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Let $X$ be a smooth complex projective variety of general type; in my applications, I work with a surface, but let me ask this question in full generality.

Assume that for some $m \geq 1$ the vector bundle $S^m \Omega_X^1$ is generated by global sections, namely, the evaluation map $$H^0(X, \, S^m \Omega_X^1) \otimes \mathcal{O}_X \to S^m \Omega^1_X$$ is surjective.

Question. Is it true that $K_X$ is ample? Otherwise, what is a counterexample?

I started working on these topics rather recently, so I apologize if this question turns out to be trivial for the experts. Any answer and/or reference to the relevant literature will be highly appreciated.

Edit (12/26/2021). Follow-up question about the base-point freeness of $|K_X|$ asked as MO412382.

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  • $\begingroup$ Is the variety smooth? If so, then you at least have that the canonical divisor class is nef. $\endgroup$
    – Jason Starr
    Commented Dec 22, 2021 at 15:54
  • $\begingroup$ @JasonStarr: oh yes, it is. I will edit the question, thanks $\endgroup$
    – Francesco Polizzi
    Commented Dec 22, 2021 at 15:55

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If $X$ is a surface it is true. In general, a smooth projective variety with $S^m\Omega ^1_X$ globally generated does not contain any smooth rational curve $C$. Indeed $\Omega ^1_C$ is a quotient of $\Omega ^1_X$, so $S^m\Omega ^1_C$ is also globally generated, which of course implies $g(C)\geq 1$.

Now if $X$ is a surface, this implies that $K_X$ is ample (in fact $K_X$ is ample if and only if $X$ does not contain any smooth rational curve with square $\,-1$ or $-2$).

Edit: As pointed out by YangMills in the comments, the result holds in all dimensions: if a smooth projective variety $X$ of general type contains no smooth rational curve, $K_X$ is ample — see Lemma 2.1 in arxiv.1606.01381.

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    $\begingroup$ Thank you for the answer. Is it also true that $|K_X|$ is base-point free (assuming $\dim X=2$)? $\endgroup$
    – Francesco Polizzi
    Commented Dec 22, 2021 at 18:21
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    $\begingroup$ I don't think so (if $m\geq 2$ of course), but I don't have an example at hand. $\endgroup$
    – abx
    Commented Dec 22, 2021 at 19:25
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    $\begingroup$ In any dimension if $X$ is smooth with $K_X$ is nef and big and $X$ contains no rational curve, then $K_X$ is ample, see e.g. Lemma 2.1 in arxiv.org/pdf/1606.01381 $\endgroup$
    – YangMills
    Commented Dec 23, 2021 at 4:03
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    $\begingroup$ so combining my comment with the answer of abx and the comment by Jason gives you an affirmative answer in all dimensions $\endgroup$
    – YangMills
    Commented Dec 23, 2021 at 4:32
  • $\begingroup$ @YangMills: Nice, thank you! I have edited my answer to make it complete. $\endgroup$
    – abx
    Commented Dec 23, 2021 at 5:14

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