0

Is it true that a constant curvature metric on a manifold $M$ is equivalent to a presentation of $M$ as a locally symmetric space $ M \cong \Gamma \backslash G/H $ where $G/H$ is a symmetric space and $ \Gamma $ a discrete subgroup of $G$?

I believe this is true based on the response of Jason DeVito to my question:

Is every surface locally symmetric?

And for a topological manifold M homeomorphic to a locally symmetric space is there any sense in which the triple $ (\Gamma, G, H) $ is unique? Or can the triple be picked to be minimal in some way? For example pick H connected and G simple or semisimple or simply connected or of minimal dimension etc...

Michael Hardy
  • 1
Ian Gershon Teixeira
  • 10,176
  • 1
    If $M$ is complete (and by constant curvature you mean constant sectional curvature), this is certainly the case, since all space forms are quotients of spheres or Euclidean/Hyperbolic spaces. – Kajelad Mar 08 '20 at 23:13
  • 2
    In dimension $>2$ there are lots of (locally) symmetric spaces that are not of constant curvature, so you'd best stick to surfaces. – Ted Shifrin Mar 08 '20 at 23:29
  • @Ted Shifrin so is it the case that Jason DeVito is wrong in claiming "Maybe I am missing something - but isn't every constant curvature space (of any dimension) locally symmetric? The universal cover is symmetric, so doesn't that give you the result you want?" in a comment on my question linked to above? Also I would love it if you could give an example of a manifold of dimension $ >2 $ that admits a constant curvature metric but is not locally symmetric! – Ian Gershon Teixeira Mar 09 '20 at 02:51
  • 1
    You didn't read my comment carefully. I'm addressing the converse statement. Your question said "equivalent" — these are only equivalent statements in dimension $2$. – Ted Shifrin Mar 09 '20 at 05:04
  • @Ted Shifrin I see, so what is an example of a manifold that can be realized as a locally symmetric space but does not admit a constant curvature metric? And is there any algebraic characterization of such spaces among all locally symmetric spaces? – Ian Gershon Teixeira Mar 09 '20 at 21:20
  • 1
    The simplest example is a Grassmannian (that is not a projective space). Look up space forms — as was already discussed, the simply connected spaces of constant curvature are very few. – Ted Shifrin Mar 09 '20 at 21:30
  • @TedShifrin $\Bbb C P^n$ endowed with the Fubini-Study metric is a simpler example than the Grassmannian in my opinion – Didier Apr 22 '24 at 19:09

1 Answers1

1

No, for dimension $ >2 $ not every locally symmetric space has constant curvature. In fact it is not even the case that every symmetric space has constant curvature. For example take $ S^2 \times S^1 $, a model for one of the eight Thurston Geometries. The $ S^2 $ is round (constant curvature $ 1 $) while the $ S^1 $ is flat (constant curvature $ 0 $). The space $ S^2 \times S^1 $ is locally symmetric (indeed it is even symmetric since it is a product of symmetric spaces). But $ S^2 \times S^1 $ cannot have constant curvature because there is no way to but the same curvature on the $ S^2 $ piece and the $ S^1 $ piece.

Although not every locally symmetric space has constant curvature, it is the case that every constant curvature space is locally symmetric. Indeed every constant curvature space has a locally isometric universal cover by $ X=G/H $ for either $ X=S^n $ if it has constant positive curvature or $ X=E^n $ if it has $ 0 $ curvature or $ X=H^n $ if it has constant negative curvature. And thus every constant curvature space can be written as a locally symmetric space as $$ \Gamma \backslash G/ H $$ where $ \Gamma $ is a discrete subgroup of isometries acting freely on $ X=G/H $ (here $ G $ is the isometry group of $ X $, and $ X $ is the simply connected symmetric space of the appropriate curvature).

And yes the canonical way to pick $ \Gamma, G, H $ for a given locally symmetric space $ M $ is to take $ X $ to be the unique simply connected symmetric space covering $ M $ then $ G $ is the isometry group of $ X $ and $ H $ is the isotropy group of the action of $ G $ on $ X $ and $ \Gamma $ is the unique subgroup of isometries acting freely on $ X $ that commutes with the universal covering map.

Ian Gershon Teixeira
  • 10,176
  • It is not even true for irreducible symmetric spaces: $\Bbb C P^n$, with the Fubini-Study metric, is symmetric, but has non-constant curvature which is $\frac{1}{4}$-pinched – Didier Apr 22 '24 at 19:07
  • 1
    Right, $ \mathbb{C}P^2 $ with Fubini-Study is a great example of an irreducible symmetric space that does not have constant curvature, in fact I think $ \mathbb{C}P^2 $ with Fubini-Study is the minimal dimensional irreducible counterexample. – Ian Gershon Teixeira Apr 22 '24 at 19:37