# Characteristic classes in $H^2(\; ;\mathbb{C})$ for complex line bundles. (Milnor and Stasheff, Charateristic Classes, page 306).

I am reading the proof of Gauss-Bonnet Theorem from the Characteristic Classes by Milnor and Stasheff.

There is a statement in the proof which says that:

"But the only characteristic class in $$H^2(\; ; \mathbb{C})$$ for complex line bundles $$\zeta$$ is the Chern class $$c_1({\zeta})=e(\zeta_{\mathbb{R}})$$ (and its multiples)."

Can anyone help me prove that why the only characteristic class in $$H^2(\; ; \mathbb{C})$$ for complex line bundles $$\zeta$$ is the Chern class $$c_1({\zeta})$$ (or a reference from the same book would be also useful).

I am familiar with the sections on the Euler class and the Chern class from the same book. But my knowledge of this area is limited. Thankyou.

Edit: I think I got the half of it. Since my manifold is closed, oriented and 2-dimensional, I have $$H^2(M;\mathbb{C}) \cong H_0(M;\mathbb{C})$$. If $$M$$ is assumed to be connected (which is not in the statement of the theorem), then $$H^2(M;\mathbb{C}) \cong \mathbb{C}$$. So I think it would be enough to show that $$c_1(\zeta)=e(\zeta_\mathbb{R})$$ is non-zero.

• Your question is more general, but in your concluding paragraph you say that your $M$ is a closed surface. Certainly there are line bundles $\zeta$ with $c_1(\zeta)=0$, but I think the point is to consider general $\zeta$. I no longer possess a copy of the book, so I cannot check the specific context. Do we not already have a list of characteristic classes and know that $c_1$ is the only one that lands in $H^2$? Jun 25, 2022 at 19:41
• @TedShifrin I have added the theorem and proof of the theorem leading to that statement for you context. Hope this helps. Jun 25, 2022 at 19:57
• Well, you have a complete list of characteristic classes. Omitting Stiefel-Whitney, since they are torsion only, you have the Euler class, Chern classes, and Pontryagin classes. Just keep track of what cohomology group each one lives in. (You can also deduce it immediately from the formulas in terms of invariant polynomials in the curvature $2$-form.) Jun 25, 2022 at 20:11

I believe the following argument is due to Serre. A characteristic class taking values in $$H^k(-, R)$$ is, more or less by definition, a natural transformation from the functor $$[-, BG]$$ of isomorphism classes of $$G$$-bundles to the functor $$H^k(-, R)$$. By the Yoneda lemma applied to, say, the weak homotopy category (to my mind this is one of the most spectacular applications of the Yoneda lemma), such characteristic classes can be identified with classes in $$H^k(BG, R)$$. These are the universal characteristic classes of $$G$$-bundles, and characteristic classes of $$G$$-bundles are given by pulling them back along classifying maps $$X \to BG$$.
Specialized to complex line bundles $$G = U(1)$$ (or $$G = \mathbb{C}^{\times}$$), the corresponding classifying space is $$BG = \mathbb{CP}^{\infty}$$, and we can compute that its integral cohomology is a polynomial ring $$\mathbb{Z}[c_1]$$ on the universal first integral Chern class $$c_1 \in H^2(\mathbb{CP}^{\infty}, \mathbb{Z})$$. In particular, by universal coefficients this gives that $$H^2(\mathbb{CP}^{\infty}, \mathbb{C})$$ is $$1$$-dimensional and spanned by $$c_1$$.
The argument you give in your edit does not suffice; a characteristic class is a function from $$G$$-bundles to cohomology classes and two such functions may a priori fail to be linearly independent even though the cohomology group they take values in is $$1$$-dimensional. What we need is to compute the corresponding cohomology group of $$BG$$ as in the above argument.