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Let $E\to M$ be a holomorphic vector bundle ($M$ is compact) and assume $s\colon M\to E$ is a non-trivial section transverse to the zero section.

Is it be possible that $s^{-1}(0)\neq \emptyset$, yet $c_{top}(E)=0$?

I exect the answer to be NO. If $M$ is Kahler, integrating the Kahler form on $s^{-1}(0)$ shows that it is homologically non-trivial. But how do you prove it for general complex manifold or without using the Kahler form.

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    $\begingroup$ The 'diagonal section of the trivial line bundle on $\mathbb C$ gives an example, no? $\endgroup$ Dec 18, 2013 at 2:01
  • $\begingroup$ I should have said M is compact. $\endgroup$ Dec 18, 2013 at 4:13
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    $\begingroup$ I suggest looking at the Hironaka twist, an example of a non-projective smooth proper threefold (Hartshorne B 3.4.1). It has two smooth disjoint curves $M_1$ and $M_2$ with the property that $M_1+M_2$ is numerically trivial. It could be possible to construct a rank 2 vector bundle $E$ with a section $s$ whose zero scheme is $M_1+M_2$. $\endgroup$ Dec 18, 2013 at 4:43
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    $\begingroup$ @user20497: Aren't the Chern classes $c_k$ defined in $H^{2k}(X,\mathbb{Z})$? Maybe the class of $D$ is a torsion class in $H^2(X,\mathbb{Z})$. $\endgroup$ Dec 19, 2013 at 13:48
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    $\begingroup$ @Jason You're right. But here is another example, hope it works: it's a primary Hopf surface $X$, Harvey and Lawson call it of class $0$ in their 1983 Inventiones paper. There exists a smooth elliptic curve $T\subset X$ such that $[T]$ (the current induced by integrating on $T$) is $d$-exact. The Hopf surface is $S^3\times S^1$ from the differential point of view, so there is no torsion in $H^*(X,{\mathbb Z})$. The line bundle $E={\mathcal O}_X(T)$ associated to $T$ satisfies $c_1(E)=0\in H^2(X,{\mathbb Z})$. Probably the non-Kahler elliptic surfaces have the same property. $\endgroup$
    – user20497
    Dec 19, 2013 at 16:24

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Take a Hopf surface $H$, projected to ${\Bbb C}P^1$ with fibers elliptic curves, and let $L=\pi^* O(1)$ be a pullback of $O(1)$ from ${\Bbb C}P^1$ to $H$. Since $H=S^3\times S^1$, all line bundles are topologically trivial, but $L$ has sections (pulled back from ${\Bbb C} P^1$) vanishing somewhere on $H$.

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