Recall for $n \in \mathbb{Z} $ that the complex projective varieties $$X_{n} = \mathbb{P}_{\mathbb{CP^{1}}}(\mathcal{O}+\mathcal{O}(n))$$ are either diffeomorphic to either $X_{0}$ or $X_{1}$. One can ask an analogous question in higher dimensions:
Question: Fix $k \geq 2$. For $n \in \mathbb{Z}$ set $$X_{n} = \mathbb{P}_{\mathbb{CP^{k}}}(\mathcal{O}+\mathcal{O}(n)).$$ Are all the $X_{n}$ diffeomorphic to either $X_{0}$ or $X_{1}$?
No. In the case $k=2$, $X_{0}$ is not diffeomorphic to $X_{2}$.
If $D_{1}$ is the divisor in $X_{n}$ which is the restriction of the $\mathbb{P}^{1}$-bundle to a line $L \subset \mathbb{CP}^{2}$, and $D_{2}$ is the section with normal bundle $\mathcal{O}(n)$. Then $$ D_{1}^{3}=0, D_{1}^{2}D_{2} = 1, D_{1}D_{2}^{2} = n , D_{2}^{3}=n^{2}. $$
Using the above formula and the fact that the Poincare duals of $D_{1},D_{2}$ generate $H^{2}(X_{n},\mathbb{Z}) \cong \mathbb{Z}^{2}$, we see that for $n=0$, this cubic form has the property that $\forall \alpha \in H^{2}(X_{0},\mathbb{Z})$, $\alpha^{3} = 3k$ for some $k \in \mathbb{Z}$.
However in $X_{2}$ by direct computation $(D_{1}+D_{2})^{3} = 13$. Hence $X_{0}$ and $X_{2}$ cannot be diffeomorphic (or even homotopy equivelant). I would guess that $X_{n}$ are pairwise non-diffeomorphic (at least for $k=2$), but I don't know how to prove it (??).