Timeline for Closed orientable 4-manifold with $H^1(M;\Bbb Z_2)=\Bbb Z_2$ and non-zero cup product $H^1\times H^1\to H^2$
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 15, 2015 at 6:41 | vote | accept | Irina | ||
| Jun 15, 2015 at 5:22 | answer | added | abx | timeline score: 11 | |
| Jun 15, 2015 at 2:33 | review | Close votes | |||
| Jun 15, 2015 at 9:08 | |||||
| Jun 15, 2015 at 2:26 | comment | added | Irina | @BenWebster Sorry, I edited the question to clarify that I mean the cup-product $H^1(M;\Bbb Z_2)\times H^1(M;\Bbb Z_2)\to H^2(M;\Bbb Z_2)$. | |
| Jun 15, 2015 at 2:24 | history | edited | Irina | CC BY-SA 3.0 |
added 61 characters in body
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| Jun 15, 2015 at 2:16 | comment | added | Zev Chonoles | Original math.SE post: math.stackexchange.com/q/1325568/264 | |
| Jun 15, 2015 at 2:15 | comment | added | Ben Webster♦ | Any closed manifold has non-zero cup product by Poincare duality. So $S^1\times S^2$ satisfies your criteria. | |
| Jun 15, 2015 at 2:08 | history | asked | Irina | CC BY-SA 3.0 |