Homotopy Theory

Peter Nelson

00:42

Isn't $X \mapsto ${free abelian group on all maps from a (whatever kind of) manifold to X} a "chain model" for (co)bordism?

I was under the impression that it was

(with dimension as grading and boundary as differential)

CPM

01:11

this answer: mathoverflow.net/questions/67169/… leads me to suspect otherwise

Peter Nelson

01:56

probably another stupid question:

Is there a difference between an R-coalgebra and an affine scheme over Spec(R)?

(apply any finiteness words you want)

Aaron Mazel-Gee

02:12

@PeterNelson i'm confused -- shouldn't an affine scheme over Spec(R) just be some Spec(A) --> Spec(R)? or do you mean affine group scheme

Peter Nelson

maybe

main reason for confusion: Ravenel-Wilson and other things by Wilson refer to Hopf rings as "ring objects in CoAlg" (and even to Hopf algebras as a group object in coalgebras)

(for my confusion, that is)

Peter Nelson

02:40

However, my brain has gotten used to replacing the words "commutative Hopf algebra" with "group scheme"

Eric Peterson

03:00

there is a difference; if R is a field, then an R-coalgebra gives a formal scheme over Spec R, but in general R-coalgebras only transmute to R-schemes if they're sufficiently nice

there is stuff on this in section 4.8 of strickland's fsfg, which is terse and difficult to make sense of unless you've also read demazure's lectures on p-divisible groups part... 1, probably?

demazure does the field case, neil points out that over not-fields things can go awry and gives a good enough condition to define a subcategory which a) does have a map to formal schemes, and b) that map preserves some colimits

Peter Nelson

fsfg is formal schemes and formal groups?

Eric Peterson

yes, sorry

Peter Nelson

and not finite subgroups of formal groups

Eric Peterson

oh goodness, yes, very sorry. i rarely look at that one

i started because functorial philosophy for formal phenomena was titled fpfp.pdf

it is definitely time to leave the office

Peter Nelson

these things both seem somewhat helpful, thanks

Jon Beardsley

03:26

@PeterNelson notice, I believe, that $R\times R\to R$ gives a coalgebra structure on $Spec(R)$

or comonoid, or whatever.

it doesn't seem immediate to me here that one needs to worry about formal phenomena

oh wait....

are you taking coalgebras over $R$?

nevermind. I'm dumb.

yeah wtf. are they talking about formal things?

Peter Nelson

@JonBeardsley who are they

Jon Beardsley

maybe one can make the same kind of formal arguments that demazure makes to go from corings to affine schemes, i dunno.

Peter Nelson

this is what eric was saying

Jon Beardsley

Ravenel and Wilson

I mean, whoever is saying that affine schemes over R are the same as R coalgebras.

Peter Nelson

@JonBeardsley no, no one was saying that. Except Demazure does, if R is a field, and Strickland does, if there are some conditions on the coalgebra

Jon Beardsley

03:36

yes.

okay. so your question is not really answered

?

Peter Nelson

I think Eric pointed me to places that answer it to my satisfaction

Jon Beardsley

But.... do those places deal with it when your not working in formal schemes?

Eric Peterson

you don't have the option of not working in formal schemes

Jon Beardsley

I mean, I do have that option. I do that all the time.

Do you mean that such a correspondence cannot exist outside of formal schemes?

Or do you mean something like anything Ravenel and Wilson do is going to be sufficiently completed or something?

Eric Peterson

yeah. if you have some coalgebra (without necessarily being part of some Hopf algebra structure), and you're tasked to build a scheme out of it, you do... what?

Jon Beardsley

03:41

Well, sure. Okay, that's fine. So you're just saying that Ravenel and Wilson must be talking about formal schemes here, because otherwise the statement doesn't make any sense.

I'll accept that.

=P

Eric Peterson

one option is to try to take linear duals to produce an algebra, and then take Spec of that. that works OK if the coalgebra is finite, but if it's infinite then duals are troublesome, and the way you get around this second hurdle is to say 'no, wait, i meant formal schemes'

yes, an alternative answer is 'this is what makes R-W sane', but i think there's more than that to say

Jon Beardsley

I just read about a cool description of the Brauer group in terms of corings that expands it so that it's always in bijection with the relevant second cohomology.

Some kind of generalization, called "Azumaya corings" or something.

Of course, then what's the point right? I mean, we use the cohomology group to try to understand something we're interested in. So if we just make the things we were originally interested in bigger, we've kind of missed the point.

Eric Peterson

idk, understanding the difference in the constructions as being a specific, analyzable source of the general discrepancy is valuable

Qiaochu Yuan

03:59

a coalgebra is a filtered colimit of finite-dimensional coalgebras, each of which is dual to some finite-dimensional coalgebra and so may be regarded as describing some scheme, and then you can think about this filtered colimit as an ind-scheme. i think?

Jon Beardsley

Yeah. That's the "formal" issue we're discussing.

In other words, formal schemes are (in my mind at least) basically just ind-schemes, and the process you describe is precisely how you get the correspondence Eric is referring to, I believe.

Qiaochu Yuan

@Peter: a commutative, cocommutative Hopf algebra describes a group in two senses. first, the multiplication describes a group object in cocommutative coalgebras. second, the comultiplication describes a group object in (commutative algebras)^{op}. it's in this latter sense that commutative Hopf algebras can be thought of as affine group schemes. this correspondence is contravariant.

cocommutative Hopf algebras, on the other hand, should be thought of as formal groups, and this correspondence is covariant. and hopf algebras which are neither commutative nor cocommutative are just some biza…

Eric Peterson

coalgebras over a field are filtered colimits of finite dimensional coalgebras

Jon Beardsley

04:39

holy crap! luc illusie is still alive. sweet!

Peter Nelson

yeah

Jon Beardsley

derp. amitsur complex is just the dual of the cech complex.

Peter Nelson

good picture

Jon Beardsley

Yeah, I really like it.

Peter Nelson

I just looked up what the amitsur complex is. The first thing I thought was "oh, that thing has a name?"

Jon Beardsley

04:51

Yeah, exactly!

It's also the ANSS, btw.

Peter Nelson

I think I would call it "the cobar construction"

Jon Beardsley

Fine.

:-)

Peter Nelson

Which I guess was callan's comment awhile ago

Jon Beardsley

Yeah.

Jon Beardsley

05:08

haha, equalizer in french = le noyau de la double fleche

Jon Beardsley