Timeline for What do cohomology operations have to do with the non-existence of commutative cochains over $\mathbb{Z}$?
Current License: CC BY-SA 2.5
16 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2016 at 16:56 | answer | added | HenrikRüping | timeline score: 3 | |
| Sep 21, 2012 at 12:26 | answer | added | Justin Young | timeline score: 5 | |
| Dec 17, 2010 at 3:01 | vote | accept | Akhil Mathew | ||
| Dec 17, 2010 at 2:35 | comment | added | Akhil Mathew | Ah. OK, this makes sense. | |
| Dec 17, 2010 at 1:56 | answer | added | Jacob Lurie | timeline score: 27 | |
| Dec 17, 2010 at 1:40 | comment | added | Tom Goodwillie | Never mind wikipedia. Look, you can consider natural cohomology operations in the broader sense of natural maps of (based) sets, and this is the (or anyway a) traditional use of the term. And it's a simple fact that a stable cohomology operation is automatically a homomorphism. | |
| Dec 17, 2010 at 1:38 | history | edited | Tom Goodwillie | CC BY-SA 2.5 |
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| Dec 17, 2010 at 0:53 | comment | added | Sean Tilson | according to wikipedia it is a natural transformation of functors into abelian groups. At every object the "value" of the natural transformation is a map of abelian groups. So according to wikipedia... | |
| Dec 17, 2010 at 0:42 | comment | added | Akhil Mathew | Not according to Wikipedia. | |
| Dec 17, 2010 at 0:32 | comment | added | Sean Tilson | Shouldn't they be homomorphisms of abelian groups? always? no matter what? | |
| Dec 17, 2010 at 0:28 | history | edited | Akhil Mathew | CC BY-SA 2.5 |
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| Dec 17, 2010 at 0:27 | comment | added | Akhil Mathew | @Sean: thanks, fixed. I don't believe cohomology operations are required to be linear in general. | |
| Dec 17, 2010 at 0:23 | comment | added | Sean Tilson | This is a cool question by the way. Also, your $C_1(X)$ is a (graded) commutative DGA right? | |
| Dec 17, 2010 at 0:19 | comment | added | Sean Tilson | The power maps are not homomorphisms of abelian groups rationally. The $p$th power map is linear mod $p$ only. | |
| Dec 17, 2010 at 0:17 | answer | added | algori | timeline score: 15 | |
| Dec 16, 2010 at 23:24 | history | asked | Akhil Mathew | CC BY-SA 2.5 |