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3:11 AM
@Drew among other things.
@Drew in fact, I think that somehow this mysterious project jack has me working on is connected to all of that. which is what i wanted all along. he's just being very mysterious about it.
I sort of get all the pieces, I just still can't really see his big picture. Or rather, his/Kathryn Hess'/John Rognes'/etc.'s big picture.
But I know for sure that that's a party I'll pay just about anything to get into.
@EricPeterson is this a very recent phenomenon? or are you talking about like, over a long period of time?
 
3:34 AM
over a long period of time, but i'm finally beginning to really notice it
 
4:33 AM
I see. Maybe it's just because you're getting smarter?
Or maybe it's because I haven't asked a question on there in a while.
 
the latter for sure
little homotopy theory these days
 
4:52 AM
:)
what a gloomy name
 
Hehehe.
As if it makes a difference that I move things.
BTW @EricPeterson I bookmarked a conversation from here, do you have any clue how to access it again? It's the one we were having about chromatic stuf.
I'll have to tell the first years at JHU about MO. There are at least one or two interested in topology. Maybe I can get them to stir up some shit.
 
which conversation about chromatic stuff
do i have other conversations here
 
hahah
all the stuff u asked about jack's paper
and then tyler and craig talking about it.
 
oh, i can find that for you
 
oh me too
i just want to understand this bookmarking stuff.
 
5:00 AM
chat.stackexchange.com/transcript/message/11703721#11703721 here. i think that link is bookmarkable
generally something labeled 'permalink' should be bookmarkable
 
no but, the chatroom has a built in bookmark function
 
oh, i hadn't noticed
 
under the room tab
i don't understand it.
i just bookmarked this conversation
but i don't know what that means....
 
ohhhh
can u see the bookmarks? or are they personal?
 
5:04 AM
i can see them, and i could see mine when i had one added
idk if non-operators can bookmark conversations
 
aha. interesting.
well that was fun. i'll have to start bookmarking conversations, for posterity.
oh shit wait a second!
it's Jack Morava night!!
Hm. I don't have anything to talk about though. And I'm pretty tired. Maybe we'll have to postpone.
 
we can appreciate what we see on this arXiv page: arxiv.org/abs/1001.0965
 
I don't even know what that stuff means, haha. Bosons....
 
i read goerss's kozmic galois notes this afternoon
 
oh nice. yeah.
i mean, u know this is based on an actual paper on the arxiv right?
Jesus. "This reformulates the Einstein-Hilbert action as the Lagrangian for a conformally invariant physical theory involving a unimodular pseudometric..."
I'm gonna need Craig's flight of stairs again.
The quotes in the beginning of the sections are pretty amazing though.
That's what's up man. We need to get crackin' on this C* homotopy theory stuff and Alain Connes and so forth.
Not really. Nevermind.
 
5:42 AM
i noticed this when it went up, i met aaron this summer
 
who/what is he?
is he a cyborg?
ohhhhh i met him at talbot like 2 years ago
 
he's not a cyborg
 
yeah, i mean, as far as we know.
 
he also recently chromotoped chromotopy.org/?p=1271
 
Wow. I think I need to talk to this guy again.
 
5:57 AM
looks like an awesome post, i can't wait to read it
 
I need to read that paper of his that he posted.
Like, ASAP. Like, that just jumped to the top of my reading list. Like, that mofo needs to come talk here.
 
what's your stake in infinity-categorical string topology
 
derived koszul duality.
i.e. kozmik galois group
it's sort of what's going on with these hopf-galois extensions
i must enter my stasis pod now. good night.
 
night
 
6:19 AM
why do you keep spelling it kozmik?
 
that's the title of the talk notes math.northwestern.edu/~pgoerss/midwest10/MidW010.pdf , koz for koszul i figure
 
 
8 hours later…
2:47 PM
i see. it's in the title, but otherwise he uses "cosmic". the k-spelling makes it feel very...soviet
 
jack's kind of soviet.
 
