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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 $$ N_g \cong G_{\mathbb{Z}} \backslash G_\mathbb{R} / H_{\mathbb{R}} $$

Ian Gershon Teixeira
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  • To whoever down voted my question, what’s wrong with it, is there some way I can improve this question? – Ian Gershon Teixeira Mar 27 '20 at 12:11
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    I did not down-vote but one reason for a down-vote would be that you have 5 questions in your post. Also, you did not tell us how you arrived to the first sentence (Every connected surface admits a complete constant curvature metric.) My questions to you would be: Are you familiar with the Gauss-Bonnet formula? If you are, what does it tell you about possible $K$ such that Klein bottle admits a metric of curvature $K$? How about connected sum of three projective planes? Do you know how to realize the full isometry group of the hyperbolic plane as a matrix group? – Moishe Kohan Mar 30 '20 at 15:00
  • I'm not very familiar with Gauss-Bonnet or Euler characteristic but having just looked it up on Wikipedia $ \frac{1}{2\pi}\int_M K dA=\chi(M) $ for a closed 2 manifold I see that the answer to my question #2 is yes. To see this let $ N_g $ be a connected sum of g+1 projective planes and let $ S_g $ be its orientable double cover, a surface of genus g. Then $ \frac{1}{2\pi}\int_{S_g} K dA=\chi(S_g)=2-2g=2(1-g)=2\chi(N_g)=2\frac{1}{2\pi}\int_{N_g} K dA $. So both $ N_g $ and $ S_g $ can only admit constant curvature metrics with the same sign as $ 1-g $ – Ian Gershon Teixeira Mar 31 '20 at 16:49
  • The isometry group of the hyperbolic plane is $ SL_2(\mathbb{R})/\pm I \cong SO_{2,1}(\mathbb{R}) $. This reminds me of the fact that the isometry group of the round sphere is $ SU(2)/\pm I \cong SO_3(\mathbb{R}) $. Is that analogy useful? – Ian Gershon Teixeira Mar 31 '20 at 17:08
  • @ Moishe Kohan Here is my logic for the first line: https://en.wikipedia.org/wiki/Uniformization_theorem#Classification_of_closed_oriented_Riemannian_2-manifolds says "every Riemann surface admits a Riemannian metric of constant curvature". And https://en.wikipedia.org/wiki/Riemann_surface says " A two-dimensional real manifold can be turned into a Riemann surface ... if and only if it is orientable and metrizable" (I define all manifolds to be metrizable). Every connected 2 manifold is orientable or has an orientable double cover and constant curvature is a local property QED . – Ian Gershon Teixeira Mar 31 '20 at 18:11
  • I didn't mean to imply that every connected 2 manifold admits a complete metric, this is true for the compact case but I doubt that the punctured plane admits any complete metric. I do not claim that every connected 2 manifold is a space form (since it may not admit a complete metric). But, if I understand correctly, you said that every connected 2 manifold admits the structure of a locally symmetric space with constant curvature in answer to my question https://math.stackexchange.com/questions/3567448/is-every-surface-locally-symmetric. – Ian Gershon Teixeira Mar 31 '20 at 18:36

1 Answers1

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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,

  1. $S$ 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.

Moishe Kohan
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