Do people care about, for a fixed ring $R$, computing things like $K(R)^\ast(X)$ for various spaces $X$? Or do people generally only care about computing $K(R)_\ast$?

For instance, do people ever compute $K(R)^*(CP^\infty)$?

@JonBeardsley I'll be somewhat perverse and note that if \ell is a prime and q a prime power coprime to \ell, then K(F_q)^{\ell} (the latter denoting \ell-adic completion) is equivalent to the \ell-adic image of J-spectrum. This is nearly the same as the K(1)-local sphere at the prime \ell.

So: everyone who computes K(1)-local homotopy groups of spectra X are also computing the \ell-adic part of K(F_q)_*(X)

Computing K(R)^*(CP^\infty) in general seems like a natural place to start; however, it's definitely not the case that if R is complex oriented, its algebraic K-theory will be as well.

@JonBeardsley Isn't this related to John Lind's work?

@AaronMazel-Gee Not sure if this is of any help, but maybe one idea would be to notice that a map in the category of arrows $X^1 \rightarrow X^0$ of simplicial sets will be a Reedy fibration if and only if $X^0 \rightarrow Y^0$ and $X^1 \rightarrow Y^1 \times _{Y^0} X^{1}$ are both fibrations, and the second condition is kinda what you're looking for..? (I think that moreover in this case the Reedy model structure will coincide with the injective one.)

Maybe you could ook at the functor that takes a simplicial set $X$ equipped with a choice of edge $p: \Delta^{1} \rightarrow X$ to the diagram $X _{/p} \rightarrow X_{/p(1)}$, from what I scribbled on a piece of paper this has a chance of being right Quillen.

(But I'm not entirely sure.)

@CraigWesterland do you have any idea of when algebraic $K$-theory spectra are complex oriented?

@SeanTilson i think so. i hadn't remembered that.

arxiv.org/pdf/1304.5676.pdf

@JonBeardsley i've been vaguely trying to advocate this for a while now, though i certainly don't have any real evidence. but my sense is that, for let's say an $E_\infty$-ring spectrum $X$, we should always think of the cotensor-object $X^{CP^\infty}$ as some sort of "formal affine line over $X$" -- complex-orientability is nothing but a sparseness condition, so that we can study the true derived algebraic geometry using mere homotopy groups.

more generally, we should think of homotopy types as giving recipes for building up new (ring) spectra out of old ones (in this way, and also using tensors -- e.g. this leads to hochschild homology)

@AaronMazel-Gee So like Ando-Hopkins-Strickland type stuff? Like what Nat and Tobias are doing some of the time? The second bit sounds like Factorization homology, but I think that is not a proper invariant in the ring spectra variable.

what do you mean by "proper invariant"?

@AaronMazel-Gee that sounds dead-on to me. but where does MU come into the picture then?

I mean, how should one think of $MU$? Because I sort of see this as something like the Witt vectors over the sphere, which means that in some sense it should parameterize all formal curves.

@JonBeardsley well of course there are some serious issues regarding just how commutative things are, but i'd guess we should expect it to be the derived lazard ring

But I agree with your characterization of complex orientability. I was just telling Romie Banerjee the other day that complex orientable theories are somewhere in between Eilenberg-Mac Lane spectra and everything else. I.e. they're sort of slightly less "algebraic" than actual rings, but still comprehensible through algebraic methods.

Hah, that's funny, he stole your dinosaur? Anyway, he's not at JHU anymore.

arxiv.org/find/all/1/all:+AND+romie+banerjee/0/1/0/all/0/1

who knows, i could've stolen his! in any case, he's adorable and i'd be happy if every single mathematician had him on their webpage as well

sites.google.com/site/romiebanerjee

right

no more dinosaur though...

But yeah, he's doing some really neat stuff. We were talking about all this.

anyways, i have no idea how to, like, really "say anything" about the DAG of more general ring spectra, but i definitely have the sense that complex orientability will look like a red herring 100 years from now

I sort of agree.

I feel like people have talked about what a "formal group" over a commutative ring spectrum is... but I don't really feel like there's a good theory.

I suspect it will have something to do with Pereira's work on truncations of the $E_\infty$-operad.

(to be clear: i'm sure romie banerjee is adorable too, but i was referring to the little dinosaur guy)

Hahaha, yeah I figured. And yeah, Romie is adorable.

