Credit to Moishe Kohan https://math.stackexchange.com/a/3604136/724711:
The Uniformization Theorem says that if $S$ is a connected Riemannian surface then it is conformally equivalent to a complete Riemannian surface of constant curvature. In particular:
Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,
$S$ is diffeomorphic (homeomorphic) to the quotient $\Gamma\backslash X$, where $X$ is either the unit sphere $ S^2 $ or the Euclidean plane $ E^2 $ or the hyperbolic plane $ H^2 $ and $\Gamma$ is a discrete subgroup of isometries of $X$ acting freely on $X$ (Note that $ S^2,E^2,H^2 $ are exactly the three simply connected symmetric spaces in dimension $ 2 $).
Part 2 does not follow from Part 1, unless you have "complete" in the statement of Part 1.
- Answering your question in a comment: It is known that every compact homogenous manifold $M$ satisfies $\chi(M)\ge 0$, this was proven by Mostow, in
Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024.
- This theorem is a consequence of a deeper structural results for homogeneous manifolds that Mostow proves in his paper. In section 4, Mostow proves that every connected homogeneous manifold (compact or not) admits an iterated fibration with fibers that are homogeneous manifolds satisfying further properties. Specializing to 3-manifolds $M$, it follows that:
i. either $M$ itself fibers, or
ii. (The exceptional case.) Either $G$ is solvable or is semisimple and in this case $H$ has finitely many connected components. Unfortunately, this exceptional case is poorly explained in the statement of the theorem in the introduction: In this case Mostow should have allowed one more subgroup in his sequence. In Mostow's notation, $k=1$, causing appearance of a subgroup $F_{-1}$ which has to be equal to $\{1\}$, while
$F_0=H$. Then the fibration in the statement of Theorem C reads:
$$
H/F_{-1}=H \to G/F_{-1}=G\to G/H.
$$
Then $G_k=G_1=G$, $\Gamma_k=\Gamma_1=H$ and we are in Mostow's case 3, where $G=G_1$ is semisimple or solvable and, in the semisimple case, $\Gamma_k$ has finitely many components. Mostow explains this better when he repeats the formulation of Theorem C in the end of section 4 (where this theorem is actually proven).
Case i. If a compact connected 3-manifold fibers, then either the fiber is a surface and the base is a circle or vice-versa.
(a) If the fiber is a surface, it has to be of nonnegative Euler characteristic (see part 1). Hyperbolic 3-manifolds do not admit such fibrations:
Namely, asphericity of $M$ implies that the fiber $F$ cannot be $S^2, P^2$. The long exact sequences of homotopy groups of a fibration implies that we have a short exact sequence
$$
1\to \pi_1(F)\to \pi_1(M)\to \pi_1(S^1)= {\mathbb Z}\to 1.
$$
If $F$ is the torus or the Klein bottle then $\pi_1(M)$ would contain a nontrivial normal virtually abelian subgroup which is known to be impossible.
(b) On the other hand, a compact hyperbolic manifold also cannot fiber over a surface $B$ with circular fibers $F$. This again follows from asphericity of $M$ which results in a short exact sequence
$$
1\to \pi_1(F)= {\mathbb Z}\to \pi_1(M)\to \pi_1(B)\to 1.
$$
Then, again we would obtain a normal infinite cyclic subgroup in $\pi_1(M)$, which is impossible.
Thus, compact hyperbolic 3-manifolds cannot admit transitive actions of Lie groups.
With more work, one can relax the compactness assumption in (2) to finiteness of volume and prove a similar theorem in higher dimensions as well.
ii. Consider now the exceptional case. If $G$ is semisimple and $H$ has finitely many components, $\pi_1(M)=\pi_1(G/H)$ is finite. This is impossible for a hyperbolic manifold. Lastly, if $G$ is solvable, so is $H$, hence, $H/H^c$ is solvable too (I am using Mostow's notation where $H^c$ denotes the identity component with respect to the Lie group topology). But then $\pi_1(M)$ is solvable, which is impossible for a hyperbolic manifold of finite volume.
Best Answer
A. Let me answer some of your questions. First of all, one should be careful using Wikipedia articles as your main source. Anybody can write/edit Wikipedia pages, regardless of how little they know about the subject. In this particular case, Wikipedia's statement "Every connected two manifold admits a constant curvature metric" is correct but incomplete: The Uniformization Theorem (UT) is much stronger than the said claim. The correct claim is:
If $S$ is a connected Riemannbian surface then it is conformally equivalent to a complete Riemannian surface of constant curvature. In particular (and this is easier than the UT):
*1. Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,
Part 2 does not follow from Part 1, unless you have "complete" in the statement of Part 1.
B. Below, I refer to $X$ from Part A as a model space. It is a basic fact that every model space is diffeomorphic to the quotient $G/K$ where $G$ is a linear Lie group and $K$ is a compact subgroup of $K$, while $G$ acts faithfully on $X=G/K$ (by left multiplication), isometrically and every isometry of $X$ comes from this action. This can be proven either by case-by-case analysis or by arguing that in dimension 2 model spaces are simply connected symmetric spaces (since each is complete and has constant curvature).
For instance, for $X={\mathbb H}^2$, $G=PO(2,1)$ (quotient of $O(2,1)$ by its center) and $K=PO(2)$ (quotient of $O(2)$ by $\pm 1$). Linearity of this group can be seen via adjoint representation of $O(2,1)$ (kernel of the adjoint representation of $O(2,1)$ is exactly the center of $O(2,1)$).
C. The projective plane does admit a Riemannian metric making it a symmetric space, namely, the quotient of the unit sphere by the antipodal map (the antipodal map is isometric with respect to the standard metric). However, $RP^2$ is not diffeomorphic to the quotient of $SU(2)$ by a subgroup.
D. Indeed, Klein bottle does admit a flat metric, but this is not because it is 2-fold covered by the 2-torus which admits a flat metric: You need a flat metric on $T^2$ invariant under a fixed-point free orientation-reversing involution. You can derive the existence of such a metric either by a direct construction or by appealing to the UT.