TITLE: Cohomology of neighborhood of $\mathbb{C}\mathbb{P}^1$ in $\mathbb{C}\mathbb{P}^n$
QUESTION [6 upvotes]: Let $\mathbb{C}\mathbb{P}^1$ be embedded linearly to $\mathbb{C}\mathbb{P}^n$ with $n>1$. (Such an embedding is given in coordinates by $[x:y]\mapsto [x:y:0:\dots: 0]$.) 

Is it true that for any open neighborhood $U$ (in the analytic topology) of $\mathbb{C}\mathbb{P}^1$ there exists a smaller neighborhood $V$ of the latter such that 
  $$H^i(V, \mathcal{O})=0 \mbox{ for any } i>0,$$ 
  where $\mathcal{O}$ is the structure sheaf?

REPLY [6 votes]: It is not hard to see that the
first cohomology is infinite-dimensional.
Take a complement $M:=CP^3 \ CP^1$
and consider a projection
$\pi:\; M \mapsto CP^1$. It is not
hard to see that $M$ is
isomorphic to the total space of
the bundle $O(1)^2$. The fibers
of $\pi$ are Stein, hence
$R^i\pi_*F=0$ for $i>0$ and any coherent sheaf
$F$, and cohomology $H^i(O_M)$ are the
same as $H^i(\pi_* O_M)$.
However, $\pi_* O_M= Sym^*(O(-1)^2)$,
because the regular functions on the
total space of $O(1)^2$ are $Sym^*(O(-1)^2)$.
However, $H^1(Sym^*(O(-1)^2))$ is
infinite-dimensional.
Same argument works for smaller
neighbourhoods $U\supset CP^1$, as
long as the fibers of $\pi:\; U \mapsto C P^1$
remain Stein.
Same is true for a neighbourhood of a 
rational line $C$ in a complex manifold 
if the normal bundle $NC$ is ample, 
but the proof is more complicated.