Let $\pi:\mathbb{C}^{n}\setminus{0}\rightarrow\mathbb{CP}^{n-1}, n\geq 3$ be the projection from affine space without the origin to the projective space. If we pull back the tangent bundle of $\mathbb{CP}^{n-1}$ we would get a nontrivial bundle over $\mathbb{C}^{n}\setminus{0}$. Now my question would be: what is $H^{1}(\mathbb{C}^{n}\setminus{0},\pi^{*}\mathcal{T}_{\mathbb{CP}^{n-1}})$?
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$\begingroup$ By $O$ do you mean the origin, i.e. $0$? $\endgroup$– Kevin CastoCommented Sep 8, 2016 at 4:02
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$\begingroup$ When $n>3$, this group is zero. When $n$ equals $3$, this group is $H^1(\mathbb{P}^2,\mathcal{T}_{\mathbb{P}^2}(-3)) = H^1(\mathbb{P}^2,\Omega_{\mathbb{P}^2})$, and this is one-dimensional. $\endgroup$– Jason StarrCommented Sep 8, 2016 at 13:53
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$\begingroup$ Isn't every vector bundle on $\mathbb{C}^n\smallsetminus 0$ trivial? (say, by Quillen-Suslin theorem and this mathoverflow.net/questions/22111/…) $\endgroup$– QfwfqCommented Sep 8, 2016 at 21:09
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$\begingroup$ Oh wait, there is THIS!! mathoverflow.net/questions/35788/… $\endgroup$– QfwfqCommented Sep 8, 2016 at 21:15
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1 Answer
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$\newcommand{tot}{\mathbb{C}^n\setminus 0}\newcommand{tan}{\mathcal{T}_{\mathbb{P}^{n-1}}}$ Since morphism $\pi$ is affine, for any quasicoherent sheaf $\mathcal{F}$ on $\mathbb{C}^n\setminus 0$ its higher direct images $R^{>0}\pi_*\mathcal{F}$ vanish, so from Leray spectral sequence we get $H^i(\tot, \mathcal{F})=H^i(\mathbb{P}^{n-1},\pi_*\mathcal{F})$. Applying this to $\pi^*\mathcal{T}_{\mathbb{P}^{n-1}}$ we get $$H^1(\tot,\pi^*\tan)=H^1(\mathbb{P}^{n-1},\pi_*\pi^*\tan)=H^1(\mathbb{P}^{n-1},\tan\otimes\pi_*\mathcal{O}_{\tot})$$ by projection formula.
Note that $\pi$ is the projection from the total space of $ \mathcal{O}_{\mathbb{P}^{n-1}}(-1)$ minus zero section.(the following is corrected thanks to Jason Starr). The direct image of the structure sheaf of the total space itself is $Sym(\mathcal{O}(-1))=\bigoplus\limits_{n\geq 0}S^n\mathcal{O}(-1)=\bigoplus\limits_{n\geq 0}\mathcal{O}(-n)$. So, if we throw away the zero section, sections of structure sheaf are allowed to have a pole along the zero section, so $\pi_*\mathcal{O}_{\tot}=\bigoplus\limits_{n\in\mathbb{Z}}\mathcal{O}(-n)$. So, the problem is reduced to computing $H^1(\mathbb{P}^{n-1}, \tan\otimes \mathcal{O}(-k))$. From Euler exact sequence we get a long exact seqeunce
$\dots\to H^1(\mathbb{P}^{n-1},\mathcal{O}(1-k)^{\oplus n})\to H^1(\mathbb{P}^{n-1},\tan\otimes \mathcal{O}(-k))\to H^2(\mathbb{P}^{n-1},\mathcal{O}(-k))\to \dots$
Both left and right groups are zero, because line bundles on $\mathbb{P}^{n-1}$ can have nonzero cohoomology only in degrees $0,n-1>2$ so $H^1(\tot,\pi^*\tan)=0$
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7$\begingroup$ For the most part this is correct, but there is a mistake. The pushforward sheaf of algebras $\pi_*\mathcal{O}_{\mathbb{C}^n\setminus 0}$ is Zariski locally isomorphic to $\mathcal{O}\otimes_{\mathbb{C}}\mathbb{C}[t,t^{-1}]$. However, the globally twisted sheaf is not $\mathcal{O}(-1)\otimes_{\mathbb{C}}\mathbb{C}[t,t^{-1}]$, but rather $\bigoplus_{d\in \mathbb{Z}} \mathcal{O}(-d)t^d$. $\endgroup$ Commented Sep 8, 2016 at 12:21
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$\begingroup$ @JasonStarr Of course, you are right, thank you! I edited the post. $\endgroup$– SashaPCommented Sep 8, 2016 at 19:25