Can I form a scalar with coordinates in general relativity? There is no position vector in general relativity. I was wondering whether a quantity like
$$k_\mu x^\mu$$
where $k_\mu$ are covariant vector components is to be treated like a scalar i.e. invariant under coordinate transformations?
A came across that problem, studying gravitational waves. In one coordinate system $x$ the wave is given by
$$h_{\mu\nu}=e_{\mu\nu}\exp(i k_\alpha x^\alpha)$$
For another coordinate system $x'$ I found the Ansatz
$$h'_{\mu\nu}=e'_{\mu\nu}\exp(i k_\alpha x^\alpha)$$
in a text book. This only makes sense to me, if $k_\alpha x^\alpha$ is invariant under coordinate transformations. Otherwise it should be
$$h'_{\mu\nu}=e'_{\mu\nu}\exp(i k'_\alpha x'^\alpha)$$
But as $x^\alpha$ are not the components of a vector, I don't understand why that would be true.
 A: It is a misnomer to denote the coordinate as $x^{\mu}$ in General Relativity. Coordinates don't transform as 4-vectors, so the upper index is misleading and incorrect.
This is not even specific to General Relativity, but is a mathematical fact from differential geometry (with is heavily used in General Relativity).
One way to see this is to consider compactified space-time. Let's say it is topologically a $d$-torus $T^d$ (just as a mathematical example). Then your Fourier integral with waves $e^{i k_{\mu} x^{\mu}}$ doesn't make any sense mathematically. Instead, $k$ take discrete values, and $x^{\mu}$ are coordinates within a single chart of the atlas, that don't have a global value.
It is, however, true, that the infinitesimal differentials $dx^{\mu}$ transform as $4$-vectors under general coordinate transformations, which partially justifies this abuse of notation. They are also atlas-independent – coordinate differentials follow the usual 4-vector transformation laws in the intersections of charts of the space time manifold's atlas.
Anticipating the question – yes, I think Spiridon's answer is incorrect and misleading.
A: In the general relativity the combination $k_\mu x^{\mu}$ is still a scalar. Performing a general change of coordinates $x \rightarrow x^{'}$, you get:
$$
k_{\mu} x^{\mu} = \frac{\partial x^{\alpha}}{\partial x^{\mu '}} \frac{\partial x^{\mu}}{\partial x^{'\beta}} k_\alpha^{'} x^{' \beta} = \delta_{\beta}^{\alpha} k_\alpha^{'} x^{' \beta} = k_\alpha^{'} x^{' \alpha}
$$
In the second equality we used the fact, that the $k_\mu$ and $x^{\mu}$ transform with the mutually inverse matrices.
Because you are considering graviational part, the polarization tensor will also transform:
$$
e_{\mu \nu}^{'} = \frac{\partial x^{\alpha}}{\partial x^{\beta}} \frac{\partial x^{\mu}}{\partial x^{\nu}} e_{\alpha \beta}
$$
So in the expression for graviational wave one sees the same exponent, but a transformed expression in front of it.
