Timeline for How many automorphisms are there of the category of filtered spectra?
Current License: CC BY-SA 4.0
10 events
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| Nov 1, 2023 at 16:57 | comment | added | Tyler Lawson | @MoriB. The bounded-belowness is used to try and determine the indicator $g(0,\infty)$ as somehow preferred among indicators; but I believe that you can still get the result for filtered spectra. (I just want to hedge my bets because I believe that I'd need to think carefully about it.) Any compact object certainly has some bound below and so most of the same arguments immediately apply. | |
| Nov 1, 2023 at 16:54 | history | edited | Tyler Lawson | CC BY-SA 4.0 |
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| Nov 1, 2023 at 13:33 | comment | added | Mori B. | thanks @Tyler, I missed that you meant non negatively filtered, tho I probably should have gathered that’s what you meant when you said “equivalent to simplicial spectra”. Is the bounded-belowness used somewhere in this argument? Do you expect unbounded filtrations to have more automorphisms? | |
| Nov 1, 2023 at 13:18 | comment | added | Tyler Lawson | @MoriB. I usually use it to denote the upward-pointing poset $(0 \to 1 \to 2 \to \dots)$. In this case, the equivalence of categories is between simplicial spectra and nonnegatively filtered spectra, rather than unbounded ones, which is why $\Bbb N$ appears rather than $\Bbb Z$. | |
| Nov 1, 2023 at 5:07 | comment | added | Mori B. | Just to clarify, in the last paragraph, \mathbb N denotes a downward pointing poset? I might have expected Z in place of N as these are unbounded filtrations, but perhaps I’m misundestanding this equivalence. | |
| Oct 31, 2023 at 23:44 | history | edited | Tyler Lawson | CC BY-SA 4.0 |
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| Oct 31, 2023 at 23:32 | vote | accept | Tim Campion | ||
| Oct 31, 2023 at 23:29 | history | edited | Tyler Lawson | CC BY-SA 4.0 |
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| Oct 31, 2023 at 22:06 | history | edited | Tyler Lawson | CC BY-SA 4.0 |
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| Oct 31, 2023 at 21:44 | history | answered | Tyler Lawson | CC BY-SA 4.0 |