I am wondering about the correspondence between the isometry group $\mathcal{I}$ and the Lie Algebra of Killing vector fields $\mathcal{K}$ on a pseudo-Riemannian manifold $(\mathcal{M}, \mathbf{g})$. Specifically
$\mathcal{L}_X \mathbf{g} = 0$
defines $\mathcal{K}$ and gives a necessary condition for $X$ to be in the Lie-Algebra $\mathcal{i}$ of $\mathcal{I}$. But as my understanding goes this is not a sufficient condition i.e not all elements of $\mathcal{K}$ generate flows $\theta_t : \mathcal{M} \rightarrow \mathcal{M}$ with $t > 0$ that are in $\mathcal{I}$. Now I have found obscure references, that only elements of $\mathcal{K}$ that generate global flows are in $\mathcal{i}$.
I guess my question boils down to something like:
Given a vector field $X$, is it possible that its maximal flow domain contains a finite interval $0 \in [a,b]$ for all $p \in \mathcal{M}$ without the flow being global? My intuition tells me: no, because if this were the case, initial conditions should be "preserved" in some sense, forcing e.g $\theta_{2b}$ to be defined as well.
The Lie algebra of the isometry group indeed consists of exactly those Killing fields, whose flows are defined for all times. The fact that this condition is neccesary is rather easy to see: In a Lie group $exp(tX)$ is always defined for all $t$, and the action of this element implements the flow of the vector field corresponding to $X$ at time $t$. Otherwise put, the fundamental vector fields of any action of a Lie group are always complete (i.e. admit a flow for all times). This plays an important role in the characterization of Lie transformation groups (which involves the condition of admitting a global flow).