2}=db\cdot n$$ $-\tau\cdot n\cdot n=-\tau$, so by the Frenet equations, $b'\cdot n=-\tau$ so $db\cdot n=-\tau\cdot
ds$.

Thus, we see that for a curve with a nowhere-vanishing curvature, $$\tau_{\phi}=\int_{S^{1}}\tau ds \label{crv}$$ which is again the total torsion of the space curve $\gamma$.

### Making sense of it all

The last step of our argument is just the Stokes’ theorem. The domain of $\overline{\Phi_{\gamma}}$ is our “cylinder compactification” of $C_{2}(S^{1})$. Pulling back the volume form 