# 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 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

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.

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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