Is the wave 4-vector a vector in general relativity?

Question

For a plane wave $f(x)=A\cos(\omega t-{\bf{k\cdot x}})$, the wave $4$-vector in special relativity is $$k^\mu=(\omega,{\bf{k}}) .$$

This is a $4$-vector in special relativity, as can be seen by checking how it transforms under a Lorentz transformation. Alternatively, the phase $\xi=\omega t-{\bf{k\cdot x}}=k_\mu x^\mu$ is plainly coordinate invariant, hence a scalar; therefore, by the quotient theorem for tensors (see Dirac's "General Theory of Relativity", Chap. 4), $k_\mu$ is a vector, since $x^\mu$ is a vector in special relativity.

Obviously, $x^\mu$ is not a vector in curved space. Therefore, I am guessing that neither is the wave vector $k_\mu$?

Here is a clearer, more detailed version of the question:

Answer 1

Actually, the wave vector is indeed a vector in curved spacetime. It is defined in generality as the gradient $k_\mu = \frac{\partial \xi(x^\mu)}{\partial x^\mu}$, where the phase function is a scalar function. So you could argue that by definition it is a four-vector (technically naturally it is a 1-form, but you can raise the index). The phase function only has the form $k_\mu x^\mu$ for a plane wave; and e.g. for light, plane wave solutions to Maxwell's solutions actually don't exist in curved spacetime.