Homotopy Theory - 2013-07-31

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12:05 AM

basic question -- is Z[G]-mod the heart of a natural t-structure on G-Spectra ?

sorry, i totally misread that.

pulled an arnav

@CPM: did that help the other day?

most definitely

thinking about simplicial commutative rings or structured spectra certainly help to get over some of those mild hang ups

I haven't read iyengar yet tho

12:32 AM

how much negative dimensional homotopy can be added to gl_1? there's the picard group, and then beneath the picard group there's the brauer group

where does it end

@ArnavTripathy depends on what you mean by "G-spectra". "naïve" G-spectra yes, "genuine" G-spectra probably not (there's a t-structure with heart the category of Mackey functors)

@eric good question.

12:54 AM

markus updated his brauer groups paper is all

Oh cool. If you have a scheme is there any reason that $H^q(X,\mathbb{G}_m) = 0$ for $q>2$?

Since q=0 is units, q=1 is the Picard group, q=2 is (almost) the Brauer group, I've always wondered what the higher things are meant to represent

1:23 AM

I don't suppose anyone has any experience with TikZ?

1:46 AM

@Drew For a general scheme probably not; I think that's the kind of identity you only expect from Spec of a local field.

Cool, thanks Tyler

2:08 AM

@Drew if X is a variety over C, you can read off these cohomology groups more or less from the long exact sequence associated to the exponential sheaf sequence 0 \to \mathbb{Z} \to \mathcal{O}_X \to \mathcal{O}_X^{\times} \to 1

in particular you can find a bunch of dudes with nonvanishing blah just by looking at a bunch of hodge diamonds

@SeanTilson damn what is an arnav

oh it is deleting comments

gotcha

sean you are a wonderful person and you should never forget that

@TylerLawson sweet thanks

Thanks Arnav - I'm showing my ignorance of algebraic geometry here

oh man you deleted the Cool

It looked funny because that was what I said to Tyler

Hodge diamond, Dieudonne crystal. How cool is maths?

I dunno on a scale of 0 to 10 maybe like a 6

Oh Arnav, I know you love maths

2:16 AM

what yeah I gave it a solid 6

better than average !!

I have this really dumb equivariant h'topy question

if G is a finite group acting on a finite contractible CW complex X, is X/G contractible ?

@ArnavTripathy $EG$ is a contractible CW; $EG/G$ is $BG$

so no

@Drew That's a problem a lot of us seem to have

oh finite. uh..

whoa

yeah even just take G = Z/2

I was unable to settle the question because ... reasons

i wanna know about the simplicial category $\cdots\to Fun(\Delta^n,C)\to\cdots\to Fun(\Delta^2,C)\to Fun(\Delta^1,C)\to C$.

@ArnavTripathy that's pretty closely related to the Sullivan conjecture

roughly, if G is finite acting on finite contractible X, is X equivariantly contractible?

2:31 AM

yeah

so like if I were asking about X^G instead of X/G

I think the question would follow in the affirmative from sullivan ?

this is equivalent to X^H contractible for all H if you're willing to move to weak equivalences

well, it would follow in the affirmative from the generalized sullivan where the group action's allowed to be nontrivial

oh well I mean the G-action to be proper discontinuous or whatever you need so that everything's still nice

the quotient should still be h'topic to CW

yeah

oh whoa I always that was called sullivan

what is that called ?

just ``generalised sullivan'' ?

if it's G-CW you don't need to weak equivalence it

something like that, sometimes still just the Sullivan conjecture

unfortunately, to make it true one generally needs to invoke some p-completion

and assume G is a p-group

2:33 AM

augh I forgot about that

yeah

so I guess there has to be some classically known reason for why you need this p-completion

and that sort of thing can give counterexamples to my question, maybe ?

yeah, there's a counterexample to generalized sullivan without it

hang on

don't have the example handy but Miller was the one who corrected it

don't know if it qualifies as a counterexample to your question, but seems likely

oho awesome

thanks !

it's probably not an example where your space is local away from |G| either, because there the homotopy theory of G-actions is much simpler

no problem

2:50 AM

so there are a lot of ways to square-zero extend a ring $R$ by a module $M$, do each of these, intuitively, constitute a different tangent vector at $R$?

