Timeline for $\operatorname{Ext}^2(O,\omega)$ as a higher extension on $\mathbb{P}^1 \times \mathbb{P}^1$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 21, 2017 at 0:30 | comment | added | solbap | Thanks Piotr! It seems plausible that there is a morphism from your sequence to each of my sequences so perhaps they are all quasi-isomorphic? This is not how I thought $Ext^i$ equivalence worked. In any case if you put your comment in the form of an answer I would accept. | |
| Jul 20, 2017 at 21:23 | comment | added | Piotr Achinger | 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. | |
| Jul 20, 2017 at 16:04 | comment | added | solbap | 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? | |
| Jul 20, 2017 at 15:55 | comment | added | solbap | 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. | |
| Jul 20, 2017 at 8:15 | history | edited | Ben McKay | CC BY-SA 3.0 |
formatting
|
| Jul 20, 2017 at 7:35 | comment | added | Piotr Achinger | (Here $[\ldots\to B\to A]$ means a complex with $A$ in degree $0$.) | |
| Jul 20, 2017 at 7:24 | comment | added | Piotr Achinger | 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 $$\mathcal{O} \cong [\mathcal{O}(-2,-2)\to \mathcal{O}(-2, -1)^2\oplus\mathcal{O}(-1, -2)^2 \to \mathcal{O}(-1,-1)^4]\to \mathcal{O}(-2, -2)[2]$$ | |
| Jul 20, 2017 at 1:25 | history | asked | solbap | CC BY-SA 3.0 |