Homotopy Theory

Jon Beardsley

4:46 AM

Certainly Gunnar did, but he left, unfortunately.

Arpon

5:21 AM

@BenLim lol hey ben

Espen Nielsen

8:55 AM

@AaronMazel-Gee Then how about $f_*$ or $f_!$ ?

Ben Lim

11:28 AM

@Arpon Hey, decided where to go for grad school yet? Are you still taking the year off in cambridge?

@Arpon I've already sent emails to MIT and Berkeley telling them I'm not coming, so it's down to two choices now

Denis Nardin

1:55 PM

Reality check: if $k$ is a field of positive characteristic, then $H(k[x])$ is not the free $E_\infty$-algebra over $Hk$, right?

Saul Glasman

2:08 PM

right

the free E_\infty-algebra over Hk has the Dyer-Lashof algebra in its homotopy

@JonBeardsley Gunnar left?

Jon Beardsley

I heard at dinner last night that he is now full time at Ayasdi.

1

Mike Catanzaro did that.

He is hitting my keyboard. 2

Ugh

There we go again.

Saul Glasman

I'm assuming you named your cat after Mike Catanzaro and this is not the human Mike Catanzaro

Jon Beardsley

No it's the human one. He's sitting right next to me.

Saul Glasman

haha

Jon Beardsley

Mike CATanzaro?

God what a great name for a cat.

Arpon

2:23 PM

@BenLim yeah i've decided on stanford, and yeah will be at cambridge next year. good luck with your choice (obviously i'm a fan of both of them)

Denis Nardin

@Saul Ok, so is anything known about the space of $E_\infty$-algebra maps $H(k[x])\to HR$ for $R$ a $k$-algebra? I have a feeling that I should know this...

Uh wait I know that: it is the same as the space of $E_\infty$-space maps from $\mathbb{N}$ to $\Omega^\infty R$

Saul Glasman

while we're on this, what's the best reference for the generators and relations way of talking about E_oo-rings over F_p?

is it Mandell's p-adic homotopy theory?

Adeel

yay european talbot

Denis Nardin

There's also something on Lurie's homepage, but I never read it so I cannot vouch for it

Saul Glasman

which thing?

Denis Nardin

2:36 PM

"Rational and p-adic homotopy theory"

Saul Glasman

oh, great

thanks

Denis Nardin

One last question on this stuff: has anyone worked a presentation as an $E_\infty$-ring (in any sense, but generator and relations would be great) for the global sections of $\bigoplus_d O(d)$ over $P^n_k$?

(when I say global sections I obviously mean derived global sections)

Denis Nardin

3:21 PM

Equivalently, what I'm after is a description of the full subcategory of $D(\mathbb{P}^n_k)$ spanned by $\{O(d)\}_{d\in Z}$ as a symmetric monoidal category. I doubt that this is actually easier, but who knows

Jon Beardsley

7:30 PM

Anyone know of a categorical description of "group-like elements" of a Hopf-algebra?

Saul Glasman

I'd like to know that too

Jon Beardsley

I was trying to think of how to construct it, like, just as some kind of limit or something?

like equalizing the diagonal composed with the multiplication and some kind of squareing map

but.... i don't know how to categorically define "squaring" except by diagonal followed by multiplication, hahah.

Saul Glasman

ah, well, it is corepresented by Z[x] with x group-like, right?

Jon Beardsley

i guess that makes sense, yeah.

maps from Z[x] as a bialgebra

or rather, maps of bialgebras out of Z[x]

Denis Nardin

Or if you prefer maps from $Z[x^{\pm1}]$ as an Hopf algebra

That is group-like elements are characters (that is group homomorphism to G_m)

Saul Glasman

7:37 PM

why does one need x^{-1}?

Denis Nardin

Well, if you want the antipode you need it

An Hopf algebra is more than just a bialgebra

Jon Beardsley

Ah, perhaps I should have said maps of coalgebras out of Z[x], with the coalgebra structure given by $x\mapsto x\otimes 1 + 1\otimes x$

Denis Nardin

Aren't those the primitives?

Jon Beardsley

sorry.... right.

I need $\Delta(x)=x\otimes x$ I guess, which is different.

@DenisNardin what's the coalgebra structure on $Z[x^{\pm}]$ making $x$ grouplike?

Denis Nardin

There is this tendency among algebraic topologists to forget to mention the antipode in the definition of Hopf algebra, but that map is actually important!

