Timeline for Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 28, 2011 at 5:15 | answer | added | Steve D | timeline score: 6 | |
| Jun 27, 2011 at 13:39 | answer | added | Dave Futer | timeline score: 13 | |
| Jun 27, 2011 at 5:43 | vote | accept | Benjamin Horowitz | ||
| Jun 27, 2011 at 0:48 | comment | added | Benjamin Horowitz | @André Henriques, Thanks, I mean non-trivial finite. | |
| Jun 27, 2011 at 0:43 | history | edited | Benjamin Horowitz | CC BY-SA 3.0 |
added 13 characters in body; added 1 characters in body
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| Jun 27, 2011 at 0:36 | history | edited | Benjamin Horowitz | CC BY-SA 3.0 |
ah, yeah, non-trivial finite
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| Jun 26, 2011 at 23:48 | comment | added | André Henriques | I'm guessing he doesn't, and is simply making a mistake: the fundamental group of $(S^2 \times S^1)_1$ # $\cdots$ # $(S^2 \times S^1)_k$ is a free group on $k$ generators (van Kampen's theorem). | |
| Jun 26, 2011 at 23:45 | comment | added | Marco Golla | When you say $\pi_1(M) = \mathbb{Z}^k$, do you mean the free product of $k$ copies of $\mathbb{Z}$? | |
| Jun 26, 2011 at 23:43 | comment | added | André Henriques | "non-finite trivial"?? Did you mean "non-trivial finite"? | |
| Jun 26, 2011 at 23:39 | answer | added | Igor Rivin | timeline score: 10 | |
| Jun 26, 2011 at 23:34 | history | asked | Benjamin Horowitz | CC BY-SA 3.0 |