(or at least, there can be a lot of ways, if i recall correctly

)

@ArnavTripathy it is when you delete a bunch of your comments

@PeterNelson are hyperbolic manifolds finite? certain 3 manifolds model EG.

@SeanTilson but then $G$ isn't finite

i think

@PeterNelson I was confused about something. it is that some manifolds are models for BG with G finite.

Almost everybody is here.

@SeanTilson yeah, like every surface of positive genus is $B$ of its $\pi_1$, but again, finiteness

there is something for other manifolds that gives you finite groups, those manifolds might not be compact though.

3:01 AM

i don't know any low dimensional topology either

@Drew i know all kinds of tikz, what's up

I feel like we should make videos like this youtube.com/watch?v=wBCmt_pJTRA

bros:

an F[t]-module is the same thing as an F-module + an endomorphism

WCATSSSSSSSS

what if i want to consider parameterized endomorphisms?

3:14 AM

@MatthewPancia yes

that is, instead of V -> V, i take V -> V \otimes_F M for some F-bimodule M

is there a similar sort of thing i can do?

whoa

where's the endomorphism come from

for an F[t]-module

applying T

you have seen this in linear algebra 101 jon

i think you should take the symmetric F-algebra on M

whoa i don't remember that class

3:15 AM

you have seen this in cats in baskets 101, jon

lol

@EricPeterson why

i sort of remember this

or... is it the tensor algebra

well, yeah, such a thing gives you an endomorphism of the tensor algebra

but not all endomorphisms of the tensor algebra are of this form

3:16 AM

so like, $tm=\phi(m)$ is the point

yes

lol, yeah don't remember this at all, though i'm relatively secure i could figure this out

if i have an operator T: V -> V, then V gets an F[T]-module structure

if my brain worked

(if V is over F)

3:17 AM

oh well sure

that seemstrue

that's all

no i mean an (F[t] = Sym(F))-module is a module w/ an endomorphism, and an TensorAlg(F^{(+)_m})-module is a module w/ several endomorphisms, all of which are composable in any order

seems like there are some thing you'd have to chec

*check

hmm

where t^2 acting used to mean apply t twice, now tuv acting means apply v, then u, then t

3:19 AM

yes, that seems right

hm that seems like a stupid and trivial fact

now that i have thought of it

thanks

well, it's kind of important

what do you mean by parameterized? a family of endomorphisms on V, indexed by M?

i mean exactly what i wrote

alright

3:21 AM

instead of having V -> V (which is like a matrix compatible with the thing defining V), i have one with coefficients in M

he means that a covector M --> F gives an endomorphism V --> V

that is true

@EricPeterson that sort of thing makes sense, but it doesn't seem germane

the issue is the the base is shifting around, so it's not really an endomorphism, nawmean

idk, that's what 'parametrized' means to me, you can pick out a piece of the parametrization which yields a single thing

oh, no, i agree with that

i was talking about the tensor algebra thing you said above

why not, i like it

3:24 AM

i like it too

okay here's a stupid question

what about an F[M]-module

does germane not mean what i think it means

it means relevant

b/c i am v. self-centered

Relevant to a subject under consideration

@JonBeardsley what do you mean

3:29 AM

well it probabyl doesn't make any sense because i don't really understand the question

i was thinking of some kind of group ring dude

like, if you want a family of endomorphisms, (T_m) for m\in M, just.... pick one

or something

yes ok i take back the symmetric algebra suggestion

a bundle of F[t]-modules over M? i dunno.

i'm proving that a square zero extension of a square zero extension of A is a cube zero extension of A. think this is true?

