Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T06:18:32Z http://mathoverflow.net/feeds/question/68888 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68888/is-a-compact-connected-orientable-3-manifold-with-mathbbzk-fundemental-gr Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? Benjamin Horowitz 2011-06-26T23:34:30Z 2011-06-28T05:15:21Z <p>According to the Kneser-Milnor prime decomposition theorem for 3-manifolds, any compact, connected, orientable 3-manifold $M$ is diffeomorphic to $S^3 / \Gamma_1$ # $\cdots$ # $S^3/ \Gamma_n$ # $(S^2 \times S^1)_1$ # $\cdots$ # $(S^2 \times S^1)_r$ # $K( \pi_1,1)$ # $\cdots$ # $K( \pi_m,1)$, where # is the connect sum, and $\Gamma_i$ is a non-trivial finite subgroup of $SO(4)$ acting orthogonally to $S^3$ (so that the result is a spherical space form).</p> <p>I suppose my question boils down to if $\pi_1(M)=\mathbb{Z} \times \cdots \times \mathbb{Z}$, k times, does that uniquely identify a manifold as a $(S^2 \times S^1)_1$ # $\cdots$ # $(S^2 \times S^1)_k$, or can the various quotient manifolds or aspherical factors create a more complicated topology without changing the fundamental group?</p> http://mathoverflow.net/questions/68888/is-a-compact-connected-orientable-3-manifold-with-mathbbzk-fundemental-gr/68889#68889 Answer by Igor Rivin for Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? Igor Rivin 2011-06-26T23:39:25Z 2011-06-27T02:06:31Z <p>NO, since the three-torus $T^3$ does not have this form.</p> <p><strong>EDIT</strong> if the OP really means a free product of $\mathbb{Z}$s, so the free group $F_k,$ then the answer is YES. It is a fact (see Hempel's book, chapter 7) that every splitting of the fundamental group of $M^3$ as a free product comes from a connected sum decomposition. On the other hand, a prime three manifold is either a $K(\pi, 1)$ or $S^2 \times S^1.$ In the former case, its fundamental group cannot be $\mathbb{Z}$</p> http://mathoverflow.net/questions/68888/is-a-compact-connected-orientable-3-manifold-with-mathbbzk-fundemental-gr/68926#68926 Answer by Dave Futer for Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? Dave Futer 2011-06-27T13:39:22Z 2011-06-27T13:57:33Z <p>Let me address a more general question:</p> <blockquote> <p>To what extent is a closed, connected $M^3$ determined by its fundamental group?</p> </blockquote> <p>Following the <a href="http://en.wikipedia.org/wiki/Geometrization_conjecture" rel="nofollow">Geometrization Theorem</a>, we have a complete answer. There are only two ways in which a closed $3$--manifold $M$ can fail to be determined by its fundamental group:</p> <ol> <li><p>$M$ is a lens space, or a connected sum of something with a lens space. It is <a href="http://en.wikipedia.org/wiki/Lens_space#Classification_of_3-dimensional_lens_spaces" rel="nofollow">well-known</a> that lens spaces are not determined up to homeomorphism by their fundamental groups. Note that in this case, you would see a ${Z}/p$ free factor in your group $G$, so it can't arise in the context of your question.</p></li> <li><p>$M = N_1 \sharp N_2$, where each $N_i$ is orientable, and each is chiral (fails to have an orientation-reversing symmetry). In this case, reversing the orientation on one factor would produce $M' = N_1 \sharp \overline{N_2}$, which is not homeomorphic to $M$ but has the same fundamental group. This also cannot arise in your context, for exactly the reasons that Igor outlined, and because $S^1 \times S^2$ <em>does</em> have an orientation-reversing symmetry.</p></li> </ol> <p>Finally, let me point out that although Geometrization may seem like an overly big hammer for this question, in fact one needs the Poincare conjecture. For, if there existed a fake $3$--sphere, one could take the connect sum with that manifold without altering the group.</p> http://mathoverflow.net/questions/68888/is-a-compact-connected-orientable-3-manifold-with-mathbbzk-fundemental-gr/69000#69000 Answer by Steve D for Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined? Steve D 2011-06-28T05:15:21Z 2011-06-28T05:15:21Z <p>If you did in fact mean the free abelian group, you can still do it (up to punctures). All references in brackets are to Hempel.</p> <p>If $\pi_1(M)\cong \mathbb{Z}$, then $M$ is one of $S^2\times S^1$, $S^2\tilde{\times} S^1$, the solid torus, or the solid Klein bottle. [5.3]</p> <p>If $\pi_1(M)\cong \mathbb{Z}^2$, then $M$ is an I-bundle over the torus. [10.6]</p> <p>If $\pi_1(M)\cong \mathbb{Z}^3$, then $M$ is the 3-torus. [11.11]</p> <p>The case $k>3$ doesn't happen. [9.13]</p>