Timeline for Can the Picard-graded homotopy of a nonzero object be nilpotent?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2022 at 21:39 | comment | added | Maxime Ramzi | yes, I was expecting this to maybe be true with shifts of spheres - if there's a counterexample though... | |
| Dec 15, 2022 at 17:23 | comment | added | Tim Campion | @MaximeRamzi Hmm... if this form argument should work, I'd expect the argument to equally show "if every map $S \to X$ is nilpotent, then $X = 0$", since I'm assuming every object is perfect. But this would be too strong: for instance, if $P$ is invertible in $Sp_{K(n)}$ but not a shift of spheres, then since $K(n)_\ast$ is 1-dimensional, every map $S \to P$ must be zero on $K(n)_\ast$. So if $P$ is perfect, then every such map is nilpotent. I'm told the "upside-down question mark" is an example of such a perfect element of $Pic$. | |
| Dec 15, 2022 at 13:07 | comment | added | Maxime Ramzi | So if $M$ is nilpotent as in Tyler's answer, and dualizable, then $M=0$. So maybe one can try to argue as follows : the collection of $P$'s that satisfy your hypothesis is thick, therefore contains $M$, and therefore $M=0$. The fact that the $P$'s are closed under cofiber sequences seems to be the hard part (if it's even true...) | |
| Dec 15, 2022 at 5:39 | answer | added | Tyler Lawson | timeline score: 4 | |
| Dec 14, 2022 at 23:36 | comment | added | Tim Campion | @TylerLawson I'm asking for both of them to simultaneously consist of nilpotents, although I guess it might already be interesting to consider just one side at a time. | |
| Dec 14, 2022 at 22:36 | comment | added | Tyler Lawson | Are you asking for one of (maps in) or (maps out) to consist of nilpotents, or are you asking for both of them to simultaneously consist of nilpotents? | |
| Dec 14, 2022 at 20:44 | history | asked | Tim Campion | CC BY-SA 4.0 |