3:07 PM
Comments: Concluding unscientific postscript & extended disco remix
 
I know
that's amazing
 
YEAH! hahaha
 
4:01 PM
i got an email from jack a few months ago that ended with: cl.ly/image/042L3F3n1x0g
speaking of soviet and generally kooky
 
H̶o̶m̶o̶t̶o̶p̶y̶ ̶t̶h̶e̶o̶r̶y̶ ̶c̶h̶a̶t̶ ̶r̶o̶o̶m̶ Jack Morava fanclub
 
4:26 PM
hahaha.
Why all those rectangles? I'm confused.
Some day I'll compile all of the hilarious e-mails I've gotten from Jack and put them in a little book.
<----- writing up a lot of examples of the chain rule (from regular calculus) :-(
 
 
2 hours later…
6:47 PM
Just read Aaron Royer's post on chromotopy. Man, what a great post!
 
7:16 PM
it is a nice post
i'm very curious what he has to say about crystalline cohomology
 
 
3 hours later…
10:03 PM
hey @TylerLawson can i ask u a question
oh wait tyler is gone.
dern.
 
Ask anyway, maybe someone else will know the answer.
(Not me though)
 
erg okay. hm. i wonder if i can still formulate it.
 
Is it about elliptic curves?
 
i kind of said that before really knowing what my question was.
no.
it's #bigpicture
 
Nice hashtag
 
10:06 PM
thankz.
i'm just not sure if i'm confused about things or not.
 
You're confused about your confusion?
 
like, given a map of "objects" (either schemes or spaces, maybe ring spectra in RngSpectra^{op} or something...)
$X\to Y$
I guess a map of sites.
then we can talk about descent.
in the correct homotopical generality, descent should be controlled by (I think!) a cosimplicial dude. something like, and well, here's my first point of confusion, either $X\otimes^L_Y X$ or $Y\otimes^L_X Y$
and i guess this depends on your variance. that is, are we using "rings" or "spaces"
 
Or descent vs 'codescent' maybe?
 
@Drew always =P
well. no. one of these objects should be thought of as a "cover"
 
Is your notation derived tensor product or something else?
 
10:09 PM
Yes.
Erm, the former.
 
Ha. I got ya
 
So, I think if we've got a map of rings, $R\to S$, we want to think of $S$ as the cover.
since this should be like the inclusion of a ring of functions on a space into a ring of functions on a larger space.
and the canonical descent coring is, if I'm remembering correctly, $S\otimes_R S$
 
That's what COCTALOS seems to say
 
Yeah, okay that's right.
And, if, e.g. we're looking at the unit map $\mathbb{S}\to MU$, we get the adams spectral sequence as our Desc-sseq
 
So everything really is a descent spectral sequence
 
10:12 PM
Well.
Not everything.
Almost everything.
Lol.
 
Everything else is a Kan extension
 
Hahah. Or THH.
Although I'm pretty sure THH is just about descent anyway.
Because descent and deformation theory are the same thing.
Whoa. Sorry.
Anyway.... back to my original business.
Well, not quite. I need to reference something real quick before I say anything else.
 
hey, how do I put two matrices side-by-side
 
Don't put them on top of eachother, that's for sure.
 
er, I am using tikz fyi
yeah, they are ending up on top of eachother :P
 
10:16 PM
Hey Jon, your tensor product above has to be over $X$ right?
 
You may want to check out the tex chat room?
 
oh shoot
wrong room :P
 
@Drew which one? did i make a mistake?
 
CPM
if you are writing tensor product then it should be algebraic and if you are writing pullback then it is geometric
so I guess Y is an X algebra and it is tensor product over X
 
The one from a map $X \to Y$. I mean $Y$ is an $X$-module but (unless its an implicit assumption) $X$ is not a $Y$ module right?
 
