Timeline for What can I say about an $E_\infty$ ring spectrum with an odd invertible element?
Current License: CC BY-SA 4.0
8 events
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| Mar 11, 2022 at 21:29 | comment | added | kiran | @MaximeRamzi good point thanks, I guess my statement is just at the level of $H\mathbb{F}_2$-modules. | |
| Mar 11, 2022 at 11:03 | comment | added | Maxime Ramzi | @kiran : this seems to require that it receives a map from its $\pi_0$, and I'm not sure how that follows - am I missing some other argument ? | |
| Mar 11, 2022 at 9:14 | comment | added | kiran | Doesn't Jeremy's comment mean that $R$ is $H(A[t^\pm])$ for some (discrete) $\mathbb{F}_2$-algebra $A$? | |
| Mar 10, 2022 at 19:18 | comment | added | Theo Johnson-Freyd | @JeremyHahn Cool. Do you want to expand your comment into an answer? Definitely $R \simeq \Sigma R$ is not enough to imply $R \simeq 0$. For example, take the strictly-commutative algebra $\mathbb{F}_2[t^{\pm 1}]$ with $t$ of degree 1. | |
| Mar 10, 2022 at 18:00 | comment | added | Maxime Ramzi | @davik : if $x$ is invertible in odd degree, then $x^2$ is invertible, and also $2$-torsion ($x^2 = (-1)^{|x|} x^2 = -x^2$), therefore $2 = 0$ | |
| Mar 10, 2022 at 14:30 | comment | added | Jeremy Hahn | MO is an E_{infinity}-algebra with homotopy groups a polynomial ring over F_2. If you invert a polynomial generator in odd degree, I think you get a non-trivial ring of the sort you are describing. What you can say if 2=0 in pi_0 of an E_{infinity} ring is that the ring receives an E_2-ring map from F_2, and in particular is an F_2 module spectrum. This means the cohomology with Steenrod action is determined from the homotopy groups, and vice versa. | |
| Mar 10, 2022 at 14:03 | comment | added | Andy Jiang | Sorry can you share the argument that it is linear over F2? I can only see that 2 can't be invertible | |
| Mar 10, 2022 at 13:51 | history | asked | Theo Johnson-Freyd | CC BY-SA 4.0 |