Birkhoff decomposition vanishing of the Chern numbers of the holomorphic line bundles of the Birkhoff-Grothendieck decomposition, is some statement I read off in One of Connes papers. Without going into detail with how it comes up in the paper, I am curious as to how this connection comes about, more or less how to consider formulating a proof about this. I must say I have a very tiny math background(struggling physics dude here) but I mean I can handle math lingo, but I need a gentle response. I am sorry if this is too trivial a question, and apologize in advance. . 



Birkhoff decomposition vanishing of the Chern numbers
The proof is straightforward if you accept some standard results on cohomology over $\mathbb{P}^1$. Let me try. We want to prove that any vector bundle $E$ on $\mathbb{P}^1$ is a direct sum of line bundles $\mathcal{O}(k)$.
The proof is by induction on the rank of $E$.
Replacing $E$ by $E(k)$ for some integer $k$ we may assume $H^0(E)\neq 0$, $H^0(E(-1))=0$. Let $s\neq 0$ in $H^0(E)$. Then $s$ is nonzero at every point (otherwise $s$ would give a nonzero section of $E(-1)$), so there is an extension
$$0\rightarrow \mathcal{O}\rightarrow E \rightarrow F\rightarrow 0\ .$$  

By induction we have $F=\sum \mathcal{O}(k\_i)$. Tensor product with $\mathcal{O}(-1)$ + cohomology exact sequence + $H^1(\mathbb{P}^1,\mathcal{O}(-1))=0$ gives 
$H^0(\mathbb{P}^1,F(-1))=0$, hence $k\_i\leq 0$ for each $i$. 


Now you have to accept that such extensions are classified by the vector space $H^1(\underline{\mathrm{Hom}}(F,\mathcal{O}))=H^1(F^\*)$. Since $H^1(\mathcal{O}(k))=0$ for $k\geq 0$, this is zero, hence the extension is trivial and you are done. 


