Timeline for Are there non-obvious finite $E_\infty$ ring spectra?
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20 events
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| Dec 9, 2021 at 20:25 | comment | added | Tim Campion | Note that if $X$ is a finite of type $n \geq 1$, then $L_{T(n)} X$ is $H\mathbb Q$-acyclic. If $X$ is $E_\infty$, then $L_{T(n)} X$ is $E_\infty$. But an $E_\infty$ ring spectrum which is $H\mathbb Q$-acyclic must vanish. So if $X$ is a finite $E_\infty$ ring spectrum, then $X$ must have type $0$. | |
| Jun 13, 2019 at 22:04 | comment | added | Ben Wieland | Oh, yeah, $X_+$ is a nonunital monoid. We should stick with unital ones. But $X_{++}$ is a unital monoid, so its reduced suspension is the unreduced suspension of $X_+$, often called $S+X$. | |
| Jun 13, 2019 at 17:00 | comment | added | Tim Campion | @BenWieland Thanks. I'm still confused by the multiplication on $X_+$ -- what's the unit? It seems to me that the only option is the basepoint $\ast$, but then the product $\ast x y z$ seems like it has to be $x$, $y$, and $z$ simultaneously if this multiplication is commutative and associative -- so it doesn't look well-defined to me. | |
| Jun 13, 2019 at 16:47 | comment | added | Tim Campion | @PhilTosteson Interesting point... I think the relevant version of Mandell's theorem says that an $E_\infty$ $H\bar F_p$-algebra is of the form $D\Sigma^\infty_+ X \wedge H\bar F_p$ for $X$ finite and simply-connected if and only if the obvious conditions on homotopy are met and the power operation $P^0$ acts injectively. But even if we have such an equivalence $R \wedge H\bar F_p \to D\Sigma^\infty_+X \wedge H\bar F_p$, I think this map need not exist before smashing with $H\bar F_p$, so there need not be an equivalence of $p$-completions as far as I can see. Still, it's something! | |
| Jun 13, 2019 at 4:23 | comment | added | Ben Wieland | "Coconnective" is wrong, but if the question were $E_\infty$ $\mathbb Z$-algebras, construction 2 would be. It seems like there is something useful to be said about objects such that when they are smashed with Z become coconnective... A monoid for the smash product is a monoid with a point that absorbs all other points; $X_+$ is a smash monoid where the product of any two elements is the basepoint... Section spaces satisfy Mayer-Vietoris, which reduces the question of finiteness to finiteness over cells, which is the usual statement that the dual of finite is finite. | |
| Jun 13, 2019 at 3:49 | comment | added | Tim Campion | @BenWieland neat ideas! Note that the dual of a finite spectrum is again a finite spectrum, and thus it is bounded below and I think unbounded above unless it's zero -- so there's not really anything coconnective going on (although the homology gets flipped around if we have a manifold I suppose). In your first comment, I think you're saying that if $X$ is a finite spectrum, then the square-zero $E_\infty$ ring spectrum $\mathbb S \vee X$ is the suspension spectrum of a finite $E_\infty$ object in $(Top_\ast, \wedge)$ -- could you explain this? In your second comment, why is the result finite? | |
| Jun 13, 2019 at 1:59 | comment | added | Ben Wieland | We can twist together the connective and coconnective(?) examples. Consider a bundle of commutative groups over X. Then instead of maps from X into the sphere, consider sections from X into the sheaf of group rings. Commutative groups only allow discrete monodromy, but smash monoids (eg, with zero multiplication) allow more interesting twisting. | |
| Jun 13, 2019 at 1:55 | comment | added | Ben Wieland | We can generalize from the unpointed suspension spectrum of a commutative product group to the pointed suspension spectrum of a smash commutative monoid. That captures the square zero extensions of the sphere spectrum. What connective examples does it miss? | |
| Jun 12, 2019 at 14:43 | comment | added | Phil Tosteson | Does Mandell's theorem imply that after $p$ completion, all examples are of type #2? | |
| Jun 11, 2019 at 19:06 | comment | added | Dylan Wilson | (This is unrelated to the “splitting” question in your post though. ‘Split square zero’ is different than ‘unit splits off’; the unit will always split off of examples built this way). | |
| Jun 11, 2019 at 19:05 | comment | added | Dylan Wilson | Square zero extensions off the sphere are all trivial (though most trivial sq zero extensions are not covered by your two examples). I bet there are lots of nontrivial 2-step extensions though. | |
| Jun 11, 2019 at 18:00 | comment | added | Tim Campion | I suppose finite $E_\infty$ ring spectra are closed under smash product too. @DylanWilson The square-zero extensions are a great source of examples -- but to be careful on the splitting question, do you actually know an example of a finite square-zero extension of a finite $E_\infty$ ring spectrum which is not split? And thanks for clarifying the nilpotence question. | |
| Jun 11, 2019 at 12:14 | comment | added | Dylan Wilson | Good point! Also, regarding nilpotence: by Mathew-Naumann-Noel, nilpotence is detected in integral homology; if R is finite then every element in nonzero degree is automatically nilpotent in homology so the same is true in homotopy. So the only thing that could go wrong is stuff in degree zero- and indeed that can happen. For example, take X to be disconnected and use a component to define an idempotent. (Note your claim in Question 3 is not quite right because of this example- the unit won’t hit this idempotent). | |
| Jun 11, 2019 at 11:31 | comment | added | Denis Nardin | @DylanWilson Duh... sorry. But at least this includes also finite commutative groups :) | |
| Jun 11, 2019 at 10:50 | comment | added | Dylan Wilson | Also, the unit map does not split off if X is empty. | |
| Jun 11, 2019 at 10:21 | comment | added | Dylan Wilson | Another class of examples are square zero extensions by finite complexes. | |
| Jun 11, 2019 at 10:20 | comment | added | Dylan Wilson | G has to be abelian to get an E_infty thing (as opposed to an E_1 thing). | |
| Jun 11, 2019 at 6:41 | comment | added | Denis Nardin | Well, as an obvious generalization of your first example there's $R=\mathbb{S}[G]$ for $G$ a compact Lie group (or more generally, a stably dualizable topological group). | |
| Jun 11, 2019 at 0:17 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Jun 10, 2019 at 23:42 | history | asked | Tim Campion | CC BY-SA 4.0 |