I've been led to believe that the Bousfield class of KO is the same as that of the 1st Johnson-Wilson theory E(1). Does anybody know where I might find a proof of this fact, or maybe provide an indication of a proof?

I assume you're at a prime? KO and K sit in a fiber sequence together and K breaks up as a wedge of E(1)s

@DylanWilson Yeah, I am at the prime p=2. Are you referring to the equivalence KO\wedge C\eta = KU?

@CWcx there's a fiber sequence KO --> KU --(psi^g - 1)--> KU, where psi^g is a topological generator for Z_p^x

@CWcx since DA(1) is finite, this would follow if you could show that DA(1) has type oo: this'd imply that <DA(1)> = <pt>, so (in terms of Bousfield classes) BP<2> = TMF wedge DA(1) = TMF.

idk how to show that though

i'm not sure of its truth, either

one question: is there any substance to the vague idea that the q-expansion/Chern character TMF --> KU((q)) has anything to do with (monstrous) moonshine?

I have two left Quillen functors $F,G:C\to D$. I have a cospan I in C where neither of the arrows are cofibrations, and I have a morphism of cospans $F(I)\to G(I)$ where the three arrows are weak equivalences. I believe this gives me a weak equivalence $F(P)\to G(P)$ where $P$ is the homotopy pushout of the cospan in $C$. Essentially because there is a weak equivalence $hocolim FI \to hocolim GI$ and left Quillen functors preserve homotopy colimits. Am I correct? I feel uneasy about it

oh, and C and D are left proper

@skd do you mean type 0? Which I think it is

@BrunoStonek The argument you give indeed shows that F(P) and G(P) are connected by a zig-zag of weak equivalences. Wether or not there is an actual map F(P)-->G(P) depends on which point-set level model of the homotopy pushout you use.

@GeoffroyHorel I would like an honest map rather than a zig-zag. Which model would give me that?

If you don't have a natural transformation between F and G but can only construct this map F(I)-->G(I) for this explicit cospan, this might be difficult.

hmm. Perhaps it would help to know that $C$ is pointed topological spaces, and that my cospan is simply $* \leftarrow S^1\to *$. And nope, I don't have such a natural transformation... in a sense, it's what I'm trying to build

put differently, I have weak equivalences $F(\ast)\to G(\ast)$ and $F(S^1)\to G(S^1)$ which are compatible with the map $S^1\to *$, and I would like to get a weak equivalence $F(D^2)\to G(D^2)$. oh, and I'd like for it to be compatible with the inclusion $S^1\to D^2$, as well...

I don't see how to do this at that level of generality.

Also I gotta go catch a train sorry :)

ok, thanks!

@Drew yeah, sorry

how do you see that it's type 0?

the way the proof goes for C(eta) is that it contains the bousfield class of anything you can find in the thick subcategory it generates, which includes (1) C(eta^n) for any n, and (2) retracts, but then eta^4 = 0 so C(eta^4) is a wedge of spheres

probably you can make a similar proof go for the 8-cell complex if you're super patient & you know some things about the nilpotence orders of the bottom several stable stems

(also: i think i learned about this in hopkins-mahowald-sadofsky, but i forget if that's right)

@EricPeterson perhaps this is too much to ask, but is there an algebraic proof of this fact using the stacks M_{bo} and M_{bu}?

here M_E is the stack associated to (MU_* E, MU_* MU (x)_{MU_} MU_ E)