For every sequence of homotopy groups is there a corresponding manifold?

Question

Some general questions about homotopy groups:

1. For any sequence $\{G_n\}_n$ does there exist a path-connected topological space $X$ with $\pi_n(X)=G_n$ for all $n$? I am aware that the answer is no if we do not impose that $G_n$ is abelian for $n\ge 2$, so I impose this.
2. Same question as above but for $X$ a manifold?

I suspect the answer is probably no for (1) and (2) - I am just unaware of general conditions on such sequences $\{\pi_n(X)\}_n$ that must be satisfied, other than the one I mentioned.

Answer 1

1. Yes. For every $n \ge 1$ we can construct an Eilenberg-MacLane space $K(G_n, n)$, which is a path-connected topological space with $\pi_n \cong G_n$ and all other homotopy groups zero (with the restriction that if $n \ge 2$ we need $G_n$ to be abelian, of course), and which is furthermore unique up to (weak) homotopy equivalence. Then we can take the product $\prod_n K(G_n, n)$.

2. No. For example, if exactly one of the $G_n$ is nonzero then the only possibility up to homotopy is an Eilenberg-MacLane space, and these generally have cohomology in arbitrarily high degrees so can't be homotopy equivalent to manifolds. E.g. if $G_2 = \mathbb{Z}$ and all the other groups are zero then we get (up to homotopy) the infinite complex projective space $\mathbb{CP}^{\infty}$, whose cohomology is $\mathbb{Z}$ in every even degree.

If you additionally restrict attention to manifolds of "finite type" (homotopy equivalent to finite CW complexes) then there are stronger restrictions. First of all, $\pi_1(X)$ must be finitely presented, and if $X$ is simply connected then all of the higher homotopy groups $\pi_n(X)$ must also be finitely presented (or equivalently, in the abelian case, finitely generated).

In the simply connected case, furthermore the dichotomy theorem asserts that the rational homotopy groups $\pi_n(X) \otimes \mathbb{Q}$ are either eventually zero or have dimensions that grow exponentially. It follows, for example, that $G_n \cong \mathbb{Z}, n \ge 2$ is impossible for any space homotopy equivalent to a finite CW complex, or even $G_n \cong \mathbb{Z}^n$ or more generally $G_n \cong \mathbb{Z}^{f(n)}$ for any sequence $f(n)$ which is not eventually zero but grows slower than exponentially, and we could even throw in arbitrary torsion. This is a consequence of a more structural theorem about how the rational homotopy groups behave as a Lie algebra (essentially the finite CW complex condition means they have to be finitely generated as a Lie algebra).

Answer 2

The easier question (without the manifold restriction) is actually not that hard: Consider $\prod_{n \geq 1} K(G_n,n)$, where $K(G_n,n)$ denotes the Eilenberg-Maclane space.