# Implications of Geometrization conjecture for fundamental group

Hempel proved that Haken manifolds have residually finite fundamental groups. With the Geometrization conjecture, this now holds for any compact and orientable 3-manifold.

How exactly does the Geometrization conjecture imply that the only non-Haken compact orientable irreducible 3-manifolds are compact hyperbolic manifolds with no cusps?

Thanks a lot.

• I guess even compact and orientable are not required for residual finiteness, only finite-generation of the fundamental group. Jan 5, 2020 at 1:00
• The statement you make is not quite correct (you can also have "small" Seifert-fibered spaces): see Theorem 1.7.6 of this book: ems-ph.org/books/book.php?proj_nr=195 Jan 5, 2020 at 2:09
• @RyleeLyman Sure, just invoking Scott's compact core theorem and that subgroups of residually finite groups are r.f, whatever the correct statement is will extend to that level of generality.
– mme
Jan 5, 2020 at 2:24
• "compact with no cusp" sounds funny: a hyperbolic manifold with a cusp can't be compact.
– YCor
Jan 6, 2020 at 18:21

There are a few ways to look at this, but let me give a synopsis of Peter Scott's discussion in Section 6 of his 8 geometries paper, because it seems to directly address the question:

Scott, Peter, The geometries of 3-manifolds, Bull. Lond. Math. Soc. 15, 401-487 (1983). ZBL0561.57001.

(Note: there is also an errata for this reference: http://www.math.lsa.umich.edu/~pscott/errata8geoms.pdf. The corrections mentioned there do not affect this summary. )

Let's set up some standard 3-manifold terminology (anything left out is clearly defined in the reference).

A compact orientable 3-manifold is irreducible if is either $$S^1 \times S^2$$ or every embedded $$S^2$$ bounds a 3-ball. From now on, assume $$M$$ is a compact, orientable, irreducible 3-manifold (unless otherwise specified).

A surface $$S$$ (smoothly) embedded in $$M$$ is incompressible if every embedded curve $$\gamma$$ in $$S$$ which bounds a disk in $$M$$ also bounds a disk in $$S$$. A compact, orientable, irreducible 3-manifold is toroidal if it contains an incompressible torus.

The most natural thing to say is that the affirmative solution to the Geometrization Conjecture implies that a compact, irreducible 3-manifold is either toroidal or is homeomorphic to a quotient of the form $$X/\Gamma$$ where $$X$$ is one of the following geometric spaces: $$S^3,S^1 \times \mathbb{R},E^3,Nil,Sol, H^2 \times \mathbb{R}, \widetilde{PSL(2,\mathbb{R})}$$ or $$H^3$$ and $$\Gamma \subset Isom^+(X)$$ acting properly and discontinuously.

If a space admits a Sol geometry it is Haken. In fact, it is both toroidal and has positive first Betti number, (see [Scott, Theorem 4.17]).

The remaining geometries are either $$H^3$$ or Seifert fibered. Of course, some Seifert fibered manifolds like $$T^3 \cong S^1 \times S^1 \times S^1$$ are both toroidal and Seifert fibered and there are other minor pathologies: for example, $$RP^3 \# RP^3$$ is Seifert fibered and reducible and in the wake of Geometrization manifolds with $$S^3$$ geometry are exactly those with finite fundamental group.

Happily, Scott gives a clean statement of what you want in the conjecture on page 484 (of course the affirmative solution to the Geometrization Conjecture implies this conjecture is now known to be true):

Conjecture (now theorem): Let $$M$$ be a closed, irreducible, non-Haken 3-manifold with infinite fundamental group. Then $$M$$ is either a Seifert fibered space or admits a hyperbolic structure.

To connect this statement to the comments mentioned above, if we assume $$M$$ is Seifert fibered, non-Haken, irreducible, and has infinite fundamental group, then it is small Seifert fibered, which implies $$M$$ that the base orbifold of $$M$$ is $$S^2(a_1,a_2,a_3)$$ with $$1/a_1 + 1/a_2 + 1/a_3 \leq 1$$.

• Is it also true that your quoted conjecture implies Geometrization pre-Perelman? Jan 7, 2020 at 3:55
• No. The conjecture (due to Scott) is only part of the Geometrization conjecture. It does not say anything about simply connected 3-manifolds for example. Jan 7, 2020 at 6:15
• The conjecture is enough to imply residual finiteness of all (finitely generated) 3-manifold groups, though.
– HJRW
Jan 7, 2020 at 21:49