Timeline for Relation between cohomological dimensions of manifolds
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Dec 4, 2021 at 7:23 | comment | added | King Khan | Thank you, this is very useful for me. | |
| Dec 4, 2021 at 7:15 | vote | accept | King Khan | ||
| Dec 3, 2021 at 18:03 | comment | added | Ben Wieland | For the Moebius band, the cohomology with coefficients in the orientation sheaf is zero in every degree, whereas the cohomology with coefficients in the trivial sheaf is not zero in dimensions 0 and 1. Real line bundles are equivalent to $\mathbb Z$-line-bundles. Given a manifold with some rank 1 local systems, you can make any of them the orient sheaf of a h eq manifold by taking the total space of an appropriate line bundle on the original, just as the Moebius band is the total space of the nontrivial line bundle on the circle. So the coh in the orientation sheaf is not much restricted. | |
| Dec 2, 2021 at 16:09 | comment | added | King Khan | So, can we say that $Ch_{\mathbb{Z}}(M)\leq Ch_{\pm\mathbb{Z}}(M)$ for open non-orientable manifold $M$? | |
| Dec 2, 2021 at 14:32 | comment | added | Mark Grant | Usually the cohomological dimension of a connected space over $\mathbb{Z}$ would mean the largest dimension in which there is non-vanishing cohomology with local coefficients in some $\mathbb{Z}\pi_1$-module. Here you seem to mean the largest dimension in which the integral cohomology (constant coefficents) is non-vanishing. If that is the case, then everything follows from standard calculations of the homology of manifolds and the Universal Coefficient Theorem. | |
| Dec 2, 2021 at 13:30 | history | edited | King Khan | CC BY-SA 4.0 |
added 557 characters in body
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| Dec 2, 2021 at 9:05 | comment | added | Nicolast | If $M$ is closed and non-orientable, then $\mathrm{Ch}_{\mathbb Z}(M) < \mathrm{Ch}_{\pm \mathbb Z}(M) = \mathrm{dim}(M)$. I don't think it is true if $M$ is not closed however: the Moebius band retracts to a circle, whose virtual cohomological dimension is $1$ anyway... | |
| Dec 2, 2021 at 7:47 | answer | added | Greg Friedman | timeline score: 2 | |
| Dec 2, 2021 at 5:47 | history | edited | YCor | CC BY-SA 4.0 |
formatting, added tag
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| Dec 2, 2021 at 4:52 | history | asked | King Khan | CC BY-SA 4.0 |