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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.)

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.)

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.)

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.)