free homotopy groups -- when do they exist? - MathOverflow

free homotopy groups -- when do they exist?

2009-10-21T07:29:41Z

Let (X,x) be a pointed space. There is an action of π₁(X,x) on π_n(X,x) -- determined by considering π_n(X,x)=π_n-1(Ω_xX,x), where Ω_xX 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 π_n(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 π_n will or won't be trivial, and does this depend on n?

Answer by Oscar Randal-Williams for free homotopy groups -- when do they exist?

2009-10-21T07:42:51Z

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 π_n = Z[t, t^-1] (= Z^infty), 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.

Answer by Charles Rezk for free homotopy groups -- when do they exist?

2009-10-21T14:16:26Z

For the last part of your question: given a group π₁ which acts on an abelian group π_n, 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(π_n,n) in such a way that realizes the action on π_n, and then use this to build a space as a fibration X-->K(π₁,1) with fiber K(π_n,n). So the only general limitation is in how the group π₁ can act on the group π_n.

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 π_n(X) is always trivial; the idea is that ΩX=Ω²Y, in which loop composition is commutative up to homotopy.

Answer by Andy Putman for free homotopy groups -- when do they exist?

2009-10-21T14:35:53Z

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.