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 ?



1 Answer 1


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.

  • $\begingroup$ Thank you for your answer. I'm trying to understand how your case is a counterexample. Can you develop on the "simply connected case" part ? $\endgroup$
    – Hypaulite
    Commented Jun 25, 2020 at 20:58
  • $\begingroup$ It might be more intuitive to think of the space (in $2+1$ dimensions) as an annular tube $\{t,x,y\in\mathbb{R}^3:1<\sqrt{x^2+y^2}<2\}$, where the vector field spirals around the annulus and the surfaces wrap around in the opposite direction. The simply connected issue is that there may need to be a "hole" in the manifold for this geometry to be possible, lest singularities form. $\endgroup$
    – Kajelad
    Commented Jun 25, 2020 at 21:10
  • $\begingroup$ Understood. I think that, precisely, if there is no hole, then this geometry might not serve as a counterexemple of my proposition. Here I am mentally thinking about helices in the whole spacetime, and trying to show that no spacetime foliation can be orthogonal to these curves, at each point. $\endgroup$
    – Hypaulite
    Commented Jun 26, 2020 at 13:23
  • $\begingroup$ I don't follow; by "hole" I don't mean missing points, so much as noncontractible loops: in this counterexample $\mathbb{R}^3\times S^1$ is the whole of spacetime; it just happens to be topologically nontrivial. $\endgroup$
    – Kajelad
    Commented Jun 26, 2020 at 15:33
  • $\begingroup$ I get your "annular tube" counter exemple. The problem is that it is "finite" in a sense, and thus you can indeed wrap the surface around, orthogonal to the helices. What I have in mind (but did not specified) is a manifold with no boundary. Take for instance in $\mathbb{R}^3$ with cartesian coordinates and picture $(x,y,z)=(R\cos u, R\sin u, u),u\in\mathbb{R}$ helices, with all radii $R\geq 0$. To me this does not seem hypersurface-orthogonal.. do you agree ? $\endgroup$
    – Hypaulite
    Commented Jun 26, 2020 at 15:50

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