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](https://mathoverflow.net/questions/412382/projective-variety-of-general-type-such-that-sm-omega-x1-is-globally-genera).



Projective variety of general type such that $S^m \Omega_X^1$ is globally generated
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](https://arxiv.org/pdf/1606.01381.pdf).


