Lemma 12.29.1.2BAAA2E519DE Let $C$ be a compact connected Riemann surface, and let $A\to C$ be a holomorphic line bundle of positive degree. Write $A^\times $ for its total space with the zero section removed. Then $A^\times $ is not holomorphically convex. More precisely, put $B=A^*$ and choose a smooth Hermitian norm on $B$. Every holomorphic function on $B^\times $ extends holomorphically to $B$, and the holomorphic hull in $B^\times $ of
\[ K_ r=\{ v\in B:|v|=r\} ,\qquad r>0, \]
contains $\{ v:0<|v|\leq r\} $ and is not compact.
Proof.
In a local holomorphic frame of $B$, a function $f\in \mathcal O(B^\times )$ has a fiberwise Laurent expansion
\[ f(z,t)=\sum _{k\in \mathbb Z}a_ k(z)t^ k. \]
By the Cauchy integral formula, each coefficient is holomorphic in $z$, and under a change of frame the coefficients $a_ k$ transform as the components of global sections of $B^{-k}$. For $k<0$ this line bundle has negative degree, so it has no nonzero section: the zero divisor of a nonzero holomorphic section has nonnegative degree. Thus all negative coefficients vanish. On a smaller base chart, Cauchy estimates on a fixed fiber circle give locally uniform convergence of the remaining power series at $t=0$. These extensions agree on overlaps and define a holomorphic function on all of $B$.
The maximum modulus principle on the disk in each fiber gives
\[ |f(v)|\leq \max _{K_ r}|f|\qquad (0<|v|\leq r). \]
The circle bundle $K_ r$ is compact because $C$ is compact. A sequence in one fiber whose norms tend to zero lies in the hull and has no convergent subsequence in $B^\times $, so the hull is not compact. Finally, fiberwise inversion identifies $A^\times $ biholomorphically with $B^\times $: a nonzero $v\in A$ corresponds to the unique $\ell \in A^*$ with $\ell (v)=1$. This proves the assertion for $A^\times $.
$\square$
Pome
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