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?
Best Answer
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 $1 - 1 = 0$.