# Hypersurface-orthogonal vector field and helicity

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

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.
• 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. Commented Jun 25, 2020 at 21:10
• 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. Commented Jun 26, 2020 at 15:33
• 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 ? Commented Jun 26, 2020 at 15:50