@SaulGlasman yeah, i think you might be right. eric explained to me that the relevant homotopy groups don't seem to line up at all, and i think he's right. i'm not super sure what you mean by nilpotent extension. i'm not sure. i know that somehow S is supposed to be a nilpotent extension of HZ, but i'm not sure how to think of this w/r/t MU since the sphere doesn't have integral generators in non-zero degrees.

Well ok, that's relieving. Because I couldn't make sense of those two things.

There doesn't seem to be a consensus on what counts as TAQ and what counts as "the cotangent complex" of a ring.

@Jon: maps X -> B^2 G classify principal BG-bundles

more generally, if you think of maps X -> BG as classifying fiber bundles with fiber F and structure group G, where G acts on F, then you can think of maps X -> B^2 G as classifying fiber bundles with fiber F and structure group BG

one way of saying what it means to have structure group BG is that F is an object in an infinity-category such that G acts on the identity automorphism of F

(so G acts as a "higher automorphism" of F)

for example, the automorphism infinity-group of Vect turns out, in a suitable sense, to be B C^{\times} (every autoequivalence Vect -> Vect of Vect as a Vect-enriched category is isomorphic to the identity, and the automorphism group of the identity functor F -> F is C^{\times}, and then we want to equip this with its natural topology), from which it follows that maps X -> B^2 C^{\times} classify Vect-bundles over X (bundles of categories isomorphic to Vect)

given a Vect-bundle over X you can now ask for sections of this bundle, which are vector bundles twisted by the map X -> B^2 C^{\times}. Giving C^{\times} its usual topology the set of homotopy classes of such maps is H^3(X, Z), and we recover the fact that complex K-theory can be twisted by classes in H^3(X, Z)

(another way to say this is that I'm thinking of Vect as a Vect-module; bundles of categories isomorphic to Vect is a model of "2-line bundles" or "line gerbes" or something)

maybe stupid question: in general what can one twist a cohomology theory by?

@Peter: in general, if F is an object in an infinity-category with automorphism infinity-group Aut(F), then F-bundles over a space X are given by homotopy classes of maps X -> BAut(F). in particular, we can take F to be a spectrum

the above action of B C^{times} on Vect gives an action of B C^{times} on the complex K-theory spectrum, but I believe the full automorphism infinity-group of the K-theory spectrum is somewhat more complicated

(maybe a lot more complicated)

this is discussed e.g. in this paper by ando-blumberg-gepner: arxiv.org/abs/1002.3004

apparently you can twist tmf by classes in H^4(X, Z). it would be cool to have a geometric interpretation of tmf that would let you see this directly...

it would be cool to have a geometric interpretation of many cohomology theories...

I should also mention that when ABG says that maps to BGL_1(R) classify twists of R as a cohomology theory they're only taking into account automorphisms of R as an R-module. there ought to be a slightly bigger automorphism group if you just want automorphisms of R as a spectrum

but I think the reason it's nice to restrict attention to thinking of R as an R-module (the way we restricted attention to thinking of Vect as a Vect-module above) is that ordinary R-cohomology then acts on twisted R-cohomology

this might be a more up-to-date reference: arxiv.org/abs/1403.4325

(is this the paper that people call ABGHR?)

ABGHR got split in two for publication purposes

this is one of them

ah. and arxiv.org/abs/1403.4320 is the other?

yes

good stuff

arxiv.org/abs/0810.4535 is the unsplit one I guess

Hey @Prasit!

@JonBeardsley Hi

@Prasit aren't you coming to talk at JHU soon?

@JonBeardsley I think everybody uses TAQ for the concept that Maria Basterra described in her thesis & first paper

@TylerLawson well, maybe, but in Basterra and Mandell's paper on cohomology of E_\infty-rings, ffor a morphism of algebras, A\to X, over B, they define LAb_A^BX to be B tensored with L\Omega_A X. In my mind, the L\Omega_AX object is "TAQ" of X over A. However, Basterra and Mandell go on to define TAQ(X,A;M) (this is cohomology) to be Ext_B(L_A^B X, M). All of which is slightly confusing to me.

Presumably TAQ-homology should be the homotopy of some TAQ-object (perhaps tensored with some module?) and TAQ-cohomology should be maps out of TAQ, but perhaps I'm being naive or just plain stupid here.

Anyway, the point is, Basterra and Mandell use $\Omega$ to denote something BEFORE they produce the left derived abelianization functor, whereas Basterra uses \Omega to denote the left derived abelianization functor.

Looking at it now, I suppose that's my confusion.

Anyway... it's really immaterial, and the way things have been going for me lately, I'm likely just being an idiot.

@JonBeardsley Right, that's TAQ-cohomology versus TAQ-homology.

@JonBeardsley Yes I will be in JHU on Oct 6.

Hi guys, I'm looking at the computation of the homology of extended powers of a spectrum X (in terms of Dyer-Lashof operations). It is pretty easy to reduce the calculation to X = some suspension of the sphere spectrum, but then for S^n there is some work to do.

I know of two proofs. An older one using the Serre spectral sequence and Kudo transgression, but I have to avoid that one.

There is a newer proof in the H-infty volume, which is rather long and an tedious, and has 25 years old now!

I was wondering if somebody knows of a newer or maybe simpler proof of the homology (mod p), p=2, of D_k(S^n) ? Maybe using some other path than in H_infty ? Thanks!

Is there a spectrum (like TMF) related to the moduli stack of cubic curves?

@JonBeardsley tmf (the connective version) is sort of what you're looking for, in the sense that its ANSS takes the form of a descent spectral sequence for M_cub.

Ah I didn't know that. Interesting.

@AndrewSenger and Tmf is the one with just the nodal curve?

Yeah, that's correct.

weird stuff.