Timeline for $\mathbb Z$-formality of spheres
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2021 at 12:57 | vote | accept | Capotasto | ||
| Jun 13, 2021 at 21:30 | answer | added | Dmitri Pavlov | timeline score: 4 | |
| Jun 13, 2021 at 18:50 | comment | added | user164898 | Sorry, I keep deleting my comments because they are not fleshed out enough to be as accurate and relevant as I would like. I will stop commenting on this question. | |
| Jun 13, 2021 at 18:45 | comment | added | mme | I agree with A.S.'s comment. (I interpreted 'trivial cohomology ring' as meaning $H^0 = H^*$; it seems like it was meant as 'all cup products are zero'. It is in general very hard to find cochains representing given cohomology classes whose products are zero on the nose, and I only know how to do this for suspensions.) | |
| Jun 13, 2021 at 18:19 | comment | added | mme | @ConnorMalin Sure, but that's easy, right? The unit gives a quasi-iso from cohomology. The best I can do with this argument seems to be that if X is a simplicial complex whose integral cohomology groups are Z-free then $C^*(\Sigma X;\Bbb Z)$ is formal. | |
| Jun 13, 2021 at 18:15 | comment | added | Connor Malin | @mme This looks right to me, but wouldn't that prove that every space with a trivial cohomology ring is formal. Is that right? | |
| Jun 13, 2021 at 17:14 | comment | added | mme | What is wrong with the following? Introduce a simplicial structure on $S^n$. We get a zig-zag of algebra quasi-isomorphisms $C^*(S^n; \Bbb Z) \to C^*_\Delta(S^n;\Bbb Z) \xleftarrow{} H^*(S^n;\Bbb Z)$, the first map given by restricting the domain of cochains to simplicial chains, the last map existing because $C^{2n}_\Delta(S^n;\Bbb Z)$ is zero on the nose. | |
| Jun 13, 2021 at 16:32 | comment | added | John Pardon | If we try to construct a map of $A_\infty$-algebras $F:H^*(S^n)\to C^*(S^n)$ by induction on $k$ (indexing the operations $F^k:H^*(S^n)^{\otimes k}\to C^*(S^n)$) whose $k=1$ component $F^1:H^*(S^n)\to C^*(S^n)$ induces the identity map on cohomology, then we run into potential obstructions when $k\equiv 2\mod n$. | |
| Jun 13, 2021 at 16:26 | comment | added | John Pardon | @A.S. (see previous comment) | |
| Jun 13, 2021 at 16:26 | comment | added | John Pardon | @abx Your reasoning shows that $H^*(S^n)$ and $C^*(S^n)$ are quasi-isomorphic as cochain complexes. Formality is the much stronger condition that they are quasi-isomorphic as differential graded algebras. | |
| Jun 13, 2021 at 15:11 | comment | added | abx | I am probably missing something here: if you map 1 to 1 and a generator $g$ of $H^n(S^n,\mathbb{Z})$ to a closed element of $C^n(S^n,\mathbb{Z})$ with cohomology class $g$, doesn't that give you a quasi-isomorphism over $\mathbb{Z}$? | |
| Jun 13, 2021 at 14:56 | review | First posts | |||
| Jun 13, 2021 at 15:08 | |||||
| Jun 13, 2021 at 14:51 | history | edited | YCor | CC BY-SA 4.0 |
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| Jun 13, 2021 at 14:49 | history | asked | Capotasto | CC BY-SA 4.0 |