Take $u=v=l_\xi$ and $w=l_\eta$ in the identity from part (e). For a <left-invariant metric>, all three scalar products are constant and $[l_\xi,l_\xi]=0$, so
$$
\langle\nabla_{l_\xi}l_\xi,l_\eta\rangle
=-\langle\xi,[\xi,\eta]\rangle.
$$
The curve $\gamma_\xi$ has velocity $l_\xi$, hence is a <geodesic> exactly when $\nabla_{l_\xi}l_\xi=0$. Nondegeneracy of the inner product now gives
$$
\gamma_\xi\text{ is geodesic}
\quad\Longleftrightarrow\quad
\langle\xi,[\xi,\eta]\rangle=0
\quad\text{for every }\eta\in\mathfrak g.
$$
This is the <geodesic-vector criterion for a left-invariant metric>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 4 / f / Solution

{title2=$\langle\xi,[\xi,\eta]\rangle=0$}

For a <left-invariant metric> on a <Lie group>, the one-parameter subgroup $\gamma_\xi(t)=\exp(t\xi)$ is a <geodesic> exactly when
$$
\langle\xi,[\xi,\eta]\rangle=0
$$
for every $\eta$ in the <Lie algebra>. The <Koszul formula> gives $\langle\nabla_\xi\xi,\eta\rangle=-\langle\xi,[\xi,\eta]\rangle$.

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