Skip to main content
MathOverflow

Return to Answer

added 31 characters in body
Source Link
Ian Agol
Ian Agol
  • 68.9k
  • 3
  • 194
  • 358

A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points out, a necessary condition is that the fundamental group is isomorphic to one of its proper finite-index subgroups, which is rather strong (in particular, implies that the group is not co-Hopfian). If such a group is also finitely presented, then it will be realized as the fundamental group of a 4-manifold (or $n$-manifold, for $n\geq 4$). However, it's not at all clear when the corresponding finite covers will be homeomorphic.

Examples comes from the non-co-Hopfian groups $BS(1,k)$. These are ascending extensions of cyclic-by-cyclic groups by cyclic groups, and unwrapping the fiber gives isomorphic finite-index subgroups: $BS(1,k)\cong \langle a, b | aba^{-1}=b^k\rangle$, with index $k-1$ subgroup $\langle a, b^{k-1}\rangle \cong BS(1,k)$. 4-manifolds with such fundamental group were classified by Hambleton-Kreck-Teichner. Their classification is (essentially) in terms of the equivariant intersection form on $\pi_2(M)$. When this intersection form is trivial (which they prove is realized in Theorem B(i)), then the covers will be homeomorphic by Theorem A (this example came out of a discussion I had with van Limbeek and Teichner).

A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points out, a necessary condition is that the fundamental group is isomorphic to one of its proper finite-index subgroups, which is rather strong (in particular, implies that the group is not co-Hopfian). If such a group is also finitely presented, then it will be realized as the fundamental group of a 4-manifold (or $n$-manifold, for $n\geq 4$). However, it's not at all clear when the corresponding finite covers will be homeomorphic.

Examples comes from the non-co-Hopfian groups $BS(1,k)$. These are cyclic-by-cyclic groups, and unwrapping the fiber gives isomorphic finite-index subgroups: $BS(1,k)\cong \langle a, b | aba^{-1}=b^k\rangle$, with index $k-1$ subgroup $\langle a, b^{k-1}\rangle \cong BS(1,k)$. 4-manifolds with such fundamental group were classified by Hambleton-Kreck-Teichner. Their classification is (essentially) in terms of the equivariant intersection form on $\pi_2(M)$. When this intersection form is trivial (which they prove is realized in Theorem B(i)), then the covers will be homeomorphic by Theorem A (this example came out of a discussion I had with van Limbeek and Teichner).

A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points out, a necessary condition is that the fundamental group is isomorphic to one of its proper finite-index subgroups, which is rather strong (in particular, implies that the group is not co-Hopfian). If such a group is also finitely presented, then it will be realized as the fundamental group of a 4-manifold (or $n$-manifold, for $n\geq 4$). However, it's not at all clear when the corresponding finite covers will be homeomorphic.

Examples comes from the non-co-Hopfian groups $BS(1,k)$. These are ascending extensions of cyclic groups by cyclic groups, and unwrapping the fiber gives isomorphic finite-index subgroups: $BS(1,k)\cong \langle a, b | aba^{-1}=b^k\rangle$, with index $k-1$ subgroup $\langle a, b^{k-1}\rangle \cong BS(1,k)$. 4-manifolds with such fundamental group were classified by Hambleton-Kreck-Teichner. Their classification is (essentially) in terms of the equivariant intersection form on $\pi_2(M)$. When this intersection form is trivial (which they prove is realized in Theorem B(i)), then the covers will be homeomorphic by Theorem A (this example came out of a discussion I had with van Limbeek and Teichner).

Source Link
Ian Agol
Ian Agol
  • 68.9k
  • 3
  • 194
  • 358

A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points out, a necessary condition is that the fundamental group is isomorphic to one of its proper finite-index subgroups, which is rather strong (in particular, implies that the group is not co-Hopfian). If such a group is also finitely presented, then it will be realized as the fundamental group of a 4-manifold (or $n$-manifold, for $n\geq 4$). However, it's not at all clear when the corresponding finite covers will be homeomorphic.

Examples comes from the non-co-Hopfian groups $BS(1,k)$. These are cyclic-by-cyclic groups, and unwrapping the fiber gives isomorphic finite-index subgroups: $BS(1,k)\cong \langle a, b | aba^{-1}=b^k\rangle$, with index $k-1$ subgroup $\langle a, b^{k-1}\rangle \cong BS(1,k)$. 4-manifolds with such fundamental group were classified by Hambleton-Kreck-Teichner. Their classification is (essentially) in terms of the equivariant intersection form on $\pi_2(M)$. When this intersection form is trivial (which they prove is realized in Theorem B(i)), then the covers will be homeomorphic by Theorem A (this example came out of a discussion I had with van Limbeek and Teichner).