Timeline for Tate Cohomology via stable categories
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2020 at 13:54 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title, added tags (the question was bumped anyway)
|
| Jul 22, 2011 at 8:56 | vote | accept | Hanno | ||
| Feb 8, 2010 at 15:17 | answer | added | Shizhuo Zhang | timeline score: 1 | |
| Jan 25, 2010 at 14:08 | answer | added | Frank | timeline score: 3 | |
| Jan 25, 2010 at 12:45 | answer | added | Mariano Suárez-Álvarez | timeline score: 3 | |
| Jan 25, 2010 at 11:18 | answer | added | Greg Stevenson | timeline score: 4 | |
| Jan 25, 2010 at 7:35 | comment | added | Hanno |
I'm sorry, but the graded commutativity is still unclear to me: In your post, you mention the graded commutativity of the "central ring" of a triangulated category. Given a morphism $\Omega^k{\mathbb Z}\stackrel{\varphi}{\to}{\mathbb Z}$, then one can construct a transformation $T_\varphi: \Omega^k\Rightarrow\text{id}$ by $T_\varphi(M): \Omega^k(M)\cong(\Omega^k{\mathbb Z})\otimes M\to{\mathbb Z}\otimes M\cong M$ Unfortunately, it's not clear to me that $T_\varphi$ is indeed in $Z^k(T)$, i.e. that $T_\varphi\Omega = (-1)^i\Omega T_\varphi$. Could you give me some details on that?
|
|
| Jan 25, 2010 at 7:30 | comment | added | Hanno | Yes, we get the long exact sequence by definition of the exact structure for those exact sequences of G-modules which split over ${\mathbb Z}$. Unfortunately, Brown for example uses non-${\mathbb Z}$-split short exact sequences in his proof, too. This is not clear to me without the use of resolutions. | |
| Jan 24, 2010 at 23:36 | comment | added | Greg Stevenson | Maybe I am misunderstanding you, but the long exact sequences should come for free in your definition as $[\mathbb{Z},-]$ sends triangles to long exact sequences and every short exact sequence gives a triangle in the stable category (with the appropriate exact Frobenius category structure on the module category). | |
| Jan 24, 2010 at 22:28 | answer | added | S. Carnahan♦ | timeline score: 7 | |
| Jan 24, 2010 at 22:15 | comment | added | Hanno | Thank you, Greg! I look for a proof using this more axiomatic approach just because I find it more elegant than the usual way using resolutions. Unfortunately, I cannot imitate the argument in Brown because (1) I'm not able to define Tate homology, and (2) I don't see how one gets long exact Tate cohomology sequences directly from the definition for any exact sequence of G-modules. For both, one really seems to need resolutions. | |
| Jan 24, 2010 at 20:17 | comment | added | Greg Stevenson | The cup product you have defined above is graded-commutative due to $\Omega$ itself being graded-commutative in some sense. One can see this from playing around with the rotation axiom for triangles (usually denoted [TR2]). You might also want to see mathoverflow.net/questions/5901/… for further details. | |
| Jan 24, 2010 at 15:33 | history | edited | Hanno | CC BY-SA 2.5 |
added 321 characters in body
|
| Jan 24, 2010 at 14:46 | history | edited | Hanno | CC BY-SA 2.5 |
added 459 characters in body; deleted 3 characters in body; added 261 characters in body
|
| Jan 24, 2010 at 14:36 | history | edited | Hanno | CC BY-SA 2.5 |
added 2296 characters in body
|
| Jan 24, 2010 at 4:48 | comment | added | Greg Stevenson | There are references for developing analogues of Tate cohomology in various triangulated settings but I am not sure if these would be helpful. I have not seen this written down anywhere for groups either. Is there any reason you expect that going via the stable category would give an abstract proof of Tate duality? Really one is still working with resolutions and it isn't immediately obvious to me that such a statement should follow from general nonsense. | |
| Jan 24, 2010 at 4:13 | comment | added | Emerton | I saw Beilinson develop Tate cohomology somewhat along these lines in a lecture at U of Chicago maybe 9 years ago, but I've not seen a written source. | |
| Jan 23, 2010 at 22:43 | comment | added | Hanno |
I mean the statement that the map $\widehat{H}^k(G;{\mathbb Z})\otimes_{\mathbb Z}\widehat{H}^{-k}(G;{\mathbb Z})\to\widehat{H}^0(G;{\mathbb Z})\cong{\mathbb Z}/|G|{\mathbb Z}$ is a duality between the finite $|G|$-torsion groups $\widehat{H}^k(G;{\mathbb Z})$ and $\widehat{H}^{-k}(G;{\mathbb Z})$ I thought I had seen it called "Tate Duality", but I'm not sure.
|
|
| Jan 23, 2010 at 22:30 | comment | added | Kevin Buzzard | Sorry to show my ignorance, but what do you mean by Tate Duality? I've heard this phrase used in the context of Galois cohomology of local and global fields but I don't think I've heard it in the context of Tate cohomology groups when $G$ is finite. | |
| Jan 23, 2010 at 22:15 | history | edited | Hanno | CC BY-SA 2.5 |
edited title
|
| Jan 23, 2010 at 22:07 | history | asked | Hanno | CC BY-SA 2.5 |