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

Fix an integer n. Can you find two finite CW-complexes X and Y which

* are both n connected,

* are not homotopy equivalent, yet

* $\pi_q X \approx \pi_q Y$ for all $q$.

In Are there two non-homotopy equivalent spaces with equal homotopy groups? some solutions are given with n=0 or 1. Along the same lines, you can get an example with n=3, as follows. If $F\to E\to B$ is a fiber bundle of connected spaces such that the inclusion $F\to E$ is null homotopic, then there is a weak equivalence $\Omega B\approx F\times \Omega E$. Thus two such fibrations with the same $F$ and $E$ have base spaces with isomorphic homotopy groups.

Let $E=S^{4m-1}\times S^{4n-1}$. Think of the spheres as unit spheres in the quaternionic vector spaces $\mathbb{H}^m, \mathbb{H}^n$, so that the group of unit quarternions $S^3\subset \mathbb{H}$ acts freely on both. Quotienting out by the action on one factor or another, we get fibrations $$ S^3 \to E \to \mathbb{HP}^{m-1} \times S^{4n-1},\qquad S^3\to E\to S^{4m-1}\times \mathbb{HP}^{n-1}.$$ The inclusions of the fibers are null homotopic if $m,n>1$, so the base spaces have the same homotopy groups and are 3-connected, but aren't homotopy equivalent if $m\neq n$.

There aren't any n-connected lie groups (or even finite loop spaces) for $n\geq 3$, so you can't push this trick any further.

Is there any way to approach this problem? Or reduce it to some well-known hard problem?

(Note: the finiteness condition is crucial; without it, you can easily build examples using fibrations of Eilenberg-MacLane spaces, for instance.)

share|improve this question

2 Answers 2

up vote 25 down vote accepted

Here is a method for constructing examples. If a fiber bundle $F \to E \to B$ has a section, the associated long exact sequence of homotopy groups splits, so the homotopy groups of $E$ are the same as for the product $F \times B$, at least when these spaces are simply-connected so the homotopy groups are all abelian. To apply this idea we need an example using highly-connected finite complexes $F$ and $B$ where $E$ is not homotopy equivalent to $F \times B$ and the bundle has a section.

The simplest thing to try would be to take $F$ and $B$ to be high-dimensional spheres. A sphere bundle with a section can be constructed by taking the fiberwise one-point compactification $E^\bullet$ of a vector bundle $E$. (This is equivalent to taking the unit sphere bundle in the direct sum of $E$ with a trivial line bundle.) The set of points added in the compactification then gives a section at infinity. For the case that the base is a sphere $S^k$, take a vector bundle $E_f$ with clutching function $f : S^{k-1} \to SO(n)$ so the fibers are $n$-dimensional. Suppose that the compactification $E_f^\bullet$ is homotopy equivalent to the product $S^n \times S^k$. If $n > k$, such a homotopy equivalence can be deformed to a homotopy equivalence between the pairs $(E_f^\bullet,S^k)$ and $(S^n \times S^k,S^k)$ where $S^k$ is identified with the image of a section in both cases. The quotients with the images of the section collapsed to a point are then homotopy equivalent. Taking the section at infinity in $E_f^\bullet$, this says that the Thom space $T(E_f)$ is homotopy equivalent to the wedge $S^n \vee S^{n+k}$. It is a classical elementary fact that $T(E_f)$ is the mapping cone of the image $Jf$ of $f$ under the $J$ homomorphism $\pi_{k-1}SO(n) \to \pi_{n+k-1}S^n$. The $J$ homomorphism is known to be nontrivial when $k = 4i$ and $n\gg k>0$. Choosing $f$ so that $Jf$ is nontrivial, it follows that the mapping cone of $Jf$ is not homotopy equivalent to the wedge $S^n \vee S^{n+k}$, using the fact that a complex obtained by attaching an $(n+k)$-cell to $S^n$ is homotopy equivalent to $S^n \vee S^{n+k}$ only when the attaching map is nullhomotopic (an exercise). Thus we have a contradiction to the assumption that $E_f^\bullet$ was homotopy equivalent to the product $S^n \times S^k$. This gives the desired examples since $k$ can be arbitrarily large.

These examples use Bott periodicity and nontriviality of the $J$ homomorphism, so they are not as elementary as one might wish. (One can use complex vector bundles and then complex Bott periodicity suffices, which is easier than in the real case.) Perhaps there are simpler examples. It might be interesting to find examples where just homology suffices to distinguish the two spaces.

share|improve this answer
    
Allen: you can avoid the J-homomorphism and Bott periodicity. All you need is a fibration $S^{k-1}\to E\to S^n$ with section given by clutching a essential map $S^{k-1} \to G_n \to F_n$ ($G_n$ = unbased homotopy automorphisms of $S^{n-1}$, $F_n$ = based homotopy automorphisms of $S^n$). Then the attaching map for the Thom complex is given by the Hopf construction of the adjoint to the clutching map $S^{k-1}\times S^{n-1} \to S^{n-1}$ and this is non-trivial because of essentialness of the clutching map. We only require non-triviality homotopy groups of spheres for this argument + Freudenthal. –  John Klein Jan 24 '11 at 13:25
    
I do realize that you are using Bott + $J$ to construct many examples for $(n,k)$ arbitrarily large. What I'm merely trying to point about is that once you know that $\pi_{k-1}^{\text{st}}(S^0)$ is non-trivial, you can cook up an example for some $n$ for that $k$. –  John Klein Jan 24 '11 at 14:09

EDIT: The two spaces I made are not n-connected; the homotopy groups up to n are trivial EXCEPT for the fundamental group.

The idea is to use the fact that a covering space for n-connected spaces ($n\ge 1$, i.e., connected with trivial fundamental group) introduces isomorphisms on homotopy groups.

Consider the CW-complexes $X=S^{n+1}, Y=S^{n+1+k}$ for some $k\ge 1$. Then $X$ and $Y$ both have free $\mathbb Z/2$ actions given by the antipodal map. So if we take $X\times Y$, then we can define $T_1$ to be the $\mathbb Z/2$-action on just the first factor with the identity on the second, and $T_2$ similarly acting only on the second factor.

If you quotient by $T_1$ you get $\mathbb {RP}^{n+1}\times S^{n+1+k}$ and if you quotient by $T_2$ you get $S^{n+1}\times \mathbb {RP}^{n+1+k}$. These will have all the same homotopy groups.

On the other hand, they can't be homotopy equivalent because they have different $\mathbb Z/2$ cohomology groups: the $\mathbb Z/2$ cohomology ring of $\mathbb {RP}^k$ is $\mathbb Z/2[x]/x^{k+1}$, and $S^k$ only has nontrivial $\mathbb Z/2$ cohomology at level $k$. So the cohomology of $\mathbb {RP}^{n+1}\times S^{n+1+k}$ has nontrivial elements at dimensions $1, \ldots n+1, n+1+k, \ldots n+2+k$, and the cohomology of $S^{n+1}\times \mathbb {RP}^{n+1+k}$ can have nontrivial elements at dimensions $1, \ldots n+2+k$.

share|improve this answer
2  
But they're not simply connected. –  Richard Kent Nov 11 '09 at 1:25
    
Right, I realized that immediately after posting. I don't know whether I should delete this answer or not, though. (Should I?) –  Elizabeth S. Q. Goodman Nov 11 '09 at 5:19
    
Please don't delete it. It's instructive nevertheless :-) –  Konrad Voelkel Nov 11 '09 at 16:09

Your Answer

 
discard

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.