Homotopy Theory

Jon Beardsley

3:00 AM

soooooo, groupoids which have the same pi_0?

David Roberts

Transitive groupoids are those such that all objects are iso.

Or, their pi_0 = 0

Jon Beardsley

oh i see what you're saying

David Roberts

or 1, depending on how you count

Howbeit in the internal case we say that a certain map has local sections rather than is just onto (which in a geometric setting makes no sense - awful maps can be onto)

Jon Beardsley

so, the kernel

David Roberts

Not sure what you mean

there

Jon Beardsley

3:01 AM

nvm

got it

David Roberts

Sorry, a better way to say it is this:

A groupoid in sets is transitive iff the map X_1 -> X_0 \times X_0 is onto

In the geometric setting we say this map has local sections

Evan Jenkins

So locally it's a split epimorphism?

Jon Beardsley

okay that makes sense

at least, i get the surjective statement

David Roberts

Yes, but locally may be in some Grothendieck topology (e.g. etale)

Evan Jenkins

Right

David Roberts

3:03 AM

OK, so now we have a groupoid X with a functor to X_0/X_1 =: pi_0(X)

Jon Beardsley

really, the point is, $\pi_0=0$

David Roberts

No, only for the fibres of the projection map

Evan Jenkins

What exactly do you mean by X_0 / X_1?

Jon Beardsley

okay.... sorry. sigh, i think i'm mixed up

oh i get what you're saying

just parsing your language wrong

David Roberts

X_0/X_1 is the coequaliser of the source and target maps X_1 -> X_0, equivalently the coarse moduli space, equivalently the space of iso classes of the groupoid etc

Jon Beardsley

3:05 AM

yes, absolutely, the fibers of that functor have trivial $pi_0$

Evan Jenkins

Ah... Does that necessarily exist?

Or are you assuming it exists?

David Roberts

In spaces, yes, for schemes, let's take the source and target maps to be smooth, or possibly etale

Jon Beardsley

i was thinking you were taking the functor $F:Groupoids\to Sets$ or something

but i get it now

David Roberts

And perhaps we want (s,t) : X_1 -> X_0 \times X_0 separated. But that is icing on the cake

Evan Jenkins

Icing is delicious.

David Roberts

3:07 AM

Alright, now we change perspective slightly, and work in spaces over X_0/X_1

Jon Beardsley

alright

David Roberts

So our groupoid has maps X_1 -> X_0/X_1 and X_0 -> X_0/X_1 compatible with all the groupoid structure

(commutes with source/target/identities/composition)

Jon Beardsley

alright, yeah, that makes sense

so they're spaces over X_0/X_1

David Roberts

Yes.

Let's let M := X_0/X_1

Jon Beardsley

k

David Roberts

3:09 AM

OK. THEN, we want to consider the map X_1 -> X_0 \times_M X_0

one moment

Jon Beardsley

the one that factors thru just regular X_0\times X_0

David Roberts

The other way around.

Jon Beardsley

ahah right

David Roberts

This is the 'relative to M' version of X_1 -> X_0^2

and NOW this is surjective, or rather, it has local sections

Jon Beardsley

so, yeah, just, coproduct in $M$-spaces

David Roberts

3:11 AM

product

Jon Beardsley

jeez sorry right

David Roberts

Alright. That's the set up

Jon Beardsley

lol ok

David Roberts

We say that X -> M is an A-bundle gerbe if X_1 -> X_0 \times_M X_0 is a principal A-bundle

Jon Beardsley

aha okay

David Roberts

3:12 AM

using the Grothendieck topology we have lying around

Jon Beardsley

sure

David Roberts

Or A-torsor or whathaveyou

Jon Beardsley

yeah, i had a feeling they were kind of like principle G-bundles in some complicated way

i mean, they are

just, over specific things

David Roberts

Another way to see this is that the groupoid X has an action by the 2-group A => 1

with quotient M

One usually starts with a space/scheme M and then takes X such that X_0/X_1 = M, though

Jon Beardsley

alright

cool!

David Roberts

3:14 AM

Note that A here must be abelian

Jon Beardsley

why?

David Roberts

For the same reason that pi_2 is

Will

because you can't deloop nonabelian groups twice

David Roberts

yep

Jon Beardsley

alright. right on. thanks for the crash course

David Roberts

3:16 AM

So there are some nice ways to generalise this just a little bit, without even going to nonabelian things, but that's probably enough for now

Jon Beardsley

i guess, intuitively, how do you think of such things?

Will

so what about Eric's question?

Jon Beardsley

what do they do?

