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Let $X$ be a normal proper variety with only rational singularities and $A$ an abelian variety. Does a rational map $X \supset U \to A$ extend to a morphism $X \to A$? If not, what is a counterexample?

For $X$ smooth, I believe any map to a group scheme defined in codimension $1$ extends to all of $X$. Can anything be said for $X$ mildly singular?

For rational maps to abelian varieties, I think extension should hold for rational surface singularities because then there exists a resolution whose exceptional fibers are trees of rational curves which are contracted in the map to the abelian variety.

In general, I would like to make the following argument work: choose a resolution $\pi : \tilde{X} \to X$ with $R^i \pi_* \mathcal{O}_{\tilde{X}} = 0$ for $i > 0$. Then any exceptional fiber $E \subset \tilde{X}$ should have $H^i(E, \mathcal{O}_E) = 0$ and therefore admit no nonconstant map to an abelian variety.

Can formal functions make the above proof work? We get that $H^i(\hat{E}, \mathcal{O}_{\hat{E}}) = 0$ where $\hat{E}$ is the formal fiber. From this, I should be able to conclude that $\mathrm{Pic}_{\hat{E}}^0 = 0$. Does this imply that maps $\tilde{X} \to A$ have to contract $E$?

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    $\begingroup$ Let $\widetilde{X}^1$ be a desingularization of the closure of the smooth locus in $\widetilde{X}\times_X \widetilde{X}$ with its projections $\text{pr}_i:\widetilde{X}^1 \to \widetilde{X}$, for $i=0,1$. By hypothesis, the natural maps $\mathcal{O}_{\widetilde{X}} \to R\text{pr}_{i,*} \mathcal{O}_{\widetilde{X}^1}$ are quasi-isomorphisms. Thus, via the Leray spectral sequence, the pullback maps $H^q(\widetilde{X},\mathcal{O}_{\widetilde{X}})\to H^q(\widetilde{X}^1,\mathcal{O}_{\widetilde{X}^1})$ maps are also isomorphisms. For $q=1$, this gives equality of the weight-$1$ Hodge structures. $\endgroup$
    – Jason Starr
    Commented May 12, 2023 at 21:54
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    $\begingroup$ The pullback map $\mathrm{Pic}^0(X) \to \mathrm{Pic}^0(\tilde{X})$ is an isomorphism since it is always injective for a resolution of a normal variety and then it suffices to know that it is surjective on tangent spaces which follows from the Leray spectral sequence. This is all that is needed to see that the map extends, since for any nonconstant map $f: Z \to A$ where $A$ is an abelian variety---we apply this to a fibre of the resolution---the pullback map on $\mathrm{Pic}^0$ is never zero (reduce to the case $Z$ is a smooth projective curve). $\endgroup$
    – naf
    Commented May 13, 2023 at 2:27
  • $\begingroup$ @JasonStarr this is a very interesting strategy,. however I must say I don't understand why the corresponding cohomological vanishing statement for X^1 holds or how you draw your confusion from this equality of hodge structures $\endgroup$
    – Ben C
    Commented May 13, 2023 at 3:23
  • $\begingroup$ I believe it is not known whether exceptional divisors for resolutions of rational singularities are always acyclic, as this can depend on singularities of the exceptional divisors themselves. What is known is that for a snc resolution, the fibers are acyclic, see Lemma 2.5 in arxiv.org/pdf/2212.06786.pdf. $\endgroup$
    – Evgeny Shinder
    Commented May 27, 2023 at 7:51

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