Timeline for Why does strong convergence of the EMSS imply that Tot commutes with suspension spectrum?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2015 at 14:46 | comment | added | Jonathan Beardsley | @CraigWesterland Oh SNAP! Thanks Craig! :-D | |
| Oct 6, 2015 at 14:28 | comment | added | Craig Westerland | A little further down the page, Tilman notes that "strong convergence" in the sense of Shipley or Bousfield is what he calls "pro-constant convergence." In Lemma 2.6, he shows this implies complete convergence, which kills the $lim^1$ term. | |
| Oct 6, 2015 at 8:18 | answer | added | Rosona | timeline score: 11 | |
| Oct 6, 2015 at 4:41 | comment | added | Jonathan Beardsley | @CraigWesterland it looks for that you still have to check the lim^1 term. Is it obvious that that vanishes? | |
| Oct 6, 2015 at 4:26 | comment | added | Craig Westerland | I don't think that the result holds for any homology theory $h$, only singular homology $H$. If you only ask this for a given $h$, I think that the result you want holds after $h$-localization. For an argument why this is true, take a look at Tilman Bauer's "On the Eilenberg-Moore spectral sequence for generalized cohomology theories," in particular Remark 2.3. This gives the result you want directly for $K=S$; I think it gives what you want for $K = H$ using the fact that an isomorphism in homology is a stable equivalence. | |
| Oct 6, 2015 at 3:12 | history | edited | Jonathan Beardsley | CC BY-SA 3.0 |
added 296 characters in body
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| Oct 6, 2015 at 2:53 | comment | added | Jonathan Beardsley | From looking at a few other references, it seems that this has something to do with the vanishing line you get in the EMSS when you have strong convergence... | |
| Oct 6, 2015 at 1:09 | history | asked | Jonathan Beardsley | CC BY-SA 3.0 |