Timeline for $k$-Disk algebras versus $E_k$ algebras
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2014 at 17:32 | vote | accept | Espen Nielsen | ||
| Nov 13, 2014 at 17:19 | comment | added | thel | The difference between framed and un-framed discs seems to be the most algebraically salient part, but you need to get some control over the topology of the space of embeddings of a disc. The clearest (though very terse) discussion of this I've seen is in John Francis's "Factorization homology of topological manifolds," section 2. | |
| Nov 13, 2014 at 15:36 | answer | added | Najib Idrissi | timeline score: 11 | |
| Sep 26, 2014 at 20:08 | comment | added | Espen Nielsen | @OmarAntolín-Camarena Maybe my terminology is nonstandard. By framed embedding, I meant here embeddings preserving a framing. An embedding $e:M\rightarrow N$ of framed manifolds is then called framed if $TM=e^*(TN)$. This is how I understand the category of framed manifolds and their structure-preserving embeddings as defiend in Ginot's notes, so it felt natural to give it such a terminology. Thank you very much for your comment about the unary operations. For some reason I did not think that the homotopies needed to preserve the endpoints, which is foolish in hindsight. | |
| Sep 26, 2014 at 0:26 | comment | added | Omar Antolín-Camarena | But ignoring that, and assuming you just mean that you allow rotations now too, I can tell you that operad is fairly different from the little cubes operad. The unary operations don't match: the space of embeddings of one cube into itself is contractible (there is "only one unary operation"), but the space of embeddings of a disk into itself, where you allow rotations, is homotopic to $SO(k)$. (Your observation that rotations are homotopic to the identity says only that $\pi_0 (SO(k)) = \ast$, which is true but doesn't make $SO(k)$ contractible.) | |
| Sep 26, 2014 at 0:23 | comment | added | Omar Antolín-Camarena | What do you mean by "framed embeddings"? I'm used to hearing that phrase for embeddings of a manifold into a higher dimensional one, where it means an embedding with a trivialization of the the normal bundle. When embedding a disk into a disk of the same dimension I don't see what you mean. | |
| Sep 25, 2014 at 23:56 | review | Close votes | |||
| Sep 26, 2014 at 20:57 | |||||
| Sep 25, 2014 at 23:37 | comment | added | Espen Nielsen | @SinanYalin That makes sense. The Morita equivalence example is nice. Thank you very much. | |
| Sep 25, 2014 at 23:36 | comment | added | Espen Nielsen | @FernandoMuro Thank you very much. I think this answers my question. | |
| Sep 25, 2014 at 22:34 | comment | added | Sinan Yalin | I would like to add a comment about your sentence "this leads me to believe that the operads themselves should be equivalent." Be careful, the fact that two categories of algebras are equivalent does not imply in general that the operads themselves are weakly equivalent. The situation is analogous to what happens in representation theory: Morita equivalence does not always imply equivalence. | |
| Sep 25, 2014 at 22:20 | comment | added | Fernando Muro | I think this is not exactly what you're asking about, but in case it might help, let me say that the little disks and the little cubes (topological) operads are weakly equivalent. Beware, however, that the little disks operad does not allow rotations. That's the framed little disks operad, which is not equivalent to the former. | |
| Sep 25, 2014 at 22:08 | history | asked | Espen Nielsen | CC BY-SA 3.0 |