Four responses to Allen Hatcher's question about large homotopy groups....DMD ___________________________________________________ Date: Tue, 15 Aug 2000 17:19:57 -0400 (EDT) From: Philip Hirschhorn Subject: Re: query abt big htpy gps On Tue, 15 Aug 2000, Allen Hatcher wrote: > Date: Tue, 15 Aug 2000 15:41:17 -0400 > From: "Allen E. Hatcher" > Subject: Query for discussion list > > Is there a finite CW complex with infinite cyclic fundamental group > and some higher homotopy group not finitely generated as a module > over the fundamental group? How about with just an abelian > fundamental group? The classical example (due, I think, to Milnor), is S^1 wedged with S^2. If we call that space X, then the universal cover of X is R^1 with an S^2 wedged on at each integer. \pi_3 of the universal cover consists of the sum of the \pi_3 of each S^3 summed with a sum of Z's, where the generators of the Z's come from taking the Whitehead product of two of the generators of the \pi_2(S^2)'s. Thus, \pi_3 has one generator for each choice of a pair of integers (corresponding to the two integers at which you've selected the (S^2)'s). The fundamental group of X acts on those (S^2)'s, but it doesn't change the distance between two chosen (S^2)'s; that is, the generator you get by choosing, e.g., the (S^2)'s at n and at k is sent to the generator you get by choosing (n+i) and (k+i) for i equal to the element of the fundamental group that is acting. Thus, even after dividing by the action of the fundamental group, there are infintely many independent generators. Phil ---------------------------------------------------------------------- Philip Hirschhorn psh@math.mit.edu psh@poincare.wellesley.edu ---------------------------------------------------------------------- Subject: Re: query abt big htpy gps Date: Tue, 15 Aug 2000 16:30:34 -0500 From: Clarence Wilkerson I think the wedge of S^1 and S^3 provides an example of the first space desired. Rationals as a homotopy group? There are examples with a homotopy group being Z[1/N] for any integer you want, but I don't know one with Q . Clarence Wilkerson ___________________________________________________________ Date: Tue, 15 Aug 2000 19:25:02 -0500 From: Bill Richter Subject: Re: query abt big htpy gps Is there a finite CW complex with infinite cyclic fundamental group and some higher homotopy group not finitely generated as a module over the fundamental group? Yes, this comes up in high-dimensional knots. Simple example is X = S^1 vee S^n for n > 1. Then pi_{2n-1}(X) is infinitely generated. Proof: Take the Ranicki CW-pi complex W = tilde X / R = vee_Z S^n Then for i > 1 pi_i(X) = pi_i(W) and the Z[Z] = Z[t, t^{-1}] action is given shifting the wedge summands. The ordinary Hilton-Milnor thm says that pi_{2n-1}(W) contains a free abelian summand with generators the Whitehead products { [x_i, x_j] : i < j } where x_i: S^n ---> W is the inclusion of the t^i factor. Clearly the Z[t, t^{-1}] action is given by t [x_i, x_j] = [x_{i+1}, x_{j+1}] So pi_{2n-1}(W) contains an infinitely generated summand, which is a free Z[t, t^{-1}] module on the generators [x_0, x_j] for j in Z - 0. \qed Explanation: If X was actually a knot complement, then the stable homotopy groups of W must be finitely generated as a module over Z[Z], since Levine showed the homology groups are finitely generated. But the EHP sequence gives us pi_*(W smash W), and there's no reason for the homology groups of W smash W to be finitely generated. Now W smash_Z W is a finitely generated Ranicki CW-pi complex, but that's not what we're dealing with. -- Bill ____________________________________________________ From: "Brian Sanderson" Subject: Re: query abt big htpy gps Date: Wed, 16 Aug 2000 12:05:30 +0100 I believe there is no compact metric space with fundamental group the rational numbers. The proof relies on exotic (my term not technical) set theory. There was an artical in the Intelligencer some time ago with this result, if memory serves. Brian Sanderson http://www.maths.warwick.ac.uk/~bjs