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Let $X$ be a (compact) homogeneous space and $V$ be a homogeneous vector bundle on $X$ of rank $n$, and such that $\operatorname{dim}X\ge n$. Suppose $V$ has a section $s$, whose zeros $s=0$ form a sub-variety of $X$ of dimension $\operatorname{dim}X-n$. Is it true, that $V$ is a semi-ample vector bundle?

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Yes, that is true. By "compact homogeneous space", I assume that you mean a smooth projective variety over an algebraically closed field $k$ (presumably $\mathbb{C}$ for you) that is homogeneous under an algebraic action of a group $k$-scheme $G$ (or holomorphic action of a complex Lie group). Let $\mathcal{F}$ be a locally free $\mathcal{O}_X$-module with an associated $G$-linearization. Then also the locally free $\mathcal{O}_X$-module $\mathcal{F}(X)\otimes_k \mathcal{O}_X$ has a $G$-linearization, and the natural $\mathcal{O}_X$-module homomorphism, $$e : \mathcal{F}(X)\otimes_k \mathcal{O}_X \to \mathcal{F},$$ is $G$-equivariant.

In particular, the cokernel of $e$ has an associated $G$-linearization. There is a maximal open subscheme of $X$ over which the cokernel is locally free -- just the open set on which the rank is minimal (this uses that $X$ is reduced). Since $\text{Coker}(e)$ is $G$-linearized, this dense open subscheme is $G$-invariant. Since $X$ is homogeneous, this dense open subscheme equals all of $X$ so that $\text{Coker}(e)$ is a locally free $\mathcal{O}_X$-module. Thus also the image of $e$ is locally free, and the short exact sequence, $$ 0 \to \text{Image}(e) \to \mathcal{F} \to \text{Coker}(e) \to 0,$$ is locally split.

Now, by hypothesis, $s$ is a global section of $\mathcal{F}$, and its zero scheme is nonempty of codimension $\text{rank}(\mathcal{F})$. However, by construction of $e$, also $s$ is a global section of $\text{Image}(e)$, so that its zero scheme has codimension no greater than $\text{rank}(\text{Image}(e))$. Hence $\text{Image}(e)$ has the same rank as $\mathcal{F}$, and $\text{Coker}(e)$ has rank $0$. Since $\text{Coker}(e)$ is locally free, this is the zero sheaf. Therefore $e$ is surjective and $\mathcal{F}$ is globally generated (hence semiample).

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  • $\begingroup$ Jason, thank you for this detailed answer! I got that you prove that the bundle is globally generated, but I don't understand details. I have some questions that you might find very silly, but let me ask them. 1) What is the difference between $\cal F$ and ${\cal F}(X)$? 2) If this is the same thing then why the homomorphism $e$ is not an isomorphism? $\endgroup$
    – aglearner
    Commented Jul 16, 2014 at 16:07
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    $\begingroup$ @aglearner. The "geometric vector bundle" associated to $\mathcal{F}$ is what you would call $V$. However, the geometric vector bundle associated to $\mathcal{F}(X)\otimes_k \mathcal{O}_X$ is not $V$, it is the constant vector bundle $\mathcal{F}(X)\times X$, where $\mathcal{F}(X)$ is the (finite dimensional) vector space of global (algebraic or holomorphic) sections of $V$. $\endgroup$ Commented Jul 16, 2014 at 17:20
  • $\begingroup$ Tanks Jason! I could not understand that ${\cal F}(X)$ are global sections :) . But now once I've got it, your answer is crystal clear :) $\endgroup$
    – aglearner
    Commented Jul 16, 2014 at 17:57

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