Is there a topological characterisation of what a (closed irreducible) hyperbolic 3manifold 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 3manifolds? 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 3manifold topology after Perelman?). Thanks!

2$\begingroup$ The criterion for knots you refer to is Perelman's geometrization theorem, and this applies to all 3manifolds. There are many surveys. A prePerelman classic is Scott's article The geometries of 3manifolds (Bulletin of the LMS). Thurston's book 'The geometry and topology of 3manifolds' is also accessible to an early stage postgraduate. PostPerelman works are rarer, but you could try arXiv:1205.0202v3 (apologies for the shameless selfadvertisement). Everyone else will have their favourites. $\endgroup$– HJRWJun 17, 2013 at 9:14

$\begingroup$ (Slight correction: the result for knots follows from the geometrization theorem for Haken 3manifolds, which was proved by Thurston. Of course, you need Perelman to deal with nonHaken 3manifolds.) $\endgroup$– HJRWJun 17, 2013 at 9:28
1 Answer
A clear statement is the following:
A compact 3manifold $M$ 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 3manifold 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 finitevolume 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 2sphere that does not bound a 3disc 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 2sphere which does not bound a 3disc (hence $M$ is reducible)
 $M$ contains an essential 2disc (hence $M$ is $\partial$reducible)
 $M$ contains an essential 2torus whose $\pi_1$ injects but which is not $\partial$parallel. Note that the 2torus 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 2disc 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.

$\begingroup$ Thanks! That's exactly what I was looking for! Just to confirm, so a compact 3manifold is not hyperbolic if either it contains an embedded essential surface of nonnegative 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? $\endgroup$ Jun 17, 2013 at 12:33

$\begingroup$ the sentence you wrote is true if you define a SSFS as a 3manifold 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. $\endgroup$ 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 BoileauPorti on orbifolds: mat.uab.es/~porti/main.pdf $\endgroup$ Jun 17, 2013 at 12:43