Consider a globally hyperbolic Lorentzian manifold $(M,g)$. Then, a well-known result of Bernal-Sánchez (see Theorem 1.1 in arXiv:gr-qc/0401112) states that it can globally be written as
$$M=\mathbb{R}\times\Sigma, \quad g=-\beta dt^{2}+h_{t}\ ,$$
where $\beta\in C^{\infty}(M,(0,\infty))$ and $h_{t}$ is a Riemannian metric on each slice $\Sigma_{t}:=\{t\}\times\Sigma$. Here, $t:M\to\mathbb{R}$ is what is called a Cauchy temporal function and has for example the property that its gradient is past-directed timelike. Now, in arXiv:2110.13672 [gr-qc] at page 9-10, the author states the following (transferred in my notation):
There are no cross terms (shift) between the $\mathbb{R}$ and $\Sigma$ parts and for any coordinates $x^{i}$, $i=1,\dots,n$ on $\Sigma$ one can choose $x^{0}=t$ so that $$g_{00}(t,x^{i})=-\beta(t,x^{i}),\quad g_{0i}(t,x^{i})=0,\quad g_{ij}(t,x^{i})=(g_{t})_{ij}(x^{i})\,.$$
Propably I thinking too much about it and makes things more complicated than they are, but I am not quite understanding this claim. I mean, properly it is more rigorous to state the Bernal/Sánchez theorem as $g=-\beta dt^{2}+\overline{g}$ where $\overline{g}$ is a tensor field on $M$ with the property that $i_{t}^{\ast}\overline{g}$ is a Riemnnian metric on $\Sigma$ with $i_{t}:\Sigma\to M, p\mapsto (t,p)$ being the obvious inclusion. So, it is not clear to me why there are not cross terms, independently of the coordinates on $\Sigma$ chosen. Since the embedding $i_{t}$ depends on time, there might be a shift vector, or am I wrong? What am I missing?
(In fact, in all these communities doing numerical relativity and the initial value formulation of general relativity, people are usually working with globally hyperbolic manifolds and they always include a shift vector field...)
