5
$\begingroup$

Let $P$ be the surface of a closed convex polyhedron in $\mathbb{R}^3$.

Q. Does every non-closed geodesic $\gamma$ fill $P$ densely?

Of course $\gamma$ cannot pass through a vertex of $P$, but it could come arbitrarily close.

Here is an informal way to phrase the question. One can have non-closed geodesics on smooth convex surfaces in $\mathbb{R}^3$ that do not fill the surface. E.g., on ellipsoids, as in the MO question "Surfaces filled densely by a geodesic":


Ellipsoid
            (Image from GeographicLib.)
So: Is there a polyhedral analog?

(One could ask the same question for nonconvex polyhedra.)

$\endgroup$
  • 2
    $\begingroup$ I expect that the words "Veech surface" and "Veech dichotomy" may be relevant here, but I am not an expert in these things and can't say anything more without spending some time thinking about it, or consulting someone who is an expert. (Unfortunately at the moment I have neither an abundance of time nor a conveniently placed expert.) $\endgroup$ – Vaughn Climenhaga Dec 13 '14 at 2:23
  • $\begingroup$ @VaughnClimenhaga: Very useful key phrases, Veech surface & V. dichotomy. Thanks! $\endgroup$ – Joseph O'Rourke Dec 13 '14 at 2:57
7
$\begingroup$

A simple triangulation of the previous example will work.

enter image description here

Suppose we have a geodesic like the blue one. The key point is to look at the green angles (better the complementary ones, those respect to the horizontal plane). Each time the geodesic crosses a black edge, this angle decreases by a fixed (this is essential!) amount, namely the angle formed by two black vertical edges if extended. If the initial angle is small enough, then we may make sure that the geodesic will cross so many black edges as we want before it touches the top square, so in fact it is not touched at all. The blue geodesic goes down and the process goes on forever.

One could answer: what happens if, when coming down, the angle is very large? For this not to happen, the polyhedron must be pointed enough. This is always feasible, as shown by this picture (which represents a periodic plane representation of the lateral faces):

enter image description here

Of course non-periodicity is a generally satisfied.

$\endgroup$
  • $\begingroup$ Nice example, convincing explanation---Thanks! $\endgroup$ – Joseph O'Rourke Dec 13 '14 at 13:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.