Homotopy Theory

Drew

00:28

Hey Eric, can Ext Chart do linear algebra over $\mathbb{F}_3$?

Eric Peterson

01:06

@Drew at the moment it just works over 2, but there's nothing special about 2. a day's work could make it go over any prime field

Drew

Oh OK. I was going to ask for a pretty picture. Its at the prime 3 though, so no stress!

Craig Westerland

01:31

@Drew, I didn't realise that I talked so much about morality... if it makes you feel any better, I picked up the use of the word "somehow" from Nora Ganter. For instance: "This should all be a fancy version of the Thom isomorphism, somehow," or, following @JonBeardsley, "Presumably, this is somehow a manifestation of Gross-Hopkins duality." Good examples abound.

Jon Beardsley

Hahaha.

Drew

Oh hi Craig! Apparently you should be careful what you say on the internet!

Craig Westerland

:-)

Jon Beardsley

@Drew I guess I'm lucky enough that Jack will definitely not ever get on here.

Craig Westerland

Unless he's secretly @EricPeterson. Has anyone ever seen the two in the same place, at the same time?

Jon Beardsley

01:36

Hm. No, but Eric had a lot of questions about things that Jack has written last night, so it seems unlikely.

Drew

No. But Craig, I've also never seen you and him in the same place, at the same time. Or you Jon (but then you'd be supervising yourself...)

Jon Beardsley

Hahah!

Hmmmmmm. Okay, so, there are enough people in here right now that it's ridiculous not to talk about some awesome math.

Who wants to get us started?

@Tyler?

Jeez, yeah, I don't think we've ever had Craig, Tyler and Nat in here all at once before.

Drew

Minnesota party!

Jon Beardsley

That's in Minneapolis right?

Craig Westerland

ha!

Drew

01:43

Have to check Google maps first Jon

Jon Beardsley

@EricPeterson Perhaps I should point out Eric's recent interesting remarks regarding Jack's note on cobordism.

Craig Westerland

Can you reference either for those of us who weren't present?

Jon Beardsley

Hmmm, shit I thought I was somehow linking to it. Ummmmm. Okay what about this.

Okay, yeah, click on the grey arrow next to my comment.

On the left side.

Tyler Lawson

What kind of awesome math are we looking for?

Drew

Or starred now anyway

Jon Beardsley

01:46

arxiv.org/pdf/0707.3216v1.pdf

Is Jack's paper.

I dunno, something chromatic I guess.

Tyler Lawson

Hrm. I remember the quotes at the beginning, but not the contents of this paper.

Jon Beardsley

Eric was referring to Jack's comments at the bottom of page 9.

Craig Westerland

Not the cats in the dark quote?

Jon Beardsley

However, I think it's related to arxiv.org/pdf/0707.3216v1.pdf as well.

Tyler Lawson

Hrm. That's a good question.

...Isn't that the same paper?

Jon Beardsley

01:49

uh

what have i done.

arxiv.org/pdf/0908.3124.pdf

There we go.

Tyler Lawson

Ah. I kind of like the idea that Jack's paper is somehow self-referential.

Jon Beardsley

Hahaha!

Yeah, it cites itself.

I dunno though, like.... I think we need to make some sense of this Tannakian formalism/ etale homotopy type of the category of spectra POV

Where HF_p is a fiber functor, so its automorphisms (the Steenrod algebra!) give us a sort of homotopy type of the category of spectra, or something.

Tyler Lawson

...I guess Jack's comment on page 9 somehow brings to mind Galois groups of local fields.

Jon Beardsley

Yeah, well, I think that's a good thing to have brought to mind.

In what way?

Tyler Lawson

If you've got a complete DVR with fraction field K and residue field k, then the absolute Galois group of k is a quotient of the absolute Galois group of K.

So there's some kind of "specialization" map from the galois group of the larger field to that of the smaller.

Jon Beardsley

01:52

hmmmm.

Tyler Lawson

This kind of happens for topological spaces too -- let's see if I can trigger that.

Anyway. His comment seems suggestive of the idea of "specialization" maps from one Morava stabilizer to the next (or, at least, from a subgroup of it).