10:18 PM
Oh sure. Yeah. Ummmmm, yes.
I phrased that wrong. I guess I should have written Spec on some of them, or somethign. don't worry about all that.
 
Ok, cool. Sorry, I know less than you on this, so I'm just along for the ride
 
No, I mean, as written, that was just nonsense.
 
CPM
but yeah Hopkins says that for a faithfully flat map R-->S you form that descent hopf algebroid (S,S\otimes_{R} S) and there is an equivalence of categories between comodules over that and the category of descent data
which I found enlightening
 
I guess ultimately my question is about the fact that $MU\wedge^{\mathbb{L}}_{\mathbb{S}}MU$ is actually (I think???) the same as $THH_{\mathbb{S}}(MU)$, i.e. it's simultaneously the thing that controls deformation data "at" the "point" MU and it's the coring which controls descent from MU to $\mathbb{S}$ along the unit.
 
Can I ask a silly question? The smash product is already derived right?
 
10:24 PM
Well, that's a good question, and one that confuses me.
The smash product though, is not homotopy invariant.
unless everything is cofibrant.
Hence we have to "derive" it, by taking a resolution
That's my (naive and ill informed) understanding of the situation.
But yeah, I just kind of put decorations on operators willy nilly. I don't really know what I'm supposed to be doing.
 
Hmm, is that before you pass to the derived category or something?
 
Yeah, the "homotopy" category.
 
Its fine. I was just wondering where you were considering your objects are lying
 
Yeah, but this whole line of thought was triggered by Aaron Royer talking about the mod-p ASS sort of... converging to things in a formal neighborhood of $p$ in Spec(S)
But this bugs me, because, in Lurie's language, the only formal neighborhoods that Spec(S) has are about arithmetic points.
And we all know that's, at least morally, unsatisfying.
 
I haven't read that post. Looks good though
 
10:30 PM
So, if we think about spectra as sheaves over a good notion of Spec(S) (i.e. something that looks like the Bousfield lattice) we can think of the various E ANSS's as telling me about, I'm not sure, something like the value of that sheaf "on" an infinitesimal neighborhood of E "in" Spec(S)
But we can also think of them as being something like.... I dunno, Tor(THH_S(E),X), I think...
lol, that says sex.
 
Haha. Nice
 
@JonBeardsley $MU \wedge_\mathbb{S} MU$ isn't THH; derived coproduct is still coproduct, so it's just $MU \wedge MU$
 
hmmmmmm. ok.
 
here's the thing I think is true: for an $E_\infty$ ring spectrum $E$, $THH(E) \cong E \wedge_{E \wedge E} E$
 
Ohhhh. hmmmm.
 
10:35 PM
with an L above the smash if you like
probably true even for $A_\infty$ if you replace the $E \wedge E$ with an $E \wedge E^{op}$
 
Right. hm.
Yeah. Weird. The THH of R over k is the relative Ext of the map k--->R\otimes R^{op}
Hmmmmmmm.
 
@SaulGlasman Why is that true?
 
the identity for THH?
 
Oh what no that's cool.
Sorry I totally misread what you said
 
In all cases you should use $E\wedge E^{op}$, but if it's commutative, you can ignore the op.
 
10:39 PM
right
but in that case I'm less confident that I know a proof in spectra
 
I see. I don't know proofs of anything. Lol.
I mean, that's just a definition, in my mind.
 
Yes, me too. Isn't it a definition?
 
probably as good a definition as any other
 
It's how they define it in EKMM
 
Hm, Jack makes this really funny statement then in one of his papers.
 
10:40 PM
in any case, when you're $E_\infty$, after taking Spec, what you have is "derived self-intersection of the diagonal"
which is a nice picture
 
But maybe he's just conflating things. I dunno.
Oh maybe Tyler can tell us.
 
