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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.
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
comment 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).