Homotopy Theory

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Clark Barwick

2:32 AM

@AaronRoyer Yeah, even for the point {0} in A^1, you're getting the k[x]-modules that are x-nilpotent.

Aaron Royer

Exactly, yeah.

Looking more closely at that example basically told me I can't do what I wanted to do, which is a shame.

It's okay, though, all it means is that D(-) is a better functor than I remember it being.

Jon Beardsley

Hey guys, do you know how cyclic sets are explicitly a subcategory of Loday's non-commutative sets?

How is the cyclic ordering data getting remembered inside of non-commutative sets?

Ah, I guess I should attribute cyclic sets to Connes, not Loday.

Oh nevermind. I'm just misunderstanding what cyclic sets are.

Jon Beardsley

3:23 AM

Wow, this paper Vigleik wrote on cyclohedra is so clear!!

13 hours later…

Jon Beardsley

4:51 PM

Does it make sense to talk about deformations of bundles of ring spectra? Or, in other words, a spectrum is a bundle over the point, but we could ask about deforming this to some nilpotent fattening of the point, and there should be some cohomology classifying these deformations, yes?

I guess one could derive the entire thing and talk about spectral sheaves over something like $Spec(K(n))$.

And then one might try to recapture some of Vigleik's results about the topological Hochschild cohomology of $K(n)$ from this point of view.

Of course then one would have to figure out what the hell one means by complete local Noetherian rings whose residue field is $K(n)$.

Denis Nardin

I think that everything you said could be made to make sense, except that $K(n)$ is not really a field, since it is not really commutative

I think there's some kind of deformation theory for noncommutative rings but I do not understand it

Jon Beardsley

Well, it's not clear to me what a nilpotent thickening of a derived scheme is in this context.

Jon Beardsley

5:58 PM

Lol, okay that didn't make any sense.

Ah, let's try this again. So, given the mapping space of two associative ring spectra $Alg(A,B)$, if $A$ is, say, the suspension spectrum of a loop space then it should be co-$E_\infty$, which makes it a cocommutative cogroup object in this category, meaning this mapping space should be an infinite loop space?

Ah but is $Alg(-,B)$ even monoidal?

Denis Nardin

Why is it a cogroup object? I think it is only a comonoid object

Jon Beardsley

That's a fair point.

I suppose I'd need more structure on $A$.

Denis Nardin

And yes, I think that $Alg(A,B)$ in this case is an E_\infty space

with operation given by sending $f,g$ to $A\to A\wedge A \to B \wedge B \to B$ where the latter is multiplication

Jon Beardsley

Any idea where I find a proof of this?

Tyler Lawson

6:23 PM

if A is the suspension spectrum of Omega X there's a nice a chain of equivalences Alg(A,B) ~= Loop-spaces(Omega X, GL_1(B)) ~= Spaces(X, BGL_1(B)); you can possibly reference ABGHR for this

if B has an E_n-structure for n > 1 then I think this should inherit a (grouplike) E_{n-1}-space structure because we can further deloop BGL_1(B)

Jon Beardsley

Hm, but you're saying that the comonoidal structure on $A$ coming from the diagonal doesn't give us a corresponding monoidal structure on the mapping spacE?

Tyler Lawson

if B is just associative, I'm not sure it's guaranteed to because B ^ B -> B might not be a map of associative algebras

Jon Beardsley

6:42 PM

Alright. I guess this is an unsuitable way of producing Hopf-algebras in associative ring spectra.

Tyler Lawson

i think it is a hopf algebra, it's just that a hopf algebra which is cocommutative but not commutative isn't necessarily a cogroup object in associative algebras

Jon Beardsley

How would you define a cocommutative hopf-algebra in the oo-category of associative algebras?

Such that the above described structure would fit into that definition?

I feel uncomfortable just saying it needs such and such maps this way and that way, because that doesn't seem to build in the necessary higher homotopical information.

In other words, I agree that the thing I described is a Hopf-algebra, I just don't know what general definition to plug it into, I guess.

This whole question is really for the purposes of trying to write something up, and working with a specific example the general case of which I cannot find existing literature.

Tyler Lawson

in these terms, so far as I'm given to understand, the definition is probably "something that has a cobar construction"

or equivalently some kind of cosimplicial object satisfying a segal-type condition

Jon Beardsley

7:04 PM

Yeah. Fair enough. Some kind of comsimplicial object w/r/t the monoidal structure on Alg(S)

satisfying the Segal condition, as you say.

Vrouvrou

7:58 PM

hello, can someone of you help me about homotopy equivalence ?

i don't understand the answer : math.stackexchange.com/questions/1019919/…

Vrouvrou

8:50 PM

@JonBeardsley can you help me please ?

1 hour later…

Denis Nardin

9:51 PM

@Vrouvrou It would be easier to help you if you explained exactly what steps you do not understand. You were presented with explicit homotopy equivalences, what is confusing you?

1 hour later…

user101036

10:53 PM

suppose I have a scheme X and let N_X be the nilpotent sheaf of ideals and let Z be a closed subscheme of X defined by a nilpotent sheaf of ideals. Suppose JN_X = 0. Why does this imply that N^2_Z \subset p(N_X) where p is the restriction?

(here N_Z is the sheaf of nilpotent ideals of Z)

this should be easy, but I am not seeing it

user101036

11:25 PM

and oh, J is the defining ideal sheaf of Z