A Non-projective Proper Surface

Before moving on to some more exotic topics let’s use these past couple of posts to construct an important example: A proper surface over an algebraically closed field of characteristic 0 that is not projective. This is important because every such curve is projective.

Using the tools we’ve developed, here is the construction. Let {X'} be a non-trivial infinitesimal extension of {X=\mathbb{P}^2_k} by {\omega :=\omega_{\mathbb{P}^2}} (the canonical sheaf). How easy was that? Such a non-trivial extension exists because we’ve already shown that iso classes of extensions are in bijective correspondence with {H^1(\mathbb{P}^2, \omega\otimes \mathcal{T})\simeq H^1(\mathbb{P}^2, \Omega^1_{\mathbb{P}^2})\neq 0}.

Here’s how we check it is not projective. We have an exact sequence by the definition of an infinitesimal extension {0\rightarrow \omega \rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X\rightarrow 0}. This induces a sequence {0\rightarrow \omega\rightarrow \mathcal{O}_{X'}^*\rightarrow \mathcal{O}_X^*\rightarrow 0} (just by the first map being {x\mapsto 1+tx} where {t} is the first map of the previous sequence).

This gives a long exact sequence of cohomology: {\rightarrow H^1(X, \omega)\rightarrow Pic X' \rightarrow Pic X \stackrel{\delta}{\rightarrow} H^2(X, \omega) \rightarrow }. Now we know the canonical sheaf is really just {\mathcal{O}(-2)}, so by knowledge of cohomology of projective space {H^1(X, \omega)=0} and {H^2(X, \omega)=k}. Also, the Picard group of projective space is {\mathbb{Z}} generated by {\mathcal{O}(1)}.

So if we can show {\delta} is injective, then {Pic X'=0}, and hence can’t be projective. Since {char k=0}, this just amounts to showing that {\delta (\mathcal{O}(1))\neq 0}. We’ll derive a contradiction if it does map to zero. If it does map to zero, then this is the zero map and hence {Pic X'\rightarrow Pic X} is surjective. i.e. there is some invertible sheaf {\mathcal{L}} on {X'} such that {\mathcal{L}\simeq \mathcal{O}(1)/\omega}, which is what it means for a sheaf to map to {\mathcal{O}(1)}.

So we examine the exact sequence {0\rightarrow \mathcal{O}(-2)\rightarrow \mathcal{L}\rightarrow \mathcal{O}(1)\rightarrow 0}. It doesn’t matter what {X'} is to know {H^1(X', \mathcal{O}(-2))=0}, so we get that {H^0(X', \mathcal{L})\rightarrow H^0(X, \mathcal{O}(1))} is surjective and hence we get a map {\mathbb{P}^2\rightarrow X'\rightarrow \mathbb{P}^2} which is the identity on {\mathbb{P}^2}. This gives us a splitting of {0\rightarrow \omega \rightarrow \mathcal{O}_{X'}\rightarrow \mathcal{O}_X\rightarrow 0}. Thus {\mathcal{O}_X'\simeq \mathcal{O}_X\oplus \omega}. Ack! This means it was the trivial extension, which we assumed it wasn’t. Thus {\delta(\mathcal{O}(1))\neq 0}, which means {X'} is not projective.

Of course this isn’t the most concrete example of one of these things, but I don’t know of any other examples that are this easy to construct and prove. If you wanted something a little bit more concrete, you could actually pick a non-zero class in {H^1(\mathbb{P}^2, \Omega^1)} and piece together your infinitesimal extension.

I’m not making any promises, but next time I might start trying to figure out what gerbes are for a few posts, and then maybe try to tie them back in with how they relate to deformation theory.

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