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?

  • $\begingroup$ When you start killing homotopy groups you may need to attach infinitely-many new cell in each step, even if $G$ is finitely presented. As a consequence of this, you may not be able to embed the resulting CW-complex in an euclidean space of any dimension. $\endgroup$ Oct 12, 2012 at 22:10
  • 1
    $\begingroup$ @Fernando: This is nota problem, you deform this complex to a locally finite one (proven by Whitehead in 1947), which then admits a proper embedding in Euclidean space. $\endgroup$
    – Misha
    Oct 13, 2012 at 3:22
  • $\begingroup$ @Misha, interesting, what paper of Whitehead is that? thanks! $\endgroup$ Oct 13, 2012 at 15:36
  • $\begingroup$ Doesn't $G$ need to be finitely presented? $\endgroup$
    – John Klein
    Oct 14, 2012 at 2:13
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    $\begingroup$ Sorry, despite my efforts to make sure I wrote "finitely presented" I ended up writing finitely generated. I'll edit the question. $\endgroup$
    – Mark Bell
    Oct 14, 2012 at 10:45

1 Answer 1


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.

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    $\begingroup$ It's always awkward when answers reach different conclusions, but I am convinced by this. I guess one could (almost) rephrase the first part by saying that if the group has infinitely generated homology in some degree $< n$, then no such compact manifold can exist. $\endgroup$ Oct 13, 2012 at 13:32
  • $\begingroup$ The (now deleted) answer answered a different question, namely if one could have a noncompact manifold without boundary. That is certainly true and not so difficult: if $G$ is countable, then the standard $BG$ has countable skeleta that can be deformed and embedded into some $R^k$ as in Mishas comment. Take an open neighborhood. $\endgroup$ Oct 13, 2012 at 17:45
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    $\begingroup$ No way if $BG$ has infinite cohomological dimension. $\endgroup$ Oct 13, 2012 at 22:38
  • $\begingroup$ @Fernando: I meant that I could embed a skeleton of $BG$, not all of it. $\endgroup$ Oct 13, 2012 at 23:31
  • $\begingroup$ Ah, I see, ok, ok. $\endgroup$ Oct 13, 2012 at 23:33

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