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Let $X = \mathbb{P}^1 \times \mathbb{P}^1$ over a field $k$ and consider $Ext^2(\mathcal{O}_X,\omega_X)\cong H^2(\omega_X) = H^2(\mathcal{O}_X(-2,-2)) = k$

Let $C = \mathbb{P}^1$. By Kunneth $H^2(\mathcal{O}_X(-2,-2))\cong H^1(\mathcal{O}_C(-2))\otimes H^1(\mathcal{O}_C(-2))$

I want to represent elements in these groups as extensions, i.e. exact sequence of sheaves upto a certain equivalence.

For example the nontrivial extension in $H^1(\mathcal{O}_C(-2))\cong Ext^1(\mathcal{O}_C,\mathcal{O}_C(-2))$ corresponds to $0\to \mathcal{O}(-2)\to \mathcal{O}(-1)^{\oplus 2} \to \mathcal{O}\to 0$

So the Kunneth isomorphism

$Ext^2(\mathcal{O}_X,\omega_X) \cong Ext^1(\mathcal{O}_C,\mathcal{O}_C(-2)) \otimes Ext^1(\mathcal{O}_C,\mathcal{O}_C(-2))$

suggests to me that one way to represent elements in the $Ext^2$ group is to pull back to two non trivial extensions to $X$ and stick them together. The issue is that there are two different ways to do this leading to

$\mathcal{O}_X(-2,-2) \to \mathcal{O}_X(-2,-1)^{\oplus 2} \to \mathcal{O}_X(-1,0)^{\oplus 2}\to \mathcal{O}_X$

or

$\mathcal{O}_X(-2,-2) \to \mathcal{O}_X(-1,-2)^{\oplus 2} \to \mathcal{O}_X(0,-1)^{\oplus 2}\to \mathcal{O}_X$

The sequences sent to each other by switching the components and they are also Serre dual sequences but they don't seem to be equivalent extensions because there aren't any non trivial homs between the middle terms (although I suspect I might be wrong about this).

QUESTION

Do the two not so short exact sequences above represent the same extension class in $Ext^2(\mathcal{O}_X,\mathcal{O}_X(-2,-2)$? If they are the same, how can we see they are equivalent?

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  • $\begingroup$ How did you get those two sequences? From tensoring the two morphisms $$\mathcal{O}\cong [\mathcal{O}(-2,0)\to \mathcal{O}(-1, 0)^2]\to \mathcal{O}(-2, 0)[1]$$ and $$\mathcal{O}\cong [\mathcal{O}(0,-2)\to \mathcal{O}(0,-1)^2]\to \mathcal{O}(0, -2)[1]$$ we get $$\mathcal{O} \cong [\mathcal{O}(-2,-2)\to \mathcal{O}(-2, -1)^2\oplus\mathcal{O}(-1, -2)^2 \to \mathcal{O}(-1,-1)^4]\to \mathcal{O}(-2, -2)[2]$$ $\endgroup$
    – Piotr Achinger
    Jul 20, 2017 at 7:24
  • $\begingroup$ (Here $[\ldots\to B\to A]$ means a complex with $A$ in degree $0$.) $\endgroup$
    – Piotr Achinger
    Jul 20, 2017 at 7:35
  • $\begingroup$ I didn't tensor the sequences, although I admit that makes more sense. If you pull back one sequence and tensor with $\mathcal{O}(0,-2)$ you get $\mathcal{O}(-2,-2)\to \mathcal{O}(-1,-2)^2 \to \mathcal{O}(0,-2)$ then the other sequence is $\mathcal{O}(0,-2) \to \mathcal{O}(0,-1)^2 \to \mathcal{O}$. So the end of the first sequence is the same as the beginning of the second sequence. Sticking them together gives one of the sequences I wrote. Reversing the roles of the sequences gives the second sequence I wrote. $\endgroup$
    – solbap
    Jul 20, 2017 at 15:55
  • $\begingroup$ I must also admit I don't totally see through your notation. I see that $\mathcal{O} \cong [\mathcal{O}(-2,0) \to \mathcal{O}(-1,0)^2$ but what is the morphism to $\mathcal{O}(-2,0)[1]$? And now I'm also wondering do the sequences I wrote down actually represent the trivial element in the $Ext^2$ group? $\endgroup$
    – solbap
    Jul 20, 2017 at 16:04
  • $\begingroup$ If I have an extension $0\to A\to B\to C\to 0$, I can regard it as a quasi-isomorphism $C\cong [A\to B]$. Here $B$ is in degree zero, so $A$ is in degree $-1$. I have the obvious map of complexes $[A\to B]\to [A\to 0]=A[1]$. The composition $C\to A[1]$ is the unique map making the triangle $A\to B\to C\to A[1]$ distinguished (or anti-distinguished, depending on the convention), and its class in ${\rm Hom}(C, A[1]) = {\rm Ext}^1(C, A)$ is the class of the extension (again, up to sign). Alternatively, I could have written $C\to [B\to C] \cong A[1]$, this gives the same map. $\endgroup$
    – Piotr Achinger
    Jul 20, 2017 at 21:23

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