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. ∎

Since $v_{1},\ldots v_{n}$ are $\mathbb{R}$-linearly independent at a point on $\partial\Sigma$, they span $TL$ at the point, so $\Omega(v_{1},\ldots v_{n})\neq 0$. By the Lemma above $v_{1},\ldots v_{n}$ are pointwise complex linearly independent as sections of the holomorphic vector bundle $TX$ over $\Sigma$, so any holomorphic first order deformation can be written as

$$v=f_{1}v_{1}+\ldots f_{n}v_{n}.$$

The functions $f_{1},\ldots f_{n}$ are holomorphic on $\Sigma$ up to boundary, and even up to corners due to $\det(a_{kj})\neq 0$. Now subtracting a constant linear combination of $v_{1},\ldots v_{n}$, we can ensure $v$ vanishes at any chosen point on $\partial\Sigma$. Then $\Omega(v,v_{2},\ldots v_{n})$ has a zero, so must be identically zero by the above Lemma, whence $f_{1}=0$ identically. Similar all $f_{k}=0$, so $v=0$. This proves that $v_{1},\ldots v_{n}$ span all first order deformations. Since the index is $n$, and the first order deformation space is $n$-dimensional, we must have vanishing obstruction space.

There is a special deformation vector field from $\mathbb{R}$ translation. The nonvanishing result then implies that the holomorphic strip is an immersion up to boundary. At the corners, the holomorphic strip is to leading order

$$(a_{1}z^{\phi_{1}/\pi}+O(z),\ldots a_{n}z^{\phi_{n}/\pi}+O(z)),\quad a_{k}\neq 0,\forall k,\quad|z|\ll 1.$$

By $\det(a_{kj})\neq 0$, this translation vector field cannot be $O(z)$ at the corner, so for at least one choice of $k$, we have $a_{k}\neq 0$. We say the failure of immersion at the corner is ‘minimal’. ∎