Does $M=(0,1)^3$ admit a foliation whose leaves, $F_{\alpha} \cong S^1\times(0,1)$ accumulate to only two points (i.e. $011$ or $100$)?
I can prove the trivial cases for $M$ in which the foliation is simply by planes, and the slightly more nontrivial case for $M$ in which the foliation is by surfaces that get arbitrarily close to $011$ and $100$ (but these don't admit closed meridian loops).
The co-dimension 1 foliation I ask about is a non-trivial generalization of a 2d foliation which partitions a semi-riemannian manifold called the Minkowski plane.
The first goal is to show that the generalization of the 2d foliation of Minkowski 2-space, is a foliation of some 3d Lorentzian manifold.
I can plot an example of a leaf in $M$ (below). And I propose that the vector field generating a flow tangent to the longitudinal meridians (black lines in the image) should be a Killing vector field, making the longitudinal foliation a metric foliation.
Take $V_+^2$ the set of all $2$-dimensional non-negative vectors $\mathbf x=(x_1,x_2)^⊤$ and $K^2$ the class of all smooth functions from $\Bbb R^2_+$ to $(0,1).$ Then there should exist some vector valued function $f(\mathbf x)$ for $k\in K^2$ which encodes the leaves of the desired foliation of $M.$ I guess that a nice form for $f(\mathbf x)$ is:
$$ f(\mathbf x):=\bigg(\int k(\mathbf x)~d\mathbf x, \int \frac{y_1 k(\mathbf x)}{c_1}~d\mathbf x,\int \frac{y_2 k(\mathbf x)}{c_2}~d\mathbf x: k\in K^2\bigg)$$
for constants $c_1$ and $c_2.$ I believe this is very close, but please let me know if this definition works and/or if there are better definitions for the leaves.
