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Let $ \pi : X \rightarrow B $ be a family of compact complex manifolds parametrized by a connected base $ B $. (By this I mean $ \pi $ is a proper holomorphic submersion.) Let $ L $ be a holomorphic line bundle on $ X $ and $ n $ a positive integer. What can be said about the locus of points $ b \in B $ such that $ L|_{X_b} $ on $ X_b $ has an $n$-th root?

I posted it originally on MSE (without a response) but I guess it's more appropriate here.

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The locus you consider is either empty, or equal to $B$.

This can be seen as follows. Line bundles on a fiber $F$ of $\pi $ are parameterized by $H^1(F,\mathscr{O}^*_F)$. This group fits into an exact sequence $$H^1(F,\mathscr{O}_F) \rightarrow H^1(F,\mathscr{O}^*_F) \xrightarrow{\ c_1\ }H^2(F,\mathbb{Z})\xrightarrow{\ i\ }H^2(F, \mathscr{O}_F)\ .$$ I claim that a line bundle $M$ on $F$ admits a $n$-th root if and only if $c_1(M)$ is divisible by $n$ in $H^2(F,\mathbb{Z})$. Indeed, assume that $c_1(M)=n\alpha $ for some $\alpha $ in $H^2(F,\mathbb{Z})$; we have $i(\alpha )=0$ since $H^2(F,\mathscr{O}_F)$ is torsion-free, hence $\alpha =c_1(N)$ for some $N$ in $H^1(F,\mathscr{O}^*_F)$. Then $M\otimes N^{-n}$ has $c_1=0$, hence comes from a class in the vector space $H^1(F,\mathscr{O}_F) $, which is of course divisible by $n$.

Now assume first that $B$ is simply connected; then $R^2\pi _*\mathbb{Z}$ is the constant sheaf $B\times H^2(F,\mathbb{Z})$ on $B$, so if $c_1(L_{|F})$ is divisible by $n$, the same holds for $c_1(L)$ restricted to any fiber. In general, suppose that $L_{|X_b}$ admits a $n$-th root for some $b\in B$; for any $c\in B$ we can choose a path from $b$ to $c$ and cover it by simply connected open subsets, so that $L_{|X_c}$ admits a $n$-th root.

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    $\begingroup$ Actually, just the second part of this argument is sufficient (and thus also applies in characteristic $p$ prime to $n$). The obstruction to infinitesimally deforming $L$ is $n$ times the obstruction to infinitesimally deforming the root invertible sheaf inside $H^2(F,\mathcal{O}_F)$. This is a vector space over the ground field. Thus, if $p$ is prime to $n$, then if one obstruction vanishes, then both obstructions vanish. $\endgroup$
    – Jason Starr
    Commented Jul 14, 2022 at 11:05
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    $\begingroup$ @Jason Starr: Right, this is more direct (and more elegant). Still the observation that having a $n$-th root can be checked on $c_1$ might be useful. $\endgroup$
    – abx
    Commented Jul 14, 2022 at 14:00
  • $\begingroup$ abx and Jason Starr, thank you. If I understand correctly, the entire argument can be repeated for schemes over a field of char not dividing n by taking the exact sequence $ 0 \rightarrow Z/nZ \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \rightarrow 0 $ on a fiber in the etale topology, where the second map is multiplying by n. Is that correct? What happens if char k divides n, out of curiosity? $\endgroup$
    – Cranium Clamp
    Commented Jul 14, 2022 at 15:05
  • $\begingroup$ In characteristic p you can have that line bundles become divisible by a power of p when specialised. This happens for supersingular K3 surfaces, but see also arxiv.org/pdf/0907.4781.pdf example 3.12. Actually Jason had also told me a few years ago of an example coming from the Moret-Bailly pencil of the Jacobian of a supersingular genus 2 curve. $\endgroup$
    – Frank
    Commented Jul 15, 2022 at 7:37

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