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Oct
21 |
comment |
Is the moduli of formal groups smooth?
(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 |
comment |
Is the moduli of formal groups smooth?
@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
20 |
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Is the moduli of formal groups smooth?
@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 |
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Is the moduli of formal groups smooth?
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. |
Sep
17 |
awarded | Nice Answer |
Sep
17 |
comment |
Do there exist “topologically significant” (and not “algebraic”) triangulated categories killed by the multiplication by $p$?
If $\mathcal{C}$ is a stable $\infty$-category, then the endomorphisms of $\iota_{\mathcal{C}}$ has the structure of an $E_2$-ring spectrum (the "Hochschild cohomology" of $\mathcal{C}$). Making $\mathcal{C}$ $R$-linear is equivalent to giving an $E_2$-map from $R$ into this endomorphism ring. And $\mathbf{F}_p$ has a very simple presentation as an $E_2$-ring spectrum: you need only the single relation ``$p=0$'' (result of Mahowald when $p=2$, Hopkins for odd primes). |
Sep
17 |
answered | Do there exist “topologically significant” (and not “algebraic”) triangulated categories killed by the multiplication by $p$? |
Aug
30 |
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A “universally non Hypercomplete” $\infty$-topos?
The localization can be described more concretely: it's sheaves with respect to the Grothendieck topology where every morphism generates a covering. Equivalently, it's functors F: {finite spaces} -> {spaces} (or you can do a pointed version if you like) with the property that for any X, F(X) is the totalization of the cosimplicial space given by applying F to the "Cech nerve" of the map X->* (in the opposite of finite spaces). This is much larger than the class of 1-excisive functors: it contains all n-excisive functors for any n, and more (such as products of n-excisive functors as n varies). |
Aug
15 |
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Schwede-Shipley theorem for monoidal categories?
True iff the unit for the monoidal structure is a compact generator. |
Aug
11 |
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Can hypercomplete objects be coreflective?
Not sure if it is explicitly stated there, but it is at least easily derived from what's there. First, reduced 1-excisive functors are spectra (Calculus III). If F is an arbitrary excisive functor then Y=F(*) is a space and $(y \in Y) \mapsto \lambda X. F(X) \times_{F(*)} \{y\}$ is a local system of reduced excisive functors on $Y$. Conversely if $\{ F_y \}_{y \in Y}$ is a local system of reduced excisive functors on a space Y, then $F(X)=\varinjlim_{y \in Y} F_y(X)$ is an excisive functor (not necessarily reduced). These constructions are homotopy inverse to one another. |
Aug
10 |
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Can hypercomplete objects be coreflective?
Tom Goodwillie, "Calculus III: Taylor Series" |
Aug
6 |
awarded | Enlightened |
Aug
6 |
awarded | Nice Answer |
Aug
6 |
revised |
Can hypercomplete objects be coreflective?
Added clarification about base points |
Aug
6 |
answered | Can hypercomplete objects be coreflective? |
Jul
18 |
awarded | Yearling |
Jul
17 |
awarded | Enlightened |
Jul
17 |
awarded | Nice Answer |
Feb
20 |
comment |
Analogues of Primitive Recursive Functions
I'm afraid that I don't follow either. What exactly are you saying you can rule out? |
Feb
19 |
awarded | Nice Question |