Timeline for Whitehead Products without Base Points?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2010 at 0:31 | vote | accept | Chris Schommer-Pries | ||
| Feb 21, 2010 at 0:31 | comment | added | Chris Schommer-Pries | @Paul & Anatoly: You are right. I made a sloppy mistake thinking the action of $\pi_1$ was the same as the Lie bracket. You can see that it is not if you look carefully at the definition of the bracket. You can also see that it is indeed functorial as you suggest. Using universal covers doesn't quite work because the universal cover is only functorial after picking a base point (you can see that from the constructions of it). Thanks! | |
| Feb 20, 2010 at 21:52 | answer | added | Anatoly Preygel | timeline score: 6 | |
| Feb 20, 2010 at 21:45 | comment | added | Paul | Did you look at how Sullivan handled this when he did rational homotopy theory for non-simply connected spaces? That would correspond to the (easier) case of $\pi_k\otimes Q$. I think of such things in terms of universal covers rather than groupoids and/or l.c. sheaves, and let the covering transformations deal with the action of $\pi_1$. What happens if you consider the homotopy groups of the universal cover based at different points? you should get another description since the homotopy groups of a cover coincide with the base, and since Whitehead products are preserved by homeos. | |
| Feb 20, 2010 at 21:31 | comment | added | Anatoly Preygel | Rather, the formula shows that bracketing with elements of $\pi_1$ gives $\pi_1$-module morphisms. I think I have a conceptual reason why things should work, so I'll post that in an answer. | |
| Feb 20, 2010 at 21:28 | comment | added | Anatoly Preygel | Chris, are you sure you that Paul's suggestion doesn't work? The action of $\pi_1$ on $\pi_k$ is not by the Lie bracket: If $g \in pi_1$ and $m \in \pi_k$, then (up to a sign) $[g,m] = gm - m$. (For instance, the usual action of $\pi_1$ on itself is by conjugation so this recovers $[g,h]=g h g^{-1} h^{-1}$.) Using this last formula, I think one gets that the Whitehead algebra is a Lie algebra in $\pi_1$-modules. | |
| Feb 20, 2010 at 19:52 | comment | added | Chris Schommer-Pries | My motivation for this comes from trying to follow up on my previous MO questions: mathoverflow.net/questions/14266/… and mathoverflow.net/questions/430/… . Basically those questions led me to try to understand the homotopy of topological commutative monoids and I need a sufficiently rich algebraic invariant to help with this. It seems very important in that world that you can't simply reduce a single component and a single base point. This question is a step towards finding that invariant. | |
| Feb 20, 2010 at 19:45 | comment | added | Chris Schommer-Pries | That was my first guess too. I played around with that a little, and the problem is that it didn't seem to give a functor to graded Lie algebras. The paths act by something which is more like a derivation. You can see this by looking at how $\pi_1$ acts. It acts by the Lie bracket [x,-]. I'm hoping someone who is more familiar with Whitehead product or graded Lie algebras will know what this structure is. | |
| Feb 20, 2010 at 19:00 | comment | added | Paul | What's wrong with just taking all higher homotopy groups as a functor $\oplus_{k>1}\pi_k:\Pi_1(X)\to $graded Lie algebras with $\pi_1$ actions? As you note, you can separate out the $\pi_1$ part. Also, could you explain why you don't want to pick base points? Whitehead products are obscure and tricky to calculate (except as commutators on $\pi_1$), and I'm curious about what context you are thinking about. | |
| Feb 20, 2010 at 16:52 | history | asked | Chris Schommer-Pries | CC BY-SA 2.5 |