Question

Is there a nice reference for the classification of closed 3-manifolds with solvable (nilpotent, abelian, etc.) fundamental group, assuming the Geometrization Conjecture?

Answer 1

Although I am late to answer this question, I wanted to put in a sales pitch for the flow charts at the end of Thurston's book (Figures 4.22 and 4.23). They do a great job for dealing with your questions for oriented manifolds.

In the first chart, Thurston uses "almost" to mean virtually, i.e. has a finite index subgroup with some property. Considering compact manifolds with abelian fundamental group, we have the following classification. (The references are to Thurston's book although older references exist for nearly all of these statements.) If $\pi_1(M)$ is finite, then $M$ is a lens space with finite cyclic fundamental group (see Theorem 4.4.14). If $\pi_1(M)$ is not finite and abelian, then $\pi_1(M)= Z, Z\times Z\times Z$ . The first case implies that $M\cong S^1\times S^2$ (see Exercise 4.7.1) and the second implies that $M \cong T^3$ (see Theorem 4.3.4).

As Richard Kent points out the paper of Peter Scott is a wonderful reference and these last two facts are discussed in greater detail there.

Finally to finish our classification for oriented manifolds, we also have to worry about the cusped manifolds. In general, the second chart has this information. The one caveat is that the unknot and Hopf link complements ($S^1\times D^2$ and $T^2 \times I$) might not neatly fit on this list, because they do not admit a cofinite $H^3$, $\widetilde{PSL(2,R)}$ or $H^2 \times R$ structures. However, unknot and Hopf link complements should be added to our list of manifolds having abelian fundamental groups as well since their fundamental groups are $Z$ and $Z\times Z$, respectively. I believe they are not considered in this chart because they would not show up in a decomposition prescribed by the Geometrization. For example, if you attach a Hopf link complement to the boundary of geometric piece via a boundary identification the resulting manifold remains geometric.

The non-orientable case, includes $S^1 \times P^2$, which has fundamental group $Z \times Z/2Z$ (see Epstein's paper Theorem 9.1).

Answer 2

Here are some references that may help answer your question:

Charles B. Thomas, Nilpotent groups and compact $3$-manifolds, Proc. Cambridge Philos. Soc. 64 (1968), 303-306; MR0233359.

Benny Evans and Louise Moser, Solvable fundamental groups of compact $3$-manifolds, Trans. Amer. Math. Soc. 168 (1972), 189–210; MR0301742.

Peter Teichner, Maximal nilpotent quotients of $3$-manifold groups, Math. Res. Lett. 4 (1997), no. 2-3, 283-293; MR1453060.

A little nugget (due to John Milnor): among the Brieskorn manifolds $\Sigma(p,q,r)$, the only nilmanifolds are $\Sigma(2,3,6)$, $\Sigma(2,4,4)$, and $\Sigma(3,3,3)$, which are circle bundles over the torus with Euler number $1$, $2$, and $3$, respectively.

Answer 3

Peter Scott, The geometries of 3-manifolds, Bull. London Math. Soc., 15 (1983), 401-487.