Timeline for Finite complexes which are not Thom spectra
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2019 at 5:11 | vote | accept | skd | ||
| Jan 14, 2019 at 3:53 | comment | added | skd | Ah, I'm sorry I got it the wrong way around. | |
| Jan 14, 2019 at 3:10 | comment | added | Dylan Wilson | I feel like the opposite is true: nowadays we like to map to pic(S) instead of BGL_1(S) | |
| Jan 13, 2019 at 23:19 | comment | added | skd | @TimCampion I think it depends on the context. In more recent papers, something seems to be called "a Thom spectrum" if it is a Thom spectrum of a rank zero stable spherical fibration, but in older papers, I would imagine that "Thom spectrum" could mean the more general notion of Thom spectrum (i.e., Z x BGL_1) that you had in mind. | |
| Jan 13, 2019 at 21:05 | comment | added | Tim Campion | Ah, I just reread, and Sanath also says he wants the bottom cell in dimension zero. Still, I'd like to know "the official word" because I don't really know what people mean by default when they say that something "is a Thom spectrum". | |
| Jan 13, 2019 at 20:58 | comment | added | Denis Nardin | @TimCampion I assume they want a Thom spectrum of a rank 0 stable spherical fibration (since they mention of $BGL_1(\mathbb{S})$ and not $BGL_1(\mathbb{S})\times \mathbb{Z}$), so you're not allowed to desuspend. | |
| Jan 13, 2019 at 20:54 | comment | added | Tim Campion | I'm probably being stupid (I don't know anything about this stuff) but if $X$ is a finite complex, then $\Sigma^n X$ is is a suspension spectrum for some $n$, and in particular the Thom spectrum of a trivial bundle. Then can't you just desuspend the bundle $n$ times to exhibit $X$ as a Thom spectrum? | |
| Jan 13, 2019 at 17:12 | answer | added | John Rognes | timeline score: 7 | |
| Jan 13, 2019 at 2:08 | comment | added | Dylan Wilson | Anyway, for the actual question: can you turn Mahowald's proof that $bo$ isn't a Thom spectrum into a proof that some skeleton of it isn't? | |
| Jan 13, 2019 at 2:05 | comment | added | Dylan Wilson | @Prasit good point :) and then I suppose that $0$ is a Thom spectrum over the empty set, so that works (or isn't allowed anyway since we restrict to complexes with a bottom $0$-cell). | |
| Jan 13, 2019 at 2:00 | comment | added | skd | Thanks for pointing that out. I guess I should really be working at some fixed prime p. (I was thinking about p=2 but the evenness of the prime shouldn't matter.) | |
| Jan 13, 2019 at 1:32 | comment | added | Prasit | @DylanWilson I do not quite get your question. what do you mean when you say mod p+1 Moore complex localized at prime p. Isn't it trivial anyway? (Since p+1 is a unit in p-adics). | |
| Jan 12, 2019 at 21:43 | comment | added | Dylan Wilson | actually even that doesn't solve the problem... yeah- if you localize at p, then the mod p+1 Moore complex seems problematic, right? | |
| Jan 12, 2019 at 20:14 | comment | added | Dylan Wilson | Presumably you've implicitly localized at a prime, otherwise you can't get the mod p Moore complex the way you describe when p>2 since 1-p is not a unit in $\pi_0S^0$. | |
| Jan 12, 2019 at 20:09 | comment | added | Prasit | Yes, indeed, I agree with your argument. Mahowald in the paper mentioned above, realizes HZ as a Thom spectrum over the Whitehead cover of $\Omega^2S^3$. Not only that he realizes integral Brown-Gitler spectra as Thom spectra over certain subcomplex of $\Omega^2S^3$ obtained from certain `operadic May filtration', viewing $\Omega^2S^3$ as universal $E_2$ space obtained from $S^1$. The upside down question mark is the first integral Brown-Gitler spectrum. I think this discussion is in section 2, see prop 2.10. But it is essentially the argument you gave above! | |
| Jan 12, 2019 at 19:39 | comment | added | skd | @Prasit I agree that my argument isn't very convincing, and is missing details, but I thought I'd add it in the body of the question anyway in case someone could provide a complete argument. Which section of Mahowald's paper is it in? Here's my idea for proving the statement about the dual question mark complex $DQ$. Identify $SO(3)$ with $\mathbf{R}P^3$, so there is a map $\mathbf{R}P^2 \to SO(3) \to O = \Omega BO$, and hence a map $\Sigma \mathbf{R}P^2 \to BO$. This map detects the nontrivial element of $\pi_2(BO)$ on the bottom cell, so it should Thomify to $DQ$. | |
| Jan 12, 2019 at 19:21 | comment | added | Prasit | The question mark complex might be an example, but I am not quite convinced with the argument. Just a fun-fact worth mentioning here: the upside down question mark complex i.e. Spanier-Whitehead dual of the question mark complex (up to a shift) is a Thom spectrum. This can be found in a paper of Mahowald ``Ring spectra which are Thom complexes." Also I think generalized Moore spectra, such as M(1,4), may not be a Thom spectrum. But cannot think of a quick argument! | |
| Jan 12, 2019 at 17:39 | history | edited | skd | CC BY-SA 4.0 |
added 724 characters in body
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| Jan 12, 2019 at 4:04 | history | asked | skd | CC BY-SA 4.0 |