$\Delta x = x\otimes x$

Jon Beardsley

7:40 PM

Okay.... And, this gives you G_m?

also - for me, I guess the objects i usually end up worrying about don't have antipodes, or at least, i haven't yet found them necessary.

Denis Nardin

Yes, it is actually easy to see that $Hom(Z[x^{\pm 1],R) = R^\times$ with the group operation

The point is that if the antipode exists it is unique, since it's basically the inverse function

Jon Beardsley

gotta close that {

Denis Nardin

But its existence is a nontrivial condition

Jon Beardsley

Right.

wow i always thought the comult on G_m came from x-->x1+1x

whups

Denis Nardin

That's G_a

Jon Beardsley

7:43 PM

I guess that makes sense.

Okay possibly a dumb question - what happens if I take Z[x^{\pm}] but with the additive group structure?

er, additive _co_group structure

does that just not work out?

Denis Nardin

That's not a coalgebra structure

Jon Beardsley

Okay, figures.

Denis Nardin

You are basically taking $G_a\smallsetminus 0$

Jon Beardsley

Right. Okay. Hmmmm....

So what goes wrong? It no longer has an identity then?

Denis Nardin

What is $\Delta x^{-1}$?

Jon Beardsley

7:45 PM

hmmmmm.

Denis Nardin

It should be $(\Delta x)^{-1}$, which does not exists

Jon Beardsley

i mean, i'm not really positing this as some kind of useful object. just trying to see how it works out.

interesting.

Denis Nardin

Basically once you remove the 0 from the additive group the operation is not defined on the whole set of pairs anymore

Jon Beardsley

unless you're working mod 2?

Denis Nardin

Even if you're working mod 2

Jon Beardsley

7:47 PM

oh shit right

b/c no zero

hahaha

very sneaky.

hmmmm okay. well that's good to know. yeah... just think of them as characters

well, too bad i don't know what my spectral G_m is supposed to be yet, lol.

primitives, it seems like, one can relatively easily obtain by cotensoring with the base ring, is that true?

(over the base ring)

Denis Nardin

Yeah they're easy (sorry I forgot what a cotensor is...)

Jon Beardsley

Sorry, yeah, sometimes I think people use cotensor to mean... like... Hom or something? hahaha.

like "this category is cotensored over spectra"

Zhen Lin

cotensor is the dual of tensor, where tensor here does not mean monoidal product

Jon Beardsley

I mean.... I dunno Zhen, you may wanna be careful with the word "is".

Denis Nardin

cotensor is used in different places of the literature as dual of the tensor and as ``like the tensor but in a dual setting''. It's very confusing

user101036

7:53 PM

Do one have examples of derived group schemes in a good sense? considering that @JonBeardsley was hoping to define spectral G_m's

Jon Beardsley

@user101036 i guess it depends on what you mean by that. we have this conversation in here once every few months, haha.

one can talk about bialgebras in E_k-ring spectra

for instance

Denis Nardin

Hopf-algebras! :P

Jon Beardsley

nope

i mean, i guess one CAN get hopf-algebras, but I'm not worried about hopf-algebras.

Denis Nardin

Wait what? Isn't that a spectral monoid scheme?

Jon Beardsley

Oh, well, okay, fine.

I just am not sure how to encode the antipode.

homotopically

user101036

7:55 PM

I suppose presumably one would want a derived G_m to take something to its "derived group of units"

But I'm not sure what that means

Jon Beardsley

Well, there's a functor like that.

Denis Nardin

We do have some sort of derived group of units

Jon Beardsley

that takes a ring spectrum R to GL_1(R)

user101036

So the question comes down to asking whether it is representable?

Jon Beardsley

I s'pose.

I guess I'm not ultimately that concerned with representability, though maybe I should be.

user101036

7:56 PM

are there hints of how such an object should look like?

Denis Nardin

@Jon When you say GL_1(R) what do you mean? The connective one is certainly representable

Jon Beardsley

What's it represented by?

Denis Nardin

S[QS^0]

Jon Beardsley

Oh okay, I didn't know that.

Denis Nardin

The group algebra on the free group on one generator :)

It's just adjunctions nonsense

user101036

7:57 PM

I suppose S here means the sphere spectrum?

Jon Beardsley

Oh. ummmmm, right. okay. \Omega^\infty\Sigma^\infty S^0

Great. Sure.

Denis Nardin

Anyway I think that the nonconnective one is better, but I don't think people know much about it

Jon Beardsley

O.... actually maybe primitives are not such a cotensor.

