Timeline for On the Hochschild cohomology of the minimal model of an $A_\infty$ algebra
Current License: CC BY-SA 4.0
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| S Aug 9, 2021 at 17:06 | history | bounty ended | CommunityBot | ||
| S Aug 9, 2021 at 17:06 | history | notice removed | CommunityBot | ||
| S Aug 1, 2021 at 15:13 | history | bounty started | Hang | ||
| S Aug 1, 2021 at 15:13 | history | notice added | Hang | Authoritative reference needed | |
| Aug 1, 2021 at 14:44 | history | edited | Hang | CC BY-SA 4.0 |
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| Jul 27, 2021 at 22:13 | comment | added | Fernando Muro | If I'm not mistaken, there is an infinity quasi isomorphism from the hochschild complex of the minimal model to that of the original algebra, constructed as in the homotopy transfer theorem | |
| Jul 27, 2021 at 20:31 | comment | added | Connor Malin | The spectral sequence is used to prove the map is a quasi-isomorphism. Whatever structure Hochschild cohomology has will be preserved since we are applying Hochschild cohomology to a map of A infinity objects. | |
| Jul 27, 2021 at 16:33 | history | edited | Hang | CC BY-SA 4.0 |
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| Jul 27, 2021 at 16:29 | comment | added | Hang | @ConnorMalin Thank you. Does this spectral sequence relation imply other additional relations than quasi-isomorphisms, such like ring homomorphisms? | |
| Jul 27, 2021 at 16:28 | history | edited | Hang | CC BY-SA 4.0 |
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| Jul 26, 2021 at 19:20 | comment | added | Connor Malin | I am not so great with references, but don't we have a spectral sequence $HH^*(H_*(A)) \implies HH^*(A)$ (coming from filtering the bar construction), so by its naturality we have a map of the spectral sequences for $A$ and for $\tilde{A}$ which is an iso on these pages, hence an isomorphism on the $E_\infty$ pages? So the answer to your question is, the complexes are quasiisomorphic. | |
| Jul 26, 2021 at 19:08 | history | asked | Hang | CC BY-SA 4.0 |