I'm not sure if you can actually make that go.

Jon Beardsley

Right.

Tyler Lawson

ah, right:

Jon Beardsley

right

Tyler Lawson

if X is a topological space, Z < X is closed and is a deformation retract of a neighborhood V, and U is the complement of Z

then there are maps of fundamental groups pi_1(X) <- pi_1(U cap V) -> pi_1(Z)

Jon Beardsley

01:57

hmmm.

It seems to me that we'd be better off working with E_n or E(n) than K(n)

If we want to specialize.

Tyler Lawson

Very possible.

Jon Beardsley

Oh, but, I guess in the analogy E_n is like the DVR

Tyler Lawson

But somehow what Jack wrote has him taking O and modding out by p

Jon Beardsley

YEah.

Tyler Lawson

that's like taking your FGL and modding out by the multiplication-by-p endomorphism

Jon Beardsley

01:59

Yeah.

Tyler Lawson

which restricts you to the action on the p-torsion part, i.e. E^*[[x]]/[p](x)

Jon Beardsley

Yeah, I don't really get why he wants to mod out by p.

Tyler Lawson

modding out by the p-series progressively starts with E_n and produces K(n)

Jon Beardsley

Hrm.

Okay. Sure.

Tyler Lawson

Well, what he says isn't really true if you don't.

There aren't really natural maps between adjacent Morava stabilizers otherwise.

Craig Westerland

02:01

There isn't really anything special about the multiplicative group here, though, right? Isn't the same construction going to always give my a map from the n'th Morava stabiliser algebra to truncated polynomials in F_{p^n}?

Jon Beardsley

Yeah. I mean, there's not really anything corresponding to E_\infty is there?

(i.e. the infty'th Morava E-theory)

Tyler Lawson

Yes, that's true.

No, there's no fancy power operation stuff here, this is just algebra on M_{fgl} so far.

Oh. Sorry, I'm incapable of reading E_infty without interpreting it a specific way.

Jon Beardsley

So somehow, if we want to end up over HF_p, which is this sort of infinite codimension closed subset of M_{fg}, we're going to have to mod out by p at some point.

Tyler Lawson

No, I don't think there's an infty'th E-theory.

Jon Beardsley

Yeah sorry, after writing that I realized it was really unclear.

lol

Tyler Lawson

02:02

Yes; and we're also going to have to mod by v_1, and v_2, and all the other coefficients of the p-series.

Jon Beardsley

Yesssss.

Tyler Lawson

@Craig: Right, this doesn't have anything to do with the multiplicative group.

In fact, that's what's confusing me: he's taking a limit of the Morava stabilizer algebras in a way that seems illegitimate.

Craig Westerland

We should be able to explicitly compute the units in F_q[F]/F^n, right?

let's see

F_q^{\times} = \mu_{q-1} if n=1

Tyler Lawson

But it also seems to be taking something close to classical (the Morava K-theories of a space only depend on their cohomologies beyong a certain stage) and asserting it as some kind of compatibility among actions of the Morava stabilizers for coherent sheaves on M_fgl.

Jon Beardsley

Right.

Tyler Lawson

02:05

Yeah, the units are just those with nonzero constant coefficient.

Craig Westerland

no, I don't think so.

for instance, F is a zero divisor

ack

i'm an idiot

you're right

Tyler Lawson

I think you were about to calculate the units the same way anyway and I interrupted you

Jon Beardsley

but wait, are we talking about F_q[[F]]/F^n or F_q[F]/F^n

Craig Westerland

same thing

Jon Beardsley

aha

of course.

lol

derrrrr

Craig Westerland

02:07

@TylerLawson: yup

Uh, so wait, is that then just going to give me F_{q^n}^\times = \mu_{q^n-1}?

Jon Beardsley

Yeah, jeez, so at each point we're basically adding a Milnor Q_i

Tyler Lawson

yep

which is cyclic

Craig Westerland

Yay characters

Jon Beardsley

(is that the group scheme?)

(\mu_{q^n-1})

Craig Westerland

I really just meant that the invertible elements in the ring F_{q^n} are the q^n-1st roots of unity

Jon Beardsley

02:09

Is he like.... talking about building the Steenrod algebra piece by piece? Is that some part of this? Or is that just really obvious?

