« first day    last day (1666 days later) » 

12:19 AM
some kind of weird version of $\mathbb{A}^1$...
 
n disjoint copies of the affine line?
 
yeah..... i guess that works, hmmmm. that somehow seems unsatisfying
 
why? as a set $R[x]/(x^n)$ is just the product of n copies of R
 
no,i mean ,absolutely
you're right
 
Jon - wouldn't it be the n-th order infinitesimals
 
12:30 AM
i guess i don't want it to be a set
well, so, right, i mean, i want something like that.... something about infinitesimals
 
Ah, no. I see what you're asking.
 
but see.... so Spec(R[x]/(x^n)) is that
but.... that's not quite qhat i want
i guess i want a scheme that gives me the n-th order infinitesimals on any ring R
 
Spec(R[x]/(x^2)) is the 'walking tangent arrow'
 
haha.... yes
 
So I guess Spec(R[x]/(x^n)) is the walking n-jet
And I mean the n-jet for schemes over R, I suppose.
 
12:33 AM
another thing i want is for the scheme $F$ such that $F(R)=R[x]/(x^n)$ to be a sub(super??)scheme of the scheme $F'$ such that $F'(R)=R[x](x^{n+1})$.
erm... should be a / on that second part
 
do you really mean that or do you mean infinitesimals
 
i don't really know what i mean. i've been doing this all in terms of algebra
and i am trying to do it in scheme
 
Just apply Spec
 
lol
welllllll
 
And turn around all the arrows
 
12:35 AM
right... i mean, if I do it for a single ring R, I apply spec, and get a nice sequence of schemes
 
because as stated the disjoint union of n affine lines is contained in the disjoint union of n+1 affine lines
 
sure, lol
 
In fact, working with affine schemes is exactly doing algebra, just one needs the geometric intuition for what the algebraic constructs are doing
 
so you are doing $F_n(R) = \{ r \in R : r^n = 1 \}$ ?
$F_n(R) = \{ r \in R : r^n = 0 \}$, sorry
?
 
well no, so, that is just $Z[x]/(x^n)$
right?
 
12:36 AM
Miles Reid has a nice book which approaches (high-level) undergrad algebra from a sneakily scheme-theoretic pov
 
erm....
 
@Jon yes
 
it's the internal hom-scheme I guess $Hom(\mathbb{Z}[x]/(x^n),-)$
or something
but no see, that's the issue
i don't want nilpotents
 
@JonBeardsley which "it's" do you mean?
 
it is
 
12:37 AM
Ah, no nilpotents, then try using varieties rather then schemes
:-)
 
@Jon so which functor do you want to represent then?
 
@Will exactly
 
let me clarify, $\mathbb{G}_a(R)=R$. I'd like some kind of.... deformation of $\mathbb{G}_a$, call it $\mathbb{G}_a'$ such that $\mathbb{G}_a'(R)=R[x]/(x^n)$
erm, let's just start with $R[x]/x^2$
 
as functors valued in sets?
 
well, i'm trying to decide that. the point is.... $\mathbb{G}_a'$ is a group scheme
 
12:40 AM
You already know G_a' is a group scheme, or you want it to be a group scheme?
 
no i mean it is
and the higher n is, the more ways it can be
 
You also need to specify what the group operation is on R[x]/x^n
Otherwise you are just giving us a (pre)sheaf of sets
 
for instance, there is a cogroup structure on $R[x]/(x^2)$ given by $f(x)\to f(x+y)$
hmmmmm okay.... so..... maybe they're cogroup schemes
 
If the group structure turns out to be just R^n,+, then you are just dealing with G_a^n
 
so, $Spec(R[x]/(x^2)$ is a group scheme
yessss
 
12:43 AM
No, not quite
 
but! there are more ways than just additivity to do it right!
oh.....
 
The Hopf algebra structure on a ring is what you get from turning around the arrows on an affine group scheme
 
This is not the same as defining a pointwise group structure on a scheme
And where by pointwise I mean functor-of-points, i.e. a lift Aff^op --> Grp of the sheaf of sets
underlying the scheme
 
i mean. there are two things going on here.
i'm pretty sure that using the unique 1-bud structure, we know that $Spec(R[x]/(x^2))$ is a group scheme, yes?
 
12:45 AM
1-bud?
 
ummmmmm, so a truncation of a formal group law
 
But Spec(R[x]/x^2) is not the same as R |-> R[x]/x^2
 
oh no absolutely not
the latter should have a cogroup structure
 
So the question is, which one do you need?
 
the latter
and i didn't entirely know that until i was forced to clarify to you what the hell i was talking about, haha
 
12:46 AM
Since I came in late, may I ask why and what for?
 
if you have a representable functor valued in cogroups you are looking at a cogroup object in the category not a group object even though the functor is contravariant
 
yeah exactly
that's what i just realized
i was getting my arrows backwards
:-)))))))
i'm super happy
 
You have lots of chins
:P
 
yeah, all i do is sit in this chatroom and eat cheezits
@davidroberts the short of it is, i'm trying to think about building formal group laws up from what are called n-buds, or formal group law k-chunks, from a deformation theory perspective
and i guess i'd like to think about it as something like.... deforming cogroup structures on a certain scheme
 
Hmm, sounds interesting!
 
