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Let $E$ be a rank $2$ vector bundle over an algebraic surface $X$, such that there exists a global section $s$, whose $0$-subscheme $Z$ is $0$-dimension.

Q

  • why $c_2(E)=\deg(Z)$.

  • How to understand the sequence $$0\to \mathscr O_X\to E\to \mathscr I_{Z}(c_1(E))\to0$$ and $$0\to\mathscr I_{Z}(c_1(E))\to\mathscr O_{X}(c_1(E))\to \mathscr O_{Z}(c_1(E))\to0.$$

PS: I do not understand the sheaf $\mathscr I_{Z}(c_1(E)$, since $c_1(E)$ is a cohomology class, how to define the sheaf?

1 Answers 1

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The $r$th Chern class of a rank $r$ vector bundle $E$ is pretty much defined to be the cycle in cohomology representing the vanishing locus of a global section of $E$ (assuming this vanishing locus has codimension $r$). In your example, $E$ has rank 2, so $c_2(E)$ is the Poincare dual of the zero-cycle $Z$ on which your global section of $E$ vanishes. Hence you can identify $c_2(E)$ with the number of points in $Z$, which I presume is the same as your "${\rm deg \ } Z$".

More generally, to define $c_k(E)$, you take $r - k + 1$ sections $\sigma_1, \dots , \sigma_{r-k+1}$ of $E$ and $c_k(E)$ is then defined to be the cycle associated to $V(\sigma_1 \wedge \dots \wedge \sigma_{r-k+1})$, assuming this cycle is of codimension $k$ as expected.

I imagine the notation $\mathcal O_X(c_1(E))$ refers to $\wedge^2 E$. This is indeed a line bundle, and it does indeed have first Chern class equal to $c_1(E)$, since $c_1(E) = c_1(\wedge^r (E))$ for a rank $r$ vector bundle $E$. The notation $\mathcal I_Z(c_1(E))$ means $\mathcal I_Z \otimes O_X(c_1(E))$. So your second SES is immediate from tensoring $0 \to \mathcal I_Z \to O_X \to O_Z \to 0$ with the line bundle $O_X(c_1(E))$.

For the first SES, follow the argument on page 306 of Eisenbud and Harris' book on intersection theory. Think of your global section of $E$ as a map $\sigma : \mathcal O_X \to E$. The statement that $Z$ is the vanishing locus of $\sigma$ is equivalent to the statement that the dual map $\sigma^\star : E^\star \to \mathcal O_X$ has image equal to the ideal sheaf $\mathcal I_Z$. So we have a Koszul exact sequence, $$ 0 \to \wedge^2 E^\star \to E^\star \to \mathcal I_Z \to 0. $$ Tensoring everything by $\wedge^2 E = O_X(c_1(E))$ gives the first of the short exact sequences in your question.

Finally, it might be worth pointing out that, in general, if $E$ is a rank $r$ vector bundle and the codimension-$r$ vanishing locus $Z$ of a global section of $E$ is smooth, then $E$ is the normal bundle to $Z$, and you have a very useful SES, $$ 0 \to T_Z \to T_X|_Z \to E|_Z \to 0$$ This is super useful for computations, though I don't think it is directly relevant to your question.

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    Since you used the statement: first Chern class of a line bundle determines the line bundle uniquely. But it is true just in smooth/continuous category. How to deduce $\wedge ^2E=\mathscr O_X(c_1(E))$ is holomorphic sense? – 2017-02-26
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    Use the standard correspondence between line bundles (i.e. Cartier divisors) and linear combinations of codimension-1 cycles (i.e. Weil divisors). Every line bundle is associated to a divisor $\sum_i a_i C_i$ in the way I described in the third paragraph. This works if your surface is smooth. I'm not sure what happens if you don't assume smoothness - I'm not an expert. – 2017-02-26
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    Actually that's not quite right is it? Linear equivalence is weaker than equivalence in homology. Take divisors on elliptic curves for example. – 2017-02-26
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    Okay then, why not just define $\mathcal O_X(c_1(E)) = \wedge^2 E$? – 2017-02-26