# Constructing the first Chern class without using the classifying map for line bundles?

I've gathered the gist of a particularly nice construction for Chern classes in a topological setting, but I can't quite figure out how to find the first class without making use of a classifying map. I've been told that this construction roughly goes through in AG without classifying maps (evidently first proposed by Grothendieck.)

Given a vector bundle $E \to B$ with fiber $V$, we form the projectivization $\mathbb P(E) \to B$, which has a tautological sub-bundle $L$, where fibers for $(x,\ell) \in P$ where $x \in B$ and $\ell \subset E_x$ is exactly $\ell$.

A formal descripton is given in $10.1.5$ here.

Now, supposing that we have a description of $\alpha \in H^2(P,\mathbb Z)$, and argue that powers of this element restrict to generators on $H^2(\mathbb CP^{n-1})$, and conclude with Leray Hirsch that $H^*(\mathbb P(E))$ is a free module over $H^*(B)$. Expressing $c_1(L)^n$ as a linear combination of the first $n-1$ powers gives the chern classes for $E$.

Question 1: How can one define $c_1(L) \in H^2(P,\mathbb Z)$ without using the classifying map $B \to \mathbb CP^{\infty}$ for line bundles?

Question 2: Can the following argument be made to work (of course by completing it?

Given the tautological bundle $L \to \mathbb P(E)$ one can use the association $Vect^1(\mathbb P(E)) \to \check{H^1}(\mathbb P(E))$ to obtain $\alpha \in H^1(P,\mathbb C^{\times})$. Is there a way to map from $H^1(\mathbb P(E),\mathbb C^{\times}) \to H^2(\mathbb P(E),\mathbf Z)$ and use this to get the chern classes?

• Sorry about the ramble below the line. I was just mostly mumbling to myself and I realized something might work (kind of.) I'll leave it in case it helps anyone else conceptually – Andres Mejia Jul 24 '18 at 2:25
• You could use the connecting morphism in the LES here: en.wikipedia.org/wiki/Exponential_sheaf_sequence (I guess you can repeat the same story on a general topological space, with $O_X$ replaced by the sheaf of continuous functions... but I don't know for sure.) – Lorenzo Najt Jul 24 '18 at 18:53
• Note that since $O_X$ (sheaf of continuous functions) is soft when $X$ is a metric space, $H^1(O_X) = H^2(O_X) = 0$, so the chern class boundary map is an isomorphism, which exactly what we expect (the chern class determines the isomorphism class of the topological line bundle). (This sanity check suggests to me that the story with the LES would work out.) Would this answer your question? I can write up some details... – Lorenzo Najt Jul 24 '18 at 19:06
• @lorenzo I think so. As long as there is a construction involved (that is * chern class referenced in your second comment.) I’d love to see how the short exact sequence story goes. – Andres Mejia Jul 24 '18 at 19:09
• It's absolutely a construction - The Chern class map appears as a boundary map in the LES associated to a certain exact sequence of sheaves. One of these sheaves has the property that its $H^1$ is naturally isomorphic to the group of line bundles, and another one of these sheaves has the property that its $H^2$ is the usual $H^2$. The boundary map connecting them is the chern class. (Have you studied sheaves and sheaf cohomology?) – Lorenzo Najt Jul 24 '18 at 19:14

The above argument can indeed be modified (and it appears that Grothendieck mentions this in his original paper.) This was also already mentioned by Lorenzo in the comments, but it took me a few days to (maybe) understand what was going on.

There is a short exact sequence of sheaves

$$A(X,\mathbb Z) \to A(X,\mathbb C) \to A(X,\mathbb C^{\times})$$

that gives rise to a long exact sequence in sheaf cohomology where the connecting homomorphism $\delta:\check{H}^1(X,\mathbb C^{\times}) \to \check{H}^2(X,\mathbb Z)$, provides the isomorphism we needed.