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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.

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    $\begingroup$ I guess even compact and orientable are not required for residual finiteness, only finite-generation of the fundamental group. $\endgroup$ Jan 5, 2020 at 1:00
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    $\begingroup$ 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 $\endgroup$
    – Ian Agol
    Jan 5, 2020 at 2:09
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    $\begingroup$ @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. $\endgroup$
    – mme
    Jan 5, 2020 at 2:24
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    $\begingroup$ "compact with no cusp" sounds funny: a hyperbolic manifold with a cusp can't be compact. $\endgroup$
    – YCor
    Jan 6, 2020 at 18:21

1 Answer 1

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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 $1/a_1 + 1/a_2 + 1/a_3 \leq 1$.

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  • $\begingroup$ Is it also true that your quoted conjecture implies Geometrization pre-Perelman? $\endgroup$ Jan 7, 2020 at 3:55
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    $\begingroup$ 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. $\endgroup$ Jan 7, 2020 at 6:15
  • $\begingroup$ The conjecture is enough to imply residual finiteness of all (finitely generated) 3-manifold groups, though. $\endgroup$
    – HJRW
    Jan 7, 2020 at 21:49

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