Timeline for Obstruction to splitting an object in derived category into a sum of two-term complexes
Current License: CC BY-SA 4.0
7 events
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| Sep 20, 2019 at 17:51 | comment | added | nikola karabatic | @Dmitry Pirozhkov: I still cannot see how it should be possible to define an obstruction without the splitting as input data. There might be many ways to split the cohomology, yielding different decompositions on homology, for example if all $\xi_i$ equal $0$. (Perhaps my fantasy is just too limited.) | |
| Sep 20, 2019 at 17:49 | comment | added | nikola karabatic | fixed the indexing and @Jeremy Rickard: you are right that this was too complicated. I upvoted your answer. | |
| Sep 20, 2019 at 17:48 | history | edited | nikola karabatic | CC BY-SA 4.0 |
added 42 characters in body
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| Sep 20, 2019 at 15:24 | comment | added | Dmitry Pirozhkov | As Jeremy Rickard said, in the case of homological dimension 2 this description becomes almost tautological: given a splitting $H^i = H^i_a \oplus H^i_b$, it produces a splitting of the complex if and only if $\xi_{i+1}$ factors through $H^i_a$ and $\xi_i$ factors through $H^i_b$. And that is more or less "the complex splits into a direct sum of two-term complexes". I would be more interested in an obstruction that doesn't start with a decomposition of a cohomology object. Intuitively the condition should be some "orthogonality" between glueing maps, but I don't know in which sense. | |
| Sep 20, 2019 at 12:10 | comment | added | Jeremy Rickard | Sorry, in the previous comment $[-i-2]$ should be $[-i+1]$. | |
| Sep 20, 2019 at 9:36 | comment | added | Jeremy Rickard | I'm a little confused by some of the indexing, but I think that the fact that $\mathcal{A}$ has homological dimension two significantly simplifies this, since then the map $H^{i+1}M[-i-1]\to\tau_{[i-1.i]}M[1]$ in your first distinguished triangle (I think you meant $H^{i+1}M$ to be shifted in degree?) is determined by a map $H^{i+1}M[-i-1]\to H^iM[-i-2]$ (which is exactly the element of $\text{Ext}^2(H^{i+1}M,H^iM)$ referred to in the question. | |
| Sep 20, 2019 at 8:53 | history | answered | nikola karabatic | CC BY-SA 4.0 |