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

As a vector space, the Lie algebra of $G$ is $\mathfrak{g}=T_{e}G$. For $\xi \in \mathfrak{g}$, definewhere $L_g$ is left translation on a Lie group. This vector field is smooth and left-invariant because $(d{L}_h{)}_g{l}_{\xi }(g)=l_{\xi }(hg)$. The assignment $\xi \mapsto l_{\xi }$ is linear and injective by evaluation at $e$. Conversely, every left-invariant vector field $V$ satisfies $V(g)=(d{L}_g{)}_{e}V(e)$, so it equals $l_{V(e)}$. Hence the map is an isomorphism.

The Lie bracket of vector fields is bilinear, alternating, satisfies the Jacobi identity, and is natural under diffeomorphisms. Therefore the bracket of two left-invariant vector fields is left-invariant. DefineThenand the inherited bilinearity, alternation, and Jacobi identity make $T_{e}G$ a Lie algebra.

The one-parameter subgroup ${\gamma }_{\xi }(t)=\exp (t\xi )$ is the integral curve through the identity of $l_{\xi }$. Its defining initial-value problem is

The Levi-Civita connection is a torsion-free connection. Its torsion form isSince $T=0$,In coordinates this is the symmetry ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ of the Christoffel symbols.

The Levi-Civita connection is also a metric connection, soand similarly for the two cyclic permutations. Add the identities with leading derivatives $u$ and $v$, subtract the one with leading derivative $w$, and useCancellation and symmetry of $g$ giveThis is the Koszul formula written with the bracket terms on the right.

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 $[l_{\xi },l_{\xi }]=0$, soThe 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 givesThis is the geodesic-vector criterion for a left-invariant metric.