I mean, that cotensor gives all $x$ that map to $x\otimes 1$

or

Denis Nardin

Everytime I think about cotensors I get confused, but I think primitives should be given by that cotensor, since no $x$ can satisfy that equation apart from 1

Jon Beardsley

crap. ugh sorry.

i'm being stupid

I feel like $H\Box Z=lim(H\otimes Z\rightrightarrows H\otimes H\otimes Z)$

equalizing the diagonal (the H-coaction on H) and the trivial H-coaction on Z given by H's unit.

I think I'm really missing something here... wtf.

Denis Nardin

8:09 PM

Weird. This does give $\Delta x = x\otimes 1$. However the only x satisfying this are the constants, because you can apply $\epsilon \otimes 1$ to both sides and you get $x=\epsilon(x)$

Jon Beardsley

Hmpf.

Denis Nardin

Which is as it should be because that contensor is computing the functions over your group invariant under left translation

Jon Beardsley

I've been calling this object "primitives" for a while.

I guess I should stop.

Denis Nardin

I think that primitives for a comodule and primitives for a coalgebra are different things

Jon Beardsley

I wonder where I got that from.

Ohhhh..... hmmmmm.

Denis Nardin

8:15 PM

I remember that terminology too, so you're not the only one.

Jon Beardsley

okay.

yes.

that's the definition of primitives for a comodule.

(according to the appendix of ravenel's green book)

I guess it's not such a surprise then that one recovers the sphere spectrum as the primitives of MU under the BU coaction.

Well... actually it's still kind of funny, since MU isn't a coalgebra.

Whatever. As long I haven't been saying something REALLY wrong for a long time.

Denis Nardin

No that is surprising. It's like saying that the fixed points of $\Spec MU$ under the action of $\Spec S[BU]$ is $\Spec S$

Jon Beardsley

Fixed points or quotient?

Denis Nardin

Uh... Maybe quotient

Jon Beardsley

I think it's the quotient.

Denis Nardin

8:21 PM

Yeah, sorry I forgot to reverse one arrow

Jon Beardsley

It's confusing, because it's supposed to be the dual of the cofixed points (in the sense of cofixed points of a coaction, haha)

I mean, in a sense.

Kathryn Hess told me yesterday to never dualize just because I'm uncomfortable with coalgebra, haha.

Denis Nardin

What I meant is that $S$ is the ring of functions over $\Spec MU$ equivariant under the $\Spec S[BU]$ action

Jon Beardsley

Right.

Yeah, it's trippy.

Adeel

anyone know whether there is a reference for vector bundles on derived schemes/stacks? locally free sheaves are easy of course, and probably so are vector bundles, but i wonder whether someone has worked out the equivalence in the derived setting already

Jon Beardsley

What would a vector bundle mean? Like, do you mean an actual bundle of vector spaces?

Denis Nardin

8:23 PM

I don't know, but I don't think that the definition of vector bundle is easy

Jon Beardsley

Or something like a locally constant sheaf?

Denis Nardin

It requires you to define $GL_n(S)$ which scares me

Jon Beardsley

Hah.

Man I should just shut up. I don't know anything anymore.

Adeel

so, I would be happy with the simplicial setting for the moment, though I am also interested in the spectral setting

Denis Nardin

Hey, I don't know anything either :)

Jon Beardsley

8:24 PM

Haha. Disagreed.

I have a friend that I went to college with. He's in Utah now doing knot theory, but he told me, after I graduated and got into grad school, that he thought I was a complete idiot because I'd ask the most basic, naive questions in class.

And I think that's still kind of the situation. I just don't seem to be able to recall basic facts very quickly.

Denis Nardin

That's how you know that you're not an idiot. I hate people who don't like when you ask stupid questions

Adeel

if you want my honest opinion, at first I thought the same thing about you, but later I realized that was a wrong impression ;)

Denis Nardin

(@Adeel I assume you want to define vector bundles as derived schemes over your scheme that locally look like the product with affine space plus some conditions on the transition functions)

Jon Beardsley

Haha. Yeah. At Hiro's talk at UIUC this morning, the last talk of the weekend, he remarked that it had been a wonderful conference because he never felt scared to ask what he thought were really stupid questions, since there were so few senior faculty present, hahah.

Somehow Adeel I'm not sure that that opinion is really something I needed to hear.

Adeel

@DenisNardin yes, that's right. and I want some equivalence of categories with locally free sheaves of course

Jon Beardsley

8:28 PM

Besides, I think it'll probably be a while before we find out whether or not I'm an idiot. Gotta get a job first. =P

Denis Nardin

@Jon Trust me, you're not :)

Jon Beardsley

Haha, thanks. :D

Adeel

@Jon sorry, I didn't mean it in a mean way. but I better shut up now

Jon Beardsley

Haha, don't worry about it. I've got thick skin (from years of practiced self-flagellation).

