Is there an atlas $\mathcal{A}$ for $\mathbb{C}P^2 \setminus \{pt\}$ such that all transition maps of this atlas are affine maps?



Can a puntured $\mathbb{C}P^2$ admit an affine structure?
The answer is 'no'. I don't have access to the right references right now, but I think an argument goes as follows. (Also, as Will points out below, you can use Stiefel-Whitney classes to get the same result.) 


If there were such an atlas on $X^4 = \mathbb{CP}^2\setminus\{p\}$, then it would induce a torsion-free flat affine structure on $X^4$. Since $X^4$ is simply connected, this affine structure would have a developing map $x:X^4\to\mathbb{R}^4$ that would be an immersion. 


In particular, it would follow that, for any embedded $2$-sphere $S^2\subset X^4$ that is smoothly homotopic through immersions to a standard linear $\mathbb{CP}^1\subset X^4=\mathbb{CP}^2\setminus\{p\}$, we would obtain an immersion $x:S^2\to \mathbb{R}^4$ whose (oriented) normal bundle would have Euler number $1$ (because two such $\mathbb{CP}^1$s intersect transversely at a single point). 


By perturbing such an $S^2$ slightly in $X^4$, we can arrange that the immersion $x:S^2\to \mathbb{R}^4$ have only transverse self-intersections, with each self-intersection consisting of two points in the domain $S^2$. However, by a theorem of Whitney (reference needed, maybe the paper of Chern and Spanier on oriented surfaces in 4-space is also relevant here), the Euler number of this normal bundle must be equal to the algebraic number of self-intersections (counted with signs) in the domain $S^2$, which is necessarily an *even* integer. 


Since we have seen that the oriented normal bundle of this $S^2$ in $X^4$ has Euler number 1, this is impossible.


