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33 votes
Accepted

Did Peter May's "The homotopical foundations of algebraic topology" ever appear?

An anonymous source told me this question is here. Dylan gave the quick answer and Tyler referred to it. I'll use the question as an excuse to give a pontificating longer answer. When I first planned ...
Peter May's user avatar
  • 30.2k
14 votes

What's with equivariant homotopy theory over a compact Lie group?

Regarding 2, there is no difficulty in defining $G$-spectra in the setting of $\infty$-categories. The only complications I can think of are that (1) the orbit category $\mathrm{Orb}^G$ is now an $\...
Marc Hoyois's user avatar
  • 8,722
13 votes
Accepted

Extending a weak version of Sullivan's generalized conjecture

The answer to the first question is negative by a result of Dror Farjoun and Zabrodsky. In Fixed points and homotopy fixed points, they prove that if a finite group $G$ is not a $p$-group, then there ...
Marc Stephan's user avatar
13 votes
Accepted

Applications of equivariant homotopy theory to representation theory

There are decades and decades of algebraic results that use techniques from equivariant homotopy theory. Some examples ... (1) Quillen's work on ring theoretic aspects of the cohomology of finite ...
Nicholas Kuhn's user avatar
12 votes
Accepted

Definition of $Fun^G( \mathcal C, \mathcal D)$ in the setting of quasicategories

A(n ∞-)category with $G$-action is just a functor $BG\to \mathrm{Cat}_∞$. Then, if $\mathcal{C},\mathcal{D}$ are (∞-)categories with $G$-action, we can get another (∞-)category with $G$ action $\...
Denis Nardin's user avatar
  • 16.3k
10 votes
Accepted

Are two equivariant maps between aspherical topological spaces homotopic?

In the language of equivariant homotopy theory, your question is as follows. You have a group homomorphism $\phi\colon G\to H$, which you use to make $EH$ into a $G$-space, and then you ask whether $[...
Neil Strickland's user avatar
9 votes
Accepted

"Oriented representation" sphere

First of all, note that right before example 3.9 they prove that $$H^G_*(S^V;\underline{\mathbb{Z}})=H_*(C^{cell}_*(S^V)^G)\,,$$ where $C^{cell}_*(S^V)$ is the cellular complex for some $G$-CW-...
Denis Nardin's user avatar
  • 16.3k
9 votes
Accepted

$p$-adic equivalence of spectra with $G$-action

I don't know if "$2$-stage nilpotent" is a standard term for this, but I'm sure that what the author means is this: a group $A$ that $G$ is acting on has a subgroup $B$ such that the action of $G$ ...
Tom Goodwillie's user avatar
9 votes
Accepted

Applications of equivariant homotopy theory in chromatic homotopy theory

The canonical answer to this question is of course the celebrated solution by Hill, Hopkins and Ravenel to the Kervaire invariant one problem Hill, Michael A., Michael J. Hopkins, and Douglas C. ...
Denis Nardin's user avatar
  • 16.3k
9 votes

Are finite $G$-spectra idempotent complete?

I think the correct setting to look at this question is that of W. Lück, "Transformation groups and algebraic K-theory". Lecture Notes in Mathematics, 1408. Mathematica Gottingensis. ...
Oscar Randal-Williams's user avatar
9 votes
Accepted

An exact sequence involving THH

Let's extract a clear question (about spectra in general) from your question, and then answer it. Let $E$ be any spectrum. There is the degree $p$ map $p:S\to S$ from the sphere spectrum to itself. ...
Tom Goodwillie's user avatar
8 votes

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

I can answer your first question in some special cases. Let $p$ be a prime and $G=C_p$ the cyclic group of order $p$. If $p=2$, the answer to your question is yes and if $p$ is odd, then it is no. ...
Justin Noel's user avatar
  • 1,738
8 votes

When do non-exact functors induce morphisms on $K$-theory?

I think the idea should be that "polynomial maps" from A to B induce maps on the K-theory spaces. Here's one possible suggestion of a way to implement this, based on Segal's proof of the Kahn-Priddy ...
Dustin Clausen's user avatar
8 votes

Homotopy fixed points of complex conjugation on $BU(n)$

I think the answer is yes, after Bousfield-Kan $2$-completion. For $n=1$, $BO(1) \to BU(1)^{hC_2}$ is an equivalence, since $BO(1) \simeq K(\mathbb{Z}/2, 1)$, while $BU(1) \simeq K(\mathbb{Z}(1), 2)$ ...
John Rognes's user avatar
  • 9,018
8 votes
Accepted

Homotopy group action and equivariant cohomology theories

From modern perspective this is much more straightforward than the "genuine" version you described above the question. Naive $G$-spaces are just functors $BG\to \cal{S}$ among infinity ...
S. carmeli's user avatar
  • 4,074
7 votes
Accepted

