What does it mean for a manifold to be geometrizable? I was reading the wikipedia article on classification of manifolds and it mentions that 4-manifolds aren't geometrizable, as opposed to in lower dimensions. It implies that this means that, in the case of 2- and 3-manifolds, they admit geometry. What is meant by this?
 A: For a good introduction to geometrization, I recommend checking out Martelli's book on geometric topology, available for free on the arXiv. The gist of it, as far as I understand, is as follows.
Any compact, connected and orientable surface (2-manifold) is homeomorphic to a $g$-holed torus $\Sigma_g$, where $\Sigma_0$ is homeomorphic to $S^2$ and $\Sigma_1$ is homeomorphic to a standard torus. This is a picture of $\Sigma_0$, $\Sigma_1$, and $\Sigma_2$:

All those spaces are geometrizable in the following sense. 


*

*The sphere $S^2$ is a very nice geometric space if we equip it with
the radial metric (or, maybe more familiarly, with the standard
Riemannian metric), because it has constant curvature. So $S^2$
could be said to be "trivially geometrizable".

*The torus $T^2$ is homeomorphic to the space $\mathbb R^2/ \mathbb
Z^2$, and again, $\mathbb R^2$ is a very nice geometric space
(arguably the most nice). We can put a metric on the torus such that
the quotient map $\mathbb R^2 \rightarrow T^2$ becomes a local
isometry, so the geometry of $\mathbb R^2$ gives us a geometry on
the torus. Thus, also the torus is geometrizable.

*For $g \geq 2$ we have that $\Sigma_g$ can be obtained by gluing
together $2g-2$ so-called
hyperbolic pairs of pants (picture below; on the left side is a pair of pants, and on the right side is a decomposition of $\Sigma_2$ into pairs of pants). Similarly, $\Sigma_g$ can be obtained by
gluing together the sides of specific hyperbolic $2g$-gons. This
gives us two ways to see that $\Sigma_g$ is locally isometric to the
hyperbolic plane $\mathbb H^2$. The second one even gives us a
quotient map $\mathbb H^2 \rightarrow \Sigma_g$, establishing the
analogy to the torus.



These examples motivate the following ad-hoc definition:

A $2$-manifold is geometrizable if it admits a Riemannian metric admitting a locally isometric universal cover isometric to either $S^2$, $\mathbb R^2$, or $\mathbb H^2$.

Note that these spaces are precisely the simply-connected Riemannian $2$-manifolds with constant curvature.
Now, for $3$-manifolds, the situation is a bit more involved, so I can only give the basic ideas here. There is a notion of prime manifolds, and with that comes a decomposition of $3$-manifolds into prime parts. Why is this interesting? Well, Thurston conjectured and Perelman later proved that any prime $3$-manifold can be cut up (JSJ decomposition) in such a way that each piece is geometrizable in the following sense:

A $3$-manifold is geometrizable, if it admits a Riemannian metric admitting a locally isometric universal cover isometric to either $S^3$, $\mathbb R^3$, $\mathbb H^3$, $S^2 \times \mathbb R$, $\mathbb H^2 \times \mathbb R$, the universal cover $\text{SL}(2, \mathbb R)$, $\text{Nil}$, or $\text{Sol}$.

In fact, these spaces satisfy a similar but more complicated characterization as in the $2$-dimensional case.
