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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?

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    $\begingroup$ There are arbitrarily small tubular neighborhoods that are disk bundles inside the normal bundle of $\mathbb{CP}^1$ relative to $\mathbb{CP}^n$. You can apply the Leray spectral sequence for projection from the disk bundle to conclude the vanishing that you want. $\endgroup$ – Jason Starr Oct 16 '18 at 16:33
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    $\begingroup$ Do I understand correctly that $E_2^{p,q}=0$ for $q>0$.? In that case how to show the vanishing of $H^1(\mathbb{C}\mathbb{P}^1)$ with coefficients in the push forward under the projection of the structure sheaf? $\endgroup$ – makt Oct 16 '18 at 17:32
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    $\begingroup$ I was wrong. I forgot to dualize the normal bundle. Please see Misha Verbitsky's post below. $\endgroup$ – Jason Starr Oct 16 '18 at 20:25
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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.

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  • $\begingroup$ To make sure: when you say that the fibers are Stein, do you refer to a version of the base change theorem? Note that the morphism $\pi$ is not proper. $\endgroup$ – makt Oct 16 '18 at 22:11
  • $\begingroup$ You don't need the base change theorem, just notice that the higher direct images vanish, because the projection is locally equivalent to a product with Stein fibers, and use it to compute the higher direct image explicitly $\endgroup$ – Misha Verbitsky Nov 2 '18 at 4:12

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