@QiaochuYuan oh, totally. good point!

@TylerLawson Thanks Tyler. What do you mean by "G acting by spectrum maps" ? If X is a spectrum, can the action be by non-spectrum maps ? I'm interested in the following case which I thing follows from what you said. If X is an E_oo spectrum, then the spectral sequence $H^{\ast}(G, THH_{\ast}(X)) \Rightarrow \pi_{\ast} THH(X)^{hG}$ for the fixed points of THH(X) is a spectral sequence of $\pi_{\ast}X$ modules, i.e., the differentials commute with multiplication by element of $\pi_{\ast}X$ ?

@Bogdan I just wanted to make sure; one could be taking a homotopy fixed-point spectral sequence of a G-action on a space instead

ah I see, yeah everything is stable here!

In the THH case you mention, though, the group doesn't act by maps of X-modules and so the differentials aren't necessarily pi_*X-linear

Ah yes. My object is pretty rigid so maybe the action maps are maps of X-modules, but I have to understand it better before I can say anything. Thank you!

Possibly silly question: is there a dg-functorial projective-cofibrant replacement functor for the dg-category of unbounded chain complexes of modules?

I admit it, I don't know Kan extensions. Before reading on them, I'd like to know if the extension Schwede has in mind in example I.5.28 of SymSpecv3 of the Eilenberg-Mac Lane functor from abelian groups to symmetric spectra, to simplicial abelian groups, is a Kan extension

@AaronMazel-Gee According to the nLab, at ncatlab.org/nlab/show/cofibrantly+generated+model+category , [C,D] is cofibrantly generated when D is cofibrantly generated and C is small. It doesn't give a precise reference to this statement, though.

The correct statement is that the projective model structure is cofibrantly generated. There is an obvious pair of generating sets for [C, D] if you are given a pair for D; you just have to check that the small object argument applies.

Orthogonal spectra are enriched over based spaces in the obvious way. What goes wrong with this definition in the case of (commutative) R-algebras when R is an orthogonal ring spectrum? I mean, why do we then have to take unbased spaces instead of based ones?

@lentic R-algebras do not have an obvious zero map, do they? Or did you mean R-modules?

ah, ok, of course. I did mean R-algebras.

here's a vague question: what can we say of that tensoring? I mean, the tensoring of orthogonal spectra over based spaces is explicit, but I find the tensoring of commutative R-algebras over unbased spaces a bit mysterious. If the space $X$ is discrete, then $A \otimes X$ is just the $X$-fold smash of $A$'s. But what if $X$ is not discrete?

this isn't necessarily very explicit, but if you can write X as a simplicial set then you can take the smash product of copies of A levelwise and then realize, since realizations of simplicial commutative ring spectra are computed in spectra

@lenticcatachresis it's interesting/unusual enough to give it a new name (topological chiral homology), and if you like it's the colimit (in commutative R-algebras) of the constant diagram on X with value A

this is how you show that THH(A) is equivalent to $A \otimes S^1$

I see. What exactly goes wrong? I mean: cotensors are created in orthogonal spectra. Why not tensors?

Because the forgetful functor from algebras to spectra is a right adjoint, not a left adjoint. Hence it preserves (homotopy) limits but not (homotopy) colimits

but on R-modules you get that both tensors and cotensors are created on spectra. is there a "coextension of scalars"?

it might help to think of the situation for limits and colimits of ordinary commutative rings and modules over them

yes, so there is a coextension of scalars for modules, and I'm guessing that it's just the same in the brave new context

yeah, it works exactly the same way

whereas we don't have such a thing for algebras. ok, that clarifies a lot. thanks everybody

@lenticcatachresis by "coextension of scalars", do you mean what also might be called "restriction of scalars"? or is this something different?

no, it's the right adjoint to restriction of scalars

coextension of scalars along a map R -> S of commutative rings is given by M |-> Hom_R(S, M)

yes, that's what I meant

ah, okay cool

yeah, @lenticcatachresis basically if you can write things out without referring to elements then it should carry over smoothly

Not sure if this questions is suited for this channel but: What are the criterions for when a functor F: Sch^{op} \to Set is representable by a scheme? It has to be a sheaf, of course, but what more is known?

these representability questions are often really hard

afaik they're most usually resolved by actually constructing the representing object

@Dedalus There are nice criteria to see if a sheaf is an algebraic space. This is one of the reasons why algebraic spaces are better behaved

What is that criteria?

I do not remember the details, it is in Artin's paper for the algebraization of formal moduli

OK! Thanks!

(I think that the paper is more interested in proving when something is an Artin stack, but of course for sheaves of sets there's no difference)

In general these results go under the umbrella name of Artin representability criteria

@Dedalus: there are some purely formal things you can say. for a category C with all colimits, a functor C -> Set is representable iff it is a right adjoint (in which case its left adjoint takes a set X to the coproduct of X copies of the representing object). so you can apply adjoint functor theorems to determine when this is the case

the non-formal statements are about when a functor CRing -> Set is representable by a scheme, since here you're not testing all schemes but only affine ones

and this is what artin representability. there's a version stated as Theorem 1 here: math.harvard.edu/~lurie/papers/DAG-XIV.pdf