Timeline for Show that the rational cohomology ring $H^*(M;\mathbb{Q})$ needs at least two generators
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34 events
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| Aug 22, 2018 at 6:25 | comment | added | Mark Grant | @Multivariablecalculus: Note that a basis for rational homotopy corresponds to a set of algebra generators of the minimal model. Now the claim follows from Proposition 1 of the paper of Vigue-Poirrier and Sullivan: projecteuclid.org/euclid.jdg/1214433729 | |
| Aug 21, 2018 at 21:36 | comment | added | Sergio Charles | Hi @MarkGrant, would you be able to explain why $H^*(M;\mathbb{Q})$ requires more than one generator iff $\dim\pi_{\text{odd}}(M)\otimes\mathbb{Q}>1$? | |
| S Jul 30, 2018 at 21:01 | history | bounty ended | CommunityBot | ||
| S Jul 30, 2018 at 21:01 | history | notice removed | CommunityBot | ||
| Jul 27, 2018 at 4:22 | comment | added | Ian Agol | Another paper on this topic (classifying biquotients with singly generated cohomology): arxiv.org/abs/math/0210231 | |
| Jul 26, 2018 at 17:03 | comment | added | Ian Agol | I also found a paper addressing the issue of polynomial quotient cohomology ring: arxiv.org/abs/1403.1801 This obstructs certain polynomial rings in certain dimensions. | |
| Jul 25, 2018 at 23:12 | comment | added | Ian Agol | In dimension at most 5, the only examples of simply-connected manifolds with polynomial cohomology ring are spheres and $\mathbb{CP}^2$ by the generalized Poincaré conjecture and Freedman's theorem. | |
| Jul 25, 2018 at 21:54 | vote | accept | Sergio Charles | ||
| Jul 25, 2018 at 5:15 | vote | accept | Sergio Charles | ||
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| Jul 25, 2018 at 3:51 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Jul 23, 2018 at 22:51 | vote | accept | Sergio Charles | ||
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| Jul 23, 2018 at 20:26 | vote | accept | Sergio Charles | ||
| Jul 23, 2018 at 22:51 | |||||
| Jul 23, 2018 at 19:56 | answer | added | Ian Agol | timeline score: 11 | |
| Jul 23, 2018 at 18:51 | comment | added | Mark Grant | @Multivariablecalculus: well, it is a condition on the non-torsion part of the homotopy groups of $M$, since tensoring with $\mathbb{Q}$ killls torsion; see rational homotopy theory. (Torsion here has little to do with torsion in the sense of differential geometry.) | |
| Jul 23, 2018 at 15:59 | comment | added | Sergio Charles | The only information known about the Riemannian manifold $M$ is that it is simply connected and closed (indeed, it is a very general case). @DylanThurston | |
| Jul 23, 2018 at 15:56 | comment | added | Dylan Thurston | Can you say more about the context? Presumably you know something about a particular manifold $M$, what sort of information do you have? | |
| Jul 23, 2018 at 15:47 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Jul 23, 2018 at 15:46 | comment | added | Sergio Charles | Is this equivalent to a condition on torsion, by tensoring with $\mathbb{Q}$ (or a divisible group)? @MarkGrant | |
| Jul 23, 2018 at 8:26 | comment | added | Mark Grant | There is an equivalent condition in terms of rational homotopy groups: $H^*(M;\mathbb{Q})$ requires more than one generator iff $\pi_{\rm odd}(M)\otimes\mathbb{Q}$ is more than one-dimensional. | |
| S Jul 22, 2018 at 19:28 | history | bounty started | Sergio Charles | ||
| S Jul 22, 2018 at 19:28 | history | notice added | Sergio Charles | Authoritative reference needed | |
| Jul 21, 2018 at 2:13 | comment | added | Sergio Charles | Coincidentally, I believe this has an analogy in Hamiltonian dynamics, but I cannot quite recall what it is exactly. | |
| Jul 20, 2018 at 22:42 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Jul 20, 2018 at 22:37 | comment | added | Sergio Charles | @Ian Agol Yes, that the sequence of Betti numbers $\{b_k(\Lambda M; \mathbb{F}_p)\}_{k\ge 0}$ is unbounded for some prime $p\ge 2$ where $\Lambda M$ is the free loop space of $M$, i.e. $H^*(M;\mathbb{Q})$ is not isomorphic to a truncated polynomial ring. | |
| Jul 20, 2018 at 22:34 | comment | added | Sergio Charles | @Igor Belegradek I see, thanks for the reference. | |
| Jul 20, 2018 at 22:18 | comment | added | Igor Belegradek | I don't think such results are known. Some examples of geometrically interesting manifolds with singly generated rational cohomology can be found in arxiv.org/abs/math/0210231 but there are many other examples and it is unclear how curvature could be relevant. | |
| Jul 20, 2018 at 22:16 | comment | added | Ian Agol | Presumably you're assuming that $dim H^k(M;\mathbb{Q})= 1$ for all $k \equiv 0 (\mod q)$ for some $1< q| dim M$? Also, I think you want to say that $H^*(M;\mathbb{Q})$ is not the quotient of a polynomial ring. | |
| Jul 20, 2018 at 22:06 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Jul 20, 2018 at 22:05 | comment | added | Sergio Charles | Yes, thank you @Tobias Shin. It does not necessarily have to be a condition on the curvature of $M$. I just need a way to force $M$ not to have a polynomial cohomology ring $H^*(M;\mathbb{Q})$. (Possibly imposing a condition on the metric of $M$?) | |
| Jul 20, 2018 at 21:53 | comment | added | Tobias Shin | I'm not sure about a necessary condition, but if your rational cohomology ring has at least two generators, then your manifold has infinitely many distinct closed geodesics. I don't know any conditions on curvature that can control the number of distinct closed geodesics however ... | |
| Jul 20, 2018 at 20:09 | review | Close votes | |||
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| Jul 20, 2018 at 19:40 | history | edited | Sergio Charles | CC BY-SA 4.0 |
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| Jul 20, 2018 at 19:29 | review | First posts | |||
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| Jul 20, 2018 at 19:26 | history | asked | Sergio Charles | CC BY-SA 4.0 |