fundamental group and complete invariant of irreducible 3-manifolds

Question

I heard that,by Perelman's work,we can get that the fundamental group is a complete invariant of irreducible 3-manifolds (except for lens spaces). can someone help explain this.Thank you!

Answer 1

Perelman has proved Thurston's geometrization conjecture, which says that every irreducible 3-manifold decomposes along its canonical decomposition along tori into pieces, each admitting a geometric structure. A "geometric structure" is a nice riemannian metric, which is in particular complete and of finite volume.

There are eight geometric structures for 3 manifolds: three structures are the constant curvature ones (spherical, flat, hyperbolic), while the other 5 structures are some kind of mixing of low-dimensional structures (for instance, a surface $\Sigma$ of genus $2$ has a hyperbolic metric, and the three-manifold $\Sigma\times S^1$ has a mixed hyperbolic $\times S^1$ structure).

The funny thing is that geometrization conjecture was already proved by Thurston when the canonical decomposition is non-trivial, i.e. when there is at least one torus in it. In that case the manifold is a Haken manifold because it contains a surface whose fundamental group injects in the 3-manifold. Haken manifolds have been studied by Haken himself (of course) and by Waldhausen, who proved in 1968 that two Haken manifolds with isomorphic fundamental groups are in fact homeomorphic.

If the canonical decomposition of our irreducible manifold $M$ is empty, now we can state by Perelman's work that $M$ admits one of these 8 nice geometries. The manifolds belonging to 7 of these geometries are well-known and have been classified some decades ago (six of these geometries actually coincide with the well-known Seifert manifolds, classified by Seifert already in 1933). From the classification one can see that the only distinct manifolds with isomorphic fundamental groups are lens spaces (which belong to the elliptic geometry, since they have finite fundamental group).

The only un-classified geometry is the hyperbolic one. However, Mostow rigidity theorem says that two hyperbolic manifolds with isomorphic fundamental group are isometric, hence we are done. Some simple considerations also show that two manifolds belonging to distinct geometries have non-isomorphic fundamental groups.

Therefore now we know that the fundamental group is a complete invariant for irreducible 3-manifolds, except lens spaces.

Answer 2

An addendum to BM's answer: you might want to start with the Kneser-Milnor theorem which gives a unique decomposition of a 3-manifold as a connected sum and hence its fundamental group as a free product to reduce the question to prime (and then irreducible) 3-manifolds. The Poincare conjecture says you don't have any summands "hidden" from $\pi_1$.

Also, you have to be careful with what kind of 3-manifolds you consider, for example if $F_1$ is a 3 punctured sphere and $F_2$ a punctured torus, then $F_1\times S^1$ and $F_2\times S^1$ (or the complement of the square and granny knots for a more subtle example) have the same fundamental group but aren't homeomorphic. So perhaps you want to first consider closed 3-manifolds. The issues for general compact 3-manifolds can get technical, but they are understood. A good starting point is Hempel's book.