Manifolds with prescribed fundamental group and finitely many trivial homotopy groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:02:20Z http://mathoverflow.net/feeds/question/109495 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109495/manifolds-with-prescribed-fundamental-group-and-finitely-many-trivial-homotopy-gr Manifolds with prescribed fundamental group and finitely many trivial homotopy groups qwerty1793 2012-10-12T21:42:45Z 2012-10-14T10:46:10Z <p>Fix $G$, a finitely <strike>generated</strike> presented group. </p> <p>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. </p> <p>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,</p> <blockquote> <p>Is it true that for each $n > 1$ there is a closed manifold $M$ such that $\pi_1(M) \cong G$ and for $1 &lt; i \leq n$, $\pi_i(M)$ is trivial?</p> </blockquote> <p>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 $1 &lt; i \leq n$) we obtain a simplical complex $X'$ with $\pi_1(X') \cong G$ and for $1 &lt; i \leq n$, $\pi_i(M)$ 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.</p> <p>Assuming that the answer to the first question is yes, can we also get manifolds of almost any dimension that we like?</p> http://mathoverflow.net/questions/109495/manifolds-with-prescribed-fundamental-group-and-finitely-many-trivial-homotopy-gr/109530#109530 Answer by Johannes Ebert for Manifolds with prescribed fundamental group and finitely many trivial homotopy groups Johannes Ebert 2012-10-13T13:14:46Z 2012-10-13T13:14:46Z <p>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 <a href="http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/" rel="nofollow">http://berstein.wordpress.com/2011/03/16/morse-theory-finiteness-properties-and-bieri-stallings-groups/</a>). 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).</p> <p>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".</p> <p>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. </p> <p>If you want to have $dim M \leq 2n$, you meet a new obstruction enforced by Poincare duality and things become really difficult. </p>