Timeline for Betti Cohomology of singular Kummer Surface
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2010 at 17:45 | vote | accept | AJ Stewart | ||
| Jun 19, 2010 at 5:36 | answer | added | Torsten Ekedahl | timeline score: 5 | |
| Jun 18, 2010 at 17:52 | comment | added | Tom Goodwillie | There is $2$-torsion; the third mod $2$ homology group is nontrivial. Quick proof: Topologically this is what you get from the product of four circle groups by identifying each element with its inverse. As such, it contains as a retract the analogous quotient space of a product of three circle groups. The latter has nontrivial third mod $2$ homology, because it is a $3$-manifold except for singularities of codimension $3>1$. | |
| Jun 18, 2010 at 16:50 | comment | added | AJ Stewart | @Jose: It is true that over Q coefficients that the cohomology of the quotient is equal to the invariants of the cohomology of the original space. However, I believe that this fact not true a priori over Z (unless I'm missing something), but ideally yes, it would just be the invariant part of the cohomology. And thank you for editing the title. | |
| Jun 18, 2010 at 16:39 | comment | added | José Figueroa-O'Farrill | Torsten, do BPvdV treat the singular quotient? | |
| Jun 18, 2010 at 16:38 | comment | added | José Figueroa-O'Farrill | I edited the title to better reflect the question, by the way. | |
| Jun 18, 2010 at 16:38 | history | edited | José Figueroa-O'Farrill | CC BY-SA 2.5 |
Edited the title to better reflect the question.
|
| Jun 18, 2010 at 16:22 | comment | added | José Figueroa-O'Farrill | In that case, why is not just the invariant part of the cohomology? (Perhaps I'm missing something obvious, though.) | |
| Jun 18, 2010 at 15:12 | comment | added | AJ Stewart | I mean the singular quotient. I suppose I really should have said singular Kummer surface in the title. | |
| Jun 18, 2010 at 14:13 | comment | added | José Figueroa-O'Farrill | By the Kummer variety do you mean the singular quotient or its resolution once the 16 singular points are blown up? | |
| Jun 18, 2010 at 2:35 | history | asked | AJ Stewart | CC BY-SA 2.5 |