Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 18 / 5 / d

Choose a positive Hermitian metric on $L$. Its curvature form $\omega =i{F}_h$ is a Kähler form on the compact complex manifold of dimension one. Part (c) gives a threshold $k_0$ with $H^1(X,\mathcal{O}(L^k))=0$ for every $k\ge k_0$. We use this to prescribe a finite principal part of a meromorphic section.

Fix a smooth cutoff $\chi$ supported inside the given coordinate chart and equal to one near $x_0$. Let $P(z)={\sum }_{j=-r}^{-1}{a}_j{z}^j$. On the punctured manifold define $t_k=\chi P(z){\zeta }^k$ inside the chart, extended by zero outside. Its Dolbeault operatoris smooth globally: it vanishes near the pole and near the boundary of the chart. It is ${\bar{\partial }}_{L^k}$-closed, either by the square-zero identity or because there are no $(0,2)$-forms in complex dimension one. The vanishing of $H^1(X,\mathcal{O}(L^k))$, together with the Dolbeault theorem, therefore gives a global smooth section $u_k$ with ${\bar{\partial }}_{L^k}{u}_k={\eta }_k$.

The section $s_k=t_k-u_k$ is holomorphic on $X\setminus \{x_0\}$. Near $x_0$, ${\eta }_k=0$, so $u_k=f_k(z){\zeta }^k$ with $f_k$ holomorphic through zero. Its Taylor series then giveswhere the second series converges near zero and its coefficients are those of $-f_k$. The threshold is determined by the fixed bundle $L$ and $X$, and is independent of the prescribed coefficients. In fact the same threshold works for every finite pole order $r$, since the Dolbeault cohomology obstruction is the same $H^1(X,L^k)$ for every such cutoff construction.