Timeline for Rational equivalence of smooth closed manifolds
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 30, 2022 at 6:02 | comment | added | Aleksandar Milivojević | To add to Denis T's comment, if the rational homotopy type in question is formal, then $X$ admits endomorphisms of arbitrarily divisible degree (see Infinitesimal Computations in Topology Theorem 12.2, and F. Manin's "Positive weights and self-maps" for a detailed discussion, Corollary 1.1 arxiv.org/abs/2108.02173v3) and so, upon precomposing with an appropriate endomorphism one can lift the map $X \to X_{Q}$ through $Y \to X_{Q}$ to get a direct rational equivalence $X \to Y$. | |
| Aug 30, 2022 at 1:42 | comment | added | Denis T | ...And SW classes are functorial along any maps which are $\Bbb Z/2$ cohomology equivalence, so one can maybe just use them instead of more complicated things like cokernel of Hurewitz map for oriented bordism provided our cohomology rings are "sturdy" enough in sense I defined above. | |
| Aug 30, 2022 at 1:37 | comment | added | Denis T | Well, yeah, that's why I'm writing it as a comment and not an answer, but I think my non-example can be promoted. By methods of Manuel Amann I think one can produce "sturdy" PD rational types, i. e. ones without self-maps of degree $\neq 1$ and retaining that property under connected sums. Then one can use some secondary obstructions (i. e. homology classes not realizable as manifolds) to distinguish two realisations of that type. Sturdiness will prevent positive degree things from killing our obstructions, which will forbid any nonzero degree maps in either direction. | |
| Aug 30, 2022 at 0:30 | comment | added | algori | .. also, the tangent Stiefel-Whitney classes are preserved under genuine homotopy equivalences of manifolds, but not in general under rational ones. | |
| Aug 30, 2022 at 0:19 | comment | added | algori | Denis: this is a good example, but rational equivalences may have degree $\neq 1$. For example, any map $S^n\to S^n$ of degree $\neq 0$ is a rational equivalence. | |
| Aug 30, 2022 at 0:05 | comment | added | Denis T | I think there's no way to connect $S^2 \times S^3$ with the unique other 5-mfd with $H_2 = \Bbb Z$ (double of nontrivial $D^3$-bundle over $S^2$) through degree 1 maps because one has trivial $w_2$ and other one does not. They are rationally equivalent, since they are both formal. | |
| Aug 29, 2022 at 20:31 | history | asked | algori | CC BY-SA 4.0 |