We know that for a curve $X$, any object $\mathcal{E}^{\bullet}$ in the derived category $D^b_{\text{coh}}(X)$ is formal, i.e. $\mathcal{E}^{\bullet}$ is quasi-isomporphic to the direct sum of its cohomology sheaves. The reason is that the cohomological dimension of $X$ is $1$. We can see Corollary 3.15 of Daniel Huybrechts' book "Fourier–Mukai transforms in algebraic geometry" for details.

Now could we find an "easy" example of object in $D^b_{\text{coh}}(\mathbb{P}^2)$ which is not formal? In particular, could we find a complex of sheaves on $\mathbb{P}^2$ of length $2$ with coherent cohomology which is not quasi-isomorphic to the direct sum of its cohomology sheaves?


Yes; whenever you have two objects in an abelian category such that $Ext^2(M,N)$ is not equal to 0, we have a nonformal object given by coning with this morphism. More down-to-earthly, the element of $Ext^2(M,N)$ is given by some complex $N \to K \to L\to M$; the non-formal complex is just $\cdots \to 0\to K \to L \to 0\to \cdots$.

EDIT: Thanks for the example below. I was too lazy to provide one, but I also think it risks camouflaging the actual point here, since there's nothing special about coherent sheaves on $\mathbb{P}^2$, this happens in any abelian category with global dimension $>1$.

  • $\begingroup$ Thank you! Maybe I need a more explicit construction. For example we know that $Ext^2_{\mathbb{P}^2}(\mathcal{O},\mathcal{O}(-3))\neq 0$ hence as you pointed out we have a complex $\mathcal{O}(-3)\rightarrow K \rightarrow L\rightarrow \mathcal{O}$. Now could we find an explicit expression of the $K$ and $L$? $\endgroup$ – Zhaoting Wei Feb 7 '15 at 16:46
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    $\begingroup$ Just take the Koszul complex associated to the surjective map $\ \mathscr{O}_{\mathbb{P}^2}(-1)^3\rightarrow \mathscr{O}_{\mathbb{P}^2}\ $ given by multiplication by $X,Y,Z$. The corresponding morphism is $\ \mathscr{O}_{\mathbb{P}^2}(-1)^3\rightarrow \mathscr{O}_{\mathbb{P}^2}^3\;$, given by the matrix $\pmatrix{0 & Z & -Y\\ -Z & 0 & X\\ Y & -X & 0}$. $\endgroup$ – abx Feb 7 '15 at 16:59
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    $\begingroup$ @abx: $O^3$ should be $O(-1)^3$. $\endgroup$ – Sasha Feb 7 '15 at 18:04
  • $\begingroup$ Oops -- right, thanks, sorry for the typo. $\endgroup$ – abx Feb 7 '15 at 18:18

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