OH, ok.

Tyler Lawson

\mu_{q^n-1} is a way to write roots of unity for those of us who have had so much category theory that we are uncomfortable identifying them with Z/(q^n-1)

Craig Westerland

and that's the cyclic group of order q^n-1

Hooray!

Jon Beardsley

Right, okaycool.

Craig Westerland

Can those of us who were at the Northwestern conference have a moment of enjoyment for Ravenel's distinction between C_2, Z/2 and F_2?

Drew

(But at least \mu_{p^n} is a group scheme anyway though right?)

Craig Westerland

02:10

actually, you didn't need to be there to enjoy that

Jon Beardsley

Well, I think it's what AGers use to denote basically the same thing, as a group scheme.

Tyler Lawson

yeah

Jon Beardsley

And like, we've got both honest groups and group schemes running around here, and I wanted to make sure I wasn't cornfused.

Tyler Lawson

in this case, it happens to (locally) be isomorphic to a constant group scheme

Jon Beardsley

Haha, sure.

Tyler Lawson

02:11

which is cyclic of order q^n-1

Drew

Phew

Craig Westerland

So Jack is telling us how to get a character from the Morava stabiliser group to the cycklic group of order q^n-1

Jon Beardsley

right.

Okay. But what's this equivariant sheaf then? I mean, what's it equivariant with respect to?

Tyler Lawson

right. so Jack appears to be telling us that for large n these are all the same

Jon Beardsley

Yes.

OH. Derrrrr it's MU(X)

Erm. BP. Or whatever.

Tyler Lawson

02:13

or, sorry: if you take a sheaf on M_fgl and look at the associated characters on its fibers, then for large n they are all the same.

Craig Westerland

@TylerLawson, by "these", do you mean for different fgl's?

I see, yess.

Tyler Lawson

I'm trying to figure out if this is obvious.

Jon Beardsley

Is there a different way to talk about the "associated characters on its fibers"?

Tyler Lawson

You have this filtration of BP by these ring spectra P(n) with homotopy Z/p[v_n, v_{n+1}, ...]

Jon Beardsley

Right.

Like, I really suck at representation theory.

Craig Westerland

02:15

interjection: let's see, if n=1, this is just the fact that Z_p^\times has \mu_{p-1} as a quotient

Tyler Lawson

and it maps to all the K(m) for n > m. It seems to me like the associated characters on the K(m) might be detected on P(n)

Jon Beardsley

can you unwind that statement?

like... what would it mean for the characters to be "detected"?

Tyler Lawson

Good question!

I think I mean the following:

We have maps P(n)_* P(n) -> K(m)_* K(m) for all m >= n.

Jon Beardsley

Yup.

Tyler Lawson

That maps hits the bottom few "Milnor primitives" Q^i, and it also hits a bunch of the Morava stabilizer algebra.

I guess the claim would be that for X finite, and n large...

Jon Beardsley

02:19

YEahhhhhh, so I was wrong, obviously we don't want to use E_n, we want to use the complement of E_n, P(n), as you say.

Craig Westerland

@Drew: is it a familiar fact that G_2 (say over F_3, since that's your favourite prime), admits a map to F_9^\times, the cyclic group of order 8? I s'pose that if we're keeping with what Jack's written here, we should be looking at G_2 over F_9, and finding a map to F_{81}^\times = Z/80...

Tyler Lawson

we have K(m)_* X = K(m)_* tensored with P(n)_*X over P(n)_*

and so P(n)_*X determines K(m)_* X, and moreover some of the "Morava stabilizer" action from K(m)_* K(m) is determined by the action of whatever lifts to P(n)_* P(n).

Jon Beardsley

hm

Tyler Lawson

And perhaps the same statements lift to M_fgl.

