Whitehead Products without Base Points? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T08:42:07Z http://mathoverflow.net/feeds/question/15904 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15904/whitehead-products-without-base-points Whitehead Products without Base Points? Chris Schommer-Pries 2010-02-20T16:52:58Z 2010-02-20T21:52:35Z <p>Let $(X, x_0)$ be a pointed space. Then we can define the homotopy groups $\pi_i(X, x_0)$ for $i \geq 1$. They are abelian groups for $i \geq 2$. It is well-known that the fundamental group $\pi_1(X, x_0)$ acts on each of the higher groups $\pi_i(X, x_0)$, and that this action generalizes to the <a href="http://en.wikipedia.org/wiki/Whitehead%5Fproduct" rel="nofollow">Whitehead Products</a> which are maps </p> <p>$$\pi_p(X, x_0) \times \pi_q(X, x_0) \to \pi_{p+q -1}(X, x_0).$$</p> <p>The details are given in the wikipedia article I linked to above. Together the Whitehead products turn the graded group $\pi_*(X, x_0)$ (for $* > 0$) into a graded (quasi-) Lie algebra over $\mathbb{Z}$, where the grading is shifted so that $\pi_i(X, x_0)$ is in degree $(i-1)$. Well, it is a little funny since the bottom group is not necessarily abelian. </p> <blockquote> <p>This is all well and good, but what if we don't want to pick base points? Is there a similar algebraic gadget in that situation?</p> </blockquote> <p>If we don't pick base points, then it seems natural to consider the fundamental groupoid $\Pi_1(X)$. Then the different homotopy groups of $X$ at different base points can be assembled into local systems on $X$. That is for each $i \geq 2$ we have a functor,</p> <p>$$\pi_i: \Pi_1 X \to AB$$</p> <p>where $AB$ is the category of abelian groups. This already incorporates the action of $\pi_1$ on the higher homotopy groups but does it in a way which doesn't depend on the choice of base point.</p> <blockquote> <p>Question: Can we enhance these local systems with a structure which generalizes the Whitehead product, and if so what precisely is this extra structure?</p> </blockquote> http://mathoverflow.net/questions/15904/whitehead-products-without-base-points/15934#15934 Answer by Anatoly Preygel for Whitehead Products without Base Points? Anatoly Preygel 2010-02-20T21:52:35Z 2010-02-20T21:52:35Z <p>As I posted in my comment, I think Paul's suggestion does work. Here's a (sloppy) description of how I think things will work:</p> <p>The local systems you describe can be obtained, by passing to homotopy groups, from a "local system of loop spaces" $$\Omega: \Pi_{\leq \infty} X \to \Omega\mathbf{Spaces}$$ One can imagine that this corresponds under the Grothendieck construction to the free loop-space fibration $\Omega X \to LX \to X$. Alternatively, if we fix a basepoint and identify $X = BG$ for a simplicial group $G$, then this is just encoding the simplicial conjugation action of $G$ on itself.</p> <p>Rather than think about (strangely-graded) Whitehead products, I prefer to think about (reasonably graded) Samelson products: We think of the structure (Whitehead product) on <code>$\pi_{*+1} X$</code> as really being a structure (Samelson product) on <code>$\pi_{*} \Omega X$</code>. I claim that Samelson products give a functor $$\pi_*: \Omega\mathbf{Spaces} \to \mathbf{grqLie}$$ so that composing with the above gives our desired "local system of graded (quasi-)Lie algebras".</p> <p>For convenience, I'll replace loop spaces with (strict) simplicial groups. Then, the Samelson product comes from noticing that the commutator map $[,]: G^2 \to G$ is trivial if one of the factors is the identity, and so factors through a pointed map $[,]: G \wedge G \to G$. This pointed map goes on to induce the (quasi-)Lie structure on homotopy. A group homomorphism $H \to G$ preserves commutators and identities, and so induces a map $H \wedge H \to G \wedge G$ compatible with the brackets, so that this construction is indeed functorial.</p>