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Suppose you have a holomorphic line bundle $L$ such that $L^{n}$ is a trivial holomorphic line bundle and the base complex manifold $M$ has no torsion cohomology classes in second degree (i.e. $H^{2}(M, \, \mathbb Z)$ is torsion free).

Then is $L$ holomorphically trivial? If yes, can we remove the restriction on cohomology?

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  • $\begingroup$ Which is your question? $\endgroup$
    – diverietti
    Commented Jun 21, 2018 at 7:27
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    $\begingroup$ What about abelian varieties? There's no torsion in integral cohomology yet there exist torsion line bundles $\endgroup$
    – Frank
    Commented Jun 21, 2018 at 8:03

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This is well-known material, so probably the question is not really suitable for MathOverflow. Anyway, since the subject can be non-trivial for someone approaching it for the first time, let me give a short answer in the case of a compact, Kähler manifold $M$.

The group $H^2(M, \, \mathbb{Z})$ classifies complex vector bundles on $M$, in other words the transition functions are assumed to be smooth and with values in $\mathsf{GL}(1, \, \mathbb{C})=\mathbb{C}^*$. In particular, if some power of $L$ is the trivial line bundle and $H^2(M, \, \mathbb{Z})$ is torsion-free, then $L$ itself is trivial in the topological sense.

Holomorphic line bundles on $M$ are instead classified by the Picard group $H^1(M, \, \mathcal{O}_M^*)$. Passing to cohomology in the exponential sequence $1 \to \mathbb Z \to \mathcal{O}_M \to \mathcal{O}_M^* \to 1$, we obtain an exact sequence $$0 \longrightarrow \mathrm{Pic}^0(M) \longrightarrow \mathrm{Pic}(M) \stackrel{c_1}{\longrightarrow} H^2(M, \, \mathbb{Z}),$$ where $\mathrm{Pic}^0(M)=H^1(M, \, \mathcal{O}_M)/H^1(M, \, \mathbb{Z})$ is a complex torus of dimension $q(M)=h^{0, \, 1}(M)= h^{1, \, 0}(M)$ and $c_1$ is the first Chern class (here we use in an essential way the Kähler assumption, see G. Elencwajg comments below).

In particular, if $q(M) >0$ the complex torus $\mathrm{Pic}^0(M)$ has strictly positive dimension, so we have a lot of non-trivial, torsion line bundles on $M$ (in the holomorphic sense) even if there is no torsion in $H^2(M, \, \mathbb{Z})$. Of course these torsion line bundles are in the kernel of $c_1$, so they are all trivial in the topological sense.

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  • $\begingroup$ Dear Francesco, aren't you using some hypothesis on $M$ in your answer like, say, compact kähler ? $\endgroup$
    – Georges Elencwajg
    Commented Jun 21, 2018 at 20:43
  • $\begingroup$ Dear Georges, I think that compact, complex manifold is enough. If not, where am I using the Kähler assumption? $\endgroup$
    – Francesco Polizzi
    Commented Jun 21, 2018 at 20:49
  • $\begingroup$ I added the assumption compact in the answer. $\endgroup$
    – Francesco Polizzi
    Commented Jun 22, 2018 at 8:26
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    $\begingroup$ Dear Francesco, here are the questions I ask myself: 1) We have $\dim H^1(M,\mathcal O)=h^{0,1.}$ (by Dolbeault) but you write $h^{1,0}$ for that number. The equality $h^{0,1.}=h^{1,0}$ requires a Kähler structure, I think. 2) Is the fact that $H^1(M, \, \mathbb{Z})$ is a full lattice in $H^1(M, , \mathcal{O}_M)$ not dependant on $M$ being Kähler? $\endgroup$
    – Georges Elencwajg
    Commented Jun 22, 2018 at 9:13
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    $\begingroup$ In their book Barth, Hulek, Peters and Van de Ven claim that indeed the image of $H^1(M,\mathbb Z)$ in $H^1(M,\mathcal O)$ is not a full lattice in the general case of a non-Kähler manifold. Here is the relevant page: books.google.fr/… $\endgroup$
    – Georges Elencwajg
    Commented Jun 22, 2018 at 9:29

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