Mmm constantly using $ hurts my eyes
What's the dagger symbol?
 
hang on, trying to figure out what the notification was
 
10:43 PM
Haha. Yeah. And then he brings in the daggers, it's nuts.
It was me asking a dopey question.
 
well, I haven't read enough to figure out the question yet
 
why use $ rather than blackboard bold S?
I mean, you'd think that the only reason to use $ is if you're trying to approximate bold S and you don't have tex
 
Yeah, I dunno. It doesn't really bother me, though it confused me at first.
 
so I think there's some confusion about THH
THH (the homology) is built out of this cyclic bar construction, and it's a simplicial object, rather than the cosimplicial one from descent
even though both are built out of smash powers
 
Oh. Yeah. That is definitely a point of confusion.
Derp.
 
10:49 PM
the philosophy you're expressing in terms of S -> E giving some kind of descent for computing a "completion" near E is legitimate
 
Yeah, but has nothing really to do with THH.
 
all things are connected
 
Okay, well then there's this other thing: there is a coring interpretation of formal groups (a la Demazure), where comodules are like formal schemes over a given formal group.
 
it's a little confusing. there's this weird relationship between simplicial and cosimplicial things that makes some of these constructions look the same
 
Shit now I gotta find my Demazure.
Yeah, Demazure gives this equivalence (for a field k) between k-corings and k-formal-schemes....
So, a coring over k "is" a formal scheme, so, in particular, if there is a k-algebra A, then A\otimes_k A is a particular formal scheme over k
but it's ALSO the descent coring associated to descending from A to k?
And Eric tells me that comodules over this coring correspond to formal curves on the formal scheme (oh and also I think algebras in the coring category are formal groups). Anyway, a space X gives me an MU\wedge MU comodule, i.e. a curve on the formal scheme associated to MU over S?
 
10:59 PM
yeah, you should be able to "dualize" the coring picture to convert it into the descent picture for rings
so far as formal curves, this I don't know very well
 
But, the coring picture is formal!
And the descent picture is not.
 
yes, it's formal -- but I'm just using my imagination here
 
=P me too
a descent datum.... is a function on a formal curve?
 
a descent datum should be a comodule
 
Whoops, yes.
 
11:01 PM
we have said 'canonical coring' enough times in here that i'm embarrassed that i still don't know what it is
 
So, just a formal curve.
 
which may be a function on a formal curve under what Eric is saying
 
Oh, it's just a complicated way of talking about the coring $R\otimes_S R$ associated to a ring map $S\to R$.
 
how is that a coring
and not to discourage, but i think (function on a) formal curve means something to me other than 'comodule'
not to be unimaginative, i mean
 
is there a situation in mind where we'd like to use this coring perspective?
 
11:05 PM
$R\otimes_S R\simeq R\otimes_S S\otimes_S R\to R\otimes_S R\otimes_S R\simeq R\otimes_S R\otimes_R R\otimes_S R$
Shit, lol
it's an R-coring
No, there's no specific situation. I just happened to notice that there is a coring associated to descent, but corings are the same thing as formal schemes, in an ideal situation.
 
oh, ok
 
sure
so related: you can do the "Adams-Novikov" cobar thing for S -> X when X isn't necessarily commutative
 
Yeah. Sure. Hm.
I guess, god sorry, maybe I'm way off the rails.
 
and on homotopy groups that should be related to (X, X^X) being a .. er, co-ring-oid
 
11:10 PM
anyway, no need to apologize, all you're interrupting is me struggling to understand commutative monoids
 
Lol.
Well, yeah, like I said it's more #bigpicture than anything specific.
 
i like little pictures
 
Haha.
Microlocal.
 
the ground object for that coring is R, not S (or "is A, not k" using your letters from earlier), which makes the corings <~~~> formal schemes correspondence less perfect. i'll bet there is still something you can say, though; maybe start by writing down what a T-point of the coring is for T an R-algebra and seeing if you find something familiar
 
Ah yeah, that's a good point.
Shit, yeah, that's a really good point.
Hm.
Derp.
Weird. So, $E\wedge_S E$ is an $E$-coring, i.e. a formal scheme over $E$, i.e. a formal neighborhood of an $E$-point in some scheme over $Spec(E)$.
Hm.
That makes me sad.
 
