Is it easy to construct examples of smooth complex projective surfaces $S$ of general type such that $h^0(\Omega_S)>1$, $alb_S:S\rightarrow Alb(S)$ is generically finite (unto its image) and $${\rm Bs}(\Omega_S):=\{x\in S,\ \omega_x(T_xS)=0 \ \forall \omega\in H^0(\Omega_S)\}$$ is non-empty?
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1$\begingroup$ If $S$ is smooth, that is impossible. By Abhyankar, the closure of the graph of the Albanese morphism is birational and projective over $S$ with uniruled fibers. Any irreducible component of any fiber of positive dimension is then a rational curve with a generically finite morphism to an Abelian variety, which is impossible. $\endgroup$– Jason StarrCommented Apr 26, 2023 at 16:16
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1$\begingroup$ Thank you very much for your answer. I have added the smoothness assumption. But I do not understand the proof. In this case the graph of the Albanese should be isomorphic to $S$ (?). The points in ${\rm Bs}(\Omega_S)$ are also the points were the differential of $alb_S$ vanishes. But I do not see the connexion to curves. May I ask for more details, please? $\endgroup$– pi_1Commented Apr 26, 2023 at 16:41
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$\begingroup$ You are right: the base locus of the space of $1$-forms is not the base locus of the rational transformation to the Albanese variety. My argument just shows that the rational transformation to the Albanese variety is everywhere regular (not everywhere smooth). $\endgroup$– Jason StarrCommented Apr 26, 2023 at 16:51
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$\begingroup$ So there are many examples by taking $S$ to be a minimal resolution of a singular surface in an Abelian variety. $\endgroup$– Jason StarrCommented Apr 26, 2023 at 16:53
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$\begingroup$ Probably, but how to make sure that the albanese of the resolution is exactly the Abelian variety you started from (not bigger)? Would not the vanishing of the differential imply that the singularity downstairs is not normal? $\endgroup$– pi_1Commented Apr 26, 2023 at 17:02
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I am just posting my comments as one answer.
Let $A$ be an Abelian variety of dimension three. Denote by $\Theta$ an ample divisor class on $A$. Let $S’$ be a smooth effective Cartier divisor on $A$ in the algebraic equivalence class of $n\Theta$ for very positive $n$. Let $f:S\to S’$ be the blowing up of $S’$ at a point $p$, precomposed with a blowing up of a point on the exceptional divisor over $p$. Denote the exceptional locus of $f$ by $E$. It is a union of two rational curves that intersect at a point $q$.
By direct computation, the pullback map is an isomorphism from $\Omega_{S’}|_p$ to $H^0(E,\Omega_S|_E)$. Thus every $1$-form on $S$ agrees on $E$ with the pullback of a unique $1$-form on $S’$. The difference is a $1$-form that is zero along $E$. This is the same as a global section of the coherent sheaf $\Omega_{S’}$ that vanishes at $p$.
By Serre vanishing and duality, both $h^1(A,\mathcal{O}(-S’))$ and $h^2(A,\mathcal{O}(-2S’))$ equal zero. Thus, the conormal sheaf of $S’$ also has vanishing $h^1$. Thus, by the fundamental short exact sequence, every global section of $\Omega_{S’}$ lifts uniquely to a global section on $S’$ of $\Omega_A|_{S’}$. Because $h^1(A,\Omega_A(-S’))$ is zero, this lifts further to a unique global section on $A$ of $\Omega_A$. Of course the only global section of this free $\mathcal{O}_A$-module that vanishes at $p$ is the zero section. Altogether, every global $1$-form on $S$ is the pullback of a unique global $1$-form on $A$. Of course all of these vanish at $q$.