Quick question that I could not find elsewhere (beeing new to the subject): Let $X$ be some complex manifold, $\mathcal{O}_X$ be the sheaf of holomorphic functions on $X$ and $\mathcal{K}_X$ the sheaf of meromorphic functions on $X$. Then there is natural a map $\mathcal{L}:\text{Div}(X) \to \text{Pic}(X)$ mapping a divisor to a line bundle (cf. Huybrechts, Cor.2.3.10). In fact, this map is up to iso simply the map $\delta : H^0(X,\mathcal{K}^\ast _X / \mathcal{O}^\ast _X) \to H^1(X,\mathcal{O}^\ast _X)$ from the long exact cohom. sequence from the ses $0 \to \mathcal{O}^\ast _X \to \mathcal{K}^\ast _X \to \mathcal{K}^\ast _X / \mathcal{O}^\ast _X \to 0$. Moreover, the first chern class $c_1 : H^1(X,\mathcal{O}^\ast _X) \simeq \text{Pic}(X) \to H^2(X,\mathbb{Z})$ is the morphism in the long exact cohom seq. from $0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}^\ast _X \to 0$. Now we know: A principal divisor $(f)=\sum \text{ord}_Y(f)[Y]$ is image of a $f \in H^0(X,\mathcal{K}^\ast _X) = \mathcal{K}^\ast _X(X)$ and therefore, I think:
$\text{im } \phi:H^0(X,\mathcal{K}^\ast _X) \to H^1(X,\mathcal{K}^\ast _X / \mathcal{O}^\ast _X) = \ker \mathcal{L}$, so any principal divisor should be mapped to the trivial bundle in $\text{Pic}(X)$.
Question: Then the chern class of it should be $c_1(\mathcal{L}(f))=0$, and also any two line bundles that come from are linearly equivalent divisors have the same chern class. Is this correct? Many thanks!
Best Answer
You are right, the first chern class of a principal divisor is $0$: in fact this statement lies in the definition of "divisor".
You can think as Chern classes as cycles (that is, the generalization of divisors to higher codimension) rather than cohomology classes. So you can think that Chern classes associate to a vector bundle $E$ on $X$ a cycle $c_i(X)$.
Considering the space of cycles is not relevant enough: you quotient your space of cycles by rational equivalence (that is, two cycles are equivalent if they differ from a "principal" cycle). You obtain a nice space called Chow group $CH^*(X)$, see https://en.wikipedia.org/wiki/Chow_group.
In the particular case of line bundles, the first Chern class associate to a line bundle $L$ the (equivalent class of) the divisor given by the zeros of some generic global section. Starting from a divisor $D$, you can define a line bundle $L(D)$: in fact you just go the opposite direction. In terms of cycle, you have $c_1(L(D))\equiv D$.
Finally, you can go from the Chow group to the cohomology of your variety $H^{2*}(X,\mathbb{Q})$ using Poincaré duality (equivalent cycles have same image in cohomology!), that's why we often say $c_1(L)\in H^2(X,\mathbb{Z})$. Principal divisor are already $0$ in the Chow group, thus so are their images in cohomology.