Killing the torsion in homotopy

Origin

This question was asked by John Baez in This Week's Finds in Mathematical Physics (Week 286). Therefore, please don't upvote this question (unless you really want to), but do upvote the answers.

Background/Motivation

For a CW complex (here for simplicity we'll have $\pi_1 = 0$), you can do the operation of "rationalizing", which will change its homotopy $\pi_n \to \pi_n \otimes \mathbb Q$. This works by attaching enough cylinders so that each original cell is killed, but its subdivisions are born instead.

Question

Does there exist a similar procedure of "killing the torsion" which would change the homotopy of 1-connected CW complex from $\pi_n$ to $\pi_n/\pi_n^{tors}$?

Thoughts

One encounters problems if one just tries to kill off the cell: the procedure might have changed higher homology (this doesn't happen in rationalizing since cylinders are simple). So I suspect the answer is "No", but how to construct a counterexample?

• I think good questions should be voted up, no matter who asks it. Commented Jan 9, 2010 at 2:11

I don't have an answer to this question, but for the analogous question for homology it looks like it can't be done. By the universal coefficient theorem, a construction like this for homology would give a construction for cohomology as well. To get a counterexample in cohomology, take an Eilenberg-MacLane space $K({\mathbb Z},n)$ with $n$ even. This has rational cohomology a polynomial ring ${\mathbb Q}[x]$ with $x$ of degree $n$. It follows from this that if you factor the torsion out of the integral cohomology ring you get a polynomial ring ${\mathbb Z}[x]$ with $x$ of degree $n$, as one can see by looking at a map from ${\mathbb C}P^\infty$ to $K({\mathbb Z},n)$ that induces an isomorphism on $H^n(--;{\mathbb Z})$, using the fact that the integral cohomology of ${\mathbb C}P^\infty$ is a polynomial ring. However, there is no space whose integral cohomology ring is a polynomial ring on a generator of degree $n$ if $n > 4$, as one sees by looking at Steenrod squares and at Steenrod powers for the prime $p=3$. (This is Corollary 4L.10 in my book.)

In the context that Baez was talking about, rationalizing the homotopy groups is equivalent to rationalizing the homology groups, so it seems to be worth knowing that one can't kill torsion in homology, at least.

No, there is no such procedure. The problem is that attaching new cells can change the nontorsion in higher homotopy degrees and make it more divisible than it used to be.

One example: If X is BSp, which appears in the Bott periodicity sequence and has homotopy groups (starting in degree 1) 0, 0, 0, ℤ, ℤ/2, ℤ/2, 0, ℤ, ... If you attach a 6-dimensional cell to kill off the ℤ/2 in degree 5, then the ℤ in dimension 8 becomes divisible by 2. Any other map that kills off this ℤ/2 factors (noncanonically) through attaching such a 6-cell and so always has divisibility of the class in degree 8.

It might go without saying, but there is a procedure for non-simply connected spaces if you're killing a perfect torsion subgroup. It's just Quillen's plus construction used in the construction of algebraic k-theory.

• Which, of course, changes homotopy groups totally: $\pi_n(BGL(R)^+) = K_n(R)$. Commented Jan 5, 2010 at 19:44
• Right! The proof that the homology doesn't change (in the general construction) and that Quillen's K-groups agree with the classical ones (in the BGL case) are beautiful.
– Josh
Commented Jan 5, 2010 at 20:09

I believe it follows from Theorem 4.4 of "Neeman, Amnon Stable homotopy as a triangulated functor, Invent. Math. 109 (1992), no. 1, 17--40" that this procedure of killing the torsion will work if you invert the prime 2.

• Can you elaborate? That theorem says that if 2 is inverted, the functor sending a spectrum to its homotopy groups factors through a triangulated functor to the derived category of chain complexes over Z[1/2]. You can certainly carry out this factorization in the derived category, but do not see how one would lift this factorization to the spectrum level based on Neeman's Thm 4.4. (This is completely aside from the spectrum vs. space level issue.) Commented Jan 27, 2010 at 13:39
• Sorry I was only thinking about the spectrum case. I thought it eliminated the possibility of non-trivial extensions, but looking more closely I can't see why it should. Thanks for pointing that out. Commented Jan 28, 2010 at 10:30
• Tyler I also made a comment on your comment on noncommutative rational homotopy type. I believe that's an open question that has been asked before (as I recall Steve Halperin asked me that questions). Commented Jan 28, 2010 at 10:38

Zabrodsky does this in a paper on phantom maps. It's not functorial, but you can do it coherently for all the spaces and maps in a diagram that is finite (in the appropriate sense).