Timeline for Intuition - difference between Moore spectrum and Eilenberg-Mac Lane spectrum
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2019 at 20:11 | comment | added | Denis Nardin | @Sasha No. Spectra in general are not $H\mathbb{Z}$-modules, they are only $\mathbb{S}$-modules. In fact the (∞-)category of $H\mathbb{Z}$-modules is equivalent to $D(\mathbb{Z})$. Moreover the Moore spectrum construction is not even functorial! If you want to continue discussing, please come to the homotopy theory chatroom so the system doesn't yell at us :) | |
| Feb 2, 2019 at 20:08 | comment | added | Sasha | Is it correct that every spectrum is an $H\bbZ$-module (where morally $n \in \pi_0 (H\bbZ)$ acts by multiplication by $n$) and the Moore spectrum construction is the left adjoint of this $Spectra \to D(\bbZ)$ functor (where $D(\bbZ)$ is the derived category of $\bbZ$-modules, which is the same, I believe, as spectrum modules over $H\bbZ$)? | |
| Feb 2, 2019 at 19:56 | comment | added | Denis Nardin | @Sasha In case it wasn't clear I was joking. But when doing homotopy theory using cohomological convention means that the homotopy group of a space are all in nonpositive degree, and that is rather annoying :) | |
| Feb 2, 2019 at 19:52 | comment | added | Sasha | I don't quite have a say, since this is so common in homological algebra (representaiton theory etc.) that I can't go against it. But perhaps it makes sense to put $\pi_1$ in negative degree, since it represents "smaller" stuff? (like the cardinality of $BG$ is $|G|^{-1}$ etc.) Anyway, probably one can find justification for each. | |
| Feb 2, 2019 at 19:49 | comment | added | Denis Nardin | @Sasha Well, as I'm fond of saying, the cohomological convention is the tool of the Devil. Don't fall for it! ;) | |
| Feb 2, 2019 at 19:46 | comment | added | Sasha | Oops, I am used to cohomological convention, so I simply calculated wrongly the long exact sequence of homotopy groups for the cofiber sequence. Thank you! | |
| Feb 2, 2019 at 19:41 | comment | added | Denis Nardin | @Sasha No, $π_0\mathbb{S}/2=H_0\mathbb{S}/2=\mathbb{Z}/2$. Maybe you're confounding it with $[\mathbb{S}/2,\mathbb{S}/2]=\mathbb{Z}/4$? (Yes, it is a bit counterintuitive but this nontrivial extension is secretly coming from $\pi_1\mathbb{S}$) | |
| Feb 2, 2019 at 19:40 | comment | added | Sasha | Thank you! This now makes perfect sense. But for p=2, it seems that the cofiber of multiplication by 2 on the sphere spectrum has $\pi_0$ with four elements, so this should differ from the Moore spectrum as defined in my question, no? | |
| Feb 2, 2019 at 19:39 | vote | accept | Sasha | ||
| Feb 2, 2019 at 11:07 | history | answered | Denis Nardin | CC BY-SA 4.0 |