yes.

certainly if x^2 =1 then x^3=0

*x^2 = 0

@PeterNelson i don't think that's quite what i'm looking for

3:35 AM

(K[e]/e^2)[f]/[f^2]

no, that's false

@TylerLawson what i'm saying?

wait did matt just give a counterexample

no i was just writing things

yeah. k[x]/x^4 = is a square zero extension of k[x]/x^2 by itself

hmmmmm yeah this is what i was worried about

it seems like.... an iterated square zero extension should give you only.... power of 2 nilpotent extensions?

is that true?

oh no, even nilpotent extensions

no nvm power of two

crap i think this means the e-mail i just sent to jacob lurie is nonsense

3:39 AM

haha

great sentence

lol

yeah. i'm sorry everyone.

?

Tywon Ronell "Ty" Lawson (born November 3, 1987) is an American professional basketball player who currently plays for the Denver Nuggets of the NBA. Lawson was drafted as the 18th overall pick in the 2009 NBA Draft, and was immediately traded to the Denver Nuggets for a future 1st round pick. Lawson played college basketball for the University of North Carolina from 2006 to 2009. Lawson was named the ACC Player of the Year, the first time a point guard had won the ACC's highest honor since fellow Tar Heel Phil Ford won the award in 1978. Middle school and high school Lawson attended si...

i'm only going to speak in messages i've already typed, by editing them

to reflect its importance

chill out JB

www.hockeyfights.com/players/4731

@TylerLawson if you get a pet hermit crab maybe you should call it Tyler Clawson

3:43 AM

Tyler the Creator

wait a second.... k[x]/x^4 is a square zero ext. of k[x]/x^2?

so k[x]/x^3 is not any kind of extension of k[x]/x^2?

it's also a square zero extension, but by a different module (k)

aha. okay. that's slightly relieving.

ohhhhhh

"by itself"

means, okay

i parsed that completely wrong

i thought you meant, i dunno, "alone" or "lonely" or something

okay. there's something very cool going on here.... i wonder if, in general, an n-nilpotent extension of an m-nilpotent extensions of k is always a... whatever, j-nilpotent extension of k

and i wonder what controls j

hrggg. gotta clean the kitchen. mmmmmm midnight cleaning is great.

(removed)

ohhhhh i get it.... square zero extend a square zero extension of k by a square zero extension of k, you get a 4-zero extension of k

so you always have to have k in mind

4:02 AM

@TylerLawson my new dream is to become wealthy/powerful enough to convince Tyler to become a hockey enforcer.

4:23 AM

@Eric, thanks I ended up just asking on the Tex stackexchange. Turns out it was easy

But do you know how to change the colour of a cell in a (tikz) matrix?

each cell in a matrix is actually a node, and you can embed nodes in nodes, so it should be ok to replace XYZ in your matrix with \node[color=red]{XYZ}

or rather \node[color=red]{XYZ};

(cf. page 172 of the tikz manual)

Yeah, that's a big manual

Oh I see, I want \node[fill=gray]{xyz}. But that leaves some empty white space which looks ugly

tikz is a complicated thing, but it's pretty regular and things are well-named enough to be able to search for reliably

maybe you don't want to be using a matrix at all then?

I've never used it. But the example people have are crazy

or maybe there's an option that controls padding / spacing

4:34 AM

Basically I just want a table, with some arrows between adjacent columns (d1's). It works now, but I was trying to be all fancy and shade some cells

But I can just not be fancy

@Drew , are you going to WCATSS or Midwest?

i think you already told me

@Drew i can get this to work without much trouble cl.ly/image/0T233p1a1431 , but i didn't set it up with a matrix to start, which means it's quite ugly

here's the underlying tikz: gist.github.com/ecpeterson/07971e7ab034adfeb73c . probably not worth it. can you just shade the text in the box instead? that would look better and be easier

