Timeline for An $E_{\infty}$-algebra is a $C_{\infty}$-algebra?
Current License: CC BY-SA 4.0
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| Nov 9, 2023 at 12:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 10, 2023 at 10:43 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 10, 2023 at 10:39 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 10, 2023 at 10:31 | comment | added | YkMz | $Com_{\infty}$ is a cofibrant operad, so we always have a lifting on any $E_{\infty}$-operad by $Com_{\infty}$ and any $E_{\infty}$-algebra is a $C_{\infty}$ algebra. | |
| Oct 9, 2023 at 9:36 | comment | added | Fernando Muro | @Walterfield the consequence is the opposite of what you say. Both notions coincide. | |
| Oct 9, 2023 at 9:05 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 9, 2023 at 9:01 | comment | added | YkMz | @FernandoMuro I misunderstood. You mean that $Com_\infty$ is just one model of $E_\infty$-operad. So it is possible to have an $E_\infty$-algebra that does not appear as an algebra over $Com_\infty$. | |
| Oct 9, 2023 at 5:45 | comment | added | Fernando Muro | @Walterfield right the contrary. | |
| Oct 9, 2023 at 2:53 | comment | added | YkMz | @FernandoMuro Thank you. You mean the answer is no because the operad $Com_\infty$ is an $E_\infty$-operad? | |
| Oct 8, 2023 at 21:54 | comment | added | Fernando Muro | @Walterfield then you can find an answer to your question in the first paragraph of 13.1.10. | |
| Oct 8, 2023 at 14:37 | history | edited | YkMz |
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| Oct 8, 2023 at 14:18 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 8, 2023 at 13:39 | comment | added | YkMz | @BertramArnold Thank you. "weakly equivalent" means quasi-isomorphic as $C_{\infty}$-algebras? | |
| Oct 8, 2023 at 13:35 | comment | added | YkMz | @FernandoMuro Thank you. I only consider characteristic 0 and the definitions of $C_{\infty} $ and $E_{\infty}$ algebras in Section 13.1.8~13.1.10 of Loday-Vallette's book. | |
| Oct 8, 2023 at 13:28 | comment | added | Fernando Muro | You should maybe clarify what you mean by $E_\infty$ and $C_\infty$. I’ve only seen the latter considered in characteristic zero, where both notions coincide for almost all possible definitions. | |
| Oct 8, 2023 at 13:23 | comment | added | Bertram Arnold | Any $C_\infty$-algebra is weakly equivalent to a cdga, namely the cobar complex of its bar complex. For $E_\infty$-algebras, Dyer-Lashov operations are an obstruction to the existence of such a weak equivalence. | |
| Oct 8, 2023 at 10:24 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 8, 2023 at 10:18 | history | edited | YkMz | CC BY-SA 4.0 |
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| Oct 8, 2023 at 10:12 | history | asked | YkMz | CC BY-SA 4.0 |