Locally Riemannian Homogeneous Surface

Question

A Riemannian manifold $ M' $ is Riemannian homogeneous if $ Iso(M') $ acts transitively.

A Riemannian manifold $ M $ is locally Riemannian homogeneous if there exists a Riemannian homogeneous manifold $ M' $ such that $$ M \cong \Gamma\backslash M' $$ where $ \Gamma $ is a discrete subgroup of $ Iso(M') $.

Every orientable connected surface admits a locally Riemannian homogeneous structure (moreover locally symmetric moreover constant curvature).

For the non orientable case I know that the projective plane is Riemannian homogeneous, while connected sums of multiple projective planes are never Riemannian homogeneous. But what about locally Riemannian homogeneous? For example

Can the connected sum of 3 projective planes be given a locally Riemannian homogeneous structure?

Answer 1

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:

1. Every smooth (topological) connected surface $S$ (oriented or not) admits a complete metric of constant curvature. Equivalently,

2. $S$ 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.

This shows that every connected surface is locally symmetric, even has constant curvature, and so, a fortiori, is locally Riemannian homogeneous.

In other words we have that every connected surface (orientable or not) admits a locally Riemannian homogeneous structure (moreover locally symmetric, moreover constant curvature).