How to show $CP^1\times CP^1$ and $CP^2$ blow up one point are not diffeomorphic?

Question

How can we prove $CP^1\times CP^1$ and $CP^2$ blow up one point is not diffeomorphic?

I tried to compute their Hodge numbers and Chern numbers but they are the same.

Answer 1

You can look at the cohomology ring structure.

More precisely, $H^2(\Bbb P^1 \times \Bbb P^1, \Bbb Z)$ is generated by $[L]$ and $[L']$, where $L = \{1\} \times \Bbb P^1$ and $L' = \Bbb P^1 \times \{1\}\}$. We have $L^2 = (L')^2 = 0$ and $LL' = 1$. In particular, for any $C = aL + b L'$, $C^2$ is even, but $E^2 = -1$ when $E$ is the exceptional divisor of the blow-up, contradiction.

In fact there are pretty close because they can be both decomposed as $$(\Bbb C^*)^2 \cup \Bbb C^* \cup \Bbb C^* \cup \Bbb C^* \cup \Bbb C^* \cup \{pt\} \cup \{pt\} \cup \{pt\} \cup \{pt\}$$ So they are equal in the Grothendieck ring of varieties, in particular have same Hodge numbers.

Answer 2

Topologically speaking, blowing up of $M$ at a point means you get $M'= M\# \bar{\mathbb CP^2}$. Signature

$\sigma(\mathbb CP^1\times \mathbb CP^1 \# \bar{\mathbb CP^2})= -1$. Where as $\sigma (\mathbb CP^2\# \bar{\mathbb CP^2})=0$. So they are not even homotopically equivalent.

Interesting fact is that if you blow up $\mathbb CP^2$ in two points then $\mathbb CP^2\# 2\bar{\mathbb CP^2}$ is actually diffeomorphic to $\mathbb CP^1\times \mathbb CP^1 \# \bar{\mathbb CP^2}$.

*EDIT I misread your original question. The reason $\mathbb CP^1\times \mathbb CP^1$ is not diffeomorphic with $\mathbb CP^2\# \bar{\mathbb CP^2}$ is because they have different co-homology ring. One way to see that in both cases $H^2$ is generated by two copies of $[\mathbb CP^1]$ namely $\alpha$ and $\beta$. But for the first case $\alpha \cup \beta \neq 0$ and $\alpha \cup \alpha = \beta \cup \beta =0$ {follows from product formula}. On the other case. i.e, $\mathbb CP^2\# \bar{\mathbb CP^2}$, $\alpha\cup \beta =0$. And $\alpha \cup \alpha \neq 0$ and $\beta \cup \beta \neq 0$. Thus they have different cohomology ring.

Answer 3

You can do this using Kirby calculus and diagrams. I'll not attempt to draw any diagrams, but since you already have two clearly articulated answers, I'll just leave this as an alternative method.

Both the spaces are $S^2$ bundles over $S^2$, and as such are characterised up to diffeomorphism by their mod 2 Euler numbers. The Kirby calculus for such objects is well understood and documented.

Each space has a handle decomposition with exactly two 2-handles. A Kirby diagram for $\mathbb{C}P^1\times\mathbb{C}P^1\cong S^2\times S^2$ consists of Hopf link of two $0$-framed unknots.

On the other hand a Kirby diagram of the blow up $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$ consists of a pair of disjoint unkots, one with framing $+1$ and one with framing $-1$. Performing a handle slide of the $-1$-unknot over the other generates a Hopf link of a $1$-framed unknot and a $-1+1=0$-framed unkot.

Further handle slides can change the $(0,0)$-Hopf link of $S^2\times S^2$ into a $(0,2n)$-framed Hopf link. Similarly, handle slides can change the $(0,1)$-framed Hopf link of $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$ into a $(0,2n+1)$-framed Hopf link, but the difference of parity cannot be reconciled by any permitted Kirby moves.

Since the handle slide operation preserves the diffeomorphism type, it's clear now that the two spaces we started with are not diffeomorphic.