on the boundary.

### Transporting the pullback of the volume form to SO(3)

Since we are now moving into SO(3), we shall convert the discussion into the language of framings. Let us break $\overline{\Phi_{\gamma}}(x,y)$ into the mapping $\phi$ from $S^{1}$ to $SO(3)$ composed on a mapping $e_{1}(x,y)$ from $SO(3)$ to $S^{2}$ . Following [@Poh], let $e_{2}$ be the unit vector normal to $e_{1}$ on the plane spanned by $e_{1}$ and the tangent, extending smoothly to the normal defined by $\phi$ on the boundary. We shall then define $e_{3}$ to be $e_{1}\times e_{2}$ at every point. The following lemma is due to William Pohl [@Poh].

$$e_{1}^{*}\omega=d(de_{3}\wedge e_{2})$$

For every $1\leq i_{3}$ in $\mathbb{N}$, $e_{i}$ defines a function $x_{i}$ by means of the relation $e_{i}^{0}\cdot
v=x_{i}(v)$ for any vector $v\in\mathbb{R}^{3}$, when $e_{i}^{0}$ denoted the restriction of $e_{i}$ to the point $\overrightarrow{0}$. The volume form in $\mathbb{R}^{3}$ is then given by the expression $$x_{1}(dx_{2}\wedge dx_{3})+ \text {
cyclic permutations.}$$

Let us pull back the volume form via $e_{1}$. $e_{i}^{*}x_{j}=
e_{i}\cdot e_{j}^{0} = \delta_{i,j}$, therefore $$e_{1}^{*}\omega= d(e_{1}^{*}x_{2})\wedge d(e_{1}^{*}x_{3})=de_{1}\cdot e_{2}^{0}
\wedge de_{1}\cdot e_{3}^{0}$$ by Leibnitz’s rule. There was nothing special about our choice of 0 as the point by which to define the functions $x_{i}$, 