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Let $\Sigma$ be a smooth convex surface in Euclidean 3-space and $\gamma$ be a unit speed minimizing geodesic in $\Sigma$. Assume that for some $a < b < c$, we have $$\gamma'(a)=\gamma'(b)=\gamma'(c).$$

Is it true that $\gamma'$ is constant on one of two intervals $[a,b]$ or $[b,c]$?

Comments

  • This is a simplification-variation of an other question I heard from Dima Burago.
  • It is not hard t construct an example of a minimizing geodesic such that $\gamma'(a)=\gamma'(b)$, but $\gamma'$ is not a constant on $[a,b]$. (See the example below.)
  • I would be also interested in the analog for $n$ points.
  • This paper: "Total curvature..." by Barany, Kuperberg, Zamfirescu is relevant.
  • Recently, in "On the total curvature..." by Nina Lebedeva and me we answered the original question of Dima Burago. The problem above remains open, likely the answer is "no".

Example. I will construct a convex polyhedron, but it is easy to smooth. Consider polyhedron defined by 5 inequlaities: $$z\ge 0,\ \ |x+\alpha{\cdot}y|\le \alpha\ \ \text{and}\ \ z\pm\beta{\cdot}y\le \beta$$ and look at the minimizing geodesic between points $(0,1-\epsilon, \beta{\cdot}\epsilon)$ and $(0,-(1-\epsilon), \beta{\cdot}\epsilon)$. For appropriately chousen $\alpha$, $\beta$ and $\epsilon$ this minimizing geodesic will pass through the faces in this order $$\{z+\beta{\cdot}y= \beta\},\ \ \{x+\alpha{\cdot}y= \alpha\},\ \ \{z=0\},\ \ \{x+\alpha{\cdot}y= -\alpha\},\ \ \{z-\beta{\cdot}y= \beta\}$$ and it will have the same velocity vector $(1,0,0)$ on both faces $\{z+\beta{\cdot}y= \beta\}$ and $\{z-\beta{\cdot}y= \beta\}$.

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  • $\begingroup$ My reaction to this is that this doesn't sound that hard but if you and Dmitri can't answer it, it's not likely I can either. Nice question, though. I will have to give it a try. $\endgroup$ – Deane Yang May 2 '11 at 16:32
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    $\begingroup$ I realize that it should be just homework for me, but could you provide the details of the example with two points with the same velocity? $\endgroup$ – Deane Yang May 2 '11 at 19:16
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    $\begingroup$ Presumably gluing your $[a,b]$ example to its reverse can only be realized on a nonconvex surface? $\endgroup$ – Joseph O'Rourke May 2 '11 at 19:53
  • $\begingroup$ @Deane: I add the example. $\endgroup$ – Anton Petrunin May 2 '11 at 22:48
  • $\begingroup$ I can see now that my initial reaction was correct. Nice question. $\endgroup$ – Deane Yang May 3 '11 at 13:03
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Just to help visualize the example, if I have interpreted the description correctly, this is one view, for $\alpha=\beta=1$, and $\epsilon=\frac{1}{16}$.
Polyhedron
In this view, the $x$-axis is horizontal, the $z$-axis vertical.

Addendum. The shortest path (yellow) from $p_1$ to $p_2$ follows the face sequence (Bk, R, Bt, L, F).
Unfolding
I tried to indicate (in the nearly invisible short dashed lines) the initial direction of the path on the (mauve) Bk and F faces, in their planar unfoldings. I should note that these initial velocities are not exactly $(1,0,0)$, which presumably is only achieved by "appropriately chosen" $\alpha, \beta, \epsilon$.

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  • $\begingroup$ On the second diagram, the angles of rotations at $p_1$ and $p_2$ should be bit more than π/2. $\endgroup$ – Anton Petrunin May 4 '11 at 17:00
  • $\begingroup$ Awesome picture, as always. $\endgroup$ – Thomas Richard Apr 19 '12 at 20:22

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