Hypersurface-orthogonality and static spacetimes I am currently learning about general relativity using these notes and I am confused by the notion of hypersurface orthogonality of a vector field.
We define a spacetime to be stationary if there exists a timelike Killing vector field K. In this case we construct a system of coordinates $(t, x^i)$ where $x^i$ are coords on a hypersurface which is nowhere tangent to $K^a$, $K = \frac{\partial}{\partial t}$ and the metric takes the form
$$ ds^2 = g_{00}(x^k)dt^2 + 2g_{0i}(x^k)dtdx^i + g_{ij}(x^k). \label{eq:metric}$$
If in addition the Killing vector $K^a$ is hypersurface-orthogonal then we say the spacetime is static. We can choose the hypersurface in the construction of the above coords to be orthogonal to $K^a$, and so the metric has $g_{0i} = 0$ in these coordinates.
My issue here is that in my head I picture that for any smooth covector field there should be some hypersurface which is orthogonal to it. This would mean that any vector field is hypersurface orthogonal and hence any stationary spacetime is automatically static, but this is clearly false. Why is it that some vector fields are not hypersurface-orthogonal?
The standard example of a stationary but not static spacetime is the Kerr metric. In Boyer-Lindquist coordinates, the metric is in the above form with $g_{0i} \neq 0$ (to be precise there is a $dtd\varphi$ cross term). This means that when lowering the index on $K^a = \left(\frac{\partial}{\partial t}\right)^a$ we get a $d\varphi$ component, ensuring that $K_a$ is not orthogonal to the constant $t$ surfaces. But why does this mean that Kerr is not static, since surely there could be a different foliation by hypersurfaces which are orthogonal to $K$?
 A: Another definition of static spacetimes is that there is a global timelike killing vector field $\xi$ and it turns out to be equivalent to the definition you have just provided. I hope this shows why Kerr is not static in another way.
You are correct when you say that every vector field determines a surface orthogonal to it. However, it is not always possible to define coordinates in spacetime that cover the whole manifold and such that keeping the first coordinate constant yields this orthogonal hypersurface. Indeed, a given set of vector fields $\{v_i\}$ define a coordinate basis if and only if their commutator vanishes, that is,
$$[v_i,v_j] = 0\:\:\:\: \forall i,j. $$
The thing is that if you have a spacetime with a given timelike vector field that is not killing, although you will be able to talk about the orthogonal surfaces to it, you will not find coordinates in these that will allow you to get coordinates in spacetime. That is, although you can find a timelike vector field, you will not be able to find vectors that are spacelike and orthogonal to it and commute. Therefore, you won't be able to find a coordinate system that covers the manifold and that areorthogonal.
