Are these 2 complex manifolds homeomorphic?

Question

This is an example give in Kodaira's Complex Manifolds and Deformation of Complex Structures, Chpt 2, Sec 2, Example 2.10

Consider $M=\mathbb P^1\times\mathbb P^1$ with $S=0\times\mathbb P^1$ where $0$ is identified as $0\in\mathbb C\subset\mathbb P^1$ and $\mathbb C$ is complex plane. Denote $W'(\epsilon)=W(\epsilon)=U_{\epsilon}\times \mathbb P^1$ with $U_{\epsilon}=\{z\in\mathbb C:|z|<\epsilon\}$. Consider the following gluing map to obtain $M_n=M-W(\epsilon)\cup W(\epsilon)$ via the following identification map. $(W(\frac{1}{2}\epsilon)-S)\to (W'(\frac{1}{2}\epsilon)-S)$ is defined by $(z,t)\to (z,t/z^n)$. (This is essentially replace $S$ by some twisted $S$.)

The book has shown $M$ not homeomorphic to $M_1$ via computing intersection number on the 2nd homology. The result is obviously extendable to $M$ not homeomorphic to $M_n$.

The book only mentions that for $n\neq m$ $M_n$'s complex structure is different from $M_m$'s complex structure.

Update: I have found in a link that intersection form $H_2(M_n)$ has some basis element with self intersection $n$.

$\textbf{Q:}$ Even from that intersection form, I cannot conclude whether $M_n$ not homeomorphic to $M_m$ for $m\neq n$. Furthermore, what is the 2 cycle that gives self intersection $n$ for $M_n$? How do I tell whether $M_n$ not homeo to $M_m$ for $m\neq n$?

Answer 1

You can conclude it because the intersection form is a topological invariant. By the way the self-intersection is $-n$ I believe (unless you took $n$ negative in your question, but it's important that the self-intersection is negative).

To explicitly identify the cycle : here is a really beautiful construction of Hirzebruch surfaces which also answer your question. It's easy to see that $M_1$ is the blow-up of $\Bbb P^1$ at a point, so a $\Bbb P^1$-bundle over $\Bbb P^1$, say with projection $\pi : X \to \Bbb P^1$. The exceptional divisor $E$ is such that $\pi_{|E} : E \to \Bbb P^1$ is an isomorphism, hence we can think of $E$ as a section (indeed it's often called the exceptional section). There is a procedure which allows you to construct $M_n$ from $M_{n+1}$ : pick a point on $p \in E$ (the exceptional divisor), blow-up at this point and contract the fiber $\pi^{-1}(\pi(p))$. Such a transformation is called a Nagata transform. This gives you explicitly the exceptional cycle (the image of $E$ by these transformation). The same construction can be reversed by first blowing-up a point outside $E$ and contracting the corresponding fiber.

Answer 2

Recall the cohomology ring is homotopy invariant, so in particular, the intersection form, up to change of basis, is homeomorphism invariant.

We have the possible self-intsersection of a 2-cycle: $$(a\cdot[\text{fibre}]+b\cdot[\text{base}])^2=-nb^2+2ab$$ where "base" is represented by the copy of $\mathbb{P}^1$ which is the point $0\in\mathbb{P}^1$ on every fibre. If $n\equiv0\pmod2$ then all self-intersection numbers are even. If $n\equiv1\pmod2$ then we get some odd numbers. So $M_n$ and $M_m$ homeomorphic $\Rightarrow m\equiv n\pmod2$.

Conversely, note that $M_n$ are $S^2$-bundles over $S^2$, and these are classified by the gluing map along an equatorial $S^1$ of $S^2=D^2\cup_{S^1}D^2$. Since $\operatorname{Aut}S^2=SO(3)$ (you could use $PGL(2,\mathbb{C})$ if you want, the calculation is the same), we have $$[\text{equatorial }S^1, SO(3)]=\pi_1(SO(3))=C_2.$$ So $m\equiv n\pmod{2}$ implies $M_n$ and $M_m$ are homeomorphic.

For complex structures, see Nicolas's answer.