Manifolds with prescribed fundamental group and finitely many trivial homotopy groups

2012-10-12T21:42:45Z

Fix $G$, a finitely ~~generated~~ presented group.

It is known that for every $k > 3$ there is a closed $k$-manifold whose fundamental group is $G$. Similarly, there is a topological space with fundamental group $G$ and all higher homotopy groups trivial.

However, even for simple examples such as when $G \cong \mathbf{Z}_2$, such a topological space is not a manifold. It seems like the problem with these spaces really lies in the infinite constructions process adding in cells of arbitrarily high dimension. So instead if we only require the first $n$ homotopy groups to be trivial can we still work with manifolds. That is,

Is it true that for each $n > 1$ there is a closed manifold $M$ such that $\pi_1(M) \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ is trivial?

Note that if we allow $M$ to be a non-compact manifold / a manifold with boundary then the answer is yes. This follows as we can always find a finite simplicial complex $X$ whose fundamental group is $G$. By correctly adding $i$-cells (for $1 < i \leq n$) we obtain a simplical complex $X'$ with $\pi_1(X') \cong G$ and for $1 < i \leq n$, $\pi_i(M)$ trivial. By embedding $X'$ in a suitably high dimensional Euclidean space and taking an closed / open regular neighbourhood we obtain $M$, a non-compact manifold / manifold with boundary with the required properties.

Assuming that the answer to the first question is yes, can we also get manifolds of almost any dimension that we like?

Answer by Johannes Ebert for Manifolds with prescribed fundamental group and finitely many trivial homotopy groups

2012-10-13T13:14:46Z

No, the answer is negative in general (if you require $M$ to be compact). $M$ comes with a map $M \to BG$ that is, by definition, $n+1$-connected (iso on $\pi_i$ for $i=0,...,n$, epi on $\pi_{n+1}$). You can turn it into a weak equivalence by attaching cells of dimension $\geq n+1$. From that you see, that there is a model for $BG$ having finite $n$-skeleton. This is a special property of a group that is called $F_n$ (for more information, see http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/). Finitely presented groups are $F_2$ and you find that a necessary condition on your $G$ is that it is of type $F_n$. The are concrete examples of groups that are $F_i$ but not $F_{i+1}$ for each $i$, which are discussed in same blog post (on page 423 in Hatcher's AT, you find the same examples in a slightly different context).

On the other hand, let $G$ be $F_n$ and let $K$ be the $n$-skeleton of $BG$; a finite complex. Then I claim there is a closed manifold $M$ with the desired properties. $M$ can be chosen of arbitrary dimension $d \geq 4,2n+1$ and to be stably parallelizable. Start with a sphere $S^d \to K$ and do surgery on $S^d$ to get rid of the homotopy groups in low dimensions. The precise formulation is for example Proposition 4 in Kreck's paper "Surgery and duality".

So we can say that a necessary and sufficient condition is that $G$ is of type $F_n$. Caveat: I might have confused $n$ and $n+1$ at various places.

If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult.

Manifolds with prescribed fundamental group and finitely many trivial homotopy groups - MathOverflow

Manifolds with prescribed fundamental group and finitely many trivial homotopy groups

Answer by Johannes Ebert for Manifolds with prescribed fundamental group and finitely many trivial homotopy groups