Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Fernando Muro Oct 12 '12 at 22:10
1  
@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. –  Misha Oct 13 '12 at 3:22
    
@Misha, interesting, what paper of Whitehead is that? thanks! –  Fernando Muro Oct 13 '12 at 15:36
    
Doesn't $G$ need to be finitely presented? –  John Klein Oct 14 '12 at 2:13
1  
Sorry, despite my efforts to make sure I wrote "finitely presented" I ended up writing finitely generated. I'll edit the question. –  Mark Bell Oct 14 '12 at 10:45
show 1 more comment

1 Answer

up vote 16 down vote accepted

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.

share|improve this answer
3  
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. –  Donu Arapura Oct 13 '12 at 13:32
    
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. –  Johannes Ebert Oct 13 '12 at 17:45
1  
No way if $BG$ has infinite cohomological dimension. –  Fernando Muro Oct 13 '12 at 22:38
    
@Fernando: I meant that I could embed a skeleton of $BG$, not all of it. –  Johannes Ebert Oct 13 '12 at 23:31
    
Ah, I see, ok, ok. –  Fernando Muro Oct 13 '12 at 23:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.