Timeline for Endomorphism ring spectrum of the Eilenberg-MacLane spectrum
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 10, 2017 at 19:10 | comment | added | Nicholas Kuhn | Right. That's what I meant to mean. | |
| May 10, 2017 at 12:30 | comment | added | Sean Tilson | I am confused, I thought $\mathrm{H}\mathbb{F}_p$ was an $MU$-algebra. I even know how to construct a map of $E_{\infty}$-ring spectra from $MU$ to $\mathrm{H}\mathbb{F}_p$. Maybe you mean the "spectral" Steenrod algebra isn't an $MU$-algebra? | |
| May 10, 2017 at 10:42 | comment | added | Nicholas Kuhn | Tyler's answer also shows that $H\mathbb F_p$ is not an $MU$--algebra, which is a bit less intuitive, as the Steenrod algebra arises when one considers the endomorphism ring of the additive formal group. | |
| May 8, 2017 at 23:04 | comment | added | Tyler Lawson | @Martin: Good question. The homological version of this is more popular because of continuity issues. It appears like this in Baker-Lazarev's "On the Adams spectral sequence for R-modules" and Carlsson's "Derived completions in stable homotopy theory". Dwyer-Greenlees-Iyengar also mention this perspective in section 5 of their duality paper. | |
| May 8, 2017 at 22:53 | comment | added | Martin Frankland | Here's a follow-up question. The displayed equivalence seems to rely on the following fact. Denote the Steenrod algebra spectrum $A = \mathrm{End}_S(H\mathbb{F}_p)$. Taking mod $p$ cohomology is the functor sending a spectrum $X$ to the left $A$-module $F_S(X,H\mathbb{F}_p) = H\mathbb{F}_p^X$. If $X$ and $Y$ are finite spectra, then the map of spectra $F_S(X,Y) \to F_{A}(H\mathbb{F}_p^Y, H\mathbb{F}_p^X)$ is a $p$-completion. I see that this is more or less the convergence of the Adams spectral sequence, but do you know a reference that states it this way? | |
| May 8, 2017 at 18:56 | vote | accept | Martin Frankland | ||
| May 8, 2017 at 18:44 | history | answered | Tyler Lawson | CC BY-SA 3.0 |