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Let $\tilde S \overset{\nu}{\to} S$ be the normalization of a projective surface $S$ over a field $k$. Assume for now that $S$ is obtained from $\tilde S$ by gluing together two disjoint curves $C_1$ and $C_2$ in $\tilde S$. Let $\sigma_1, \sigma_2 : C \to \tilde S$ be parametrizations of $C_1$ and $C_2$ used for this gluing (e.g. one can take $C = \nu(C_1)=\nu(C_2)$).

If a line bundle $\mathcal{L}$ on $\tilde S$ is the pullback of a line bundle $L$ on $S$ then we get an isomorphism $\alpha: \sigma_1^* \mathcal{L} \to \sigma_2^* \mathcal{L}$, here $\alpha$ is essentially the gluing data on $\mathcal{L}$ to get $L$. More specifically, we use that $\nu \circ \sigma_1 = \nu \circ \sigma_2$ as well as $\mathcal{L} = \nu^* L$ to get the canonical isomorphism $\alpha$.

How about the reverse, given a pair $(\mathcal{L},\alpha)$ as before, can ve reconstruct a line bundle on $S$ corresponding to this pair?

I suspect this could be done as follows. One could define the kernel $L$ of the following exact sequence $0 \to L \to \nu_* \mathcal{L} \overset{s}{\to} \sigma_2^* \mathcal{L} \to 0$ where the latter map is defined as $s(u) = \sigma_2^*(u) - \alpha(\sigma_1^*(u))$. However, I have a hard time proving that $L$ is a line bundle.

Is there an easy way to see that $L$ defined in this way is a line bundle?

Generalizations to more complicated normalizations are welcome!

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  • $\begingroup$ The question is local, so it is enough to show that there is an exact sequence $0 \to O_S \to \nu_*O_{\tilde{S}} \to \sigma_{2*}O_{C_2} \to 0$. $\endgroup$
    – Sasha
    Commented Jul 29, 2016 at 20:18
  • $\begingroup$ There is such an exact sequence, but we tweaked the quotient map. $O_S$ itself gives the obvious isomorphism between the two pullbacks $\sigma_i^* O_{\tilde S}$, taking $1 \mapsto 1$. I am essentially asking if I glue using $1 \mapsto c$ for some $c \in k$ will I still get a line bundle. $\endgroup$
    – Emre
    Commented Jul 29, 2016 at 20:43
  • $\begingroup$ Imagine you defined sheaf $L$ by your sequence. Since locally $\mathcal{L} \cong O_{\tilde{S}}$, it follows that locally $L \cong O_S$, hence is a line bundle. $\endgroup$
    – Sasha
    Commented Jul 30, 2016 at 5:36
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    $\begingroup$ See Th 2.2 in D. Ferrand, Bull. Soc. Math. France 131 (4) 2003, 553-585. $\endgroup$
    – Laurent Moret-Bailly
    Commented Jul 30, 2016 at 7:36

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