(wave hands)

Jon Beardsley

Hahaha,

Drew

02:23

Craig, that doesn't appear obvious. But then I've only ever read about maximal finite subgroups. Then at n=2,p=3 there are only 2 conjugacy classes of max finite subgroups; C_2 and the (non-trivial) semi-direct product of C_3 and C_4

(That's for S_2 anyway)

Tyler Lawson

hrm. that's somehow unsatisfying, but it seems in the vein of what Jack is saying ... that there's some "master" information behind the scenes that governs both the Morava stabilizer action mod-p and the Steenrod coaction.

Jon Beardsley

Tyler, possibly related: have you ever thought about a spectral sequence coming from a sequence of generalized Moore spectra: S-->M(p^i)-->M(p^{i},v_1^{j})-->...

Tyler Lawson

Hrm.

So the layers there are also suspensions of Moore spectra

and the target would be the homotopy of the colimit of these Moore spectra. right?

Jon Beardsley

hmmmm, yeah, okay.... maybe that's not quite what i want. those layers don't sound right.

okay. back to our regular programming

Craig Westerland

Huh, maybe this C_8 is the (not so semi-)direct product of the Galois C_2 with the C_4 in the maximal finite. But that only works if that C_4 splits off of S_2...

Drew

02:29

So I think the Galois C_2 and the C_4 give you Q_8

Craig Westerland

right

Jon Beardsley

Whoa guys. This isn't the group theory chat room.

Tyler Lawson

@Craig: so any quotient of G_2 that's prime-to-3 factors through the quotient by the ideal generated by F, which is (I think!) GL_2(Z/3)

Drew

ALGEBRAIC topology right Jon?

Eric Peterson

@JonBeardsley are you sure

Jon Beardsley

02:30

No. I'm not sure of anything anymore.

We'll take what we can get.

Drew

What is F? I wasn't paying enough attention

Jon Beardsley

I think it's the Frobenius?.

Craig Westerland

Yep

Jon Beardsley

If we're still talking about the Morava stabilizer algebra.

Craig Westerland

it's what Ravenel calls S

Tyler Lawson

02:31

yeah, it's the element that is sometimes called S in the division algebra

too slow

Drew

Ah OK. Good old S

Craig Westerland

go @TylerLawson

Drew

Jon, maybe you just need to convince Jack to come on here!

Tyler Lawson

damn group theory. is C_8 a quotient of GL_2(Z/3)?

Craig Westerland

I think it's a field of def'n issue

GL_2(F_9) will act on F_9[F]/F^2

Jon Beardsley

02:33

Haha. He might, if I tell him Tyler and Craig are on here wilin' out.

But I already tried to get him to get a twitter acct.

Craig Westerland

= F_9 \oplus F_9

Jon Beardsley

He wasn't having it.

Craig Westerland

omigod, Jack on twitter, best thing ever

Jon Beardsley

I KNO!

But, yeah, he refused. I think he worries about having even less time than he already has.

Tyler Lawson

OH. Damn it. Yes, I'm making mistakes.

Craig Westerland

02:35

So: this group is not \mu_8; I think it's a subgroup of GL_2(F_9) of order 8

Tyler Lawson

So the quotient of S_2 by the 3-sylow is F_9^x, which "is Z/8" generated by an 8th root of unity in F_9

and in G_2 you get a semidirect product of this Z/8 by Z/2

Jon Beardsley

Wait.... am i being stupid. Is the composition $M(p^{i_0})\to M(p^{i_1},v_1^{i_1})\to M(p^{i_0},v_1^{i_1},v_2^{i_2})$, where the maps are the cofibers, a trivial composition?

Tyler Lawson

nope, it's not trivial

Jon Beardsley

Okay, good.

Tyler Lawson

craig, now I'm confused. where did GL_2(F_9) come in?

Jon Beardsley

02:37

@Drew your buddy travis is making trades like a wild man in fantasy

or attempting to

Drew

@JonBeardsley, I think he is losing pretty badly

Jon Beardsley

Yeah. He's offered me one. I dunno yet.

He really wants Giovanni Bernard.

Is there any meaningful interpretation of K(n)_\ast(K(n)) as THH?

Tyler Lawson

Do you want it to be THH?

Jon Beardsley

I dunno.

I don't know what I want anymore.

Craig Westerland

Doesn't (F_9[F] / F^2)^\times act on the rank 2 F_9 vector space F_9[F] / F^2? This gives a map to GL_2(F_9)...

