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All varieties will be smooth when necessary.

Earlier i learned that the first Chern class of a line bundle on an algebraic variety does not determine the bundle up to algebraic isomorphism, i.e. the map $$ \alpha: \quad\operatorname{Pic}(X) \longrightarrow A^1(X) $$ sending a line bundle to its first Chern class can have a nontrivial kernel. (isn't this kernel nonzero whenever the albanese variety is nontrivial?)

(see the comments on Are vector bundles on $\mathbb{P}_{\mathbb{C}}^n$ of any rank completely classified? (main interest $n=3$))

However, i quote from Gathmann's notes "The top Chern class $c_r(F)$ has the additional geometric interpretation as the zero locus of a section of $F$". Here $F$ is a bundle of rank $r$. In the case of line bundles this is the first Chern class.

I found the same statement in Zach Teitlers notes:, see fact 10.

The problem i am having is that these statements seem contradictory: it is known that a line bundle on a smooth variety is completely determined up to isomorphism by the divisor of zeroes of a section. But according to Gathmann, that is exactly its first Chern class, so Gathmann is basically saying that $c_1(\mathcal{L})$ does determine $\mathcal{L}$ up to isomorphism.

Question 1: Can someone clear this contradiction up? Do Gathmann and Teitler both make the assumption that $\alpha$ is injective? Or am i missing something? (i am new to this subject!)

Question 2 (probably related): Teitler explains the method of degeneracy loci to compute the Chern classes of a bundle. I have the feeling that this method needs special assumptions on the variety in order to work, that Teitler does not explain. Is this true? He does assumes projectivity, smoothness and irreducibility, are these necessary?


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By the way, why do you think the method of degeneracy loci should need more assumptions, beyond projectivity and smoothness? – user64687 Apr 16 '13 at 16:22
@AsalBeagDubh, well if the method of degeneracy loci would always work, then Gathmann's and Teitler's claim would be true, but my because of what i thought to be a contradiction i assumed that they made assumptions on the variety. Anyway, your crystal clear answers solves that one as well, so thanks! – Joachim Apr 16 '13 at 17:10
Ok, I understand the motivation for the second question now. Anyway, graag gedaan! – user64687 Apr 16 '13 at 17:12
up vote 8 down vote accepted

To answer question 1: yes, this is confusing! The explanation is that for (say) complex algebraic varieties, there are different "flavours" of Chern class, taking values in different groups: topological Chern classes, which take values in cohomology $H^*(X,\mathbf{Z})$, and algebraic Chern classes, which live in the Chow groups $A^*(X)$. There is a so-called cycle map $A^*(X) \rightarrow H^*(X,\mathbf{Z})$ linking the two.

In my comment to the linked question, what I was saying was that the topological first Chern class $c_1^{top}(L)$, which is an element of $H^2(X,\mathbf{Z})$, does not determine the line bundle $L$.

However, the statements by Gathmann and Teitler both refer to the (more refined) algebraic first Chern class, $c_1^{alg}(L)$. This takes values in $A^1(X)$, which is exactly the same thing as $Pic(X)$.

In much fewer words, your map $\alpha$ is always an isomorphism. The map that can have a nontrivial kernel is the cycle map $A^1(X) \rightarrow H^2(X,\mathbf{Z})$.

A good example to keep in mind is an elliptic curve $E$. On such a curve, every point $p$ has an associated line bundle $O(p)$, and these are not isomorphic for different points. On the other hand, any two points clearly give the same class in $H^2(X)$.

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Off-topic comment: nice match of posting name and avatar! – Brian M. Scott Apr 17 '13 at 5:33
Dear @ Brian: what do you mean ? – Georges Elencwajg Apr 17 '13 at 6:34
@Georges: I think Brian is just referring to my user picture, which indeed shows an "asal beag dubh". – user64687 Apr 17 '13 at 13:31

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