We have a section of $\Lambda^{n}TX|_{\Sigma}$ given by $v_{1}\wedge\ldots v_{n}$, which takes boundary value in $\Lambda^{n}TL$ on $\partial\Sigma$. Since $v_{1},\ldots v_{n}$ are all holomorphic, so must be the function $\Omega(v_{1},\ldots v_{n})$. Assume this function is not identically zero. By holomorphicity, the zeros are isolated. We claim that the order of zeros must be nonnegative everywhere. This is clear for the interior and the boundary points. We analyze the ends of the strip as the origin in the upper half plane model with holomorphic coordinate $z$, putting $TL_{\pm}$ in the standard form at the corner point. The deformation vector field has the leading asymptote

$$v_{k}=(a_{k1}z^{\phi_{1}/\pi},\ldots,a_{kn}z^{\phi_{n}/\pi})+O(z),\quad k=1,2,\ldots n,$$

hence

$$\Omega(v_{1},\ldots v_{n})=z^{(\sum\phi_{k})/\pi}(\det(a_{kj})+o(1)),$$

and the excess vanishing order is nonnegative. By the index formula (23), the index $\deg q-\deg p\geq n$, and when equality is achieved all vanishing orders must be zero. In particular $\det(a_{kj})\neq 0$ at the corners. ∎