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

I'm starting to study at the elementary level the relationship between topology and geometry of a Riemannian manifold of negative curvature. The first two theorems, simple and interesting in this direction are:

$\bf{Preissmann~~ Theorem}:$ Let be $M$ a Riemannian manifold with sectional curvature $K<0$, then every non trivial Abelian subgroup of the fundamental group $\pi_1(M)$ is cyclic infinite.

$\bf{Byers~~ Theorem}:$ Let be $M$ a Riemannian manifold with sectional curvature $K<0$, then every no trivial solvable subgroup of the fundamental group $\pi_1(M)$ is cyclic infinite, and $\pi_1(M)$ have no cyclic subgroup of finite index.

$\bf{My ~~Question}:$ I'm looking for nontrivial examples (counterexamples) for Byers Theorem, i.e, non trivial examples of a Riemannian manifold that:

  1. Has a no solvable fundamental group.

EDIT:

2 Has a cyclic $\bf{infinite}$ subgroup of finite index of the fundamental group.(In this case the trivial examples are welcome.)

share|improve this question
    
simultaneous post at math.stackexchange.com/questions/156571/… –  Will Jagy Jun 10 '12 at 18:08
    
I'm just trying to learn mathematics. –  Juan Valdez Jun 10 '12 at 19:20
    
You need to add the hypothesis that $M$ is compact in the statement of these theorems. –  Ian Agol Jun 10 '12 at 21:43
    
of course, thanks. –  Juan Valdez Jun 11 '12 at 15:13
add comment

1 Answer

up vote 4 down vote accepted

1) To construct a manifold with no solvable fundamental group, take for example a finite unsolvable group $G$ and embed it into $SU(n)$ as a discrete subgroup. This embedding is obtained by realizing $G$ as a subgroup of the permutations $S(G)$ of the set $G$, then note $S(G)$ is isomorphic to a subgroup of $SU(n)$ via monomial matrices ($n=|G|$). The quotient $SU(n)/G$ carries a natural manifold structure with fundamental group $G$. Note these examples are compact. Taking product with $R$, one obtains non-compact examples.

2) For manifolds with cyclic subgroups of finite index, consider products of real projective spaces $RP^n$ or lens spaces $S^{2n+1}/Z_p$ with manifolds with finite fundamental group.

To combine 1) and 2), take products.

share|improve this answer
    
Hi Malte, I apologize if I was not clear enough to write my question but I wish that the subgroup of the second item to be infinite as well as cyclical –  Juan Valdez Jun 10 '12 at 21:00
    
Then take an $S^1$ in place of $RP^n$ or the lens space. –  Malte Jun 10 '12 at 21:13
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.