 Solution

For the positive-frequency branch and $K^2=k^2+m^2$, the wavefront-normal phase velocity and the group velocity areTheir scalar product vanishes since $k(k{m}^2)+m(-m{k}^2)=0$. Equivalently, the frequency is homogeneous of degree zero in the wave vector, and hence $\mathbf{k}\cdot {\nabla }_{\mathbf{k}}\omega =0$.

The vertical sign relation isIt is negative for the usual geophysical ordering $N>|f|$ and nonzero $k,m$. Thus vertical phase and energy propagation are opposite under that ordering; negative-frequency waves obey the same product relation. The PDF does not state this ordering, so its unconditional direction assertion needs qualification. For example, $N=1$, $f=2$, $k=m=1$ gives positive vertical components on the positive-frequency branch. If $N=|f|$ the group velocity vanishes, and in axial limiting directions one vertical component can be zero. Orthogonality is valid throughout, while strictly opposite nonzero vertical components need the stated nondegeneracy and $N>|f|$.