I am currently (trying) to learn more about orbit spaces generated from an isometry group of a manifold. I cannot quite pinpoint what I (don't) understand, so I will try to lay out what I could gather:

Suppose we consider pseudo-Riemannian manifold $M$. Let us assume that this manifold has an isometry group $G$. Now, from I understand, I can/should really understand this $M$ as being a $G$-manifold. For example, if $G=SO(3)$, then its action on a point $p \in M$ would be the usual "matrix multiplication". Anyway, this means that we also have an associated Lie algebra $\mathfrak{g}$. From what I could gather, is it correct to say that a Killing vector $\xi$ generated from the isometry in fact lives in $\mathfrak{g}$? In which case, we can use the exponential map for : $exp(t \xi) \in G$. Now,I could see from here that $d/dt \ exp(t \xi) * p$ is a tangent vector? Intuitively, I can see why: the one -parameter group $exp(t \xi)$ is in $G$ and acts on $p$, thus giving another point in $M$. Differentiation then gives a tangent vector I suppose, but I cannot see this more precisely, especially when viewing tangent vector as directional derivatives? So, if I wanted to act $d/dt \ exp(t \xi) * p$ on some function $F \in C^\infty(M)$, how is this defined? This then brings me to the main problem: the induced metric, following again a previous post, would be: $$g\left(d/dt \ exp(t \xi) * p,d/dt \ exp(t \xi) * p\right)$$ I cannot see how one can compute this? As an example, suppose we have the metric: $$g = dr^2+r^2 d\theta^2$$ It has an evident Killing vector $\xi = \frac{\partial}{\partial \theta}$. but I cannot see how to use the above to calculate the induced metric? I mean, very naively, I would suspect the metric "perpendicular" to the orbits to be of the form: $$g^{perp}_{ab} = g_{ab} - r^{-2} \xi_a \xi_b$$ because $g^{induced}_{ab}\xi^b=0$. The above does not really make sense to me, insofar that the orbit space really should not depend on the $\theta$ coordinate anymore, whereas the indices $a,b$ still run over the full manifold! It is just that we made sure that $g^{induced}_{\theta \theta}=0$, and so naively, we can "forget" about "this part" of the metric?? Reversing it, the induced metric of the orbits would then be: $$g^{induced}_{ab} = r^{-2} \xi_a \xi_b?$$


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