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I heard that there is a theorem due to Rosenlicht which says the following:

Theorem. Let $X$ be a complex projective manifold and $V$ a non-trivial holomorphic vector field on $X$. Then $X$ is uniruled,ie,can be covered by rational curves,if $V$ has a zero.

I have thought for a few days and failed to give myself a proof. Can somebody give me the reference or say something about the idea of proof?

Thanks in advance.

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1 Answer 1

up vote 3 down vote accepted

You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds.

His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component of the identity of the automorphism group of $X$, is an algebraic group which acts algebraically on $X$.

Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian.

If $p \in X$ is a zero of your vector field then $p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves.


Added later: For an alternative proof see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.

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Thank you very much! –  Jun Li Jul 7 '11 at 3:18
    
I'm glad to hear that there's a proof more "analytical".It seems more easy for me.Thank you again. –  Jun Li Jul 11 '11 at 7:02

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