5:04 AM
So, the last room I created with this topic was inadvertently linked to math.se rather than MO. Hopefully I did this one right.
Anyway, here's something I've been thinking about lately--is there an easy (read: algorithmic) way to compute Mumford-Tate groups?

5:45 AM
So this is a new room!

Yep!

I was confused that the room of homotopy theory occurs quite early, but actually algebraic geometry is the most popular tag at MO.
I guess this is because there are not many users start to use chat in this new version.

Yeah the homotopy-theorists are pretty active. I figured I'd give this a shot though, see what happens.
I think in general the higher-category people tend to be early adopters of web technology, e.g. the nLab and n-category cafe etc.

Are you focusing on things related to the Hodge conjecture on abelian varieties?

6:02 AM
Could be quite cool if this one gets going. The representation theory room hasn't really worked out so far.

@YuchenLiu Not really--I've been wondering about the following sort of question
(as of like two days ago, not really what I'm working on.)
Namely, suppose X is a smooth projective surface; then the Hodge index theorem tells us which quadratic space NS(X)_R is
There are certain natural questions in algebraic geometry where one wants to know whether quadratic forms defined in terms of the intersection pairing on NS(X) represent certain integers
and it would be nice to be able to apply the Hasse principle to such questions.
So I want to know if one can identify the quadratic space NS(X)_{Q_p}.
It's too much to hope that the answer is determined by e.g. numerical invariants of X, as in the case of NS(X)_R
as you'll see if you think about CM Abelian surfaces
(which all have the same numerical invariants, but whose intersection form depends on the totally real subfield of the CM field acting on the variety.)
But I'm wondering if one can figure out the form if one knows the mumford-tate group...
And I'd also like to know how to compute some more examples, just to know what's up.
@GregStevenson Yeah, I figure I'll recruit a bit also...

@DanielLitt Will NS(X)_{Q_p} be determined by NS(X)_Z?

Yes, of course; it's just NS(X)_{Z} \otimes Q_p by definition...

6:19 AM
So the quadratic form on NS(X)_Z is of type (1, \rho-1), and the structure of indefinite unimodular quadratic forms is simple, as for odd form it can be diagonalized, for even form it splits to direct sum of hyperbolic form and E_8 form, which integer do you want?

Well, for example, the discriminant and \epsilon-invariant...

6:37 AM
@DanielLitt OK... I realize that I made a mistake -- the intersection form on NS(X)_Z is not unimodular in general... Thanks for your explanation.

No worries! It is sometimes, I think, whence your comment is pretty helpful.

So this numerical invariants has lot to do with the field in CM Abelian surfaces?
It is interesting, I haven't thought about these invariants

Yeah, if I've worked things out correctly (no guarantees) the discriminants depend only on the underlying totally real field...

8 hours later…
2:35 PM
oh man i've been waiting for such a room to start! i would have started it myself, but I didn't want to take people away from the htpy room! great idea @danielLitt !

2:47 PM
so maybe the people in here know, if I take an augmented R-algebra A to the group I/I^2, where I is the augmentation ideal, is this some kind of cotangent complex?

1 hour later…
4:03 PM
@JonBeardsley It seems to me that if R is regular, then that's the conormal bundle of the embedding Spec (R)-->Spec(A) induced by the augmentation--hence a shift of the cotangent complex.

4:17 PM
Sorry, I should have required that both R and A are regular.

2 hours later…
6:09 PM
@DanielLitt aha! fantastic!

I think it should also suffice if the map Spec(A)-->Spec(R) is smooth, by the so-called "transitivity triangle"
Just curious, did you have some heuristic reason for thinking that should be true beforehand? I have no intuition for the cotangent complex, so I'd be curious to know how you reached that guess.
@JonBeardsley Also, have you read the unpublished version of Quillen's "On the (co-)homology of commutative rings"? It's where I learned the little cotangent complex stuff I know, and is pretty readable; here's a djvu: chromotopy.org/paste/quillen.djvu

i just read it! i'll tell you about why i'm interested in it in a second!
(someone is teaching me how to make a website at the moment)
@danielLitt so i ran into this thing while dealing with formal group laws
you may know that one builds a formal group law out of "formal n-buds"
where an n-bud is an equivalence class of symmetric power series such that f(f(x,y),z)-f(x,f(y,z))=0 modulo degree n+1 terms
and eventually what you get is a group structure on an infinitesimal neighborhood of the origin on your group scheme
but it seems that the first step of this thing is basically looking at the functor valued in groups which takes an augmented ring R to I(R)/(I(R))^2
which you can think of as the tangent space of that group scheme, in some sense, and the obstructions to extending the unique 1-bud (the additive one f(x,y)=x+y) live in its derived functors

6:43 PM
Ah hah. Is this a fruitful point of view on formal groups?

