 Solution

A Riemannian metric is a smooth positive definite symmetric bilinear form $g_q$ on each tangent space. The inverse form on the cotangent space defines the kinetic HamiltonianIt is smooth and satisfies $H(q,ap)=a^{2}H(q,p)$. With the canonical symplectic form $\Omega ={\sum }_{i}d{q}^i\wedge d{p}_i$, define ${\iota }_{X_H}\Omega =dH$. This gives Hamilton's equationsThus $p_i=g_{ij}{\dot{q}}^j$. Differentiating the inverse metric yields ${\dot{p}}_i=\frac{1}{2}{\partial }_i{g}_{jk}{\dot{q}}^j{\dot{q}}^k$. Differentiating $p_i=g_{ij}{\dot{q}}^j$ and substituting givesTherefore the projections of the Hamiltonian integral curves are affinely parametrized geodesics, and lifting a geodesic by its metric covector gives a Hamiltonian integral curve.

For the Killing vector field define the fibrewise linear function $f(q,p)=p(Y_q)=p_i{Y}^i(q)$. Along the lifted geodesic it equals $g(\dot{q},Y)$. Metric compatibility and the geodesic equation giveThus $f$ is conserved. Equivalently the cotangent lift of the isometry flow preserves $H$, and its generator has Hamiltonian $f$, so their Poisson bracket vanishes.

For the final two metrics write $s=\ell (x)+m(y)$, restricting to an open set where the printed tensors are smooth nondegenerate metrics. Their inverse forms giveAlong the $H_1$ flow, $H_1$ is constant, whileHenceThis is the Liouville metric geodesic integral, a quadratic conserved quantity. It does not require a Killing vector field; it arises from the separated metric coefficients. When both printed metrics are Riemannian, their positivity in particular requires $s>0$, $m>0$ and $\ell <0$; the calculation also applies wherever the corresponding inverse tensors are defined.