David Roberts

The trick is, as usual, what it means for two of these to be equivalent

Will

for me the fact that you can't deloop GLn/U(n) more than once for n > 1 is a total show stopper

David Roberts

3:17 AM

What you need to do is take more general 2-groups, rather than just trying to deloop a nonabelian group more times.

Will

sure

but what 2-group has the right to replace a twice delooped U(n)?

David Roberts

Giraud got stuck at this point, because he defined nonabelian gerbes using the crossed module (same thing as a 2-group) Ad: G -> Aut(G)

And nothing was functorial in homomorphisms G-> H

Ah, so here's the key example from higher gauge theory

Eric

omg, i have to leave and now you start talking about this

Will

transcripts, bro

David Roberts

Consider the group Omega G of smooth loops in a compact simple simply-connected Lie group G

Email me, @Eric

Omega G has a canonical extension hat(Omega G) by U(1)

It turns out this is 2-connected (just to keep the homotopy theory up)

In fact the 2-connected cover of Omega G.

I've never seen this fact mentioned in print before

Then recall we have the space of connections on the trivial G-bundle on a circle A_G

This can be identified with a certain space of smooth paths [0,1] -> G (periodic derivatives and something like that), which send 0 to e \in G

Evan Jenkins

3:22 AM

Sorry, catching up, is A here the same as $\pi_1(X)$?

David Roberts

Sorry, getting ahead of myself

Not quite - changed gears completely.

Evan Jenkins

I'm confusing letters.

David Roberts

I'm describing a 2-group that we can consider gerbes/2-bundles for

Evan Jenkins

I'm talking about way back.

David Roberts

Ah, that was X

Yes, A = pi_1(X) = Aut(x) for any x \in X_0

For the ordinary bundle gerbe, all the automorphism groups are abstractly isomorphic, but the groupoid X is not transitive

I'll push on, but do stop me if need be

Now Omega G sits inside A_G, as periodic paths certainly have periodic derivatives

Evan Jenkins

3:26 AM

Is anybody else still here?

Will

yes

Evan Jenkins

Just checking.

David Roberts

Glad I'm not talking to myself (and let's not start that again)

We then consider the homomorphism hat(Omega G) -> Omega G -> A_G

There is a conjugation action of A_G on Omega_G

And this lifts to a 'conjugation' action on hat(Omega G)

And this makes hat(Omega G) -> A_G a very interesting crossed module.

Will

I'm not too familiar with crossed modules

David Roberts

As it represents the next step in the Whitehead tower of G

Will

3:29 AM

but with this you've constructed a 2-group, right?

David Roberts

yes

We take the groupoid of hat(Omega G) acting on A_G via the given map

Then the action of the latter on the former means we have a monoidal groupoid

which is a 2-group

I'm afraid I don't know any interesting examples of 2-group schemes

Aside from the obvious ones

Will

so wait

what is the Whitehead tower of G?

David Roberts

Ah, going back a step, if we assume G = Spin(n)

Will

then this is String?

David Roberts

Then we get Spin(n) -> SO(n) -> O(n)

Yes

And String_G for any old 1-connected G

Evan Jenkins

3:32 AM

I heard a talk by Christoph Wockel once that sounded a lot like this.

Is this related?

David Roberts

Yes, he's all over this

Evan Jenkins

OK

David Roberts

It certainly is. It's the Baez-Crans-Schreiber-Stevenson model

Evan Jenkins

I'm peeking back at my notes: math.uchicago.edu/~ejenkins/notes/nwtft/cw.pdf

Will

I have a related silly question

is there a sensible way of doing this modulo 8 for O(n)?

Jon Beardsley

3:35 AM

lol

David Roberts

Hmm, not sure what that would mean

Do you mean like Clifford algebras

Or over Z/8?

Will

well

I know doing a Whitehead tower relies on you having killed off lower groups before proceding higher

David Roberts

Or Bott periodicity

Yes

Will

but is there a way of killing off groups mod 8 here

yes because of Bott periodicity

David Roberts

Ah. Hmm

Not that could even imagine

So you want to kill off pi_i(O(n)) (n >>0) for i = 3mod 8 or similar

Will

3:38 AM

that's the idea

although fixing n there is going to cause troubles

David Roberts

Man, that's pretty crazt

crazy

Will

I said it was silly

David Roberts

So you want O, rather than O(n)

Will

yes

David Roberts

Hot dog

plus.google.com/photos/…

Will

3:41 AM

I don't think that link works for people other than yourself

David Roberts

hmm, it should

Never mind

Let's just say I don't feel optimistic

One more try: plus.google.com/103404025783539237119/posts/EbWRn3wJwCP

Will

that worked

David Roberts

:-)

So a question I would be interested in is what sort of interesting 2-groups one can get in schemes

Hi Jon

Will

I'll pretend to be competent and say: won't String be such an example?

