Timeline for homology and cohomology of a quotient manifold
Current License: CC BY-SA 2.5
14 events
when toggle format | what | by | license | comment | |
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Dec 23, 2010 at 18:10 | comment | added | Sergey Melikhov | "How to compute the (co)homology of orbit spaces (when the action is not free)?" mathoverflow.net/questions/1080 | |
S Dec 14, 2010 at 12:35 | vote | accept | HKSHLZW | ||
Dec 14, 2010 at 12:35 | vote | accept | HKSHLZW | ||
S Dec 14, 2010 at 12:35 | |||||
Dec 14, 2010 at 12:35 | vote | accept | HKSHLZW | ||
Dec 14, 2010 at 12:35 | |||||
Dec 10, 2010 at 1:00 | comment | added | user2464 | Let us try to use a sequence of cutting and you get that you want. | |
Dec 9, 2010 at 15:34 | comment | added | Dan Ramras | For a finite group acting on a reasonable space, the rational cohomology of the quotient is the fixed points in the rational cohomology of the total space (this works with any field coefficients). See my answer here: mathoverflow.net/questions/18898/… | |
Dec 9, 2010 at 14:17 | answer | added | Mark Grant | timeline score: 3 | |
Dec 9, 2010 at 12:43 | answer | added | Daniel Loughran | timeline score: 4 | |
Dec 9, 2010 at 12:30 | history | edited | HKSHLZW | CC BY-SA 2.5 |
added 123 characters in body
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Dec 9, 2010 at 10:33 | answer | added | Kostya | timeline score: 7 | |
Dec 9, 2010 at 10:33 | comment | added | HKSHLZW | Or, if essentially my assertion about surface is wrong in any way , so what i concern is just the question: can we analys the cohomology of $N$ from the one of the $M$? | |
Dec 9, 2010 at 10:31 | comment | added | HKSHLZW | sorry, i didn't know what the wiki is ,then i think it as a good choice . So , if adding a condition that if $N $ is oriented ,so the situation works as i concern works? | |
Dec 9, 2010 at 10:26 | comment | added | José Figueroa-O'Farrill | 1) why is this community wiki? Don't you expect a definite answer? 2) Are you sure about your claim concerning surfaces? $H_1(\mathbb{RP}^2) \neq H_1(S^2)$. | |
Dec 9, 2010 at 10:02 | history | asked | HKSHLZW | CC BY-SA 2.5 |