11:20 PM
i won't be sad until you write out the T-points and find nothing
 
Hahah. By T-points you mean, Hom(R\otimes_S R,T)?
I don't think I understand.
I guess I'm not sure what there is to "find"?
(sorry, now I'm being unimaginative)
 
the T-points of a formal scheme presented as a coring C are something else; they're elements of C (x) T (thought of as Hom(C^*, T)) satisfying some properties (thought of as the conditions to be a Hom)
 
Aha, I see.
Hm, yeah, I mean, in that case the T-points are basically the free C-comodule?
Hm, but wait, T is an R-algebra.
So, $R\otimes_S R\otimes_R T\cong R\otimes_S T$
And that's kind of weird. I mean, that's like regarding $T$ as an $S$-module and then bumping it back up to $R$.
i.e. going once around the descent comonad.
Freaky.
Where's my Hess? I need to look at my Hess.
 
"Professor Hess" makes me think of American Astronaut before hopf-galois hess
 
Yeah, an $R$-algebra $T$ is not naturally an $R\otimes_S R$ comodule, I don't think. But let's see, $R\otimes_S T\cong R\otimes_S S\otimes_S T\to R\otimes_S R\otimes_S T$
oh but fuck. that's over the wrong base ring.
wtf. i dunno. screw it. lol.
I have some hideous homologies to compute.
haha I had never heard of this: imdb.com/title/tt0243759
 
11:32 PM
it's pretty nutty and also pretty good
 
weird geometric interpretation of the homology of BSO inside of that of BO
 
What's the story with 'communicated by...' for PAMS papers?
 
Well, at least in the good old days, you had to have someone like, present your work to the Royal Academy or whatever. Maybe that's still the case?
 
It just seems kind of archaic
 
Haha. I used to get archaic and arcane mixed up.
I guess I still do a little bit.
 
11:43 PM
Some journals published by learned societies or national academies require that “communications” be presented (or sponsored) by a member of the society. This was the case, for example, of the PNAS (Proceedings of the National Academy of Sciences) until July 2010; the top of an article looked like this:
 
it marks the editor who was responsible for handling the article
 
I cannot believe that nobody out there has computed the map H_*(BSp)-->H_*(BSO). It must be around here somewhere.
(2-locally)
 
Or rather, in Z/2 coeffs...
Anyway, I'm going to have to do it, it looks like.
 
Pengelley is a name that comes to mind...maybe
 
11:48 PM
Yeah, I have at least one of his papers around here.
Hahaha, I just found low-cost flights from Missoula (MSO) to Minneapolis (MSP)
(and vice-versa)
 
Where is Missoula?
 
i've thought about this some before, i can stumble around and maybe get somewhere
 
I still can't believe jousting is the state sport of Maryland
 
Montana.
Whatever bro, at least it's not didgeridooing.
Relatedly - you know what the national sport of NZ is? I don't think we can say it in public.
 
Ha. Of course I do
Definitely NSFW
 
11:51 PM
Lol.
Do countries have national sports? I can't seem to find one for Australia. I guess we'd probably have to go with footy tho?
The state sport of Alaska is "mushing."
 
Not officially I guess. In Victoria/SA it's definitely Australian rules football
NSW/QLD it would be rugby (league or union)
 
@EricPeterson well, I have a map from H_*(BSp) to H_*(BU) from an old Sem. Cartan.
given in terms of duals of characteristic classes.
Jack has recommended that I work with the homology of SO and Sp instead, since the exterior algebras make things easier, but then how will I ever determine the homology of MSO\wedge_{MSp} MSO from the homology of SO\wedge_{Sp} SO???
that wouldn't work, would it??
 

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