@TomBachmann It's not necessarily the R/G-cohomology of BG. For example, let's say G is a finite subgroup of a monoid Δ and R = HQ[Δ], with the action of G (on the right). Then R/G = HQ[Δ/G], and [R/G, R/G]_{R} = (HQ[Δ/G])^hG is the Hecke algebra of double coset functions for G in Δ. On the other hand, the R/G-cohomology of BG is just R/G again because R/G is rational. E.g. the two are additively different for G an order two subgroup of Δ = the symmetric group on three letters.

You do always get an identification of it as (R/G)^{hG}.

But you have to figure out the action, and this always confuses me a little. The R-linear (right) action is a group map ρ:G -> GL_1(R) which acts by r * g = r ρ(g), and then you get a compatible left action by g * r = ρ(g^{-1}) r.

The action that you're taking homotopy fixed-points for is the latter one, I think.

Nope, there's no inverse and I can't edit it anymore. Should be ρ(g) r. (Forgot it was a right action.)

Are n-truncated spaces closed under filtered colimits?

@TylerLawson So this problem goes away if I assume R to be commutative?

@RuneHaugseng I feel I must be overlooking something obvious, but doesn't "take homotopy groups" commute with filtered colimits, and n-truncated = no homotopy groups above n, so ... yes?

@RuneHaugseng doesn't this follow from the fact that the boundary of a simplex is a compact object?

X is a filtered colimit colim F where F(i) is n-truncated. Given a map ∂Δ^k -> X, it factors through some F(i) since ∂Δ^k is compact, and if k>n, then a lift to Δ^k exists to F(i) and ergo to X, ergo all k-boundaries for k>n admit fillers, so we're done. I think this works

@RuneHaugseng Yes. One cheap way of seeing directly (beyond what Harry and Tom already mentioned) is that a space $X$ is $n$-truncated iff the diagonal $X→X^{S^{n+1}}$ is an equivalence and $S^{n+1}$ is a compact object (HTT.5.5.6.17)

This also shows that the result is true whenever you are in an ∞-category where finite limits commute with filtered colimits (e.g. an ∞-topos)

@DenisNardin Frobeniua is not possible: fourth declension neuters have the nominative in -u, not -us

@MatheinBoulomenos That's not right: en.m.wiktionary.org/wiki/Appendix:Latin_fourth_declension Anyway I'm going to use Frobeniusse

What have I done ^^

Only a good thing:)

@DenisNardin the link supports what I said

Ah because it is neuter. Oh well, it doesn't matter anyway:)

Let R -> S be a morphism of E_oo-ring spectra. Then A = S smash S and B = S smash_R S form "Hopf algebroids over R" in an appropriate sense, right? Moreover I should be able to form the "derived cotensor" of A and S over B. Does this give me S smash R? Moreover, I feel like this whole "hopf algebroid in spectra" business, including the answer to the above question, should just be some kind of stupid thing about cosimplicial objects. Can I read about this somewhere?

@JonathanBeardsley is the man to ask about these things

A long time ago in this very chat I learned from @EricPeterson that Segal proved that the map BU(1) --> BU of infinite loop spaces is a rational equivalence. Paraphrasing a remark of Eric, since the splitting lemma for vector bundles tells you that any such bundle is decomposed as a sum of line bundles, it would be interesting to understand why BU(1) --> BU is not an equivalence mod p. i.e. to have a geometric explanation for the difference that appears mod p.

I haven't learned anything in that direction since that moment but I'm still intrigued, wondering if any of you here have anything to add

Are you giving the multiplicative E_∞-structure to BU here?

I don't think so. But let me check

Otherwise I don't see how to give an infinite loop space map BU(1)→BU, but maybe I'm being dumb

no, I'm saying something wrong

the map BU(1) --> BU gives you a spectrum map S[CP^\infty] --> ku

this is the one that's a rational equivalence

Ok, this makes sense

I wonder what cohomology theory is represented by $\mathbb{S}[CP^∞]$

As in, does it have a geometric description?

good question

it should be related to line bundles, right?

Dunno. My intuition for this is basically zero

It receives a multiplicative map from the Picard group, but it's quite clearly not an isomorphism

You could say it's the universal repository of the first Chern class

Could it be true that this map S[CP^infty]--->BU is n connective after inverting n! or something of the sort that its connectuvity is improved when you invert hiegjt and higher primes?