Timeline for Homotopy transfer in the opposite direction
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2015 at 22:11 | comment | added | Vladimir Dotsenko | Yes indeed... you are absolutely right. | |
| Aug 14, 2015 at 22:10 | history | edited | Vladimir Dotsenko | CC BY-SA 3.0 |
deleted 4 characters in body
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| Aug 14, 2015 at 12:56 | vote | accept | Fernando Muro | ||
| Aug 14, 2015 at 12:56 | comment | added | Fernando Muro | $\prod$ instead of $\bigoplus$? | |
| Aug 14, 2015 at 12:47 | comment | added | Vladimir Dotsenko | @FernandoMuro: done! | |
| Aug 14, 2015 at 12:47 | history | edited | Vladimir Dotsenko | CC BY-SA 3.0 |
moved the correct argument from comments
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| Aug 14, 2015 at 8:27 | comment | added | Vladimir Dotsenko | @FernandoMuro I suppose I prefer to view infinity structures as Maurer-Cartan elements of the convolution pre-Lie algebra between the coaugmentation coideal of the Koszul dual cooperad and the endomorphism operad; the inclusion of endomorphism operads induces an inclusion of the corresponding convolution algebras. | |
| Aug 13, 2015 at 22:47 | comment | added | Fernando Muro | Trying to look for a more conceptual explanation, it looks like if it works because $\mathcal O_\infty$ is augmented and the endomorphism operad of $X$ is a non-unital suboperad of the endomorphism operad of $Y$, right? The inclusion of endomorphism operads being defined via the splitting. | |
| Aug 13, 2015 at 16:36 | comment | added | Vladimir Dotsenko | @FernandoMuro Is it actually much simpler than we (I) think? In your picture the map $i\colon X\to Y$ must be injective and is an isomorphism with its image. Let us now use the projector $\pi=i\circ p$ to split $Y$ as the direct sum of its kernel and its image. (The latter coincides with the image of $i$. The former, we do not care.) Then let us argue like in the minimal model theorem: if we transfer the structure that we have on $X$ to the image of $\pi$ via the isomorphism $i$, and let all yet undefined structure maps be zero, we get an O-infinity structure on $Y$. Or something fails? | |
| Aug 13, 2015 at 16:09 | comment | added | Vladimir Dotsenko | Oops. You are right. Did you inform Bruno about it? | |
| Aug 13, 2015 at 15:57 | comment | added | Fernando Muro | Ok, that's theorem 10.3.7 in the published version. The proof is mistaken, if the argument were true any quasi-isomorphism should be injective! Actually, what I'm asking about would yield a proof of that theorem as a corollary (at least over a field), since any quasi-isomorphism between cofibrant complexes can be replaced with a roof of strong deformation retractions. | |
| Aug 13, 2015 at 15:21 | comment | added | Vladimir Dotsenko | Hi Fernando, I refer to Theorem 10.3.12 (p.310) in the electronic version. (math.unice.fr/~brunov/Operads.pdf) | |
| Aug 13, 2015 at 13:23 | comment | added | Fernando Muro | Hi Vladimir, I can't find any 10.3.12. I've tried to figure out what result you mean in that chapter without much success, maybe theorem 10.3.7? | |
| Aug 13, 2015 at 7:40 | history | answered | Vladimir Dotsenko | CC BY-SA 3.0 |