Timeline for Homology of the universal cover
Current License: CC BY-SA 4.0
9 events
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| Jun 16, 2019 at 19:56 | history | edited | John Klein | CC BY-SA 4.0 |
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| Jun 14, 2019 at 17:49 | comment | added | John Klein | @DannyRuberman I misphrased my previous comment. The statement about primes was rubbish. I meant to write: 1) The universal cover, which is 1-connected, is finite dimensional. 2) The total homology with Z coefficients is not finitely generated. 3) The total homology with any field coefficients is finitely generated. so we could get an example wit, say, a copy of $\Bbb Q$ in its homology and 2 and 3 will hold. | |
| Jun 13, 2019 at 1:30 | history | edited | John Klein | CC BY-SA 4.0 |
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| Jun 12, 2019 at 2:06 | comment | added | John Klein | @DannyRuberman Yes, I had forgotten that. So these examples are not finitely generated over $\Bbb Z$ in some degree but are so over any field, implying that there is torsion at in infinite number of primes in that degree. | |
| Jun 12, 2019 at 0:59 | comment | added | Danny Ruberman | Milnor (Infinite Cyclic Coverings, Assertion 5) shows that with field coefficients, the homology of the infinite cyclic covering of any knot complement is finitely generated. The proof is a nice application of the Milnor exact sequence in homology derived from the SES $0 \to C_{*}(\tilde{X}) \overset{(t_*-1)}{\to} C_{*}(\tilde{X}) \overset{p_*}{\to} C_{*}(X) \to 0$. | |
| Jun 11, 2019 at 23:51 | comment | added | John Klein | For any knot the map $X\to S^1$ is a homology isomorphism by Alexander duality. | |
| Jun 11, 2019 at 22:36 | comment | added | GSM | In this case $X\rightarrow S^1$ is a homology isomorphism ? | |
| Jun 11, 2019 at 21:57 | history | edited | John Klein | CC BY-SA 4.0 |
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| Jun 11, 2019 at 21:50 | history | answered | John Klein | CC BY-SA 4.0 |