Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


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.


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.


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


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?

share|improve this question
I think good questions should be voted up, no matter who asks it. –  Sam Derbyshire Jan 9 '10 at 2:11

5 Answers 5

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
Which, of course, changes homotopy groups totally: $\pi_n(BGL(R)^+) = K_n(R)$. –  Ilya Nikokoshev Jan 5 '10 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. –  jd.r Jan 5 '10 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.

share|improve this answer
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.) –  Tyler Lawson Jan 27 '10 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. –  Don Stanley Jan 28 '10 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). –  Don Stanley Jan 28 '10 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).

share|improve this answer
Is this the paper you're talking about? springerlink.com/content/127q151u7v86h4w4 –  j.c. Jan 29 '10 at 20:48
Please correct me if I'm wrong, but it looks like what Zabrodsky does is a little weaker than what I interpreted the original question to ask, which was for a map that induced a surjection on homotopy groups with kernel the torsion subgroups. Zabrodsky only asks for a map that induces an isomorphism on rational homotopy groups and kills the torsion, so after factoring out torsion in the domain it would be injective but not necessarily surjective. Tyler's example from January 5 is a counterexample to the stronger requirement but not to Zabrodsky's weaker requirement. –  Allen Hatcher Jan 30 '10 at 5:20
Yes, that is the paper, and no, it doesn't do the full job. –  Jeff Strom Jan 31 '10 at 0:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.