= Solution
Choose the positive-frequency branch with $k>0$, let $K=|\mathbf k|$, and write $\mathbf k=K(\cos\theta,\sin\theta)$. The <internal gravity wave> <dispersion relation> is $\omega=Nk/K=N\cos\theta$. The <phase velocity> and <group velocity> are
$$
\boxed{\mathbf c_p=\frac{N}{K}\bigl(\cos^2\theta,\cos\theta\sin\theta\bigr),\qquad
\mathbf c_g=\frac{N}{K}\bigl(\sin^2\theta,-\cos\theta\sin\theta\bigr).}
$$
They are perpendicular: the phase advances normal to the crests while energy travels along them. For a general signed $k$, the positive frequency is $N|k|/K$ and the horizontal component of group velocity changes sign with $k$.
If $\omega>N$, there is no real wave vector satisfying this relation. For prescribed real horizontal <wavenumber>, $m^2=k^2(N^2/\omega^2-1)<0$, so a bounded boundary response is an <evanescent wave>, rather than a freely propagating plane wave.
For a single incompressible plane wave, <internal-wave polarization> gives $\mathbf u_1\cdot\mathbf k=0$. Its oscillatory displacement $\boldsymbol\xi_1$ is parallel to $\mathbf u_1$ and therefore also perpendicular to $\mathbf k$. Every spatial derivative of the plane-wave field is proportional to a component of $\mathbf k$, so
$$
\boxed{(\boldsymbol\xi_1\cdot\nabla)\mathbf u_1=0.}
$$
There is no leading <Stokes drift> for that single wave. The fluid parcels oscillate along straight lines within phase planes and never sample a spatial change of the wave along their displacement. This conclusion does not prohibit cross terms between different wave vectors.
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