Chern Class = Degree of Divisor?

Solution 1:

If $M$ is a complex manifold of dimension $n$, the exact sequence of sheaves $$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$ yields a morphism of groups in cohomology (part of a long exact sequence) $$c_1:H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)$$
Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the morphism $$c_1:Pic(M)\to H^2(M,\mathbb Z)$$
This morphism coincides with the first Chern class defined (for $C^\infty$line bundles) in differential geometry in terms of curvature of a connection.

If now $M$ is a compact Riemann surface (= of dimension $1$), we may evaluate a cohomology class $c\in H^2(M,\mathbb Z) $ on the fundamental class$[M]$ of the Riemann surface $M$, thereby obtaining an isomorphism $I: H^2(M,\mathbb Z) \xrightarrow {\cong} \mathbb Z: c\mapsto \langle c,[M] \rangle$, which composed with the first Chern class yields the degree for line bundles: $$deg=I\circ c_1:Pic(M)\to \mathbb Z: L \mapsto deg(L)=\langle c_1(L),[M] \rangle$$

Finally, if $L$ is associated to the divisor$D$ i.e. $ L=\mathcal O(D)$, we have the pleasantly down to earth (but highly non tautological!) formula $$deg(\mathcal O(D))=deg (D)$$
which says that the degree on the left, obtained by sophisticated differential geometry, can be computed in a childishly simply way by computing the degree of the divisor $D\in \mathbb Z^{(X)}$ seen purely formally as an element of the free abelian group on $X$.

All this is explained in Griffiths-Harris, Principles of algebraic geometry, Chapter 1.