Timeline for If $A\in \mbox{Rings}\subset E_\infty\mbox{-rings}$, what is the equivalence between objects of $\mathcal{D}(\mbox{Mod}_A)$ and $A$-module spectra?
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2012 at 2:52 | vote | accept | Aaron Mazel-Gee | ||
| Jan 31, 2012 at 2:24 | comment | added | Aaron Mazel-Gee | Ah, I must have been forgetting that he passes from $E_\infty$-ring spaces to $E_\infty$-rings, since he makes the jump without actually doing anything. Clearly a single set couldn't possibly be a spectrum! | |
| Jan 30, 2012 at 19:48 | answer | added | Justin Noel | timeline score: 8 | |
| Jan 30, 2012 at 14:11 | comment | added | David White | I think the early comments of Fernando and Dylan should answer your first paragraph. I note that you're still concerned about why we consider $A$ as an $E_\infty$-ring. At first glance this seems to lose information, but it retains enough for our purposes. Here's a moral reason why. Because the derived category is a homotopy category (i.e. you get there after inverting quasi-isomorphisms), that category doesn't know the difference between a strict commutative ring object and an $E_\infty$ ring object. The theory is better developed for $E_\infty$, so that's all we need to know about $A$. | |
| Jan 30, 2012 at 4:48 | comment | added | Moosbrugger | It might be clarifying to note that the $0$-space of $HA$ is $A$ considered as a discrete space, and that there's not much choice other than Eilenberg-Maclane for how to turn $A$ into a spectrum. | |
| Jan 30, 2012 at 0:26 | comment | added | Dylan Wilson | @Aaron: Sean is correct. The way one "considers" $A$ as an $E_\infty$-ring is precisely by declaring it to be $HA$. | |
| Jan 30, 2012 at 0:23 | comment | added | Fernando Muro | I dare say that the only way of regarding a ring as a ring spectrum is via its Eilenberg-MacLane ring spectrum | |
| Jan 29, 2012 at 23:22 | comment | added | Aaron Mazel-Gee | @Sean: I don't think so. $HA$ may be an $E_\infty$-ring, but he certainly means to consider $A$ itself as an $E_\infty$-ring. This is outlined in the preceding page or so. | |
| Jan 29, 2012 at 23:00 | comment | added | Sean Tilson | So what he means by module spectra over a discrete ring $A$ are module spectra over $HA$. There are interesting maps between these as spectra, but not as $HA$ modules. | |
| Jan 29, 2012 at 20:50 | comment | added | Dylan Wilson | And theorem 7.1.2.13 for something more explicit... Okay I'll stop! | |
| Jan 29, 2012 at 20:49 | comment | added | Aaron Mazel-Gee | @Dylan: My previous comment was in response to your first. In response to your second comment: titcr (urbandictionary.com/define.php?term=titcr) | |
| Jan 29, 2012 at 20:48 | comment | added | Fernando Muro | The answer to your question is rather technical, but you can find all details in the following nice paper by Shipley: math.uic.edu/~bshipley/zdga17.pdf | |
| Jan 29, 2012 at 20:47 | comment | added | Aaron Mazel-Gee | @Dylan: That's possible, but since the homology of the (cellular/singular/whatever) chain complex only gives you singular homology, that might not be sufficiently faithful. | |
| Jan 29, 2012 at 20:47 | comment | added | Dylan Wilson | Alternatively you can use the universal property of the derived category (for $\infty$-categories). Also you can look at the proof of this claim in Lurie's Higher Algebra p.681 (Prop. 7.1.1.15) | |
| Jan 29, 2012 at 20:41 | comment | added | Dylan Wilson | Maybe just the cellular chain complex will give you a map one way? Also remember that we're dealing with $\infty$-categories here, so you'd have to then take homotopy categories to get the classical derived category. | |
| Jan 29, 2012 at 20:41 | history | edited | Aaron Mazel-Gee | CC BY-SA 3.0 |
nobody ever asserted an equivalence of categories!; edited title
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| Jan 29, 2012 at 20:34 | history | asked | Aaron Mazel-Gee | CC BY-SA 3.0 |