@theL not an expert, but it seems that there is no obvious reason for an algebraic geometer to care about spectral AG. As far as I know, the 'smallest' natural extension is with simplicial commutative rings, not E_\infty algebras. See for example, mathoverflow.net/questions/226082/…

@theL one reason that I've heard that might actually interest some people (namely I've heard this about the people doing p-adic Hodge theory) is that there is a natural "Frobenius" in any characteristic.

namely, for any spectrum M there is a map from M to the Tate spectrum on the p-fold smash product of M. if M = R for some commutative ring spectrum R then we can compose with the multiplication and get a map from R to the Tate spectrum of R.

if R is an Eilenberg-Mac Lane spectrum then in degree zero this realizes the p'th power map R -> R/(p)

if R is the sphere then this turns out to be equivalent to the map from S to the p-completed sphere

but even if R came from a simplicial ring, this is usually not a map you can realize by a map of simplicial rings or even of algebras over Z, only over the sphere spectrum

@MingcongZeng well there's the homotopy orbit sseq, and then you also have to contend with the boundary map (for which you may need to understand the Tate spectrum and homotopy fixed points as well)... but yeah, it's not easy.

(So at the very minimum you need to know how G, or the Weyl group of H, acts on the homotopy of X as an H spectrum)

Hmm, i see. It looks like something I can handle in my case.i will definitely give a try tomorrow. Thank you!

Thanks @TylerLawson, that's sound interesting.

Fix spaces $A$, $B$, $X$. Let $E$ be the set of pairs $(f,g)$ where $f\colon A\to X$ and $g\colon B\to X\cup_f CA$ are continuous maps, where the latter is the usual mapping cone construction.

Is it possible to topologize $E$, so that $E\to Map(A,X)$ is a fibration with fibers $Map(B,X\cup_f CA)$?

how does one define the natural transformation from genuine to geometric fixedpoints in the language of spectral mackey functors?

geometric fixed points of X are themselves the genuine fixed points of a spectrum: the localization of X into spectra whose identity fixed points are zero

so the natural transformation you want is the composite of that localization with evaluation at G/G

say we write $i=triv : Fin \rightleftarrows GFin : (-)^G = j$, then $\Phi^G = j_! : Mack(GFin) \to Mack(Fin)$ while $(-)^G = i^* : Mack(GFin) \to Mack(Fin)$

(here G = C/p, I probably shouldn't reflexively assume that)

no, what you said is true for any finite group i think

but okay, so how do i write down that localization in mackey functor language?

only if I change "identity fixed points are zero" to "fixed points for any proper subgroup are zero"

oh right, sure

but so that localization should be easy enough in mackey functors

so I'm not sure precisely what you're asking, but the fiber of that localization map is the restriction and left Kan extension of X from A^{eff}(G) to A^{eff}(nontrivial G-orbits) and back

ah, i was figuring it itself would come from a left kan extension

you can also view it as the left Kan extension you wrote down above

you mean that there is a unit map $id \to j^* j_!$, and i apply $i^*$ to this?

that works

okay, cool. i guess the point is that $i^*j^* = id$, probably just because $ji = id$

oh yeah, that's true

and I guess it's clear at this point, but $j^* j_!$ really is exactly the localization I first talked about

beautiful

thanks!