Let $M$ be a compact K\"ahler manifold, which is assumed to be projective, i.e. there exists an ample line bundle over $M$ giving an embedding into $\mathbb{C}P^n$.

Let $\mathcal{L}$ be a smooth line bundle over $M$. Can $\mathcal{L}$ always be endowed with a holomorphic structure? Is such a structure necessarily unique, or can there exist more than one?


The group of isomorphism classes of complex line bundles on $M$ is isomorphic to $H^2(M,\mathbb{Z})$, via the map $L\mapsto c_1(L)$. The analogous group $\operatorname{Pic}(M) $ for holomorphic line bundles fits into an exact sequence $$0\rightarrow T\rightarrow \operatorname{Pic}(M)\xrightarrow{\ c_1\ } H^2(M,\mathbb{Z})\cap H^{1,1} \rightarrow 0$$ where $T$ is a complex torus (usually denoted $\operatorname{Pic}^{\mathrm{o}}(M)$), of dimension $h^{1,0}$ (both statements follow from the cohomology exact sequence associated to the exact sequence of sheaves $0\rightarrow \mathbb{Z}\rightarrow \mathcal{O}_M\xrightarrow{\ \mathbf{e}\ }\mathcal{O}_M^*\rightarrow 1$, where $\mathcal{O}_M$ is the sheaf of $C^{\infty}$ or holomorphic functions, and $\mathbf{e}(f)=\exp(2\pi if)$). Thus, if $h^{1,0}$ and $h^{2,0}$ are $\neq 0$, there are complex line bundles which do not admit holomorphic structures, and those which do admit a whole family, parametrized by $T$.

  • $\begingroup$ And for a reference: this is, I believe, originally the content of Kodaira-Spencer (1953). $\endgroup$ Oct 11 '19 at 1:18
  • $\begingroup$ So if $h^{(1,0)} = h^{(2,0)} = 0$, then every line bundle admits one, and only one, holomorphic structure? $\endgroup$ Oct 11 '19 at 5:54
  • 1
    $\begingroup$ Yes, that's right. $\endgroup$
    – abx
    Oct 11 '19 at 6:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.