5:02 AM

Jon, I'll be at midewest

Eric, that's nice. I'll probably just shade the text, but I'll save your code for future reference

oh i see craig will be speaking

is he talking about something interesting

maybe he's solved the chromatic splitting conjecture while i was busy with this silly programming thing

5:46 AM

Ermm, I'm not sure what he's talking about. But if he has solved the chromatic splitting conjecture, he hasn't told me

6:25 AM

@TylerLawson thanks for the answer, and it is late, so i'll read it carefully tomorrow. about the last paragraph though: i know about the ABGHR construction, and i find it hugely aesthetically pleasing, but i was (and, i guess, am) pretty unconvinced that it would formulate this issue in a computationally accessible way

maybe that was misguided, idk

if it was misguided, and that's both much cleaner and still computationally plain, then maybe it's worth the trouble of writing about it too. not to ask for too much

5 hours later…

11:42 AM

Does Matt Ando have some secret document about Dieudonné crystals?

yes, I'll send you a copy

3 hours later…

2:50 PM

hahaha i really love that last interaction between drew and callan

3:15 PM

wait so isn't a stable spherical bundle over a finite CW complex actually just a sphere bundle?

yes

and its thom spectrum is just a suspension spectrum

i think that's what tyler is saying, in more words?

it's a finite spectrum, at least; it may have been desuspended or twisted in an unfriendly way

ahah. okay.

that is part of what tyler is saying; i still haven't read it, i just got up a bit ago

3:21 PM

okay. yeah, the twisting and desuspending is what i wasn't thinking about right

yes right, because it's not just a trivial bundle. cool.

yeahhhhh i get lost at clutching map

4:02 PM

@Drew also available on the dropbox, under 'unpublished or unshareable'

2 hours later…

6:04 PM

so there should be like, a monoidal structure on nilpotent extensions of a ring R. i think it's related to "joins" of simplicial sets

1 hour later…

Christian Nassau

7:13 PM

@Sean I never really warmed up to the Dyer Lashhof algebra [embarrassed grin], so I don't know how much my programs could help.

I'm sure that if you'd seriously think about writing an Ext program for that algebra there would be something in my code that you could copy.

On the one hand you'd need a good implememntation of the algebra: efficient enumeration of a basis, a good multiplication routine, etc.

On the other hand, what makes my programs fast is a very efficient "trick" to compute minimal resolutions, that uses vanishing lines for sub algebras of the Steenrod algebra. It's quite possible that this trick could also apply to the Dyer Lashhof algebra (but keep in mind my ignorance here.)

I suppose you would be looking at $\mathrm{Ext}(M,N)$ where $M$ and $N$ are still close to being free? My programs are currently better equipped at the opposite extreme...

In any case, I'd probably start by writing an implementation of the algebra in Sage and see how far that gets me. I'm still occasionally working on the Sage interface to my programs. Although it will probably still take another year or two before I have something that I can release to the public...

Aaron Mazel-Gee

7:50 PM

question: so in non-symmetric operads you have the terminal operad Ass, and in symmetric operads you have the terminal operad Comm. but you can consider Ass as a symmetric operad (that i'll call Asssym), too. then, are non-symmetric operads equivalent to symmetric operads over Asssym? i feel like the usual process of making a symmetric operad out of a non-symmetric one should be inverse to taking the fiber over the appropriate sub-thing of Asssym...

8:11 PM

That seems like it is the case, the construction I think is just taking the disjoint union of Sigma_n copies of C(n) where C is a non-symetric operad and then all the operads over Asssym are like this because the action of Sigma_n gives equivalences of connected components and those components lie over Asssym because the map is equivariant

perhaps I am missing a subtlety

8:28 PM

hey @user36938 remember that question of mine you commented on? about cocycles? do you happen to know how to prove that?

