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15 votes
6 answers
2k views

Why are equivariant homotopy groups not RO(G)-graded?

I know very little about the fancy equivariant stable homotopy category, so I apologize if this question is silly for one reason or another, but: I think that stable homotopy, in the non-equivariant ...
Dylan Wilson's user avatar
  • 13.1k
15 votes
2 answers
1k views

$RO(G)$-graded homotopy groups vs. Mackey functors

Everything here is model-independent: either take co/fibrant replacements wherever appropriate, or work $\infty$-categorically. Also, I've looked through other similar MO questions, but I didn't find ...
Aaron Mazel-Gee's user avatar
11 votes
0 answers
250 views

Trouble with Stable Equivariant Profinite Homotopy Theory

I've heard that there are some problems in developing a good formalism for stable equivariant homotopy theory (either from the spectral mackey functors perspective or from the orthogonal spectra ...
user95456's user avatar
  • 111
11 votes
0 answers
371 views

How does the HHR Norm functor interact with the cotensor over $G$-spaces?

Let $N_H^G$ be the norm functor from orthogonal $H$-spectra to orthogonal $G$-spectra. We know the category of orthogonal $G$-spectra $\mathcal{S}_G$ is enriched over the category of based $G$-spaces $...
Jack Davies's user avatar
10 votes
2 answers
463 views

Are finite $G$-spectra idempotent complete?

Question: Let $G$ be a compact Lie group (you can assume that $G$ is finite if you like). Is the category of finite $G$-spectra idempotent complete? Here, by "finite $G$-spectra", I mean ...
Tim Campion's user avatar
  • 59k
9 votes
2 answers
1k views

Genuine equivariant ambidexterity

A particular case of Lurie and Hopkins' ambidexterity theory is that if $G$ is a finite group acting on a $K(n)$-local spectrum $X$ then the norm map $$ X_{hG} \to X^{hG} $$ is a $K(n)$-local ...
Yonatan Harpaz's user avatar
9 votes
1 answer
1k views

A heart for stable equivariant homotopy theory

Let $G$ be a finite group. I wonder whether the following statement is true, known and written down: There is a t-structure on the stable $G$-equivariant homotopy category such that the associated ...
user2146's user avatar
  • 1,233
9 votes
0 answers
427 views

Stable homotopy type of $BG^{\wedge}_p$ in algebraic terms

In the mid 90's, Martino- Priddy proved that given two finite groups $G, H$, the following are equivalent: $\mathbb{F}_p\mathrm{Inj}(P,G)\cong \mathbb{F}_p\mathrm{Inj}(P,H)$ as $\mathbb{F}_p\mathrm{...
Victor TC's user avatar
  • 795
8 votes
2 answers
735 views

Models for equivariant genuine commutative ring spectra

The following question is an attempt at understanding various flavours of equivariant commutative ring spectra; it may not be suitable level for this forum. Let $\mathcal{C}(G)$ be a symmetric ...
Tom Bachmann's user avatar
  • 1,951
8 votes
1 answer
349 views

(Non)-equivariant equivalence in $G$-spectra

In HHR, an important part is the periodicity theorem. For proving the theorem, they invert a carefully defined class $D \in \pi^{C_8}_{19\rho_8}(N^8_2MU_{\mathbb{R}})$ and they can find an element in $...
Mingcong Zeng's user avatar
8 votes
0 answers
207 views

Fibrations of orthogonal G-spectra and fixed points

There are at least two fixed point functors that characterize stable equivalences of orthogonal G-spectra: the geometric fixed points and the naive fixed points of a fibrant replacement. Is this true ...
Emanuele Dotto's user avatar
7 votes
1 answer
455 views

Naive G-spectrum representing geometric equivariant cobordism

Let $G$ be a finite group. By the transversality results of Wasserman $G$-equivariant bordism (say real) should be a naive homology theory, and as such it should be represented by a naive G-spectrum. ...
Emanuele Dotto's user avatar
7 votes
0 answers
188 views

A confusion about geometric fixed points via spectral Mackey functors and smashing localisations

Let $G$ be a finite group and $N$ a normal subgroup. One of the modern ways to construct the $\infty$-category of $G$-spectra is as product-preserving spectral presheaves $\text{Sp}^G = \text{Fun}^{\...
user99383532's user avatar
6 votes
2 answers
407 views

What are the naive fixed points of a non-naive smash product of a spectrum with itself?

Let $X$ be an object in one of the well-known symmetric monoidal model categories of spectra. E.g., an $\mathtt S$-module in the sense of EKMM, or an orthogonal spectrum, or a $\Gamma$-space, etc. One ...
Gregory Arone's user avatar
6 votes
1 answer
594 views

Where can I find basic "computations" of equivariant stable homotopy groups?

