Timeline for Is the moduli of formal groups smooth?
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| Oct 21, 2015 at 14:25 | comment | added | Jacob Lurie | (continued) d) The moduli stack of formal groups of height exactly n (regarded as a reduced locally closed substack of the entire moduli stack) has trivial cotangent complex over $\mathbf{F}_p$ (if $n > 0$), so any square-zero extension of that has a unique splitting (provided that the extension lives in simplicial $\mathbf{F}_p$-algebras). | |
| Oct 21, 2015 at 14:24 | comment | added | Jacob Lurie | @Tomer a) The moduli stack of formal groups (no need to mention height) is formally smooth: that is, it satisfies the usual infinitesimal lifting criterion. This follows immediately from the fact that the Lazard ring is polynomial. b) The infinitesimal lifting criterion only applies in the affine case, so it doesn't address your problem. c) The moduli stack of formal groups has a well-defined cotangent complex which will let you rephrase your question in homological terms, provided that by "CDGA" you really mean "simplicial commutative ring"... | |
| Oct 21, 2015 at 9:34 | history | edited | Lennart Meier | CC BY-SA 3.0 |
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| Oct 21, 2015 at 8:59 | comment | added | Tomer Schlank | @JacobLurie, My situation is that I have a square-zero extension of the formal stack MFG^{n} by some suspension of some coherent sheaf on MFG^{n} and (presented as a sheaf of square -zero CDGA's) I want to show that it splits. I found a theory of cotangent complexes for stacks in HAG2 (section 1.4). But it is stated in a very general way, and was hoping for a more direct reference useful for this case. | |
| Oct 20, 2015 at 15:34 | comment | added | Jacob Lurie | @Tomer, I'm not sure what definition you have in mind. The moduli stack admits an pro-etale surjection from the spectrum of a finite field, as you point out. If this qualifies for you as "smooth", then the answer is yes. If "smooth" means "admits a smooth surjection from something smooth" (as in Lennart's answer), then the answer is no. | |
| Oct 20, 2015 at 14:39 | comment | added | Tomer Schlank | @JacobLurie , Since I'm basically interested in formal smoothness, it seems your answer settles (1) in the negative but it also seems to settle (2) in the positive. Is this correct? | |
| Oct 20, 2015 at 12:52 | comment | added | Jacob Lurie | The map $g$ is not smooth. I also don't know what the OP has in mind by smoothness, but the usual definition fails for $X = \mathcal{M}^{\leq n}$. If $X$ were smooth over $\mathbb{Z}_{(p)}$ (by the usual definition), then it would have a perfect cotangent complex. But the cotangent complex of $X$ isn't perfect: if you take the fiber at a point corresponding to a formal group of height $m > 0$, then you get an $(m-1)$-dimensional vector space. For a perfect complex, the Euler characteristic of the fibers is locally constant. | |
| Oct 20, 2015 at 11:16 | comment | added | Neil Strickland | Also, what exactly is the definition of "surjective" here? I couldn't find it in the stacks project. At least under some definitions, the image of $\text{spec}(E(k)_*)$ (for $k<n$) will be contained in $\mathcal{M}_{FG}^{\leq n}$ but will not be covered by $\text{spec}(E(n)_*)$. | |
| Oct 20, 2015 at 11:08 | comment | added | Neil Strickland | How do you know that $g$ is smooth here? Does that not involve consideration of $U\times_XU=\text{spec}(E(n)_*E(n))$? | |
| Oct 20, 2015 at 9:51 | history | edited | Lennart Meier | CC BY-SA 3.0 |
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| Oct 20, 2015 at 7:35 | history | answered | Lennart Meier | CC BY-SA 3.0 |