Homotopy groups of smooth manifolds?

Question

For a fixed $d$, is there a relationship between the homotopy groups of smooth $d$-manifolds?

The $d=1$ case is trivial, but I already don't know how to approach $d=3$ (I should have said that the case of $d=2$ is simple as well, since there is only a sphere to consider, but I don't know how to formulate the property of "having the same homotopy groups as $S^2$" in a simpler way).

Note about the discussion on the comments: it's unreasonable to expect an easy complete characterization of homotopy groups of $S^2$, even less for other manifolds. But I think one could try some partial relations. An interesting relationship would be: for some $d$, the groups $\pi_n$ can be determined from groups $\pi_m$ for $m<N<n$ (this is unlikely to be true though).

Answer 1

For $d=3$ the homotopy groups can be pretty elaborate. Consider the connect-sum of some lens spaces. The universal cover embeds in $S^3$ as the complement of a cantor set (except for a few degenerate cases where you have $\mathbb RP^3$ summands). So the homotopy-groups are pretty complicated ($\pi_2$ is finitely generated over $\pi_1$). You could probably make an argument that this is about the worst thing that can happen for the homotopy-groups of 3-manifolds.

You might want to phrase your question as a question about the Postnikov towers of manifolds. Eilenberg-Maclane spaces are rarely compact boundaryless manifolds.

edit: I guess another spin on your question could go like this. We know the fundamental groups of compact manifolds are all possible finitely presentable groups provided $n \geq 4$. So is there a sense in which the homotopy-algebras of manifolds can be anything finitely presentable? Say, for example, $\pi_2$. As a module over the group-ring of $\pi_1$, are there any restrictions beyond being finitely generated? I suppose you could construct a compact $6$-manifold with $\pi_2$ (almost) any finitely-presented thing over any finitely-presented $\pi_1$ pretty much the exact same way $4$-manifolds with any finitely presented $\pi_1$ are constructed. I think if $H_2(\pi_1)$ is non-trivial you might run into problems following the analogous construction, in that $\pi_2$ might strictly contain the $\pi_2$ you're trying to create.

2nd edit: So regarding 3-manifolds I think your question has something of an answer now, right? $\pi_n M$ is $\pi_n$ of the universal cover provided $n > 1$. The universal cover of a geometric 3-manfold is homeomorphic to $\mathbb R^3$ or $S^3$. So by climbing up the JSJ and connect sum decomposition of a 3-manifold, the universal cover is diffeomorphic to a punctured $S^3$ -- the number of punctures is either $0$, $1$, $2$ or a Cantor set's worth of punctures. In the Cantor set case we're giving this complement the compactly generated topology induced from the Cantor set complement's subspace topology. So among other things, $\pi_2 M$ is a direct sum of copies of $\mathbb Z$, similarly $\pi_3 M$, torsion first appears in $\pi_4 M$. The complement you think of as a directed system of wedges of $S^2$'s so the Hilton-Milnor theorem tells you the homotopy groups.

Answer 2

The case for $d=4$ also runs into complications, pretty much like it happens for $d=3$. One can show that the homotopy groups of a simply connected smooth closed oriented four manifold $M$ is completely determined by its second Betti number $b_2$. More precisely, if $b_2=k$ then the homotopy groups of $M$ are given by the homotopy groups of $\sharp^{k-1}S^2\times S^3$. Again, these groups are very hard to compute.

Answer 3

Arapura wrote "a brief guide to some recent work on fundamental groups of varieties". Pridham gave new new restrictions on etale fundamental groups of smooth varieties. Perhaps Sullivan's work relates to the question.