I am just sort of trusting Lazard for now, because I spent what seemed like an inordinate amount of time trying to prove it, but I'd love, I dunno, a hint, or something, haha.

this question: mathoverflow.net/questions/137288/lazards-gamma-nf-as-cocycle

8:41 PM

@EricPeterson sorry for leaving it in a half-formed state.

roughly I'd like to say: let G be the based homotopy automorphism monoid of S^n. then your spherical fibration gives you a principal G-bundle P -> X, and the cellular filtration turns into a filtration of Sigma^infty P_+ by free S[G]-modules.

the suspension spectrum of X is the quotient by the action of G (smashing over S[G] with S, with the trivial action), whereas the Thom spectrum is a different quotient (smashing over S[G] with S, using "the natural action of G on spheres).

so you could view the cellular filtrations of X and the Thom spectrum as two specializations of this filtration by S[G]-modules.

I'm not sure how much of this I have the time to raise to even a low bar of MO-quality

Wow, that's awesome. Is it lame to talk about $\infty$-torsors or something? Is that what this business is?

@TylerLawson is G in this case an honest spectrum?　ｈｏｗ　ｗｏｕｌｄ　ｏｎｅ　ｗｒｉｔｅ　ｉｔ？

ｗｈａｔ　ｉｓ　ｇｏｉｎｇ　ｏｎ．　ｗｈａｔ　ｈａｐｐｅｎｅｄ　ｔｏ　ｍｙ　ｔｅｘｔ．

ｗｅｌｌ　ｔｈａｔ＇ｓ　ｓｔｒａｎｇｅ

there we go

8:58 PM

G is a topological monoid. it's probably some kind of infinity-torsor, yes.

A "real" torsor for a monoid always factors through a torsor for the grouplike elements, so you do need to be a little careful.

hm those are closer to the sorts of sentences i was hoping to hear, i'll have to think about it

nat left for germany today, i'm leaving soon, we still haven't finished this wretched paper, and all the other things that come with leaving soon have all suddenly cropped up this past week. i'm distracted too, definitely don't worry too much about beefing up the answer

OK.

maybe it would help to indicate what vague sort of thing i'd like to try to see with this: ignoring the usual construction of CP^n and taking X_1 = CP^1, i would like to try to show 1) that X_n appears as either a quotient complex or maybe a subcomplex of X_n^L =: X_{n+1}, 2) L extends to X_{n+1} in some inductive way, 3) some interpretation about what happens when studying mL instead of L

given enough knowledge of the stable homotopy of spheres, can you define these spaces in a way that make things like james periodicity evident

i realized i was saying 'compute' a bunch and that wasn't super helpful, i mean 'compute' along the lines of shuffling elements of pi_* S

mostly a curiosity anyway

that's all independent of the question, but it informs the answer

seeing the bottom few cells of X(n): also desirable

9:55 PM

haha yeah that'd be nice.

dang ol' X(n)

10:15 PM

dang ol' X(n) i tell you h-what

Exactly.

So I don't think my email to Jacob was complete nonsense.

i have a hard time arguing how good that show was to others

Hahah, yeah, it was fun. I used to watch it all the time, before I even thought critically about TV shows, and I think that's a good indicator to me that it was indeed a quality show.

Just taught my last online Calc III lecture of the summer!

haha i talked to vitaly about his online course

Now I get like, a month to just go nuts with my own stuff. So happy.

Yeah, he likes teaching.

10:18 PM

he described it as 'lecturing into the abyss' b/c essentially no one showed up in person to the lectures

sounds pretty awful

That can be true. I was having a few. But still, they wouldn't interact at all.

And you can't see any faces, or anything. One person asked a question today, which was really nice.

I get these weird moments of self-awareness in which I notice that I'm actually gesturing and being very active physically as I'm speaking, even though nobody can see me.

i do that on the phone, not when just chatting but when trying to uhhh communicate an idea

when lecturing, essentially

Yeah exactly.

10:42 PM

poor vitaly

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