TITLE: Hypersurface-orthogonal vector field and helicity
QUESTION [0 upvotes]: In a (3+1)-dimensional Lorentzian manifold equipped with a metric $g_{ab}$ (context: general relativity), I define a vectof field $k^a$ to be a Helical Killing Vector (HKV) if
i. it is a Killing vector field, i.e. $\mathcal{L}_k g_{ab} = 0$
ii. it is of the form $k^a := t^a + \Omega\, \phi^a$,
where $t^a$ is a timelike vector field ($g_{ab}t^at^b<0$ everywhere), $\Omega\neq 0$ is a constant, and $\phi^a$ is a spacelike vector field ($g_{ab}\phi^a\phi^b>0$ everywhere) with closed integral curves.
The integral curves of an HKV depict helices in spherical-like system of coordinates. It seems to me that such a vector cannot be hypersurface orthogonal, because of its helical nature (think of the twisted fibers in a rope). However I am having troubles proving this rigorously, even with the many formulations of the Frobenius theorem.
Does anybody know how to do this, or have any ideas/sources as to how to do it ?
Thanks

REPLY [1 votes]: Such a vector field can be hypersurface orthogonal, but perhaps not in the simply connected case. Here's a potentially illustrative counterexample:
Consider the space $\mathbb{R}^3\times S^1$, with coordinates $t,x,y,\theta$ (with $\theta$ defined modulo $2\pi$, or locally) and the Minkowski metric $g=-dt^2+dx^2+dy^2+d\theta^2$. Choosing $\epsilon\in(0,1)$, we can choose a timelike foliation given by $t=\epsilon(\theta+2\pi n+c)$ with $c\in[0,2\pi)$ labeling the leaves. These sheets are everywhere orthogonal to $\epsilon^{-1}\partial_t+\partial_\theta$ which is a killing vector field fitting your description.
This approach might rely on the existence of noncontractible spacelike loops. If these are absent, something might go wrong when shrinking these loops to a point.
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