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09:00
@S.carmeli I agree that this is a counterexample to the proof. You should email Jacob about it!
09:13
@MarcHoyois I did but apparently to the wrong email and I tried again now so hopefully he will answer soon.
 
6 hours later…
14:56
If $\Gamma$ is a height $n < \infty$ formal group over a perfect field $k$, canonically associated to $\Gamma$ we have the $(n-1)$-dimensional vector space $Ext^{1}(\Gamma, \mathbb{G}_{a})$ or, equivalently, $Def(k[\epsilon]/\epsilon^{2})$ (the groupoid of deformations and $*$-isomorphisms).
(Now that I think of it, I don't see an immediate canonical identification between the two. Maybe they're actually dual, or even not related to begin with?)
Does this vector space have a different description in terms of the geometry of $\Gamma$ alone?
@PiotrPstrągowski what does “geometry alone” mean? Certainly it depends on the actual group structure, right?
15:16
@DylanWilson Right, that's what I mean. I guess I'm looking for any other descriptions of this vector space.
15:26
I've never really understood how it fits in, but another (n-1)-dimensional vector space floating around is the cotangent space of the second exterior power of Gamma.
16:02
@WilliamBalderrama it sounds pretty related. A deformation of a group G to k[e]/e^2 is given by taking the set of elements (g+eh) and take the multiplication to be (g+eh)(a+eb)=gh+e(gb+ah + \phi(g,h)) and then you have a biliniar form \phi showing up right? so something related to the tangent space of the alternating power or so.
@PiotrPstrągowski is the summary of the kodaira-spencer map in section 1.6 of swc.math.arizona.edu/aws/2019/2019HopkinsNotes.pdf helpful?
I was probably confused from something related because it doesn't make sense sorry again :-(
 
2 hours later…
18:12
@EricPeterson This is very interesting, and I'd like to understand it better. Let's say $\Gamma$ is a height $n$ formal group over a perfect field, then if I understand correctly $Lie(\Gamma)$ is a $1$-dimensional affine space with the additive group structure, and trivial action of $\Gamma$.
Since $\Gamma$ is formal, I assume that for the purpose of computing cohomology of $\Gamma$ we can replace $Lie(\Gamma)$ with the formal additive group, so that is what Hopkins denotes $H^{2}(\Gamma, Lie(\Gamma))$ the same as $H^{2}(\Gamma, \mathbb{G}_{a})$ in your book?
But then, if $E_{0}$ is the associated Lubin-Tate ring with maximal ideal $m$, isn't $m/m^{2}$ an $n$-dimensional $k$-vector space rather than $(n-1)$-dimensional?
@WilliamBalderrama That is very interesting. It wasn't even clear to me that this would have the right dimension, but I just found a reference!
skd
skd
18:50
this could be a useful description (that you probably already know, and which is related to the kodaira-spencer formalism): Ext^1(G, G_a) = Hom(omega_{G^}, G_a), where omega_{G^} is the invariant differentials in the Cartier dual G^ to G. there's more about this in mazur-messing's "Universal Extensions and One Dimensional Crystalline Cohomology".
 
4 hours later…
23:19
@PiotrPstrągowski i also have gotten confused about n vs (n-1) and how the arithmetic deformation direction gets tracked through all this. i thought i'd gotten it straight in the book—it produces a bunch of classes beta_j in H^2(Ga; Ga), j ≥ 0, and claims that H^2(G; Ga) is spanned by beta_j for 0 ≤ j < n, so has dimension n—but i also see that i claim that "beta_0" in H^2(Ga; Ga) has a representative given by c_{p^0}(x, y) = (x + y) - x - y, so something must still be broken.
and welcome back @EricPeterson
mjh's account of this in coctalos (section 17, drawing from section 3) suffers from the same confusion
@skullpatrol ty. fwiw, i still lurk the transcripts as much as anyone else
How is this coming along?
Nov 1 '17 at 4:41, by Eric Peterson
https://github.com/ecpeterson/FormalGeomNotes/raw/master/main.pdf this got sent off to the publisher today 🎉🎉 but there's still lots of time to take advice, should any of you like to try to read it
which still holds the record with 24 stars
23:40
all done cambridge.org/us/academic/subjects/mathematics/… but this discussion ^^ portends to uncover a mistake in 3.4
the 24 star link seems to be broken
I have to click the original link to get access

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