Is $S^2\times S^2$ homeomorphic to $\mathbb{CP}^2\#\mathbb{CP}^2$?

Question

Is $S^2\times S^2$ homeomorphic to $\mathbb{CP}^2\#\mathbb{CP}^2$?

My idea: by the product formula for the Euler characteristic, we have $\chi(S^2\times S^2)=\chi(S^2)^2=4$. By the sum formula for Euler characteristic we have $\chi(\mathbb{CP}^2\#\mathbb{CP}^2)=2\chi(\mathbb{CP}^2)-\chi(S^4)=4$. So both manifolds have the same Euler characteristic.

By the sum formula for the signature, $\sigma(\mathbb{CP}^2\#\mathbb{CP}^2)=2$. Is the signature of $S^2\times S^2$ also 2? I'm not sure how to compute it.

If so, then I think we can conclude that the two manifolds must be homeomorphic by Friedman's classification of simply connected manifolds.

Is this reasoning correct?

Answer 1

The intersection form on $\mathbb{CP}^2\# \mathbb{CP^2}$ is $\pmatrix{1 & 0 \\ 0 & 1}$, while that of $S^2 \times S^2$ is $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$. The former has signature $1 + 1 = 2$; the latter has signature $1 - 1 = 0$.

Answer 2

It is not even true that $S^2\times S^2$ is homotopy equivalent to $\mathbb{C}P^2\#\mathbb{C}P^2$.

Pinching to one factor gives a map $\mathbb{C}P^2\#\mathbb{C}P^2\rightarrow \mathbb{C}P^2$ which can be used to detect a non-trivial Steerod square $Sq^2:H^2(\mathbb{C}P^2\#\mathbb{C}P^2;\mathbb{Z}_2)\rightarrow H^4(\mathbb{C}P^2\#\mathbb{C}P^2;\mathbb{Z}_2)$. On the other hand all non-identity cohomology operations on $H^*(S^2\times S^2)$ vanish identically.

In fact it is not difficult to show that

$$\Sigma(\mathbb{C}P^2\#\mathbb{C}P^2)\simeq \Sigma\mathbb{C}P^2\vee S^3$$

and

$$\Sigma(S^2\times S^2)\simeq S^3\vee S^3\vee S^5$$

which clearly demonstrates the non-existence of even a homotopy equivalence between the spaces in question.