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Lubin and Tate show in their paper Formal moduli for one-parameter formal Lie groups that for any formal group over a field $k$ of characteristic $p>0$ with height $h<\infty$, the functor of deformations is represented by a formal scheme isomorphic to $\mbox{Spf } \mathbb{W}(k)[[u_1,\ldots,u_{h-1}]]$. Modulo lower terms, $u_i$ is the coefficient of $x^{p^i}$ in the $p$-series of the universal deformation. (We take $p=u_0$.)

Does this carry over for the additive group? Certainly there is an evident deformation to $\mbox{Spf } \mathbb{W}(k)[[u_1,u_2,\ldots]]$. In the paper, the finite height assumption is present in various results that they cite from elsewhere, so without being intimately familiar with the whole theory it's kind of hard to tell if this assumption is essential.

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Cross-posted to here from math.se due to lack of activity: math.stackexchange.com/questions/242061/… –  Aaron Mazel-Gee Nov 27 '12 at 10:54
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One thing to note is that the formal additive group has infinitesimal automorphisms (the dual Steenrod algebra), while a formal group of height $n$ does not. –  Akhil Mathew Nov 27 '12 at 12:45
    
@Akhil: Rather, you mean that a formal group of height $n$ doesn't admit infinitesimal homomorphisms to the formal additive group. –  Eric Peterson Nov 27 '12 at 19:47

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$\DeclareMathOperator{\Ext}{Ext} \newcommand{\G}{\hat{\mathbb{G}}} \DeclareMathOperator{\Maps}{Maps} \renewcommand{\phi}{\varphi}$ The analysis of the infinitesimal deformation space of the Honda formal groups $H_n$ uses three calculations which govern the existence of square-zero deformations. These can be phrased in terms of certain $\Ext$ groups, and here's how I think of them: we want to study the deformation of a formal group $G_0$ over a ground ring $R_0$ along a square-zero infinitesimal deformation $I \to R \to R/I = R_0$ to a formal group $G$ over $R$. Because $I$ is square-zero and the only formal group law truncated at degree $2$ is the additive formal group, we think of $G$ as sitting in an extension $I \otimes \G_a \to G \to G_0$ sort of "over" the original extension of $R$-modules. So, to study these extensions (their existence, their uniqueness, so on), we want to calculate some kind of $\Ext$ groups of the form $\Ext^*(G_0; \G_a)$.

This is exactly what Lubin and Tate do in their paper. Fix a pair of formal groups $F$ and $G$; then there's a simplicial object $BF = B_*(*, F, *)$ associated to $F$ via its group structure, and the cosimplicial object $\operatorname{Maps}(BF, G)$ should be a thing whose cohomology $H^*_{LT}$ computes $\Ext^{*-1}(F; G)$. Trying to work out exactly what this means, you'll find the following descriptions of the first few groups (written in coordinates, but this is inessential):

$$H^1_{LT}(F; G) = \{\phi: F \to G \mid \phi(x) - \phi(x + y) + \phi(y) = 0\} = \operatorname{Hom}(F, G),$$ $$H^2_{LT}(F; G) = \frac{\{\phi: F^2 \to G \mid \phi(x, y) - \phi(x, y + z) + \phi(x + y, z) - \phi(y, z) = 0\}}{\{\delta^1 \phi \mid \phi : F \to G, \delta^1 \phi(x, y) = \phi(x) - \phi(x + y) + \phi(x)\}},$$ $$H^3_{LT}(F; G) = \frac{\left\{\phi: F^3 \to G \middle| \begin{array}{c}\phi(x, y, z) - \phi(x, y, z + w) + \phi(x, y + z, w) - \\ \phi(x + y, z, w) + \phi(y, z, w) = 0\end{array}\right\}}{\{\delta^2 \phi \mid \phi : F^2 \to G, \delta^1 \phi(x, y) = \phi(x, y) - \phi(x, y + z) + \phi(x + y, z) - \phi(y, z)\}}.$$

General facts about infinitesimal deformation theory then tell you the utility of these groups: $H^1_{LT}$ tracks automorphisms of the square-zero deformation space, so it tells you whether you should expect the deformation space to be a scheme or some sort of stack; $H^3_{LT}$ tracks the obstruction to the existence of square-zero deformations; and $H^2_{LT}$ tracks the available square-zero extensions, in the sense that when $H^1_{LT} = 0$, its dimension will tell you the dimension of the tangent space of the deformation space.

So, Lubin and Tate go about computing these three things in the case that $G_0 = H_n$ is the height $n$ Honda formal group. They know that $\operatorname{Hom}(H_n, \G_a) = 0$, and so $H^1_{LT} = 0$. They also compute that $H^3_{LT} = 0$, so they know that the deformation space they're computing is actually a formal variety --- there are never any kind of conditions imposed on their generators $u_i$ to ensure the existence of a deformation. The only thing left is to determine the dimension of the formal variety, and they compute $\dim \Ext^1(H_n, \G_a) = n-1$.

You ask what happens when we replace $H_n$ with $\G_a$ and instead study $H^*_{LT}(\G_a; \G_a)$. The computation of these cohomology groups is worked out in the COCTALOS notes, and also in Theorem 4.3 and Section 8.4 of a project I worked on: [link]. Grinding through the homological algebra, you'll find that the group $H^n_{LT}$ consists of the polynomials of homogeneous degree $n$ in the dual Steenrod algebra, after assigning the degrees $|\xi_*| = 1$, $|\tau_*| = 1$, and $|P_*| = 2$.

All of these things are woefully $\infty$-dimensional. I don't actually know what that means for the existence of your object --- and so this isn't a proper answer --- but I do know that it isn't anywhere as simple as the case Lubin and Tate analyze.

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Actually, I am a little surprised that you can make this characterization in terms of binomials and trinomials in the Steenrod algebra. The statement is true, but I fear it may be an accidental isomorphism rather than an instance of a natural construction. Caveat lector. –  Eric Peterson Dec 2 '12 at 1:41
    
Thanks, Eric! I'd say this certainly means there isn't a deformation space; if anything, there might be an Artin stack. If we could present it as $X/\!/G$ carrying a universal deformation $\widetilde{\mathbb{G}}$, then perhaps "Morava $E_n$ at $n=\infty$" could be reasonably taken to mean the Lubin-Tate spectrum associated to $\widetilde{\mathbb{G}}(X)$ along with its group action... or maybe the fixed points thereof. –  Aaron Mazel-Gee Dec 2 '12 at 16:48

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