Feeds

Jon Beardsley has added Eric to the list of this room's owners.

Jon Beardsley

3:47 AM

oh hai

Will

Spin is an algebraic group, seems like String should be a 2-group scheme

or is that nonsense?

Jon Beardsley

:(

Will

did Mathematica not give you the polynomials you wanted?

David Roberts

Perhaps we can take formal loops in G, and get an Ind-scheme

Jon Beardsley

no it did. they're just big

plus i've been really silly about typesetting all of them really nicely

David Roberts

3:49 AM

I've seen worse

Will

I guess it's at this point that my lack of understanding of cohesion betrays me

Jon Beardsley

yeah. it won't take too long

sorry for interrupting, continue

David Roberts

Lemma 5 page 15 of math.sunysb.edu/~claude/hem10.pdf

Will

good performance

David Roberts

@Will - is that to Jon or myself

(re: lack of understanding)

Will

3:51 AM

to you

David Roberts

Ah, about String

Will

I understand his polynomials pretty well

David Roberts

ok, some history there....

Will

yes, of cyberbullying

he told me not to do it again

David Roberts

Ah, that was just before I came in

So can we replace Omega G by the formal loops ind-scheme, when G is a 1-connected (reductive, if that makes sense) group scheme?

Jon Beardsley

3:53 AM

cyberbullies

Will

so you mean Hom(Spec(k((t))),G) ?

sorry I don't know what formal loops are

David Roberts

That's something like it

Perhaps I'm thinking along the lines of polynomial loops

Just looking it up

OK, something like R |-> G( R((t)) ), but that's probably what you wrote

(source: arxiv.org/abs/math/0607130, Pappas-Rapoport)

Will

ok

well that certainly seems like something you can do but I have no idea if it leads anywhere

David Roberts

Then we'd need something like the transgression of the canonical 3-cocycle (assume everything nice, so this exists) on G to this loop group

Yeah, I'm just talking off the top of my head. I'm not an algebraic geometer, after all

Other than that, perhaps something like the automorphism 2-group of a stack, although that's not going to be a scheme except in special cases

sorry, 2-group in schemes

Will

well that's where cohesive structures come in right, but I don't know how you do it really

David Roberts

4:04 AM

Well, I suppose it's probably a group stack

Cohesion is probably overkill just to get one simple example :-)

OK, so here's an idea: what's the automorphism 2-group of the stack of elliptic curves?

Will

do we expect to be able to name it?

David Roberts

No, compute it

It would be nice if the action lifted to the sheaf of E_oo ring spectra which is elliptic cohomology

Perhaps people already know bits and pieces of this already

Jon Beardsley

dang ol mathematica won't solve my equations

David Roberts

Shalosh B. Ekhad would be able to.

Jon Beardsley

i don't know what that means

Will

4:12 AM

so you mean you don't know who to call?

Jon Beardsley

gerst bersters

David Roberts

arxiv.org/find/all/1/all:+ekhad/0/1/0/all/0/1

Will

I enjoy that some of the papers have no coauthors

David Roberts

+1 Will

Jon Beardsley

Another Hanukkah Miracle: The Gaps Between Consecutive Christmas-in-Hanukkah Years is ALWAYS a Fibonacci Number!

David Roberts

4:16 AM

Anyway, you were mentioning, @JonBeardsley about the groupoid of fgls

Will

yes, use mathematica to compute an automorphism 2-group Jon

Jon Beardsley

dear mathematica, what is the homotopy of the 893-sphere? localized at 17

@DavidRoberts yeah, that's the only one i've ever significantly bumped into

well, i'm sort of interested in descent groupoids, but i don't really know what that means yet

i gotta get this book by janelidze from the library

David Roberts

Don't laugh, but cs.ox.ac.uk/people/jamie.vicary/twovect/guide.pdf

Ah, the one on Galois theory?

Jon Beardsley

haha

Will

oh god 2-vector spaces

David Roberts

4:20 AM

Why?

Will

so re:Eric's question, what is a 2-vector bundle?

David Roberts

Know anything bad?

Well, let's just wait and see if I get my research grant

And then I'll be able to work on them for 3 years

Seriously, though...

Jon Beardsley

i want to understand Kontsevich's stuff about polyvector fields and Hochschild homology

Will

so what is the 2-group corresponding rank n 2-vector bundles?

Jon Beardsley

yeah the one on galois theory

sorry, just saw that

David Roberts

4:21 AM

Well, that's a slightly tricky question...

