Global conformal Killing vector fields in a Riemannian manifold

Question

Suppose $t$ is a globally smooth Killing coordinate function in $(M,g)$ such that $\partial_t$ is a Killing vector field. This gives rise to an embedding $\mathbb R \times \Omega$ for the manifold $M$. Under what conditions can we deduce that the metric must take the form: $$ g(t,x) = \,dt^2+ 2\,dt\otimes \eta + h $$ where $\eta$ is a one-form and $h$ is a Riemannian metric on $\Omega$ with both $\eta$ and $h$ independent of $t$.

Answer 1

Here is the general form of metric tensor that is invariant with respect to vertical shifts: $$ g(t,x) = f(x){\cdot} dt^2+ 2{\cdot} dt\circ \eta + h.$$

Evidently, $g(\partial_t,\partial_t)=f(x)$. So it is sufficient to assume $g(\partial_t,\partial_t)=1$.