@AaronMazel-Gee I guess I mean that I have been informed that it is not suitably functorial in $E_n$-algebras. I could be misremembering this. Mostly, factorization homology is not, in general, a homology theory for rings except when it agrees with the ones that we already know (the manifold in question is a (smash) product of $S^1$'s). This is what I mean.

Say I consider Spec k[x] as a group scheme and dualize it to get a formal group scheme, I think I should be getting the formal additive group scheme, is there a clean way to see this?

this is something that you can simply write down

if you look at the comultiplication map k[x] -> k[x] tensor k[x] giving the group scheme structure, which takes x to x tensor 1 + 1 tensor x, the linear dual is clearly the addition map

and conversely

I thought we got some divided power structure. For example, taking x^o to be the dual to x, then x^o * x^o = 2x^{2,0}

@SaulGlasman but maybe I misunderstood something, if we're taking ring homomorphisms k[x] -> k, then I agree that the coaddition structure turns into ordinary addition

@QiaochuYuan Of course, I'm thinking of viewing k[x] with the ''common'' comultiplication (the one that Saul wrote down) . I want Spec k[x] to be G_a, the additive group scheme

or what it's called

@Dedalus: yeah, sorry, misread your question. what's the dual of a group scheme?

@JonBeardsley I don't really know any criteria that ensures it. To be honest, I think that the only example I know is when R = C, so that KR = ku. Of course, even that doesn't work -- you need to replace ku with KU.

there's a few ways to do it if I recall correctly (but they all amount to the same thing of course). The easiest way is probably to look at the underlying Hopf algebra A, consider Hom_k(A,k) and note that this reverses all the arrows in the relevant diagrams (if your algebra is not finite-dimensional, you need to use that any coalgebra is the filtered colimit of its finite dimensional coalgebras) you get a formal group scheme

(and if the algebra was finite, this of course gives a group scheme again)

The hopf algebra needs to be cocommutative as well

If I recall correctly, in this way you get an equivalence of categories between formal schemes and cocommutative coalgebras; formal commutative groups and affine commutative group schemes

algebraic K-theory of a (connective) ring spectrum is "almost never" complex orientable, i think

it would, for instance, need to see \eta trivially. i remember andrew salch giving me some conditions on when this occurs...

Now that I think about it, I can't think of many examples of connective complex orientable ring spectra, but maybe that's me focusing almost exclusively on periodic cases...

@AaronMazel-Gee I was thinking something like this. Suppose you have a functor $p: X \rightarrow S$ of $\infty$-categories, together with a choice of an edge $f: \Delta^{1} \rightarrow X$ that has the equivalence property you need.

You can now treat this as an arrow in simplicial sets over $\Delta^{1}$ equipped with the Joyal model structure and find a factorization $X \rightarrow _{s} X^{\prime} \rightarrow _{q} S$ into an acyclic cofibration followed by a fibration.

Wouldn't $s: X \rightarrow X^\prime$ be the categorical equivalence you're looking after..? I think $s(f): \Delta^{1} \rightarrow X^{\prime}$ will both have this "equivalence property" you had, because this should be invariant under categorical equivalences over $S$...

… as well as property that the induced map $X^{\prime} _{/s(f)} \rightarrow X^{\prime} _{/s(f)(1)} \times _{S _{/p(f)(1)}} S_{/p(f)}$ is a fibration, because it's a matching map of a Reedy fibration by what you verified.

if R is commutative enough to make KR a ring, eta is the image of -1 in the unit group of R, so the only time KR doesn't possess eta is when 2=0 in pi_0 R or R is nonconnective. sometimes it goes away when you do a completion, though

@Dedalus: yeah, okay, in that case i'm pretty sure you just get the formal additive group. shouldn't this just be a calculation?

@TylerLawson isn't the next obstruction in the AHSS related to $\nu$ or something??

I can't remember.

But I mean, one can obviously write down obstructions. I guess that's not a very satisfying theoretical answer though.

So, if I have an associative $\mathbb{S}$-algebra $A$ (a la EKMM), and a right $A$-module $M$ and a left $A$-module $N$ it's not true that $M\otimes_AN$ is an $A$-module again correct?

But somehow this is true if we assume that $A$ is $E_2$?

But this doesn't really fit into the EKMM set-up, does it?

Or wait... I guess that's not what I'm trying to get at... I want to think about smashing $A$-algebras over $A$, i.e. $M$ and $N$ are $A$-algebras...then I don't have that their tensor is still an algebra...

but if $M$ is a left $R$-module and $N$ is a right $R'$-module, then $M\otimes_A N$ is an $(R, R')$-bimodule.