Foliating R^3 with straight lines

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.

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By complete characterization you mean a classification up to to diffeomorphisms $\mathbb{R}^3 \to \mathbb{R}^3$ preserving such foliations? –  Michael Bächtold Apr 27 '11 at 19:56
... or do you want a method to produce all such foliations. And do you mean smooth foliations or arbitrary partitions of space into lines. Concerning the latter I think there was a related question on MO. –  Michael Bächtold Apr 27 '11 at 20:07
The foliation by hyperboloids is described here: mathoverflow.net/questions/1194/… –  Ian Agol Apr 27 '11 at 20:50
This shouldn't be too tricky. Since the leaves of the foliation are regular, the foliation will induce a submersion from \R^3→X for some surface X (namely, the space of leaves of the foliation). π_1(X) and π_2(X) are both trivial. So, X is diffeomorphic to \R^2. From here, it's a matter of classifying submersions over the plane with 1-dimensional leaves with a given affine structure. If we can find a surface embedded in R^3 that intersects each leaf exactly once, then this can be used to give the submersion a line bundle structure which means it's trivial. –  Brendan Foreman Apr 28 '11 at 2:07
@Brendan Foreman: Nice comment. Wouldn't you develop it as an answer? –  Giuseppe Tortorella Apr 28 '11 at 7:37

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 $X.$

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The nearest point to the origin is a continuous function of a line, so you get a continuous section in any case. Note that the tangent space to the foliation as well as its normal bundle is automatically oriented because R^3 is simply-connected -- you don't need to detour through embeddability of surfaces. But here's a question: given an embedding of $\mathbb R^2$ in $\mathbb R^3$, can you give a criterion whether it is transverse to such a foliaiton? the surface obtained from a foliation by this construction? –  Bill Thurston Apr 28 '11 at 14:42
I guess you can also make all the lines have exactly the same fixed nonzero angle with the $xy$ plane, by considering the corresponding cone of lines through the desired point. Or one can make all the angles different, but contained in a very tight interval of possible angles; etc. –  Joel David Hamkins Apr 28 '11 at 13:24
You can ensure that every line in the partition passes within $\epsilon$ of the origin. –  Joel David Hamkins Apr 28 '11 at 14:16