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Mar
26 |
awarded | Self-Learner |
Mar
19 |
comment |
N-periodic derived categories
You can think of $N$-periodic complexes as modules over the differential graded algebra $R[x^{\pm 1}]$, where the differential is trivial and $x$ lies in degree $N$. This gives an answer to $(1)$ (there's a model structure on modules over any differential graded algebra, where the fibrations and weak equivalences are maps that are fibrations and weak equivalences on the underlying chain complex). |
Feb
24 |
awarded | Revival |
Feb
21 |
answered | Variant of Conceptual Completeness |
Jan
16 |
comment |
Genuine equivariant ambidexterity
That's another way to describe the problem. The genuine fixed point construction is right adjoint to a functor from spectra to $G$-spectra, given informally by "let $G$ act trivially". But the "trivial action" functor doesn't preserve homotopy limits, and therefore doesn't have a left adjoint. (Though if you work in the $K(n)$-local setting and $G$ is finite, then this issue goes away: the right adjoint will also be a left adjoint.) |
Jan
16 |
revised |
Genuine equivariant ambidexterity
deleted 1 character in body |
Jan
16 |
answered | Genuine equivariant ambidexterity |
Jan
16 |
comment |
Genuine equivariant ambidexterity
@Yonatan Not to my knowledge. I believe that a $G$-spectrum (for $G$ cyclic of prime order) is determined by the data described above. |
Jan
16 |
comment |
Genuine equivariant ambidexterity
@Yonatan I'm not sure what you mean by the "genuine norm map", unless you mean the map $f$. |
Jan
16 |
comment |
Genuine equivariant ambidexterity
Let $G$ be cyclic of order $p$. To specify a genuine $G$-spectrum $X$, you need to specify a spectrum $Y$ with a "naive" $G$-action together with a factorization of the norm map of $Y$ as a composition $f: Y_{hG} \rightarrow Z$ with $g: Z \rightarrow Y^{hG}$; here $Z$ is the genuine fixed point spectrum. If $Y$ is $K(n)$-local, then the norm map $g \circ f$ exhibits $Y^{hG}$ as the $K(n)$-localization of $Y_{hG}$. I'm not sure if I understand the question: is it "if $Z$ is also $K(n)$-local, then does $f$ exhibit $Z$ as the $K(n)$-localization of $Y_{hG}$"? If so, the answer is no. |
Jan
11 |
awarded | Popular Question |
Dec
26 |
awarded | Nice Question |
Dec
25 |
awarded | Curious |
Dec
24 |
asked | Variant of Conceptual Completeness |
Dec
22 |
awarded | Nice Answer |
Dec
22 |
answered | elliptic curves and group cohomology |
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 |
comment |
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 |
comment |
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. |