Let (X,x) be a pointed space. There is an action of π1(X,x) on πn(X,x) — determined by considering πn(X,x)=πn-1(ΩxX,x), where ΩxX denotes the space of loops in X based at x, and x denotes the constant loop — given simply by conjugation. We can speak unambiguously of πn(X), the free (i.e., not necessarily basepoint-preserving) homotopy group exactly when this action is trivial.
On an algebraic level I'm fine with this, but I'm having trouble envisioning how a homotopy class might be conjugated to a different homotopy class in this way. Besides my admittedly small collection of toy examples, my issue could also be that I'm mainly thinking about π1, in which case it might (???) be that the action is trivial. (I seem to recall that before learning about general homotopy theory, I heard a statement along the lines of "for path-connected spaces, you may as well ignore basepoints". Certainly the groups are all isomorphic, but I'm not certain whether there is a unique natural isomorphism.)
Also, are there (necessary and/or sufficient) conditions for when the π1 action on πn will or won't be trivial, and does this depend on n?
Best Answer
For the last part of your question: given a group π1 which acts on an abelian group πn, there is always as space X with these homotopy groups with this action, and you can manufacture one using Eilenberg-MacLane spaces. You can make the group π1 act on the Eilenberg-MacLane space K(πn,n) in such a way that realizes the action on πn, and then use this to build a space as a fibration X-->K(π1,1) with fiber K(πn,n). So the only general limitation is in how the group π1 can act on the group πn.
There is one condition on a space X which implies the action is trivial: if X is a loop space (i.e., X is homotopy equivalent to ΩY for some Y), then the action of π1(X) on πn(X) is always trivial; the idea is that ΩX=Ω2Y, in which loop composition is commutative up to homotopy.