Timeline for A question about cohomology with local coefficient
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 6, 2023 at 19:22 | comment | added | Mark Grant | You'd just need to check if the 0th Alexander-Spanier cohomology is functions on the connected components. | |
| Dec 6, 2023 at 13:54 | comment | added | Mehmet Onat | @MarkGrant No, $H^0(F;\mathbb{Q})=\mathbb{Q}$, but $F$ is not path-connected, just connected, then is it trivial? | |
| Dec 6, 2023 at 13:39 | comment | added | Mark Grant | I see what you're asking. If $H^0(F;\mathbb{Q})\cong\mathbb{Q}$ then $F$ is path-connected, and since the action of $\pi_1(B)$ on $H^0(F;\mathbb{Q})$ is induced by a permutation of path components, this action is trivial. | |
| Dec 6, 2023 at 12:18 | comment | added | Mehmet Onat | @MarkGrant if the space $F$ is not nice, but $H^0(F;\mathbb {Q})=\mathbb{Q}$, then still, is the system of local coefficients simple?, that is the homomorphism $\pi_1(F) \rightarrow Aut(H^0(F;\mathbb{Q}))$ is trivial? | |
| Dec 5, 2023 at 12:36 | comment | added | Mark Grant | You are right. Most treatments of spectral sequences and local coefficients I've seen use singular or cellular cohomology, and/or restrict to nice spaces (so that path connected and connected coincide). | |
| Dec 5, 2023 at 8:49 | comment | added | Mehmet Onat | @MarkGrant The local coefficients system is given by $\pi_1 (B) \rightarrow Aut (H^0(F))$. So It seems to me that path connectedness of $F$ is related to the cohomology used. Am I wrong? | |
| Dec 1, 2023 at 10:11 | comment | added | Mark Grant | It is induced by the standard action map $\Omega B\times F\to F$ which comes from the homotopy lifting property, as described in many textbooks e.g. Hatcher's Algebraic Topology, Chapter 4. | |
| Nov 30, 2023 at 19:13 | comment | added | Mehmet Onat | @MarkGrant Can you write more clearly this action? | |
| Nov 28, 2023 at 13:40 | comment | added | Mark Grant | I don't know about Alexander-Spanier cohomology, but presumably replace $\pi_0(F)$ with the set of connected components in the above comment. | |
| Nov 28, 2023 at 13:40 | comment | added | Mark Grant | You can think of $H^0(F;R)$ as functions $\pi_0(F)\to R$, and $\pi_1(B)$ acts by pre-composing with the action of $\pi_1(B)$ on $\pi_0(F)$. If $F$ is path-connected this action is clearly trivial. | |
| Nov 28, 2023 at 12:00 | history | edited | Mehmet Onat | CC BY-SA 4.0 |
added 105 characters in body
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| Nov 28, 2023 at 11:43 | history | asked | Mehmet Onat | CC BY-SA 4.0 |