Skip to main content
11 events
when toggle format what by license comment
Jan 21, 2021 at 21:44 vote accept Tim Campion
Jan 21, 2021 at 21:43 answer added Tim Campion timeline score: 4
Jan 21, 2021 at 5:31 history became hot network question
Jan 21, 2021 at 4:28 comment added Jonathan Beardsley The same statement holds for any spectrum $E$ such that $\langle E\rangle < \langle H\mathbb{F}_p\rangle$ (these are the Bousfield classes) I believe. So you have a whole slew of spectra that satisfy the equation $x^2=0$, basically by taking the Brown-Comenetz dual of any connective spectrum with finitely generated homotopy groups.
Jan 21, 2021 at 4:26 comment added Jonathan Beardsley Doesn't the Brown-Comenetz dual of the sphere, $I$, have the property that $I\wedge I\simeq 0$? So I guess it's integral?
Jan 20, 2021 at 22:12 answer added Nicholas Kuhn timeline score: 17
Jan 20, 2021 at 21:54 comment added Tim Campion No worries. I think that ultimately varying the subring would be an interesting thing to do.
Jan 20, 2021 at 21:53 comment added Fernando Muro Sorry, I missed that.
Jan 20, 2021 at 21:53 comment added Tim Campion @FernandoMuro over $\mathbb Z[\Sigma,\Sigma^{-1}]$. I'm sure there are other reasonable choices, but this seems like the "minimal reasonable one"
Jan 20, 2021 at 21:52 comment added Fernando Muro Integral over what subring?
Jan 20, 2021 at 21:29 history asked Tim Campion CC BY-SA 4.0