First Chern class $c_1$ of $U(1)$ bundle over $S^2$ v.s. $\pi_1(S^1)=\mathbb{Z}$

Question

Let us compare the four statements:

1. Consider the $U(1)$ fiber over the $S^2$. Such that this $S^1$ fiber over $S^2$ gives a first Chern class $c_1$ on the $S^2$ with $c_1=1$.

2. Consider the $S^1$ fiber over $S^2$ as a Hopf fibration, so that we have a $S^3$ constructed out of $S^1 \hookrightarrow S^3 \to S^2$.

3. Consider the classification of the homotopy map between the $U(1) =S^1$ and the target space as the $S^1$ boundary of the flat $\mathbb{R}^2$, as $\pi_1(S^1)=\mathbb{Z}$. To compare with the first two statements, we take the unit integer class 1 in $\pi_1(S^1)=\mathbb{Z}$. I think there is a correspondence of this map to the above two constructions.

4. Consider the homotopy class map between $M=S^2$ and $BU(1)=\mathbf{P}^\infty$, so that $[M, BU(1)]=[S^2,\mathbf{P}^\infty] = [S^2, K(\mathbb{Z},2)]$ which is $H^2(S^2,\mathbb{Z})=\mathbb{Z}$. To compare with the first three statements, we take the unit integer class 1 in $H^2(S^2,\mathbb{Z})=\mathbb{Z}$. I think there is a correspondence of this map to the above two constructions.

How to show or prove that the above four statements are the same or related constructions? Especially the first two statements?

Answer 1

First, it should be noted that the Hopf bundle has Chern class $c_1(L)=-1$, not $1$. Having said that, let's proceed.

For $1\iff 2$, we look at sheaf cohomology. There is an exact sequence of sheaves $$0\to \mathbb{Z}\to C^\infty_\mathbb{C}\xrightarrow{\exp(2\pi i\cdot)}(C^\infty_\mathbb{C})^\times\to 0$$ which results in a long exact sequence of cohomology groups. In particular, because $C^\infty_\mathbb{C}$ is soft, its higher cohomology vanishes, and this gives an isomorphism $$H^1(X,(C^\infty_\mathbb{C})^\times)\cong H^2(X,\mathbb{Z})$$ whenever $X$ is a smooth manifold. The first group classifies line bundles up to isomorphism, by associating a line bundle to an open cover together with its associated transition functions. This isomorphism sends a line bundle to its first Chern class (as an integral cohomology class). Since $H^2(S^2,\mathbb{Z})$ is torsion free, however, we can just view these Chern classes as sitting inside the de Rham cohomology. So if two line bundles have the same first Chern class, they are isomorphic. This proves $1\iff 2$, once you choose a connection on the Hopf bundle and integrate it.

Exercise: Consider the natural connection on the Hopf bundle, defined by the $1$-form $\omega=\bar{z}_1dz_1+\bar{z_2}dz_2$ on $S^3\subset\mathbb{C}^2$. Show that its curvature integrates to $-2\pi i$.

Suppose we want to classify line bundles over $S^2$. This is done via the clutching construction. We take the standard open cover for $S^2$ corresponding to projection from the respective poles $N$ and $S$, i.e. $U=S^2\setminus\{N\}$ and $V=S^2\setminus\{S\}$. We have $U\cong\mathbb{R}^2\cong V$. Both of these are contractible topological spaces, and hence any vector bundle over $U$ or $V$ must be trivial. Therefore, a (unitary) line bundle is completely determined by the gluing map $$\varphi:U\cap V\to U(1)$$ Denote the resulting line bundle by $L_\varphi$. Then we know that homotopic maps give isomorphic line bundles, i.e. $\varphi_1\simeq\varphi_2\iff L_{\varphi_1}\cong L_{\varphi_2}$, so isomorphism classes of line bundles are in bijective correspondence with $[U\cap V,U(1)]$, i.e. homotopy classes of maps from $U\cap V$ to $U(1)\cong S^1$. Now, $U\cap V=S^2\setminus\{N,S\}\cong S^1\times\mathbb{R}\simeq S^1$. That is, $S^1\times\mathbb{R}$ is homotopy equivalent to $S^1$. Therefore, $[U\cap V,U(1)]\cong [S^1,S^1]=\pi_1(S^1)\cong \mathbb{Z}$. In conclusion, the line bundle is classified by an integer $n$, which is the degree of the map $\varphi:U\cap V\to S^1$.

Exercise: pick a small disk $U_0$ around $N$ (the north pole). Show that, for an arbitrary line bundle $L$ and trivialisation $L|_{U_0}\cong U_0\times\mathbb{C}$, it is possible to pick a connection $\nabla$ on $L$ such that $\nabla|_{U_0}=d$ under this trivialisation.

Algebraic topology tells you that for a map $\varphi:S^1\to S^1$, we have $$\text{deg}(\varphi)=\frac{1}{2\pi i}\int_{S^1}\varphi^{-1}d\varphi$$ Using this equation and the exercise above, you can show that $$\int_{S^2}\frac{F_\nabla}{2\pi i}=\int_{S^2}c_1(L)=\text{deg}(\varphi)$$ In particular, the transition function for the Hopf bundle is given by $z\mapsto 1/z$, which has degree $-1$, and so $c_1(L)=-1$ where $L$ is the line bundle associated to the Hopf bundle under the natural representation. This shows $2\iff 3$.

Finally, there is $\mathbb{CP}^\infty$ and we have $\pi_2(\mathbb{CP}^\infty)\cong\mathbb{Z}$. By the Hurewicz theorem, this implies $H^2(\mathbb{CP}^\infty,\mathbb{Z})\cong\mathbb{Z}$. There is a canonical generator for this group, denote it $\xi$. Since $\mathbb{CP}^\infty$ is the classifying space for line bundles, every line bundle $L\to S^2$ defines a map $f:S^2\to \mathbb{CP}^\infty$, and homotopic maps give isomorphic bundles. As such, every line bundle determines a cohomology class $$f^*\xi\in H^2(S^2,\mathbb{Z})$$ It turns out that $f^*\xi=c_1(L)$. I would assume the details can be found in Hatcher's book on vector bundles, although I am not sure.