3
$\begingroup$

Let $X$ be a minimal surface of general type. Recall that a vector bundle $\mathscr{E}$ on $X$ is called globally generated if the evaluation map of global sections $$e \colon H^0(X, \, \mathscr{E}) \otimes \mathcal{O}_X \to \mathscr{E}$$ is surjective. Instead, $\mathscr{E}$ is called ample if the line bundle $\mathcal{O}_{\mathbb{P}(\mathscr{E})}(1)$ is ample on $\mathbb{P}(\mathscr{E})$. By the cohomological characterization of ampleness, if $\mathscr{E}$ is ample then the symmetric power $\operatorname{Sym}^n \mathscr{E}$ is globally generated (and ample) for $n \gg 0$.

Now let us take $\mathscr{E}=\Omega_X$, the cotangent bundle. I'm looking for examples of $X$ such that:

  1. $\Omega_X$ is neither globally generated nor ample;
  2. $\operatorname{Sym}^n \Omega_X$ is globally generated for $n \geq n_0$, where $n_0 \geq 2$ is an explicit constant (I would be already happy with $n_0=2$).

Note that these conditions provide several restrictions. For instance, 1. tells us that the Albanese map $a_X \colon X \to \operatorname{Alb}(X)$ is not a local immersion, since the non-surjectivity of the evaluation map for $\Omega_X$ at a point $x \in X$ is equivalent to the non-injectivity of the differential $da_X(x)$. Moreover, 2. implies that $K_X$ is ample: indeed, by my previous question MO412306, if $\operatorname{Sym}^n \mathscr{E}$ is globally generated then $X$ contains no rational curves at all; as a consequence, the failure of $a_X$ to be locally immersive is not related to the contraction of rational curves.

I have looked for such examples, so far without success. Probably, one of the reasons is that I do not know a geometrical characterization of the global generation of $\operatorname{Sym}^n \Omega_X$ in terms of the Albanese map. So let me ask the following

Question. What are examples of minimal surfaces of general type that satisfy 1. and 2. above? More generally, how can I check in general whether $\operatorname{Sym}^n \Omega_X$ is globally generated?

Edit. Will Sawin's nice answers provide examples $X$ which are étale quotients $Y \to X$, where $Y$ is a suitable hypersurface in a reducible abelian threefold. Then $X$ has zero second Segre class, namely, $s_2(X)=c_1(X)^2-c_2(X)=0$, because $Y$ has the same property and this is preserved by étale maps.

Question 2. Can we obtain examples with positive $s_2$?

My guess is that this should be possible, by generalizing Will's construction to étale quotients of suitable complete intersections in reducible Abelian fourfolds.

$\endgroup$
5
  • 1
    $\begingroup$ An aside comment: your globally generated condition implies that $\mathcal O_{\mathbb P(\Omega^1)}(1)$ is nef. Then, the second Segre number positive gives you bigness of $\mathcal O_{\mathbb P(\Omega^1)}(1)$. So you are looking at surfaces with big&nef cotangent sheaf, right? $\endgroup$
    – diverietti
    Jan 23, 2022 at 11:32
  • $\begingroup$ @diverietti: yes, in particular, the examples I'm looking for will have big and nef $\Omega_X$ (but neither globally generated nor ample). In fact, as you correctly remark, the generation condition on $\mathrm{Sym}^n \Omega_X$ implies that $\mathcal{O}_{\mathbb{P}(\Omega)}(n)$ is globally generated, hence nef. Is it the converse true? $\endgroup$ Jan 23, 2022 at 12:30
  • 1
    $\begingroup$ Unfortunately not! You cannot infer globally generation of the vector bundle from globally generation of the tautological line bundle on the projectivized bundle. Ernesto Mistretta made me remark this once! $\endgroup$
    – diverietti
    Jan 23, 2022 at 13:41
  • $\begingroup$ A typographical comment: ample in place of $ample$. $\endgroup$
    – Z. M
    Jan 24, 2022 at 19:44
  • $\begingroup$ Typo fixed, thanks. $\endgroup$ Jan 25, 2022 at 7:12

1 Answer 1

5
$\begingroup$

Let $E$ be an elliptic curve, $A$ an abelian surface, and $x\in A$ a point that's not 2-torsion.

Let $Y \subset A \times E$ be a surface of high degree, stable under inversion, and containing $x \times E$, that is general among hypersurfaces satisfying these conditions.

If the degree is sufficiently high then (by a dimension-counting argument) a general such $Y$ will be smooth and not contain any two-torsion points of $A \times E$.

So the quotient of $Y$ by inversion will be smooth. Call this $X$.

I claim $X$ is general type. Indeed, its unramified double cover $Y$ is an ample hypersurface in an abelian variety, hence of general type.

I claim $\Omega_X$ is not globally generated, and indeed has no global sections - they would pull back to involution-invariant global sections of $\Omega_Y$, which by Lefschetz would give involution-invariant global sections of $\Omega_{A \times E}$.

I claim $\Omega_X$ is not ample. Restricted to the elliptic curve $x \times E$, it has the rank one trivial quotient $\Omega_E$, and thus cannot be ample.

I claim $\operatorname{Sym}^n \Omega_X$ is globally generated for all even $n$. Indeed, every section of $\operatorname{Sym}^n ( H^0(A \times E, \Omega_{A \times E}))$ for even $n$ is involution-invariant and thus, restricted to $Y$, descends to $X$, and these sections locally generate $\operatorname{Sym}^n \Omega_{A \times E}$, hence they also locally generate $\operatorname{Sym}^n \Omega_Y$ and $\operatorname{Sym}^n \Omega_X$.

So $X$ is an example.

$\endgroup$
2
  • $\begingroup$ Thank you for your nice answer. These surfaces $X$ are étale quotients of divisors in Abelian threefolds, hence (if my computations are correct) their second Segre class is zero, namely, $s_2(X) = c_1(X)^2-c_2(X)=0$. For technical reasons, I would like to have examples with positive $s_2$. Perhaps we could also obtain them by means of your construction, by taking as $Y$ a suitably complete intersection in $A \times E$, with $A$ Abelian threefold? $\endgroup$ Jan 23, 2022 at 8:39
  • 1
    $\begingroup$ @FrancescoPolizzi All my arguments should work in this context, so I think so. $\endgroup$
    – Will Sawin
    Jan 23, 2022 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.