Timeline for Detecting the Brown-Comenetz dualizing spectrum
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 4, 2018 at 17:12 | vote | accept | skd | ||
| Jun 4, 2018 at 8:43 | answer | added | Drew Heard | timeline score: 12 | |
| Jun 4, 2018 at 6:13 | comment | added | skd | In fact, since $\bigvee_n \langle K(n) \rangle \geq \langle I \rangle$ (all finite spectra are harmonic), this shows that $\langle E \rangle\geq \langle I \rangle$ on finite spectra. I think that this is true on all spectra, and in fact, I believe that the question is equivalent to the following: if $E$ is a spectrum with no finite acyclics, then $\langle E\rangle \geq \langle I\rangle$. (Note that $\langle E\rangle \geq \langle I\rangle$ implies that $E$ has a finite local, namely $S/p$.) I don't know how to prove this equivalence, though. | |
| Jun 4, 2018 at 5:59 | comment | added | skd | Note that if such an $E$ with a finite acyclic existed, it would be a counterexample. Indeed, $\langle E \rangle \geq \langle X \rangle \geq \langle T(n) \rangle$ for every $n\geq N$, where $X$ has type $N$. As $\langle T(n) \rangle \geq \langle K(n) \rangle$, this implies that $\langle E \rangle \geq \langle K(n) \rangle$ for every $n\geq N$. If $E$ had a finite acyclic $F$, then $K(n)_\ast F=0$ for every $n$, i.e., $F$ would be contractible (as finite spectra are harmonic). | |
| Jun 4, 2018 at 5:59 | comment | added | skd | Here is a failed attempt at answering this question. Suppose $E\wedge I \neq 0$. Then the same argument as in Strickland's answer, combined with the self-duality of the generalized Moore spectra, shows that $[E, M(i_0, \cdots, i_n)] \neq 0$. If $E$ had a finite acyclic $F$, then $[E,F] = 0$. If $F$ has type $N$, then every type $N$ spectrum (in particular, $M(i_0, \cdots, i_N)$) can be obtained from $F$ via a finite number of retracts and cofiber sequences, so we would run into a contradiction. Therefore $E$ cannot have a finite acyclic. | |
| Jun 3, 2018 at 22:39 | answer | added | Neil Strickland | timeline score: 10 | |
| Jun 3, 2018 at 16:03 | history | edited | skd | CC BY-SA 4.0 |
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| Jun 3, 2018 at 16:02 | comment | added | skd | You're right, of course. I'll fix this. | |
| Jun 3, 2018 at 8:29 | comment | added | Jonathan Beardsley | A silly comment, but X(1) is the sphere spectrum, so you probably don't want to include that in your list of examples of spectra that don't detect. | |
| Jun 3, 2018 at 4:06 | history | asked | skd | CC BY-SA 4.0 |