Timeline for Covers of a 4-manifold pull back a cohomology class to any algebraic multiple
Current License: CC BY-SA 4.0
11 events
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| Sep 12, 2020 at 14:04 | comment | added | Will Sawin | @MikeMiller It seems Moishe gave a simpler argument, but there is a Leray spectral sequence associated to any morphism, and not just a fibration, and it also works for orbifolds, and even for nastier spaces like algebraic stacks - why wouldn't it? | |
| Sep 12, 2020 at 14:01 | comment | added | Moishe Kohan | @WillSawin: I do not have a reference but this is indeed the case that $b_1(SFS)= b_1(B)$ or $b_1(SFS)=b_1(B)+1$ for every Seifert manifolds SFS woth the base $B$; It is easy to prove by looking at fundamental group presentations (which one can find pretty much in any source of Seifert manifolds). However, since the question changed, this is all mostly irrelevant. | |
| Sep 12, 2020 at 12:05 | comment | added | mme | @WillSawin Seifert manifolds are not fibrations, but rather singular fibrations. If you like, you can think of them as fiber bundles over an orbifold. But there is no Serre SS available here. | |
| Sep 12, 2020 at 11:35 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 9, 2020 at 16:56 | comment | added | Will Sawin | @MoisheKohan Unless I made a mistake, it's easy to calculate using the spectral sequence associated to the fibration by circles. | |
| Sep 9, 2020 at 16:35 | comment | added | Moishe Kohan | @WillSawin: Reference for $H^1$ of Seifert manifolds? | |
| Sep 9, 2020 at 15:00 | comment | added | Will Sawin | @MoisheKohan $H^1$ of a Seifert manifold is either $H^1$ of the underlying surface or $H^1$ of the surface + $H^1$ of the fiber. So maybe you only get units again for Seifert manifolds ( algebraic integers of degree at most 3 from the 3-torus). Do you know a reference? | |
| Sep 8, 2020 at 17:59 | comment | added | Moishe Kohan | Very few closed 3-manifolds admit nontrivial self coverings: I think you only get Seifert manifolds (and not all of them). | |
| Sep 8, 2020 at 17:18 | history | edited | user164740 | CC BY-SA 4.0 |
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| Sep 8, 2020 at 17:17 | comment | added | Will Sawin | We need $x$ an algebraic integer (and not just an algebraic number). If $x$ is an algebraic unit (i.e. $x^{-1}$ is an algebraic integer) then $x$ is an eigenvalue of a matrix in $Sp_{2g}(\mathbb Z)$ and so $x$ is an eigenvalue of a map from a surface to itself (or a $3$-manifold by taking the product with $S^1$.) So the difficulty is for algebraic integers that are not units. | |
| Sep 8, 2020 at 17:05 | history | asked | user164740 | CC BY-SA 4.0 |