Is there a topological characterisation of what a (closed irreducible) hyperbolic 3-manifold is? I don't know any Riemannian geometry and still want to understand what an exceptional Dehn surgery is. For definition of a hyperbolic knot we can avoid the real understanding of what a hyperbolic manifold is by simply saying that it is not an unknot, not a torus knot and not a satellite. Is there a similar characterisation for hyperbolic 3-manifolds? Cameron Gordon's Park City lecture notes seem to suggest that such a criterion would be that the manifold is simple. But I couldn't find a clear statement of this. If there is such a criterion, a reference for a survey discussing these things would be very desirable(and in general, is there a survey - suitable for early stage PhD students - about the 3-manifold topology after Perelman?). Thanks!
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ The criterion for knots you refer to is Perelman's geometrization theorem, and this applies to all 3-manifolds. There are many surveys. A pre-Perelman classic is Scott's article The geometries of 3-manifolds (Bulletin of the LMS). Thurston's book 'The geometry and topology of 3-manifolds' is also accessible to an early stage postgraduate. Post-Perelman works are rarer, but you could try arXiv:1205.0202v3 (apologies for the shameless self-advertisement). Everyone else will have their favourites. $\endgroup$– HJRWCommented Jun 17, 2013 at 9:14
-
$\begingroup$ (Slight correction: the result for knots follows from the geometrization theorem for Haken 3-manifolds, which was proved by Thurston. Of course, you need Perelman to deal with non-Haken 3-manifolds.) $\endgroup$– HJRWCommented Jun 17, 2013 at 9:28
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
A clear statement is the following:
A compact 3-manifold $M$ is hyperbolic if and only if it has infinite fundamental group and does not contain any essential surface with $\chi \geqslant 0$.
You may remember that by saying that $M$ is hyperbolic unless there is some clear obstruction, and the obstrucion is $\pi_1$ finite or the existence of some surface with $\chi \geqslant 0$. This statement is pretty general and works for any compact 3-manifold with any kind of (possibly empty) boundary, using the appropriate notions of "hyperbolic" and "essential".
- Hyperbolic means that, after removing tori and Klein bottles from $\partial M$, the manifold admits a complete finite-volume metric with constant curvature $-1$ and with geodesic boundary.
- Essential means that the surface is immersed in $M$ with a map that is $\pi_1$-injective and not homotopic to some map whose image is in some component of $\partial M$. In addition, a 2-sphere that does not bound a 3-disc is essential.
More concretely, an orientable manifold $M$ is hyperbolic unless one of the following obstructions occurs:
- $\pi_1(M)$ is finite,
- $M$ contains a 2-sphere which does not bound a 3-disc (hence $M$ is reducible)
- $M$ contains an essential 2-disc (hence $M$ is $\partial$-reducible)
- $M$ contains an essential 2-torus whose $\pi_1$ injects but which is not $\partial$-parallel. Note that the 2-torus may be immersed and not embedded: this occurs precisely in the small Seifert spaces
- $M$ contains an essential annulus (for instance in $\Sigma \times [0,1]$ for any surface $\Sigma$)
This characterization implies the one for knots that you mentioned, in virtue of the following:
- every knot complement has infinite $\pi_1(M)$ and is irreducible,
- a knot complement contains an essential 2-disc if and only if it is trivial,
- a knot complement contains an essential annulus or torus if and only if it is not prime, a torus knot, or a satellite.
answered Jun 17, 2013 at 11:34
-
$\begingroup$ Thanks! That's exactly what I was looking for! Just to confirm, so a compact 3-manifold is not hyperbolic if either it contains an embedded essential surface of non-negative Euler charasteristic or it is a Small Seifert Fibred space? By SSFS do you mean base surface $S^2$ with at most 3 exceptional fibres or do you also include the $P^2$ with at most one exceptional fibre as well? Do you know a text that I could cite making this statement? $\endgroup$ Commented Jun 17, 2013 at 12:33
-
$\begingroup$ the sentence you wrote is true if you define a SSFS as a 3-manifold that has a Seifert fibration on S^2 with at most 3 singular fibers: such a "weak" notion of SSFS includes everything that has finite pi_1, and everything which contains "immersed tori" but not embedded ones. $\endgroup$ Commented Jun 17, 2013 at 12:41
-
$\begingroup$ A manifold fibering over P^2 with one singular fiber also fibers over S^2 with three singular fibers. This is due to the fact that the orientable fibration over the Mobius strip also fibers over the disc with two singular points of order 2. You can find this in the book of Matveev and Fomenko, with lots of pictures. Concerning the general statement, I suppose that some of the surveys of Cameron Gordon should contain it. See also the paper of Boileau-Porti on orbifolds: mat.uab.es/~porti/main.pdf $\endgroup$ Commented Jun 17, 2013 at 12:43