Dolbeault groups of the negative twistor-fibre twists

A global section of $\mathcal{O}(-k)$, $k=1,2$, would have an entire coefficient $f_0(t)$ with $f_0(t)=t^{-k}{f}_{\mathrm{\infty }}(1\mathrm{/}t)$. By the normally convergent expansions in [F2], the left side has only powers $t^n$ with $n\ge 0$, whereas the right side has only powers $t^n$ with $n\le -k$. Uniqueness of Laurent coefficients forces every coefficient to vanish. Thus both degree-zero groups vanish.

A one-cochain on the two-chart overlap has a normally convergent Laurent expansion $g(t)={\sum }_{n\in \mathbb{Z}}{a}_n{t}^n$. Coboundaries subtract the entire $t$-series with $n\ge 0$ and the series $t^{-k}{f}_{\mathrm{\infty }}(1\mathrm{/}t)$ with $n\le -k$. These two tails converge on their respective entire charts by [F2]; the cover is acyclic for both twists because $r=1$ and $k=1,2$. For $k=1$ they exhaust every integer exponent, so the analytic first cohomology is zero. For $k=2$ the sole remaining exponent is $n=-1$; its coefficient cannot be a coboundary, and it identifies the analytic first cohomology with $\mathbb{C}$.

Apply [F3] to the two holomorphic line bundles on $L$. Its hypotheses hold for $r=1$ and the full AC assumption [A1]. It converts precisely the analytic sheaf groups computed in steps 1.1–2.1 to the smooth Dolbeault groups in the Statement. The transition calculation in [F1] gives $K_L\cong \mathcal{O}(-2)$. No scheme cohomology or unproved analytic/algebraic comparison is used.