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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-1xX,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?

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up vote 5 down vote accepted

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.

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More generally, the action of \pi_1 on \pi_n will be trivial if X is an H-space. – Eric Wofsey Oct 21 '09 at 15:04
Assuming that the H-space is connected. – Tyler Lawson Oct 21 '09 at 16:07

If you are thinking about π1, the action is just that by conjugation in this group, so is trivial iff the fundamental group is abelian.

I think the easist place to see the nontriviality of this action in higher dimensions is for the space S1 v Sn. By considering its universal cover and using Hurewicz's theorem, one can see that πn = Z[t, t-1] (= Zinfty), the ring of Laurent polynomials in a single variable t (as an additive group). You must then exercise the imagination to convince yourself that 1 in Z = π1 acts as multiplication by t.

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You don't need imagination to see that π_1 acts by t -- this is exactly what you are proving when you compute π_n = Z[t,t^-1]. Just go to the definition of the action of π_1 on π_n. – Tom Church Oct 21 '09 at 18:21

Just to make explicit what was implicit in the above answers -- while pi_1(X,v) is isomorphic to pi_1(X,w) for all v and w, the isomorphism depends on a choice of path from v to w. This isomorphism will be natural in the sense that it is independent of this choice of path if and only if pi_1(X) is abelian.

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Right, and the isomorphism only depends on the homotopy class of the chosen path. In other words, we can always interpret pi_n(X) as a local system of groups on X. The OP's question then is about when this local system is trivial. – Reid Barton Oct 21 '09 at 16:35

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