Jon Beardsley

because i'm interested in descent groupoids associated to galois and hopf-galois extensions of ring spectra

Will

yeah that's why I haven't been convinced to like 2-vector spaces

Jon Beardsley

but i don't know that stuff too well w/r/t dang ol regler rings

David Roberts

Just a moment... brb

The main problem, to my mind, is that we have all these flavours of 2-vector spaces flying around, and then a massive general version due to Urs Schreiber

Will

good old good old

David Roberts

4:27 AM

And it's not entirely clear which applications they are good for.

Will

so do I need to just accept there is no one such thing as "2-vector bundle" and that different notions of 2-vector space give different notions of 2-vector bundles?

David Roberts

For instance, Kapranov-Voevodsky ones are used for 2-K-theory by Baas-Dundas-Rognes - but only at the beginning

then it's all spectra calculations

Yes - that is right

Will

hrmpf

David Roberts

Well, there is not one sort of 1-vector space, when you think of it

Will

?

David Roberts

4:29 AM

Fin. dim, separable Hilbert, Banach, Frechet, etc

And then we have all sorts of different fields

Things just bifuricate

Will

ok sure

David Roberts

OK - the REAL problem: what is a 2-field?

Jon Beardsley

lol

Will

but if you settle on finite dimensional, and settle on a field

are there still bifurcations in the 2-theory?

David Roberts

Well, what's a 2-field?

Will

4:31 AM

yeah fair enough, I see

David Roberts

It seems like R and C, which are the easy ones, are very special anyway

KV 2-vector spaces take Vect as the base ring, and right there you have lost the topology on the objects

A given vector space doesn't admit any deformations, say

Man, I've been hogging the airwaves.

Will

I've enjoyed it

David Roberts

Don't get to talk about this stuff much at my department

Thanks @Will !

Jon Beardsley

no worries

glad there's some math going on around here

David Roberts

And it's all even vaguely relevant :-)

I might even ask a question on MO about Aut(Ell)

imagine that!

Will

4:35 AM

could turn out quite interestingly

Jon Beardsley

good!

quote this chatroom!

haha

David Roberts

Ok, let me thrash out a question here

What is Eq(Ell)?

Will

What is Eq?

David Roberts

The stack of elliptic curves is a very concrete, nice stack [Q on which site?]

OK, that's why I wanted to do it here

Aut(Ell)

What is Aut(Ell)?

The stack of elliptic curves is a very concrete, nice stack [on X site]

and as an object of the 2-category of stacks, its endofunctors which are equivalences are the objects of a 2-group. That is, a monoidal groupoid with weakly invertible monoidal product.

[I think I need a link to the description of Ell.]

Will

you might as well do it over the complex numbers

and draw a punctured sphere with two cone points as an explanation :P

Jon Beardsley

4:41 AM

alright, have fun guys, these stupid polynomials have made me want to put my eyes out with an icepick. i gotta go to bed

Will

careful with the icepick

David Roberts

ok, so assuming 6 is invertible, what's that ring again?

Yes, g'night @JonBeardsley

Jon Beardsley

i always sleep with an ice pick. later.

Will

which ring are you asking about, sorry?

David Roberts

Ah, there's a very simple ring from which one can get a lot of information about M_{1,1}.

I'm trying to find it

OK, so there's a Hopf algebra (=groupoid in Aff) which presents the moduli stack of elliptic curves

Evan Jenkins

4:45 AM

There are some lecture notes by Mike Hopkins that describe this nicely.

math.rochester.edu/people/faculty/doug/otherpapers/coctalos.pdf

Section 20

Page 70

Will

ah so you mean the formula for an elliptic curve in Weierstrass form with j-invariant j, j not 0 or 1728?

David Roberts

That's the one, @EvanJenkins

And yes, @Will

ok, so I now have my affine groupoid scheme

(cont.) ... So as the stack is so concrete, we can try to calculate this automorphism 2-group.

My first question is, has this been done?

At least, away from the primes 2 and 3 there shouldn't be massive problems.

Then, if we have this 2-group, one can ask whether the action lifts to objects related to Ell.

For instance, we have the universal elliptic curve over Ell, which we can view as the total space of a stack over Ell

[I think? Does this sound reasonable? Perhaps it's not a stack over Ell, but over the base site, but with a map to Ell...]

Then we also have the sheaf of ring spectra from which we get elliptic cohomology theories: is this an equivariant sheaf? (in the appropriately weak way)

Evan Jenkins

Lurie does something like this in his survey of elliptic cohomology.

David Roberts

using Aut(Ell)? I know there's some equivariant stuff in there, but it's not in front of me

In fact I'd be mildly surprised if Aut(Ell) acted on the sheaf of spectra, but there's no harm asking.

Evan Jenkins

Ah, no, probably not.

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