12:50 AM
yeah. i'm sort of pleased with it. Michel Lazard basically proved that any n-bud can be deformed into an n+1-bud, but he didn't do it from this perspective
 
So it's working from the bottom up, rather than starting from the Lazard ring?
 
yeah exactly
and Lazard actually did that originally
but.... i guess he didn't know deformation theory, haha
well actually, the cohomology theory that the obstructions live in was developed later by lubin and tate
and they used it to do some very cool stuff
 
So my question is, given an elliptic curve, can you deform it to give an explicit fgl one level up?
And I mean a specific elliptic curve from Weierstrass form
 
well, ummmmmm, i think such things always give you formal group laws, right?
when you say one level up, are you referring to height?
 
I mean chromatically
 
12:53 AM
yeah, so, that's an awesome question. I think that other people have worked out that elliptic curves always give you FGL's of either height 1 or 2
 
yes according to ordinary or supersingular
 
I ask because people, when talking about cohomologies from fgls say: G_a -> HR, G_m -> ku (or similar, can't remember connective or otherwise)
 
@will so are elliptic curves which are ordinary, are they... smooth?
 
Then E -> elliptic cohomologies
 
12:54 AM
all elliptic curves are smooth by definition
 
for E a curve
 
oh okay, good!
 
And then what next?
 
It's all quotients of BP and so on.
 
12:54 AM
it's just that in characteristic p the p-torsion of an elliptic curve can be either (Z/pZ) or (Z/pZ)^2
(as sets)
 
Shimura varieties or something
aha. okay, that's very helpful, thankyou
 
sorry
 
I would like to see an example of a cohomology theory induced by a different fgl
 
0 or Z/pZ
 
@Will that makes more sense
 
12:55 AM
and so, an elliptic curve is an affine group scheme, and i get an FGL by look at an infinitesimal neighborhood of the identity?
 
the formal group essentially only sees what is RIGHT BY the origin
 
Er, yes
 
in the case where it's Z/pZ
you have it spread out a bit
so in a sense you only have on Z/pZ factor that has collapsed
if you have 0
 
hmmmmmm, okay
 
12:56 AM
the scheme still has degree p^2
but it's all concentrated around the origin
 
right
 
and the formal neighbourhood of the origin sees all of that
 
interesting
 
so the height is 2
 
12:57 AM
Nice
 
so @DavidRoberts, there are for instance the Honda formal group laws
which are related to, if i'm not mistaken (and i often am), MoravaK-theory
 
Ah, ok.
How far up the chromatic tower are they?
 
arbitrary height
 
Hmm ok
 
is that right @will? you probably know better than i
 
12:58 AM
But say you're given a curve, can you deform it to a Honda fgl at height 3?
 
i don't think so
 
I don't know, are they associated to higher rank L functions or what?
(that was for you @Jon)
 
Have a look at www.math.uiuc.edu/~rezk/hopkins-miller-thm.pdf
 
yeah i know. considering that these chats are saved, i'm not sure i want to say a lot more
hahaha
but yeah, that's the place to look @davidroberts
 
Example 3.8
 
1:02 AM
i'm sorry, Honda FGLS give you Morava E-theory
 
apologies, but I've got to go now
 
later
 
1:34 AM
the thing with everything being on record is that you can't trick people into thinking the chat was active and they need to join in
 
speak for yourself
i am v. tricky
 
wait what
i don't understand the trick i might polay
*play
 
people can tell whether or not you were lying
b/c they can see when the last person who talked was
 
you can't just start a conversation mid-way through to get a reaction of someone who just joined
 
1:36 AM
yeah, beardsley
 
yeah, i guess i haven't spent enough times in chat rooms to understand this particular trick
 
i guess
 
now would have been the perfect opportunity for instance
 
like when eric joined?
 
instead he can alleviate the confusion by scrolling up
 
1:42 AM
hah hah!
we were just discussing our favorite homotopies, eric
 
i see
 
homotopoii
 
yeah. and cogroup schemes
 
when david comes back: you can't perform an p-primary infinitesimal deformation to change the height of a formal group law
you can do noninfinitesimal things, but then examples are obvious
 
is the Honda formal group law something obvious
 
1:45 AM
no, it certainly is not
there are papers by charles rezk and a student of tyler lawson which give explicit models for the honda formal group laws in characteristics 2 and 3 and at height 2 using accidentally available models of elliptic curves
the general case is abysmal
i forget tyler's student's name. yi fei? idk
 
ok thanks
 
it exists
technically
 
haha, good stasheff email
 
yeah totally
i responded
but he didn't respond to me
 
well i hope he arrives
 
1:47 AM
me too
 
it's uncomfortably hot in berkeley :(
 
which means returning to boston is going to be really miserable
 
is it
 
it's miserably hot in Baltimore
and probably in Boston too
 
1:48 AM
i just won't go outdoors i guess
 
yeah, use your tunnel system
 
high of 77 here
 
you've been burrowing a lot in Boston right?
 
someone has
grad students of the past
 
yeah. hm. maybe it was Michael Andrews.
probably Nitu
that would explain all the tunnels at JHU too
 
1:59 AM
@JonBeardsley !!!!! An elliptic curve is not affine!!!
 
@JonBeardsley Are you working with Nitu?
 

« first day    last day (1666 days later) »