Tyler Lawson

02:40

Ah, yes. Good point.

I think that's even upper triangular if you give it the basis {1,F}.

Craig Westerland

rightyo

Jon Beardsley

Jack's given me a thesis project that's very geometrical. Which is cool. But I'm kind of bummed I'm getting away from chromatic stuff.

(Hence not knowing what I want, nor being sure of anything)

Drew

Just find some link to chromatic homotopy theory!

Tyler Lawson

Define "geometrical".

Jon Beardsley

Ahaha, well it's linked to "odd formal group laws" which are linked to MU, sort of.

Looking at some connections between MSp, MU, MSO, Hopf-Galois extensions, stuff like that.

Well, perhaps "stuff like that" doesn't really help explain anything.

Tyler Lawson

02:44

AH. Err, I might contend that this sounds relatively chromatic to me. ;)

Jon Beardsley

Haha. That's true. I just miss out on all the Morava E-theory fun.

Nat Stapleton

:)

Tyler Lawson

E-theory even only looks at one chromatic layer, you're doing something MORE chromatic

Jon Beardsley

Lol.

Tyler Lawson

though, of course, E-theory is fun too

Jon Beardsley

02:48

I'm really rackin' up the stars tonight.

Wednesday nights should be Jack Morava night in the homotopy chat room.

Okay.... that's enough.

You guys are lucky I like seeing my name on the internet, otherwise I'd be deleting all those stars.

Maybe I'll just send the transcript of this chatroom along with all my job applications.

Eric Peterson

publications, MO rep, star count

Jon Beardsley

Yeah.

It'd be more valuable I think if I could say who the stars were from.

Like, 10 stars from a tenured researcher = 1 recommendation letter.

@NatStapleton where are you going to get a job!??

Please for the love of christ say Johns Hopkins.

Eric Peterson

so what's the consensus on page 9

Jon Beardsley

A resounding maybe.

Tyler Lawson

it leads into page 10

Craig Westerland

03:00

I think that it gives us a homomorphism from G_n to GL_n(F_q) whose image has order at most q^n-1

Eric Peterson

i had some other questions in the brick from last night that weren't literally on page 9, like: On / p is like End (Ga taken to order p^n), which sounds like the story involving A(n)

Jon Beardsley

(Presumably)

Eric Peterson

can that be made solid

being of height n means you have a logarithm defined beneath order p^n, probably not an accident

hm i've never actually thought about P(n), maybe i should learn some basic facts

Tyler Lawson

unfortunately P(n) isn't commutative at p=2

Eric Peterson

well that's what you get when you go quotienting things

Jon Beardsley

03:09

Yeah, don't mess with stuff. Just let it be. Just accept what you've got.

Craig Westerland

Two things: first O_n/ p is an algebra, and so acts on itself, giving a map O(n) / p ---> End(O(n)/p). But second: I think the fact that O_n/p is rank n over F_q, and not rank n^2 (the rank of End(O(n)/p) means that this map only realises O(n)/p as a subalgebra of End(O(n)/p) = Mat_n(F_q), and not the whole thing (unless n=1)

Jon Beardsley

Well that was a good chat. Now I have to grade like 50 Calc I homeworks.

Craig Westerland

Got a staircase handy?

Jon Beardsley

I live on the 13th floor. I think these screens come out pretty easily.

(If I misunderstood your statement, that comment probably seems really strange)

Eric Peterson

there's a joke about grading papers by giving As to all the ones that land on the first step, Bs to all the ones that land on the second step, e t c

Craig Westerland

03:13

I meant something of this persuasion: joannejacobs.com/2009/12/grading-exams-the-staircase-method

Eric Peterson

your students all get Fs, i guess

Jon Beardsley

Oh. I thought I was throwing myself down the staircase.

Eric Peterson

i'm assisting with Discrete Mathematics this semester, which has turned out to be a small mess

Jon Beardsley

@EricPeterson hahaha. that was a pretty entertaining interchange.

I apologize for my dark humor.

@Drew that's "humour," in English.

Drew

Thanks Jon.

