# Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalent

A standard example of two CW complexes which have isomorphic homotopy groups but are not homotopy equivalent is $RP^2 \times S^3$ and $RP^3 \times S^2$.

The easiest way to see that they are not homotopy equivalent is by looking at their homology (i.e. ala the Kunneth formula).

What is an example of two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalent?

It seems like an obvious place to look would be to have non-isomorphic cohomology (as rings). If there is such an example, then the next question would be for an example of two non-homotopy equivalent CW complexes with isomorphic homotopy groups, and isomorphic homology (as graded groups) and cohomology (as rings). Do such examples exist? I suspect they do, but I don't know any.

Update: there are good answers below to the first question, with links. But it seems to still beg the question about cohomology: does anyone know an example of two non-homotopy equivalent CW complexes with isomorphic homotopy groups, homology groups, and cohomology rings?

• – Grigory M Dec 6 '11 at 18:02
• (Re: update) See Johannes Ebert's answer @ MO – Grigory M Dec 6 '11 at 20:27

$S^2\times S^2$ and $\mathbb CP^2\#\overline{\mathbb CP^2}$ have the same homotopy and homology groups but not homotopy equivalent since they have non-isomorphic cohomology rings (link).

• You can also add $\mathbb{C}P^2 \#\mathbb{C}P^2$ to that list. – Jason DeVito Jun 3 '12 at 0:56

Sorry to be late to the party, but here's an example of 2 compact simply connected manifolds which have the same homotopy groups, same homology groups, same cohomology ring, and yet are not homotopy equivalent. The examples are motivated by Grigory M's examples:$S^2\times S^2$ and $\mathbb{C}P^2\#\overline{\mathbb{C}P^2}$. His examples are both $S^2$ bundles over $S^2$.

If we extend this further, it turns out there are precisely two $S^3$ bundles over $S^2$. Of course, one is the product $S^3\times S^2$, while another doesn't have a more common name, so I'll just denote it $S^3\hat{\times} S^2$.

Both of these spaces are diffeomorphic to quotients of free linear $S^1$ actions on $S^3\times S^3$. Letting $X$ denote either bundle, we have a long exact sequence of homotopy groups $$...\pi_k(S^1)\rightarrow \pi_k(S^3\times S^3)\rightarrow \pi_k(X)\rightarrow \pi_{k-1}(S^1)\rightarrow ...$$

which can be used to show that $\pi_k(X) \cong \pi_k(S^3\times S^3)$ for $k\geq 2$ and $\pi_1(X) = \{e\}$ and $\pi_2(X) \cong \mathbb{Z}$.

The Hurewicz theorem together with Universal coefficients theorem implies $H^1(X) = 0$ and $H^2(X) \cong \mathbb{Z}$. Poincare duality then forces the rest of the cohomology rings to agree.

Finally, to see $S^3\times S^2$ and $S^3\hat{\times} S^2$ are different, one computes the Stiefel-Whitney classes of their tangent bundles. It turns out $w_2(S^3\times S^2) = 0$ while $w_2(S^3\hat{\times}S^2)\neq 0$. (And all other Stiefel-Whitney classes are $0$ for both spaces). Since the Stiefel-Whitney classes can be defined in terms of Steenrod powers, they are homotopy invariants, so $S^2\times S^3$ and $S^3\hat{\times}S^2$ are not homotopy equivalent.

• Looking back on this example years later, I feel I should add that there is nothing special about $S^3$ bundles over $S^2$. In fact, there are always precisely two linear $S^n$ bundles over $S^2$ (for $n\geq 2$). For $n\geq 3$, they have isomorphic homotopy groups and cohomology rings, but $w_2$ distinguishes them. – Jason DeVito Sep 6 '15 at 14:30

The lens spaces $L(p,q_1)$ and $L(p,q_2)$ are homotopy equivalent if and only if $q_1q_2 \equiv \pm n^2 \mod p$ for some $n\in \mathbb{Z}$ (and homeomorphic if and only if $q_1 = \pm q_2^{\pm} \mod p$.) The homology groups and cohomology ring only depend on $p$.

The homotopy groups are isomorphic as well, since each $L(p,q)$ is the quotient of $S^3$ by a $\mathbb{Z}/p$ action. Off the top of my head I'm not sure how $\pi_1$ acts on the higher groups.

I am not sure about the homotopy groups, but the standard example is to look at $S^2\vee S^4$ and $\mathbb{C}P^2$. They have the same homology groups and the same cohomology groups, but the cohomology rings are different. Now suspend each space once and now they have the same ring structure, all products are zero. However, they are different modules over the Steenrod algebra.

Again, I don't know if the homotopy groups are the same. In fact, there stable homotopy groups must be different because of an Adams Spectral Sequence computation.

• $\pi_3(S^2\vee S^4)=\pi_3(S^2)=\mathbb Z$, $\pi_3(\mathbb CP^2)=\pi_3(S^5)=0$. – Grigory M Dec 7 '11 at 5:03