Timeline for Generalized de Rham cohomology on product bundle giving specified cohomology
Current License: CC BY-SA 4.0
8 events
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Sep 9, 2019 at 14:13 | comment | added | user126256 | I think I understand my issue now; on page 147 of the book, Fuks says that the first page differential $d_1^{p,q}$ is induced by the de Rham differential. However, if you track the calculations back a bit, this differential would have to come from a de Rham differential on the space $\Gamma(\Lambda^\bullet T^* M \otimes E)$ for some other vector bundle $E$. He does not mention (but other papers by Gelfand, Fuks and Losik regarding this matter do), that this bundle $Q$ is flat, and even trivializable. Thus my question becomes redundant for my purposes. Thank you regardless! | |
Sep 6, 2019 at 11:44 | comment | added | mme | I don't immediately see where Fuks' book uses what you are looking for. Can you point to the precise step you are thinking of? | |
Sep 3, 2019 at 14:14 | history | edited | user126256 | CC BY-SA 4.0 |
Added requirement on differential
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Sep 3, 2019 at 14:11 | comment | added | user126256 | Yeah, that is a good point. I think a reasonable requirement in my context should be that the differential is a differential operator of order 1. I will add this to the post. Thank you! | |
Sep 3, 2019 at 14:06 | comment | added | mme | You should put some conditions on these differentials, or it must be completely useless to you. Because all of the individual terms $C^k$ are infinite dimensional, you can cook up a differential with essentially arbitrary homology, so long as it is concentrated in degrees $0$ through $\dim M$. | |
S Sep 3, 2019 at 14:01 | history | suggested | Ali Taghavi |
I add the tag "Cohomology" since as it is indicated in the question it is some thing different from the standard de Rham cohomology.
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Sep 3, 2019 at 14:00 | review | Suggested edits | |||
S Sep 3, 2019 at 14:01 | |||||
Sep 3, 2019 at 13:50 | history | asked | user126256 | CC BY-SA 4.0 |