For holomorphic line bundle we define its first Chern class by exponential sequence $$0\to \mathbb Z \to \mathcal O \to \mathcal O^* \to 0 $$ and we can similarly define Chern class for smooth line bundle by the short exact sequence $$0\to \mathbb Z \to \mathcal C^\infty \to (\mathcal C^\infty)^*\to 0$$

Then there is a natural morphism from the first short exact sequence to the second one, so there is a natural map $H^2(\mathbb Z)\to H^2(\mathbb Z)$. Is this map isomorphic? Similarly, is the map $H^1(\mathcal O^*)\to H^1((\mathcal C^\infty)^*)$ just the natural map of on equivalent classes of line bundles?

In fact I am almost sure this is true (because they looks natural), but I do not know how to show this formally?


The map $\Bbb Z \to \Bbb Z$ in your morphism of short exact sequences is just the identity, and so induces an isomorphism on $H^*(X;\Bbb Z)$.

The second and third maps are inclusion, and think about what this says at the level of building up a vector bundle from transition functions: given a representative for an element of $H^1(X;\mathcal O^*)$, we have charts $U_i \subset X$ and holomorphic maps $U_i \cap U_j \to \Bbb C^\times$ that we think of as transition functions, the way to glue $U_i \times \Bbb C$ to $U_j \times \Bbb C$ over the overlap. The corresponding element of $H^1(X;(\mathcal C^\infty)^*)$ is the same maps $U_i \cap U_j \to \Bbb C^\times$, but we have forgotten that they are holomorphic (and just remembered that they are smooth). Thus the line bundle we construct is the same smooth manifold (it has the same charts and transition maps!), but we have forgotten the holomorphic structure it naturally inherits.

So the conclusion you get is that the map $c_1: H^1(X;\mathcal O^*) = \text{HolLineBun} \to H^2(X;\Bbb Z)$ factors through $H^1(X;(\mathcal C^\infty)^*) = \text{SmLineBun}_{\Bbb C}$.

(That is, the first Chern class is really an invariant of smooth complex line bundles, and to take the first Chern class of a holomorphic line bundle, just forget that it's holomorphic!)

  • $\begingroup$ Maybe stupid but why identity map must induce isomorphism on cohomology? I think this is not directly followed from definition. $\endgroup$ – User X Aug 6 '18 at 7:26
  • $\begingroup$ @UserX I think it is direct: the induced map on the level of cocycles is the identity, as it is on the level of coboundaries. It induces the identity on cohomology. $\endgroup$ – user98602 Aug 6 '18 at 15:18

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