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Sep 23, 2022 at 17:18 history edited Francesco Polizzi CC BY-SA 4.0
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Sep 23, 2022 at 16:52 history edited Francesco Polizzi CC BY-SA 4.0
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Sep 23, 2022 at 5:22 comment added Francesco Polizzi @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?
Sep 23, 2022 at 4:54 comment added abx "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.
Sep 23, 2022 at 0:26 comment added Jason Starr 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 $\text{pr}_C,\text{pr}_D$.
Sep 22, 2022 at 22:58 history edited LSpice CC BY-SA 4.0
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Sep 22, 2022 at 22:36 comment added Francesco Polizzi Thanks for the comment. Could you please give more details?
Sep 22, 2022 at 20:50 history edited Francesco Polizzi CC BY-SA 4.0
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Sep 22, 2022 at 10:09 comment added Jason Starr I think that fails when $X$ is a product of two hyperbolic curves.
Sep 21, 2022 at 16:08 history edited Francesco Polizzi CC BY-SA 4.0
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Sep 21, 2022 at 15:54 history edited Francesco Polizzi CC BY-SA 4.0
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Sep 21, 2022 at 15:41 history asked Francesco Polizzi CC BY-SA 4.0