TITLE: Foliating R^3 with straight lines
QUESTION [11 upvotes]: Is there a complete characterization of ways one can foliation the 3-dimensional Euclidean space with straight lines?
For example, one can partition R^3 into parallel planes and fill up each plane with parallel lines. Slightly more involved examples should be possible, like filling space with hyperboloids.
I would like to know what conditions one can say about an arbitrary foliation of R^3 by straight lines.

REPLY [4 votes]: The leaves of the foliation are regular.  So, the space of leaves $X$ induces a submersion $\pi: R^3\rightarrow X$ with leaves of $R.$  Using the standard homotopy sequence, we see that $X$ is simply-connected.  Furthermore, $X$ is path-connected since any two leaves can be connected by a path of leaves in the foliation.
We can actually define an embedding $\phi: X\rightarrow R^3$ by letting $\phi(l)$ be the point on the leaf $l\in X$ that is closest to the origin in $R^3.$  This gives us two things:  a vector space structure on each leaf by using $\phi(l)$ as the origin and an orientation of $X$ (since only orientable simply-connected surfaces are embeddable in $R^3$).  We use the orientation of $X$ to give an orientation on each leaf.
Thus, we have a 1-dimensional orientable vector bundle over a simply-connected space $X$, which means that the vector bundle is trivial.  Hence, all smooth foliations with leaves of lines on $R^3$ are diffeomorphic.
I rather wanted to say all continuous foliations of such sort are homeomorphic, but I'm nervous about the embedding of $X.$

REPLY [3 votes]: Probably you don't, but if you mean to include arbitrary partitions of space into lines, then a tidy classification will be impossible, since with the axiom of choice one can construct extremely bizarre partitions. Just enumerate the points in space in order type continuum, and iteratively pick a line through the next point missing the previous lines. There are continuum many such lines, since only fewer than continuum many lines were chosen so far. So one can arrange, for example, that no two of the lines in the partition are parallel, that no two of them have the same angle with the axis planes, or each other, and so on. This is because at each step of the recursion, such requirements only rule out fewer than continuum many of the continuum many lines through the desired point.