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Let $X$ be a projective surface of general type over $\mathbb{C}$, and assume that $\Omega_X$ is globally generated. Then the Albanese map $a \colon X \to \operatorname{Alb}(X)$ is a local embedding namely, a finite, étale cover with image $a(X) \subset \operatorname{Alb}(X)$. It is also well known that $a(X)$ generate $\operatorname{Alb}(X)$.

Question. Is the normal bundle of $a(X)$ an ample vector bundle?

By a result of Hartshorne (see 1), in order to have a positive answer, it suffices to check that every curve in $a(X)$ generates $\operatorname{Alb}(X)$, see also MO187149 On a proposition in Hartshorne's paper "Ample vector bundles on curves". In particular, this shows that the answer to the previous question is "yes" when $\operatorname{Alb}(X)$ is simple. But what can happen in general?

References.

1. Hartshorne, R., Ample vector bundles on curves, Nagoya Math. J. 43, 73-89 (1971). ZBL0218.14018.

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    $\begingroup$ I think that fails when $X$ is a product of two hyperbolic curves. $\endgroup$
    – Jason Starr
    Commented Sep 22, 2022 at 10:09
  • $\begingroup$ Thanks for the comment. Could you please give more details? $\endgroup$
    – Francesco Polizzi
    Commented Sep 22, 2022 at 22:36
  • $\begingroup$ If $X$ equals a product $C\times D$, then the Albanese morphism for $X$ is the product of the Albanese morphisms of the factors, $\text{alb}_C\times\text{alb}_D:C\times D \to \text{Alb}(C)\times \text{Alb}(D)$. Thus, the pullback of the tangent bundle by the Albanese morphism is the direct sum, $\text{pr}_C^*\text{alb}_C^*T_{\text{Alb}(C)}\oplus \text{pr}_C^*\text{alb}_D^*T_{\text{Alb}(D)}$. Therefore the normal bundle is also a direct sum $\text{pr}_C^* N_{\text{Alb}(C)/C}\oplus \text{pr}_D^*N_{\text{Alb}(D)/D}$. These are not ample on curves in fibers of $\text{pr}_C,\text{pr}_D$. $\endgroup$
    – Jason Starr
    Commented Sep 23, 2022 at 0:26
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    $\begingroup$ "The Albanese map is a finite, étale cover onto its image": not necessarily. You only know that it is unramified, the image could very well be singular. $\endgroup$
    – abx
    Commented Sep 23, 2022 at 4:54
  • $\begingroup$ @abx: ok, thanks. In fact, I only know that the differential of the Albanese map is everywhere of maximal rank. Do you know any example of surface such that $\Omega_X$ is globally generated and $a(X)$ is singular? $\endgroup$
    – Francesco Polizzi
    Commented Sep 23, 2022 at 5:22

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