Eric Peterson

03:16

we covered e.g. cantor's diagonalization proof, where students were free to use things like decimal representations of real numbers without any discussion, and now today they had some homework problem where the professor got real fussy about whether or not it was ok for them to say 'pair' without being precise about what they meant

Jon Beardsley

As in an element of the direct sum or something?

That sounds pretty rough.

Eric Peterson

something else; they were supposed to check that the mod p residue classes in {2, 3, ..., p-2} all pair up with their multiplicative inverses (which are distinct from themselves and from each others' inverses)

literally that the equivalence classes in the closure of a ~ b := 'b = a^-1 mod p' are all of size 2

Jon Beardsley

Aha. I see. Right, so nobody can say the word "isomorphism" or something.

Or bijection. Or whatever.

Eric Peterson

idk, it's unclear what they can say

Jon Beardsley

Hahah.

Eric Peterson

03:19

that's the frustration

Jon Beardsley

Yeah, that's terrible.

We've got a ripe batch ourselves. Very argumentative, yet very poorly informed about high school algebra.

Craig Westerland

Undergrad set theory is kind of like learning to drive a car with both hands tied behind your back; it's hard to figure out when you're going to allow yourself to use a knee or not to steer.

Jon Beardsley

Haha! Yeah.

Eric Peterson

i imagine it'll get progressively better as the semester goes on and what they're allowed to take for granted gets larger and larger

Jon Beardsley

I dunno. My undergrad set theory was really good. Like, we started with functions, sets, bijections, equivalence classes, and everything was simple, but really rigorous.

And we just did tons of basic proofs.

It was actually a lot of fun (this was back when math was easy)

It bums me out that we don't teach anything like it at JHU. Like, no set theory, no basic logic, you have to teach math without really ever saying what a proof is. It's rough.

Well, lower level math.

Hi @PaulFabel!

Jon Beardsley

04:42

My students are really having a hard time with the intermediate value theorem. :-(

Jon Beardsley

14:56

→ 10 messages moved to The Graveyard

→ 1 message moved to The Graveyard

→ 2 messages moved to The Graveyard

Hmmmm.

Aaron Mazel-Gee

gaaaaaah. mike hopkins speaking on brown--gitler spectra and E_2 algebras, or sutherland speaking on zhang's progress on twin primes and the subsequent polymath project?

Jon Beardsley

hopkins

Besides, what a delightful dilemma.

Drew

15:22

Also vote for Hopkins. I liked a story Tobi told at Detroit about Hopkins giving a 60 minute talk, and then not getting to his conjecture. And then asking for more time, speaking for another 20 minutes, and getting distracted and still never actually getting to the conjecture

Christian Nassau

well, you can read about twin primes in the new york times, but brown gitler spectra are more rare; not to mention that hopkin's talks are usually superb - even if he runs out of time

what conference is that, by the way?

Aaron Mazel-Gee

haha he'll be speaking for 2 hours, so he better not have more to say than that

oh, these are just coincident seminars that are both happening today

Christian Nassau

ah, I see

I thought it was a pity that they didn't give Hopkins an extra 2 hours at the Simons conference

he had so many slides - but the audience was loszt

Jon Beardsley

btw @CraigWesterland, @TylerLawson, @EricPeterson @Drew: Jack says: "I'll try to write up some explanation re the p 9 stuff but I may not have
a chance to get at it till later this weekend.

If X is a finite complex then for some n >> 0, the AH sseq's for K(n)^*(X) will degenerate, so all the higher K(n)'s will look like H^*(X) (with suitably funny gradings). So the question arises, how are their S_n - actions related to the Steenrod actions on H^*(X)? That's really what
I was alluding to..."

Tyler Lawson

Great!

Eric Peterson

16:14

@JonBeardsley wonderful

Jon Beardsley

23:29

so, is there a fibration $U/SO\to BSO\to BU$? anyone know anything about it?

Eric Peterson

there is, and not me

Jon Beardsley

hah. alright. yeah. looks like I'm diving into some serious French mathematics from the 60's again.

thanks tho. just wanted to make sure i was actually thinking about the right fibration

Transcript for