I don't know yet! I think so.
but I believe that in general, the obstruction to extending an n-bud to an n+1-bud live in the derived functors of R/R^{n+1}

Are there obstructed n-buds?

not in the smooth case!
Lazard showed originally that the obstruction is always a coboundary

Sorry, what are you asking to be smooth?

oh sorry
so, in this case, a formal group law describes a group structure on spf(Z[[x]])

6:46 PM
right
which is smooth as a formal scheme

right
and it's the completion of the affine line

oh I suppose you might be interested in extending group structures on Artin schemes to bigger artin schemes

moreover, one could take another abelian variety

and these are often obstructed
sure

but with another abelian variety, the tangent space will sort of always be n-dimensional affine space

6:47 PM
ya

so yeah, a simple example that i plan on trying to look at is just a curve with a node or a cusp
you can repeat the process, and probably get obstructions
i.e. complete at that node, and your "tangent space" is gonna be nasty
or, well, that's not quite what i mean to say, to be precise, but i think the idea is clear

in any case i think you might need a bit more technique than quillen's approach to the cotangent complex

yeah, well, it's not purely categorical

deformations of group schemes are much more subtle than deformations of schemes

hmmmm

6:49 PM
e.g. you're deforming a diagram

aha right
interesting

you should look in illusie

his thesis?

the big cotangent complex book

oh okay, i'll try to get that

6:50 PM
was that his thesis? no idea

i dunno, i think he wrote his thesis on cotangent complex stuff

(but honestly this case you can probably do "by hand")

yeah
you totally can. i mean, i have been.
all this is just kind of... fancy pants language

yeah

mostly you actually have to do computations in cohomology
hahah

6:51 PM
cohomology of what?

well, i mean, so that's what the derived functors are

sure

cohomology of this certain cochain complex

yep

basically, i replace R[x] with the cobar complex

6:52 PM
oh wow this is still pretty fancypants

well, it's not the smartest resolution to use\
but it always works, which is nice
so yeah, i take the cobar complex on R[x], then i apply this functor (R|-->I(R)/(I(R))^n)
levelwise
and the obstruction lives in the third cohomology of this guy

hmmmmmm

this is kind of the same thing as taking the bar complex on a group scheme
which is basically doing group scheme cohomology
so i get this complex.... R[x]-->R[x,y]/(n-degree things)-->R[x,y,z]/(n-degree things) etc.
this is something like an iterated cotangent complex
or iterated TAQ, or AQ, or TQ, or whatever

hmm interesting
need to think about it a bit more/grab lunch

anyway, i probably shouldn't be blabbing about this so much
lol

6:56 PM
nah it's pretty interesting

yeah, i agree, but i'd like to write a paper about it hahah
and blabbing about new ideas in public forums is perhaps ill advised

eh i think one always gains more by talking than one loses
anyway i gotta run--later dude

cya!

7:10 PM
Illusie's part II is probably the more relevant bit
(and it was, indeed, his thesis)

1 hour later…
8:32 PM
@JonBeardsley BTW, are there examples of a non-smooth formal group schemes in char 0?

nope
ummmmm, actually i shouldn't be so sure
but i don't think so

i doubt it too
but i don't see the proof

i mean certainly every formal group law is, over Q, isomorphic to the additive one
but saying the words "formal group law" sort of... implies smoothness to start

um but a formal group law is the same as a smooth formal group scheme
right
so i'm asking about group objects in the category of formal schemes
which a priori need not be smooth
group schemes are smooth in char 0 (they're automatically reduced by a result of Oort, whence the result follows from generic smoothness + translating around)

oy. i'm out of my element trying to prove this in any way

8:36 PM
i'm just wondering about the existence of the objects you're trying to study

aha
let me add one thing
actually.... hrm... no i don't think that would make any sense. nevermind, haha.

ah found it
looks like all group objects in formal schemes are smooth in char 0
but there should be non-smooth guys in char p
oh i guess this is kind of obvious, e.g. \mu_p works and if you want a higher dim'l example just multiply a \mu_p on

okay, that makes sense

cool cool
well if you do end up working out the def-obs theory for infinitesimal group schemes, i'm curious
when you have a writeup, i'd be pretty interested

okay cool
ur at stanford? are u a graduate student?

8:41 PM
yep
working with ravi vakil

oh great. i know arnav pretty well.
never met ravi, though i'd like to at some point.

yeah arnav's my academic bro

hopefully all the stuff i'm talking about here can be extended to the derived context
i mean, i'm supposed to be doing homotopy theory after all

cool cool
applications aimed at chromatic htpy theory or something?

ummmmmm sort of
there's a sort of urban legend that S is a nilpotent thickening of Z or something
or HZ, rather
so that's something i'd like to make precise
also, there are spectra which determine n-buds on the coordinate rings of cohomology theories, and they're kind of like truncated versions of MU (which determines formal group laws on such) so i'd like to realize that in the framework of some kind of TAQ or something
similar to how i'm writing this stuff down as TQ of group schemes

8:49 PM
oh cool

or i guess i'm actually sort of writing it down as something like TQ of derived group stacks, or some nonsense

S is a nilpotent thickening of Z -- is this different from saying \pi_0(S)=Z?

i mean.... haha, i dunno

fair nuff :P

i think it's supposed to have some DAG sense to it

8:50 PM
gotcha--cool cool, don't know much about DAG

i dunno, maybe S is like, the universal nilpotent extension of HZ or something
like, if S is a thickening of HZ, this should show up in some TAQ, or something