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 $\text{dim}(X)$.

  • $\begingroup$ I honestly don't understand the definitions well enough to be sure, but other Grassmannians seem like a good target; their tangent bundles are the tensor product of two bundles which I think are ample. $\endgroup$
    – Ben Webster
    Aug 9, 2015 at 12:58
  • 1
    $\begingroup$ @BenWebster The tensor product of two ample vector bundles is again ample, yet the only variety with ample tangent bundle is $P^n$. For the Grassmannian, the tangent bundle is however globally generated, for the reason you mentioned. $\endgroup$
    – J.C. Ottem
    Aug 9, 2015 at 13:59
  • $\begingroup$ @J.C.Ottem Thanks for the clarification. If Grassmannians don't work, it seems hard to imagine anything else would. $\endgroup$
    – Ben Webster
    Aug 10, 2015 at 0:23
  • $\begingroup$ Can you point to a place in the article of Andreatta-Wisniewski where it is necessary for the sheaf homomorphism to be generically injective (as opposed to generically nonzero)? I realize that they state their result only for generically injective homomorphisms, but I do not immediately see a place where this is necessary. $\endgroup$ Aug 10, 2015 at 11:30
  • $\begingroup$ Also, you might look at the articles of S'andor Kov'acs, Carolina Araujo, Stephane Druel, et al. They generalize Andreatta-Wisniewski in various ways. They may address your question. $\endgroup$ Aug 10, 2015 at 11:31

3 Answers 3


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.

  • $\begingroup$ (By the way, I think your question is a priori slightly stronger than mine--they would be equivalent if every ample subsheaf of $T_X$ was a quotient of an ample vector bundle.) $\endgroup$ Aug 12, 2015 at 17:51

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


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.

  • 2
    $\begingroup$ No, it isn't. That morphism is only on $\mathbb P^1$ and it maps to the restriction of $T_V$ to $\mathbb P^1$. So, you really just get a morpohism $T_{\mathbb{P}^1}\rightarrow T_V\left|_{\mathbb P^1}\right.$. It also contradicts the Andreatta and Wiśniewski result mentioned, so if this were correct, that theorem would be toast. $\endgroup$ Aug 22, 2015 at 17:03
  • $\begingroup$ Sándor is correct of course; this doesn't work. Thanks though! $\endgroup$ Aug 23, 2015 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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