# Geodesics of manifold with $n$ Killing vector fields

I'm trying to compute rotations of a generalized rigid body, namely where rotations correspond to geodesics on Lie group with left-invariant metric. In this context one always has for every group dimension a Killing vector field. Inspired by this physical problem, I was wondering about the following:

Given an $n$-dimensional Riemannian manifold with $n$ Killing vector fields $X^{(i)}$ that are everywhere linearly independent, I'm interested in finding the geodesic $\gamma(t)$ through the point $\gamma(0)$ with fixed tangent vector $v_0=\dot{\gamma}(0)$.

Does the following simplification work?

• Along the geodesic $\gamma$, the inner products $C^{(i)}=\langle\dot{\gamma}(t),X^{i}_{\gamma(t)}\rangle$ is constant, because $X^{(i)}$ is a Killing vector field.
• Knowing all the local inner products $\langle X^{(i)},X^{(j)}\rangle$ together with $C^{(i)}$, should allow us to construct at every point $p$ of the manifold a vector $Y_p$, such that $C^{(i)}=\langle Y_p,X^{(i)}_p\rangle$.
• I believe that if $\gamma$ goes through a point $p$, then it must have tangent vector $Y_p$.

This means I can compute the geodesics $\gamma$ as integral curve of the flow $Y$ starting at the point $\gamma(0)$.

Is my reasoning correct? Is this a well-known construction?

• Is there a particular reason you asked this question here and not on Mathematics, since you gave no physical context for your question? – ACuriousMind May 25 '17 at 16:31
• Yes, the context of my question comes from physics and I included it above. – LFH May 25 '17 at 17:58
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