I am new to this subject; so please correct me if I will say something wrong or if you don't like my notation. In particular, I don't know whether it is reasonable to consider an infinite group $G$ (...
Mikhail Bondarko's user avatar
5 votes
1 answer
182 views

Rational G-spectrum and geometric fixed points

For a finite group $G$, how is a rational $G$-spectrum $X$ detected by the geometric fixed point functor $\phi^H$ where we consider the conjugacy class of $H\leq G$? I tried finding a reference for ...
Steven Patrak's user avatar
4 votes
1 answer
158 views

$E^G_\ast(E)$ tensored with the rationals

Lemma 17.19 of Switzer's "Algebraic topology - Homology and Homotopy" states that $E_\ast(F)\otimes\mathbb{Q}$ is isomorphic to $\pi_\ast(E)\otimes\pi_\ast(F)\otimes\mathbb{Q}$. I wanted to ...
user avatar
4 votes
1 answer
198 views

Equivariant complex $K$-theory of a real representation sphere

Consider the one-point compactification of a $U(n)$-representation $V$, denoted by $S^V$. I want to caclulate $\tilde{K}_\ast^{U(n)}(S^V)$. When $V$ is a complex $U(n)$-representation, we can use the ...
user avatar
4 votes
1 answer
262 views

Calculate homotopy groups of $\mathbb{Z}_2$-equivariant loop spaces of "complex" topological spaces

Let $X$ be a topological space such that complex conjugation is defined (e.g. $\mathbb{C}^n$) and let us define the set of maps $$S_d:= \left\{f: (I^d,\partial I^d)\to (X,x_0)\mid \overline{f(k)} = f(...
Mathematics enthusiast's user avatar
4 votes
0 answers
119 views

Equivariant phantom maps

In the stable homotopy category a map of spectra $f\colon X \rightarrow Y$ is called phantom is the induced map between the associated homology theories $X_* \rightarrow Y_*$ is zero, it is know that ...
N.B.'s user avatar
  • 757
3 votes
0 answers
127 views

Equivariant classifying space and manifold models

The classifying space $BS^1$ for $S^1$-bundles can be taken to be the colimit of $\mathbb{CP}^n$ which are smooth manifolds and the inclusions $\mathbb{CP}^n \hookrightarrow \mathbb{CP}^{n+1}$ are ...
UVIR's user avatar
  • 933
3 votes
0 answers
117 views

Equivariant spectra with coefficients

In “The localization of spectra with respect to homology”, Bousfield describes localizations with respect to Moore Spectra. Given a spectrum $E$, and a group $M$, he describes the spectrum with ...
user avatar
3 votes
0 answers
81 views

Twists of equivariant spectra

Let $A$ be a spectrum, defined by deloopings $A_n$ (n an integer). Then the identity $A = S^1\wedge A_1$ together with antipodal equivariant spectrum structure on $S^1$ gives genuine $\mathbb{Z}/2$-...
Dmitry Vaintrob's user avatar
3 votes
0 answers
139 views

Maps between equivariant loop spaces

I have an elementary question about equivariant loop spaces that I feel it should be well known. Given a finite group $G$ and a finite $G$-set $J$ let $S^J=\mathbb{R}[J]^+$ be the permutation ...
Emanuele Dotto's user avatar
2 votes
1 answer
152 views

Double coset decomposition for compact Lie groups

The starting point of my question is the following fact: suppose $G$ is a finite group and let $H,K \leq G$ be arbitrary subgroups, then there exists an isomorphism of $G$-sets as follows \begin{...
N.B.'s user avatar
  • 757
2 votes
0 answers
194 views

Is the equivariant Steenrod algebra useful?

I am a newbie to the field, so please excuse any potential obvious gaps in knowledge. I have been wondering of late about the equivariant (dual) Steenrod algebra in the context of genuine $G = C_p$ ...
abelian_cat's user avatar
2 votes
0 answers
128 views

Geometric fixed points of induction spectrum

I was reading the paper "The Balmer spectrum of rational equivariant cohomology theories" of J.P.C. Greenlees and I found the following interesting fact, expressed in Lemma 4.2 and Remark 4....
N.B.'s user avatar
  • 757
1 vote
0 answers
78 views

Equivariant spectrum with coefficients

I am curious to know whether spectra with coefficients as defined in Adams's Blue book be defined to an equivariant setting. In the non-equivariant case, for a spectrum $E$ and an abelian group $A$, ...
anon's user avatar
  • 19