# 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.