# Projective variety of general type such that $S^m \Omega_X^1$ is globally generated

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.

• Is the variety smooth? If so, then you at least have that the canonical divisor class is nef. Dec 22, 2021 at 15:54
• @JasonStarr: oh yes, it is. I will edit the question, thanks Dec 22, 2021 at 15:55

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.

• Thank you for the answer. Is it also true that $|K_X|$ is base-point free (assuming $\dim X=2$)? Dec 22, 2021 at 18:21
• I don't think so (if $m\geq 2$ of course), but I don't have an example at hand.
– abx
Dec 22, 2021 at 19:25
• 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 Dec 23, 2021 at 4:03
• so combining my comment with the answer of abx and the comment by Jason gives you an affirmative answer in all dimensions Dec 23, 2021 at 4:32
• @YangMills: Nice, thank you! I have edited my answer to make it complete.
– abx
Dec 23, 2021 at 5:14