Varieties with an ample vector bundle mapping to their tangent bundle

Question

A well-known result of Andreatta and Wisniewski says: Let $X$ be a projective complex manifold whose tangent bundle $T_X$ contains an ample sub-bundle $\mathscr{E}$. Then $X$ is isomorphic to projective space $\mathbb{P}^n$ for some $n$. Furthermore, the bundle $\mathscr{E}$ is either $\mathscr{O}(1)^{\oplus r}$ or $T_{\mathbb{P}^n}$ itself.

I would like to know if a slightly stronger statement is known/true.

Suppose there is an ample vector bundle $\mathscr{E}$ and a non-zero map $$f: \mathscr{E}\to T_X.$$ Is $X$ necessarily isomorphic to $\mathbb{P}^n$ for some $n$?

Of course if $\text{im}(f)$ is a vector bundle, the result follows by Andreatta-Wisniewski, as quotients of ample vector bundles are ample.

Remarks (Added later). The statement is true if $\mathscr{E}$ is a line bundle, or if $X$ is a curve or surface. I can also show that the existence of a non-zero map $\mathscr{E}\to T_X$ with $\mathscr{E}$ ample implies $X$ is birational to an (etale) $\mathbb{P}^k$-bundle over a lower-dimensional variety (and in particular, is uniruled). The answer of pgraf gives a reference showing the statement is true if $\text{Pic}(X)$ has rank $1$.

The paper "Galois coverings and endomorphisms of projective varieties" by Aprodu, Kebekus, and Peternell, which pgraf has referenced and which covers the case of Picard rank $1$ only uses the hypothesis in Corollaries 4.9 and 4.11, as far as I can tell. In particular, many of the other arguments in Section 4 of that paper hold true; for example, one knows that if $T$ is a family of rational curves in $X$ of minimal degree, any split curves must land in the singular locus of $\text{im}(f)$. Analyzing the arguments of that paper seem to give several other cases, e.g. if $f$ has generic rank equal to $\text{dim}(X)$.

Answer 1

This is not a complete answer, but it's way too long for a comment.

You can define ampleness for arbitrary coherent sheaves. This is done for example in V. Ancona, "Faisceaux amples sur les espaces analytiques". The definition is the same as in the locally free case: A coherent sheaf $\mathscr E$ on a compact complex space $X$ (say, nonzero and torsion-free) is said to be ample if for any coherent sheaf $\mathscr F$ on $X$ there exists $n_0$ such that for $n \geq n_0$ the sheaf $\mathscr F \otimes S^n \mathscr E$ is globally generated.

This is equivalent to $\mathscr F \otimes S^n \mathscr E$ having vanishing higher cohomology groups for $n \gg 0$, and to the line bundle $\mathscr O_{\mathbb P(\mathscr E)}(1)$ on $\mathbb P_X(\mathscr E)$ being ample (Proposition 2.5 in Ancona's paper).

Now a nonzero quotient of an ample vector bundle will be ample in this general sense. Hence your question becomes: if $T_X$ contains an ample subsheaf $\mathscr F$, then is $X$ isomorphic to a projective space? In case the Picard number $\rho(X) = 1$, this is Corollary 4.3 in "Galois coverings and endomorphisms of projective varieties" by Aprodu, Kebekus, and Peternell. In fact they show that in this case, $\mathscr F$ is already locally free and then they apply the result of Andreatta and Wiśniewski.

Answer 2

The answer to this question is now known to be yes -- it is Corollary 1.2 of this paper of Jie Liu.

Answer 3

I hope this is correct: Consider $V:=\mathbb{P}^1\times \mathbb{P}^1$. It is clear that the rank of $\mathrm{Pic}(V)$ is greater than $1$. Hence $V$ is not isomorphic to $\mathbb{P}^n$. Let $\Delta:\mathbb{P}^1\rightarrow V$ be the diagonal morphism. The map $\Delta$ induces a non-zero sheaf-morphism $T_{\mathbb{P}^1}\rightarrow T_V$. But $T_{\mathbb{P}^1}$ is an ample tangent bundle and therefore we get a counterexample.