2
$\begingroup$

Let S be the surface of a 3D convex region (a 'convex surface'). Let S' be a subset of S. We shall refer to S' as geodetically convex in S if the following condition holds: If A and B are two points in S', (at least one of ) the shortest path(s) on S between A and B contains no point that is not in S'.

Question: It is easy to see that if S is the surface of a sphere and n is any integer, S can be partitioned into n equal area pieces that are also geodetically convex in S - these pieces are separated by great circle arcs between any 2 antipodal points on S. Prolate and oblate spheroids (and perhaps all surfaces got by rotating a curve about an axis) too seem to allow such equipartitions. Are there any other convex surfaces for which such an equipartition exists for any n? If so, how does one characterize such surfaces?

Note 1: If S is the surface of a cube, I suspect it cannot be cut into n geodetically convex equal area pieces for all n. One could also ask which convex surfaces are worst for such equipartitions - ie if there are convex surfaces which do not allow such an equipartition for any n.

Note 2: Even if we drop the equal area requirement and need the n pieces only to be geodetically convex in S, the question appears tricky for general n. For n=2, since a closed convex surface always has a closed geodesic (as per a theorem of Poincare), such a geodesic cuts a convex surface S into 2 pieces that are geodetically convex in S.

$\endgroup$
4
  • 2
    $\begingroup$ Concerning the cube, which $n$ do you know about? 2 and 6 are easily "yes", and 4 can be done with each piece consisting of three isosceles right triangles. What about 3 and 5? $\endgroup$ Commented Jan 24, 2023 at 10:42
  • 1
    $\begingroup$ @JukkaKohonen: For $n=3$ on cube, this is natural but fails: Slice with planes parallel to the base at heights $\frac{1}{4}$ and $\frac{3}{4}$. The middle band fails to be geodetically convex. Another reason that an alternative definition might be preferred. $\endgroup$ Commented Jan 24, 2023 at 11:53
  • 1
    $\begingroup$ @Joseph, that was my first attempt as well. IMHO it just shows that the problem has two possible formulations, a "local" and a "global" one. Both might be interesting in their own right. Perhaps the "local" one that you are proposing would have better title to the name "geodesic" since usually being a geodesic is understood as a local condition (not "globally shortest path"). $\endgroup$ Commented Jan 24, 2023 at 13:42
  • $\begingroup$ Yes, I couldn't find an answer for n=3 or 5 for the cube. Maybe there aren't. $\endgroup$ Commented Jan 25, 2023 at 8:31

1 Answer 1

1
$\begingroup$

Let $C$ be a cone with lateral side $A$ and base $B$. So $C = A \cup B$. The base is geodetically convex. But if I'm interpreting the definition correctly, $A$ is not geodetically convex: For any two points $a$ and $b$ on the rim, the unique shortest path from $a$ to $b$ lies in $B$ (blue in the figure). The geodesic segment on $A$ (red) is always longer.

Cone

Perhaps this is the behavior you want. But if not, you might change the definition to:

A set $S$ is convex if, for every pair of points in $S$, there exists a geodesic segment $\gamma$ connecting them which remains entirely within $S$.

I.e., drop the "shortest path(s)" condition.

Even with this definition, there are what might be considered anomalies. Consider the cone $C$ minus the apex $x$: $C \setminus \{x\}$. Then I believe this surface with a point-hole is convex, because no geodesic passes through $x$. Again, maybe this is fine, e.g., for smooth surfaces.

$\endgroup$
2
  • $\begingroup$ Thank you. Yes, as per our definition, the curved surface of the cone 'A' that you showed cannot be called a geodetically convex subset of the full cone C ('full cone' means the closed surface formed by this curved part and base). And as far as I can make out, C minus the apex is indeed a geodetically convex subset of C. $\endgroup$ Commented Jan 24, 2023 at 4:15
  • 1
    $\begingroup$ And if the full cone surfaced C is cut into 2 finite pieces by a plane parallel to its base, it appears the lower portion (containing the base) will not be geodetically convex in C. $\endgroup$ Commented Jan 24, 2023 at 9:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .