# Riemannian symmetric spaces and Lie algebras

1. Jul 14, 2012

### TrickyDicky

I'm interested in the crossover of Lie groups/differential geometry and I'm a bit confused about the relation of Lie algebras with symmetric spaces.
Take for instance the Lie group G=SL(2,R), we take the quotient by K=SO(2) as isotropic group(maximal compact subgroup) and get the symmetric space G/K= H2(hyperbolic plane).
How is it then the tangent vector space of the hyperbolic plane exactly related to the Lie algebra sl(2,R) of G (if at all)?
Thanks in advance, I would also be interested in references on introductory textbooks that treat Lie groups from the geometrical side rather than the purely abstract algebraic one.

2. Jul 14, 2012

### Ben Niehoff

The Lie algebra sl(2,R) is generated by the three Killing vectors of the hyperbolic plane. Should be a simple calculation to do if you want to see for yourself.

3. Jul 14, 2012

### TrickyDicky

Hi Ben, thanks for that hint.
I'm trying to make some conceptual connections to get an intuitive understanding of Lie algebras wrt Riemannian manifolds, I'm trying to link the Lie algebras in Riemannian manifolds with the group of isometries of the manifold, so what you say makes sense since Killing fields are generators of isometries and form a Lie subalgebra of vector fields on the manifold. So I gues in the case of the hyperbolic plane, sl(2,R) can be considered the Lie algebra of the isometry group (with the isometry group being the Lie group SL(2,R)? of the hyperbolic plane, is this correct?

4. Jul 14, 2012

### Ben Niehoff

Right.

For hyperbolic planes, there is a trick that makes this easy to visualize. A hyperbolic n-plane embeds isometrically in $R^{n,1}$ as one sheet of a two-sheeted hyperboloid inside the lightcone. All the symmetries of H^n then correspond to point symmetries of the origin in $R^{n,1}$; i.e., the Lorentz group SO(n,1). For n = 2, we have so(2,1) = sl(2,R).

(Incidentally, the one-sheeted hyperboloid outside the lightcone is de Sitter space.)

In fact, ALL the SO(p,q) groups can be realized as isometry groups on quadric surfaces in $R^{p,q}$ in a similar fashion. For other Lie groups, I don't know of a simple trick.

However, given the standard Killing form on the Lie group (which gives a metric on the group manifold itself), I think there is a straightforward way to find a metric on the manifold whose isometry group is said Lie group (but I don't know what it is).

5. Jul 16, 2012

### TrickyDicky

Thanks, this helps me figure ot what I had in mind in the OP:
A way to relate the Lie algebra of the group of isometries from a given Riemannian manifold, to the inner products (metric tensor) of each manifold point's tangent space.
For semisimple Lie algebras like sl(2,R) in which the Killing form is nondegenerate it serves as metric tensor of the manifold according to wikipedia.
Would this Killing form of sl(2,R) be the metric tensor of the manifold (in this case the hyperbolic plane) whose group of isometries is the Lie group SL(2,R)?

6. Jul 17, 2012

### TrickyDicky

To formulate my question better as I obviously implied something wrong in my previous post question. The nondegenerate Killing form of a semisimple Lie algebra is the metric tensor of the corresponding Lie group, does anyone know the above mentioned way to find a metric on the manifold whose isometry group is said Lie group?