Space form structure for non orientable surfaces

The projective plane is homogeneous $$ N_0=P^2 \cong SO_3/O_2 $$ The connected sum of two projective planes is the Klein bottle $ K= N_1 $ which is also homogeneous (for the construction of this homogeneous space see for example Is it possible to realize the Klein bottle as a linear group orbit? ). For $ g\geq 2 $ the Euler characteristic is negative so by a theorem of Mostow the surfaces are not homogeneous. But orientable surfaces can be presented as double coset spaces $$ \Gamma \backslash SL_2 /SO_2 $$ where $ \Gamma $ is a surface group. Is it possible to present the sum of $ g+1 $ projective planes as an arithmetic manifold for $ g \geq 2 $? In other words does there exists $ G $ a linear algebraic group, and $ H $ a subgroup such that $$ N_g \cong G_{\mathbb{Z}} \backslash G_\mathbb{R} / H_{\mathbb{R}} $$

Solution 1:

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,

2. $S$ is diffeomorphic (homeomorphic) to the quotient $\Gamma\backslash X$, where $X$ is either unit sphere or the Euclidean plane or the hyperbolic plane and $\Gamma$ is a discrete subgroup of isometries of $X$ acting freely on $X$.*

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.