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

In both the orthogonal world and the EKMM world one can set things up so that $G$-spectra are just spectra with an action of $G$, and the naive and genuine equivariant stable categories are the ...
Neil Strickland's user avatar
7 votes

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

In the context of functors with smash product (FSP), or symmetric spectra, or orthogonal spectra, the spectrum with $\Sigma_2$-action $X \wedge X$ prolongs essentially uniquely to a $\Sigma_2$-...
John Rognes's user avatar
  • 9,018
7 votes

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

For any finite abelian $G$ and $H\leq G$ we have a geometric fixed-point functor $\phi^H\colon\text{Sp}_G\to\text{Sp}$ which preserves smash products and sends the equivariant sphere $S^0_G$ to $S^0$. ...
Neil Strickland's user avatar
7 votes

Rational G-spectrum and geometric fixed points

This is implied by Theorem 3.10 in Wimmer's "A model for genuine equivariant commutative ring spectra away from the group order". Taking $R = \Bbb Q$ and $\mathcal F$ to be the full family ...
Tyler Lawson's user avatar
  • 51.8k
6 votes
Accepted

$RO(Q)$-graded homotopy fixed point spectral sequence

For a based $G$-space or $G$-spectrum $X$, the homotopy fixed point object $X^{hG}$ is by definition $F_G(EG_+,X)$. Suppose we write $EG$ as the colimit of a sequence of $G$-subspaces $A_k$. This ...
Neil Strickland's user avatar
6 votes
Accepted

When do non-exact functors induce morphisms on $K$-theory?

As suggested by Dustin Clausen in his answer, polynomial functors induce maps on $K$-theory. In the setting of stable $\infty$-categories, you proved this in your joint work with Barwick, Glasman, and ...
6 votes
Accepted

Universal space for the family of subgroups of a finite cyclic group

Let $X$ be a space with an action of $G$ such that $X^G=\emptyset$, and for every proper subgroup $H$, $X^H\ne \emptyset$. Then the infinite join of $X$ with itself is a universal space for the family ...
Gregory Arone's user avatar
6 votes

K-theory of free $G$-sets and the classifying space, and generalization

The general point is just that if $\mathcal{U}$ is equivalent to the free symmetric monoidal category $F\mathcal{C}$ generated by $\mathcal{C}$ then $K(\mathcal{U})\simeq\Sigma^\infty_+\mathcal{C}$. ...
Neil Strickland's user avatar
6 votes

How does one rigorously lift a map $Sp \rightarrow Sp$ of spectra to equivariant spectra?

Let $C$ be a complete $\infty$-category. Let $U:Fun(BC_n,C)\to C$ denote the forgetful functor, $\mathrm{CoInd}$ its right adjoint, and $(-)^{triv}$ the functor given by precomposition along $BC_n\to *...
Maxime Ramzi's user avatar
6 votes

The adjoint representation of a Lie group

For Question 2, there are only two kinds of such examples: the case where $G$ is commutative and even-dimensional, and the case where all simple factors of $G$ come in isomorphic pairs. (In the ...
LSpice's user avatar
  • 11.6k
6 votes
Accepted

$\operatorname{Spaces}/BG$ $\sim$ $\operatorname{Spaces}^G$ $\sim$ $??(\Omega G)$

If $A$ is a braided ∞-group, the delooping $\def\B{{\sf B}}\B A$ is an ∞-group. Consider the ∞-category of spaces equipped with an action of the ∞-group $\B A$. Since $\B Ω G≃G$, this ∞-category is ...
Dmitri Pavlov's user avatar
5 votes

Is the Milnor construction contractible

Oliver Straser is correct in that Milnor himself in 1956 [1] only showed his model for $EG$ is weakly contractible (that is, all its homotopy groups vanish) with his coarse topology on the join. (For ...
Qayum Khan's user avatar
5 votes

Homotopy group action and equivariant cohomology theories

Much has already been said in the other answers and comments, but let me summarize a few points. One way to obtain from a category a 'homotopy theory' (aka an $\infty$-category) is to specify a notion ...
Lennart Meier's user avatar
5 votes

Are finite $G$-spectra idempotent complete?

To complement Oscar's more systematic answer, let me expand my comment about the case $G = \mathbf{Z}/p\mathbf{Z}$ for a prime number $p$, where the answer is no when $\tilde{K}_0(\mathbf{Z}[G]) \neq ...
5 votes

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

Here is one possible approach, which is specific to the adjoint representation. Let $X$ denote $U(n)$, regarded as a $U(n)$-space by the rule $g.x=gxg^{-1}$. Put $X_k=\{x\in X:\text{rank}(x-1)\leq k\...
Neil Strickland's user avatar

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