You can interpret this physically. Essentially, the Schrödinger group can be seen as modelling quantum nonrelativistic particles whereas the wave group models ultra relativistic particles. Recall that in general, the Hamiltonian is given by: $$ H = \sqrt{m^2+g(p,p)} $$ After quantisation you recover the curved analogue the Klein-Gordon equation: $$ (\partial_t^2+m^2-\Delta)\psi = 0 $$ with $m$ the mass.

With $m\to0$ you get the ultra relativistic limit (wave group): $$ H = \sqrt{g(p,p)} $$ After quantisation you recover the curved analogue the D'Alembert equation: $$ (\partial_t^2-\Delta)\psi = 0 $$ For $m\to\infty$ you get the nonrelativistic limit (Schrödinger group): $$ H = m+\frac1{2m}g(p,p) $$ After quantisation you recover the curved analogue the Schrödinger equation: $$ (i\partial_t+\frac12\Delta)\psi = 0 $$ In all cases, you get geodesic flow i.e. the rays in the eikonal limit follow geodesics. The difference is in the dispersion relation. For the wave group, it is dispersionless, the speed of the "particles" is the "speed of light" ($1$ according to the normalisation) hence the constant speed propagating wave front. This is to be contrasted with the Schrödinger group which is dispersive with arbitrarily large group velocities.

You can interpret this physically. Essentially, the Schrödinger group can be seen as modelling quantum nonrelativistic particles whereas the wave group models ultra relativistic particles. Recall that in general, the Hamiltonian is given by: $$ H = \sqrt{m^2+g(p,p)} $$ After quantisation you recover the curved analogue the Klein-Gordon equation: $$ (\partial_t^2+m^2-\Delta)\psi = 0 $$ with $m$ the mass.

With $m\to0$ you get the ultra relativistic limit (wave group): $$ H = \sqrt{g(p,p)} $$ After quantisation you recover the curved analogue the D'Alembert equation: $$ (\partial_t^2-\Delta)\psi = 0 $$ For $m\to\infty$ you get the nonrelativistic limit (Schrödinger group): $$ H = m+\frac1{2m}g(p,p) $$ After quantisation you recover the curved analogue the Schrödinger equation: $$ (i\partial_t+\frac12\Delta)\psi = 0 $$ In all cases, you get geodesic flow i.e. the rays in the eikonal/WKB limit (short wavelength) follow geodesics. Classically, this is because the particles evolve according to Hamiltonian mechanics with the corresponding Hamiltonian. In general, if the Hamiltonian is of the form $$ H = h(g(p,p)) $$ From the equations of motion: $$ \begin{align} \dot q^k &= h'g^{kl}p_l & \dot p_k &= h'\frac{\partial g^{ij}}{\partial q^k}p_ip_j \end{align} $$ you recognise the geodesic equation: $$ \frac1{h'}\frac d{dt}\left(\frac1{h'}\dot q^k\right)+\frac1{(h')^2}\Gamma_{ij}^k\dot q^i\dot q^j = 0 $$ so that the time parameter $t$ is related to the arc length $s$ by: $$ \frac{ds}{dt} = h' $$ The difference is in the dispersion relation (wave analogue of the time parametrization). For the wave group, it is dispersionless, the speed of the "particles" is the "speed of light" ($1$ according to the normalisation) hence the constant speed propagating wave front. This is to be contrasted with the Schrödinger group which is dispersive with arbitrarily large group velocities.