TITLE: Geodesic flows and Curvature
QUESTION [6 upvotes]: I have some conceptual questions related to geodesic flows and cuvature.

Suppose you have one parameter group of isometries from your manifold to itself. Since isometry preserves metric then it preserves Levi-Civita connection and curvature. How would one tie this to geodesic flows*. Is there a way to understand whether if a manifold has constant curvature by its geodesics (besides the criteria I gave below). For instance given a point $p$ on $M$, if $p$ can be connected to any other point on the manifold by a geodesic (as in sphere) then does the manifold have constant curvature? I would assume that if you have a "neighbourhood of geodesic flows" then its pullback preserves metric on that nbd. However it is not a global isometry. 

*-I know one theorem where if every geodesic circle has constant curvature then the manifold has constant curvature. 

My second question is where can I get some information about the set of all isometries of a manifold as a space itself? Is there a good geometry book on this topic as a reference?

REPLY [2 votes]: Some clarifications:
Every complete Riemannian manifold has the property that for any two points $p, q \in M$, there exists a geodesic $\sigma_{pq}$ connecting $p$ and $q$. Moreover, the geodesic can be chosen to be minimizing, that is, there are no other curves $\alpha : I \rightarrow M$, geodesics or not, with length strictly less than the length of $\sigma_{pq}$. This is part of a basic result known as Hopf-Rinow's Theorem.
Thus, if it were true that being able to connect any $p \in M$ with any other $q \in M$ by a geodesic implies constant curvature, then every complete Riemannian manifold would have constant curvature, which is obviously false.
A related property which does imply constant curvature is homogeneity. A Riemannian manifold $(M, g)$ is homogeneous provided that for any $p, q \in M$, there exists an isometry $\Phi_{pq} : M \rightarrow M$ sending $p$ to $q$. The intuition behind this is that the metric looks the same at every point, and thus everything metric-related (like curvature) must be constant.
About the isometry group: the result is that if $(M, g)$ is Riemannian (finite-dimensional, I'm assuming), the its isometry group is a Lie group. There are no additional conditions. A lot is known about isometry groups of Riemannian manifolds, and since you like Kobayashi, you can take a look at another one of his books, called Transformation groups in Riemannian geometry, which has a nice exposition about this topic and many others.