Adeel

@DenisNardin but actually, what I'm more interested in at the moment just a way to associate to a locally free sheaf a derived scheme. That should be easy, right?

you just need a global Sym functor and a relative affine spectrum I guess

Denis Nardin

8:32 PM

Yeah that should be easy. You just need derived spec

Exactly

(possibly you need to dualize before you take the free algebra, depending on your preference)

Adeel

haha right

but so I guess there is no reference for this at present, then

Denis Nardin

Unless it's buried in some part of DAG, I would doubt it

also you need to be careful with what you mean by a "locally free sheaf", is the suspension of a free module free?

Adeel

for me, no, I think...

Denis Nardin

Ok, then you're good :). It is going to be annoying that the Picard group doesn't classify locally free modules of rank 1 anymore though :P

Jon Beardsley

Yeah.... Ugh... graded things.

Denis Nardin

8:37 PM

Everything would be a lot easier if we knew how to take the Spec of nonconnective rings. Until someone comes up with that definition we're stuck with these problems

Jon Beardsley

Haha.

I dunno. No reason to think there IS a good definition.

I tend to think we should just try to solve outstanding problems at this point, and the necessary structure will reveal itself.

Adeel

so, what is the problem with nonconnective guys? is there a decent Zariski/etale topology?

Denis Nardin

I don't know, nor does anyone else as far as I know

Adeel

so backing up, do we have cotangent complexes?

Jon Beardsley

I still wonder about what happens with making your covers "maps which are of effective descent."

Denis Nardin

8:40 PM

@Adeel In what sense? As modules, sure.

Adeel

but then we can define etale morphisms, right

Denis Nardin

There's the whole of deformation theory worked out for derived schemes I think

Yes, we have the notion of étale morphisms I think. But it is very restrictive

I don't really know this stuff, so I'm expecting someone to contradict me anytime now :)

Adeel

i mean, give me a subcanonical Grothendieck topology and I'm ready to start doing geometry (in theory)

Denis Nardin

These are called TAQ-étale maps, and if I recall correctly the set of TAQ-étale covers of a scheme depend only on the $\pi_0$ of the base

Adeel

That is true in the connective setting at least

Denis Nardin

8:44 PM

Yep, and that's why I don't like it

I want thick subcategories of E-modules to be seen from the topology of $Spec E$

Adeel

right, I remember you mentioning that before

Jon Beardsley

Yup.

I don't know that anyone's ever seriously thought about taking an $\infty$-categorical analog of the Bousfield lattice.

And trying to get sheaves of localized categories thereon.

I.e. saying that the topology shouldn't be based so much on the rings, but rather at the various localization functors.

Denis Nardin

That's complicated, because thick subcategories should correspond to arbitrary intersections of quasicompact opens

Jon Beardsley

Hm.

Denis Nardin

And I don't know how to get "true opens"

For example the generic stalk is a localization functor

At least. say in Spec Z

Zhen Lin

9:28 PM

it's a lot easier if you work in the category of schemes and know what the closed immersions are – then the open immersions are their complements, in an appropriate sense

but working with just the categories of quasicoherent sheaves makes it a lot harder, I imagine

Denis Nardin

It's not that I want to work with quasicoherent sheaves. But I don't know what closed immersions are in the derived world, either. (yes, yes you can define them as effective epimorphisms. You still get too few closed subsets)

Zhen Lin

yeah, I don't expect any of that to work in the derived setting

I mean, I already find it disturbing that effective epimorphisms are not actually epimorphisms in general

Adeel

epimorphisms in infinity-categories are so weird :(

Zhen Lin

epimorphisms in stable categories moreso

Jon Beardsley

11:11 PM

Hey y'all, here's a question about torsors: what's the qualitative difference between the two following "torsor conditions" for some topos, a group object $G$, a potential $G$-torsor $P$ and an object $B$ in that topos:
on one hand, we can ask for a relative torsor condition $P\times_B P\simeq P\times _B G$ versus $P\times_B P \simeq P\times G$?

I mean, I guess I have to have a map $G\to B$ in the first case.

But what does this latter condition really mean, I guess, if I assume that $G$ isn't over $B$, but is global.

I suppose if we have the latter condition, we can restrict $G$ to $B$ to get the former condtion?

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