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

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

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’. ∎