Timeline for Homotopy groups of filtered homotopy limits
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 15, 2021 at 18:29 | vote | accept | Mikhail Bondarko | ||
| Dec 24, 2013 at 19:20 | comment | added | Ricardo Andrade | Regarding examples with $\lim^2$ non-zero. It appears there is such an example (which is not very transparent, I am afraid) of an uncountable limit of abelian groups in theorem 6 of Martin Ziegler's Higher inverse limits which I referenced above. You can also look at chapter 6 (for example, proposition 6.1) of the Lecture Notes in Mathematics volume 254, Les foncteurs dérivés de lim et leurs applications en théorie des modules by Christian Jensen. I look forward to hearing your thoughts on this. Merry Christmas! | |
| Dec 24, 2013 at 18:13 | comment | added | Ricardo Andrade | Dear @David: No worries! Just to clarify my first objection, it was essentially the following. If you apply your homotopy equalizer construction to a limit along the ordinal sum $\omega + \omega$, none of the maps you consider connects the first copy of $\omega$ to the second one. So the homotopy equalizer you get is actually the product of the two homotopy limits along each copy of $\omega$. In the case of a countable limit, this objection can certainly be circumvented (by taking a cofinal copy of $\omega$), but I fail to see how to avoid this problem for uncountable limits. | |
| Dec 24, 2013 at 12:38 | comment | added | David White | Hi Ricardo. Thanks for your comments; you raise some interesting points. Sorry for the delayed response. Christmas and all that. First, I agree that one step in what I wrote seemed sketchy, and that was taking an uncountable product of maps. Still, I'm not sure if I buy your argument about limit ordinals or not; morally, it feels like a limit containing all the $f_\alpha$ will get the limit maps for free. More concerning to me is your comment about lim$^i$. Can you think of an explicit example of an uncountable sequence of maps with a nonzero lim$^2$ term? That would definitely convince me. | |
| Dec 21, 2013 at 17:50 | comment | added | Ricardo Andrade | To the best of my limited knowledge, all we can do in the general situation of the question is to try applying the fringed Bousfield-Kan spectral sequence for the homotopy groups of the homotopy limit. The $E_2$ term of that spectral sequence involves the higher derived functors of limit. Thus, the above cited work on the derived functors of limit might be helpful. | |
| Dec 21, 2013 at 17:50 | comment | added | Ricardo Andrade | For such results on higher derived functors of limit, see for example: Christian Jensen, Les foncteurs dérivés de lim et leurs applications en théorie des modules, volume 254, Springer Verlag, 1972. Some of the relevant results there are also summarized in the preprint Higher inverse limits by Martin Ziegler. There was some more work done on higher derived functors of uncountable limits, for example by Dana Latch in On derived functors of limit. | |
| Dec 21, 2013 at 17:47 | comment | added | Ricardo Andrade | I think what you state in the first paragraph will not work, and is in fact very specific to countable limits. For limits along uncountable ordinals, you will probably need to consider more maps than $f_\alpha:X_{\alpha+1}\to X_\alpha$, as there are elements without a predecessor. As such, you might not be able to express them as homotopy equalizers. In fact, that would be rather surprising, as the higher derived functors $\operatorname{lim}^i$, $i>1$, of the limit of abelian groups along an uncountable ordinal are in general non-zero. This is unlike the countable case, where they are zero. | |
| Dec 21, 2013 at 15:02 | history | answered | David White | CC BY-SA 3.0 |