There are probably many ways of talking about a second (integral) cohomology class of a smooth, closed, orientable manifold $M$ of dimension $n$. Here are a few, with $\alpha\in H^2(M,\mathbb{Z})$:
- Using Poincare duality, obtain a homology class $\hat{\alpha}\in H_{n-2}(M,\mathbb{Z})$. This can be represented by a submanifold $A$.
- $\alpha$ is represented by (the homotopy class of) a map $f:\ M\rightarrow \mathbb{CP}^\infty$, which can be homotoped to have target $\mathbb{CP}^N$, where $2N\ge n$. Taking the preimage of a (transverse) $\mathbb{CP}^{N-1}$ gives a submanifold $B$.
- $\alpha$ is the first Chern class of a complex line bundle $\mathbb{C}\rightarrow E\rightarrow M$. If $s:\ M\rightarrow E$ is a generic section, then the zero set of $s$ is a submanifold $C$.
What's kind of amazing is that $A$, $B$, and $C$ are all homotopic (the same)! I have used this many times, but don't think I've ever seen a detailed proof of all this before. I understand the basic setup (the construcion of $f$ in 2., the line bundle setup in 3., ...), but I don't know why these three ideas give essentially the same result: that is, why $B$ and $C$ are indeed Poincare dual to $\alpha$. I would really enjoy some references, outlines of proofs, etc.
OK, I have tracked down some references, etc., for all of this.
To show that 2. works, I refer to theorem VI.11.16 in Bredon's Topology and Geometry, where a complete proof is given.
To show that 3. works, I refer to theorem 5.2 of these notes; all of the necessary material is contained in Bredon's book, but I couldn't find so explicit a statement as that in the notes. Plus, the notes are nice anyway.
The only thing to do is show how 2. and 3. are connected.
Now if $\alpha\in H^2(M,\mathbb{Z})$, then it is represented by some (smooth) map $f:\ M\rightarrow \mathbb{CP}^N$ for high enough $N$; this is the map mentioned in 2. The representation is by the pullback of the fundamental class $u\in H^2(\mathbb{CP}^N,\mathbb{Z})$: $\alpha=f^\ast(u)$. Now $u$ itself is the first Chern class of the canonical line bundle (call it $E$) over $\mathbb{CP}^N$: $u=c_1(E)$. By naturality, we have $\alpha=f^\ast(c_1(E))=c_1(f^\ast(E))$. $f^\ast(E)$ is the complex line bundle mentioned in 3.