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visits | member for | 4 years, 5 months |
seen | Nov 28 at 21:22 | |
stats | profile views | 16,926 |
Nov 28 |
comment |
Failure of “equivariant triangulation” for finite complexes equipped with a $G$-action
Actually wait: I've misunderstood the question again, and what I said doesn't necessarily apply. (It applies when X comes from a finite object in the "genuine" equivariant category. But your finiteness condition looks like it might be weaker than that, even without allowing retracts.) |
Nov 28 |
comment |
Failure of “equivariant triangulation” for finite complexes equipped with a $G$-action
Ah, misunderstood you. I don't have very much helpful to suggest, then. One comment: in the special case G = Z/p, if a space X with an action of G is to have the form you describe, then it satisfies the Sullivan conjecture. In particular, the homotopy fixed points for the action of G on the p-adic completion X_p must itself arise as the p-adic completion of a finite space. Possibly this could give you a way to test if something is a counterexample? (A finite X w/G action such that (X_p)^hG is not the p-adic completion of anything finite?) |
Nov 28 |
comment |
Failure of “equivariant triangulation” for finite complexes equipped with a $G$-action
In Fun(BG,S), the orbits G/H generally won't be compact when H is nontrivial. But G itself generates the compact objects under finite colimits and retracts (this follows formally from the fact that it is a compact object which corepresents a conservative functor: in this case, the functor which forgets the G-action). |
Oct 18 |
awarded | at.algebraic-topology |
Oct 17 |
comment |
Associative Ring Spectra and Derived Completion
Let I be the fiber of the map A -> B. Then the fiber of the map A -> Tot^n is the (n+1)st smash power of I over A. For these fibers to vanish in the limit, it suffices that they are getting more connected. Since A is connective, it suffices for I to be connected, which is immediate from your hypothesis. |
Oct 17 |
answered | Associative Ring Spectra and Derived Completion |
Oct 13 |
awarded | Good Answer |
Oct 11 |
awarded | Enlightened |
Oct 11 |
comment |
Are there noncommutative extensions of $\alpha_p$ by $\mathbb{G}_m$?
Any line bundle over the spectrum of a local ring is trivial. |
Oct 11 |
awarded | Nice Answer |
Oct 10 |
answered | Are there noncommutative extensions of $\alpha_p$ by $\mathbb{G}_m$? |
Oct 7 |
awarded | Enlightened |
Sep 18 |
awarded | Pundit |
Sep 18 |
comment |
Commutation of simplicial homotopy colimits and homotopy products in spaces
For a counterexample in spaces, let X_* be the simplicial set with vertices the integers and edges joining each integer n to n+1. Regard X_* as a simplicial object of spaces whose geometric realization is contractible. Then a product of infinitely many copies of |X_*| is contractible. But the geometric realization of a product of infinitely many copies of X_* is not connected: for example, there's no path joining the identity map Z -> Z with the constant map joining Z -> {0} -> Z. (Contradicting your "elementary computation".) |
Sep 18 |
comment |
Commutation of simplicial homotopy colimits and homotopy products in spaces
Ah, I misunderstood you. Your original statement is false, even in the infty-category of spaces. (Also, it wouldn't follow from a general infty-topos from there, because left exact functors only preserve finite products. Also, sifted colimits don't commute with finite products in a general presentable infty-category.) |
Sep 18 |
comment |
Commutation of simplicial homotopy colimits and homotopy products in spaces
The statement that colimits commute with products (in each variable) is a special case of the assumption that colimits are universal (namely, that they are preserved by pullback along the map X -> *), which is one of the axioms on your list. |
Jul 18 |
awarded | Yearling |
Jul 17 |
awarded | Enlightened |
Jul 17 |
awarded | Nice Answer |
Mar 21 |
awarded | Notable Question |