Timeline for G-equivariant Whitehead's Theorem
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 23, 2013 at 12:41 | vote | accept | Sean Lawton | ||
| May 23, 2013 at 12:11 | answer | added | Tom Goodwillie | timeline score: 10 | |
| Jul 18, 2012 at 4:51 | comment | added | Andy Putman | @Sean Lawton : It is true that smooth $G$-manifolds and $G$-semialgebraic sets admit $G$-CW structures when $G$ is a compact Lie group. See the following paper and the references therein : MR1770606 (2001j:57032) Illman, Sören(FIN-HELS) Existence and uniqueness of equivariant triangulations of smooth proper G-manifolds with some applications to equivariant Whitehead torsion. J. Reine Angew. Math. 524 (2000), 129–183. | |
| Jul 18, 2012 at 3:31 | history | edited | Sean Lawton | CC BY-SA 3.0 |
added 255 characters in body
|
| Jul 18, 2012 at 3:24 | comment | added | Sean Lawton | @Peter: Just out of curiosity, I guess smooth $G$-manifolds admit $G$-CW structures. What about (semi-)algebraic $G$-spaces (maybe stratify by orbit type, triangulate strata, and reassemble)? @Tom: Brilliant! Exactly what I needed. OK, so now further assume that the fixed locus is contained in the sub-complex $Y$ (or more weakly, the fixed locus $G$-equivariantly SDRs to a subspace of $Y$). | |
| Jul 18, 2012 at 2:53 | comment | added | Tom Goodwillie | Yes. Here is an example showing that orbits instead of fixed points isn't enough. Take a CW space $Z$ that is acyclic but not contractible. Let $X$ be the suspension of $Z$, and let $G$ of order $2$ act on it by switching the two cones, with $Z$ as fixed point set. Then both $X$ and the orbit space are contractible, but $X$ is not equivariantly contractible because the fixed point set is not contractible. | |
| Jul 18, 2012 at 2:18 | comment | added | Peter May |
Technical care is needed here: $G$-CW complex has a precise meaning, just like CW complexes but cells of the form $G/H \times D^n$'. So when $G$ is compact Lie, $n$-skeleta do not have geometric dimension $n$. Subcomplex must be taken in this equivariant sense, a union of equivariant cells. Then if $X^H \to Y^H$` is a weak homotopy equivalence, $X\to Y$ is the inclusion of a SDR. All bets are off for other guesses as to what a $G$-CW complex means, and for orbits replacing fixed points, and for merely nonequivariant SDR's. Just not in the cards.
|
|
| Jul 18, 2012 at 1:09 | comment | added | Sean Lawton | Great! That definitely answers the first question Tom. Thanks! But in that example, there is no SDR from $X$ to $Y$ at all. So what about the second question, with the addendum that I have a SDR from $X$ to $Y$ too (just not necessarily equivariant)? I had added that thinking that the answer to the first question was no. And I meant by "$G$-stable" simply that $G$ sends cells to cells, but I am happy assuming that it preserves the $n$-skeletons (i.e. acts by cellular maps). | |
| Jul 18, 2012 at 0:49 | comment | added | Tom Goodwillie | The usual statement is that if $X\to Y$ is an equivariant map of $G$-CW complexes and if for every closed subgroup $H$ the induced map of fixed-point spaces $X^H\to Y^H$ is a homotopy equivalence then in fact the map has an inverse up to equivariant homotopy. A reasonable question is, is the analogous statement true with orbits instead of fixed points? I presume not. | |
| Jul 18, 2012 at 0:41 | comment | added | Tom Goodwillie | I am guessing that by $G$-stable you mean that $G$ acts by cellular maps? Which is more or less what they mean by $G$-CW complex. If so, then no. Think of $G$ of order $2$ acting on a circle $X$ by reflection, and $Y$ one of the two fixed points. | |
| Jul 18, 2012 at 0:15 | history | asked | Sean Lawton | CC BY-SA 3.0 |