# Isometry group, one parameters sub-group geodesics

I have general questions about the group of isometries of a metric space. -When is the isometry group of a space a lie group? -when the isometry group is a Lie group, is there a relation between the one parameter sub-group of the isometry group and the geodesics of the metric space?

I am relatively new to these concepts, any piece of answers will help me.

Thanks

I'm sure there are several cases I could give to answer this, but the two cases I would consider to occur the most commonly are the following:

• Any finite metric space, since all finite groups are (compact) Lie groups
• By a theorem of Myers and Steenrod, Riemannian manifolds have Lie groups as isometry groups. This appears as the first part of Theorem 1.2 of Chapter II of Kobayashi's Transformation Groups in Differential Geometry. Thus, if the Riemannian manifold is connected, and thus admits a global distance function, then the isometry group in the standard sense is a Lie group.

Every compact Lie group $G$ admits a bi-invariant Riemannian metric, and under this metric, the one-parameter subgroups of $G$ are precisely the geodesics through the identity. However, I do not know of any general connections between geodesics on the spaces acted upon by isometry groups and the one-parameter subgroups on that group.

One very nice case occurs when $G$ is a non compact semi-simple Lie group (meaning its Lie algebra is a direct sum of simple ideals) with finite center. In this case, there is a maximal compact subgroup $K < G$ (the fixed subgroup of a Cartan involution) such that $G/K$ admits a Riemannian metric with respect to which $G/K$ is a globally symmetric space and the obvious left action of $G$ on $G/K$ is isometric.

Also, the one parameter subgroups of $G$ (whose generating vectors lie in a particular subspace, $\frak{p}$= the $-1$ eigenspace of the Cartan involution) push down to geodesics in $G/K$ and there is a close correspondence between left multiplication by these one parameter subgroups and the geodesic flow of $G/K$.

Finally, I believe that if $Ad(K)$ acts transitively on the unit sphere in $\frak{p}$, $G$ will be the full isometry group of $G/K$ (Maybe you also have to throw in the Cartan involution, which pushes down to the local symmetry at $K \in G/K$, if you want to flip the orientation).

So for instance think about $G= SL(2,\mathbb{R})$ acting on the upper half plane by hyperbolic isometries. The stabilizer of $i$ is $SO(2, \mathbb{R})=K$, so $G/K$ is diffeomorphic to $\mathbb{H}^2$. $SL(2, \mathbb{R})$ is the full (orientation preserving) isometry group (up to quotienting by the finite center) since $SO(2,\mathbb{R})$ acts transitively on the unit circle in $T_i(\mathbb{H}^2)$ (it's just rotations!). This action corresponds (via the derivative of the map $G\rightarrow G/K$ at $e$) to the action of $Ad(SO(2,\mathbb{R}))$ on $\frak{p}=$ the space of symmetric matrices in $\frak{sl}(2,\mathbb{R})$, as you can check.

I've been told that every symmetric Riemannian manifold is a quotient in this way, so it seems that this is what happens more generally, which is neat. Also, I think the compactness/ non compactness of $G$ may determine the sign on the curvature of $G/K$, but here I'm out of my depth. All of this is in the book on symmetric spaces by Sigurdur Helgason.

• $SL_2(\mathbb{R})$ is not the full isometry group, even after quotienting by its center. Consider the transformation $z\mapsto\frac{1}{\bar{z}}$. This is an isometry for the upper half plane model. Other than that, a nice answer. – Robin Goodfellow Sep 29 '14 at 2:15
• @RobinGoodfellow Ah, ok, the full group of orientation preserving isometries! :) – Tim kinsella Sep 29 '14 at 2:17
• There you go! :D – Robin Goodfellow Sep 29 '14 at 2:17
• Thanks for the answer, i will need a few days to digest this... – Chevallier Sep 30 '14 at 9:12
• Tim Kinsella, when you say "there is a close correspondence between left multiplication by these one parameter subgroups and the geodesic flow of G/K", the close correspondance is simply the action or is it more subtle? – Chevallier Oct 1 '14 at 12:30