Timeline for Degrees of self-maps of aspherical manifolds
Current License: CC BY-SA 3.0
9 events
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May 23, 2015 at 4:48 | history | edited | Ian Agol | CC BY-SA 3.0 |
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Oct 15, 2010 at 19:11 | history | edited | Ian Agol | CC BY-SA 2.5 |
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Oct 15, 2010 at 18:00 | comment | added | Ian Agol | I found a paper which for $n=4$ implies that the classifying map is degree one if and only if an invariant of the Postnikov tower $k_3(M)$ vanishes. ams.org/mathscinet-getitem?mr=1809909 The paper doesn't give any examples where $k_3(M)$ is non-vanishing. In any case, maybe surgery is not the right approach. | |
Oct 15, 2010 at 17:32 | history | edited | Ian Agol | CC BY-SA 2.5 |
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Oct 15, 2010 at 17:30 | comment | added | Ian Agol | @ Dmitri: since $\pi_1(M)$ is finitely presented, we only have to kill finitely many elements of $ker(\phi)$ before we get a presentation for $\phi(\pi_1(M))\cong \pi_1(M)$ (just enough to get one of each relation for $\pi_1(M)$). As for your second question, this points out a gap in my argument. The classifying map of a manifold to an aspherical manifold with the same fundamental group need not be degree one. So I don't know how to complete the argument. | |
Oct 15, 2010 at 9:13 | comment | added | Dmitri Panov | Great answer! Could you please expend a bit the last two lines? Why "$\phi'$ is homotopic to the classifying map" implies $deg(\phi')=\pm 1$? Also, why $Ker(\phi_{\#})$ is finitely normally generated? | |
Oct 15, 2010 at 6:21 | history | edited | Ian Agol | CC BY-SA 2.5 |
added a complete argument
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Oct 15, 2010 at 5:08 | history | edited | Ian Agol | CC BY-SA 2.5 |
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Oct 15, 2010 at 3:42 | history | answered | Ian Agol | CC BY-SA 2.5 |