Timeline for Comparing stabilization of stable category modulo injectives and a Verdier localization
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jan 23 at 2:01 | history | bounty ended | CommunityBot | ||
| S Jan 23 at 2:01 | history | notice removed | CommunityBot | ||
| Jan 15 at 8:08 | history | edited | Jeremy Rickard | CC BY-SA 4.0 |
spelling
|
| Jan 15 at 7:36 | answer | added | Jeremy Rickard | timeline score: 3 | |
| S Jan 15 at 0:18 | history | bounty started | Snake Eyes | ||
| S Jan 15 at 0:18 | history | notice added | Snake Eyes | Draw attention | |
| Jan 14 at 12:34 | comment | added | Snake Eyes | @LeoAlonso: that is fine ... thank you for taking an interest :) | |
| Jan 13 at 21:10 | comment | added | Leo Alonso | Oh, I see your $S$ denotes stabilization, sorry for the misunderstanding. I am afraid I do not have intuition about this. | |
| Jan 12 at 23:04 | comment | added | Snake Eyes | @LeoAlonso: No I do not .... If $\mathcal A$ were Frobenius, then $S(\mathcal A/\mathcal I)\cong \mathcal A/\mathcal I$ which is then just $D_{sg}(\mathcal A)$ I think ... | |
| Jan 12 at 17:48 | comment | added | Leo Alonso | Do you assume your category $\mathcal{A}$ is Frobenius? | |
| Jan 12 at 0:06 | comment | added | Snake Eyes | @LeoAlonso: The stabilization of any right/left triangulated category is triangulated ... | |
| Jan 11 at 22:10 | comment | added | Leo Alonso | I guess the answer is no because $D^b(\mathcal A)/K^b(\mathcal I)$ is triangulated while $S(\mathcal A/\mathcal I)$ is merely right triangulated. | |
| Jan 11 at 8:39 | history | edited | Snake Eyes |
edited tags
|
|
| Jan 11 at 8:12 | history | edited | YCor | CC BY-SA 4.0 |
formatting, included top-level tag
|
| Jan 11 at 6:54 | history | asked | Snake Eyes | CC BY-SA 4.0 |