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While pondering a problem on separating objects in $R^n$ by hyperplanes, I came across the following puzzle:

Suppose we are given a convex subset of $S^n$ which we call $M$.

For each point $p \in S^n$, let $C(p)$ define the great circle obtained by considering the point $p$ as the north pole and $C(p)$ as the equator. (If we imagine $p$ to correspond to a unit vector $v(p)$ in the obvious manner, then $C(p)$ corresponds to the set of all unit vectors orthogonal to $v(p)$).

Now, define the set $X = \{x \in S^n$ s.t $C(x) \cap M \neq \phi \}$

Is $X$ necessarily convex? The above definition is equivalent to saying that $X$ is the set of all unit vectors orthogonal to at least one unit vector in $M$.

Edit: Sasho has clarified in his answer that $X$ need not be necessarily convex. I am wondering if we may say something weaker about the topology of $X$. I will conjecture that $X$ would be decomposable into $2$ separate convex components?

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    $\begingroup$ $S^n$ itself is not convex (it usually is defined to have unit vectors only, and that is the definition the question seems to be using), and its only convex subsets are singletons. Am I missing something here? $\endgroup$ – Piyush Nov 6 '11 at 9:55
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    $\begingroup$ Piyush, convexity on the surface of $S^n$ can be defined as follows. For any two points $x$ and $y$ in convex set $M \subset S^n$, then either the minor or major arc between them should also lie in $M$. This is analogous to the definition of convexity in the Euclidean space, that for any two points in the convex set, the line segment between them should be contained in the set. $\endgroup$ – Amir Nov 6 '11 at 15:46
  • $\begingroup$ Suppose we take $M$ to be the $90^o$ arc between $(1,0,0)$ and $(0,1,0)$ in the $xy$ plane. Consider the points $x_1 = (1, 0, 1/2)$ and $x_2 = (0,1,1/2)$, which are both in $X$. But the "convex hull" of $x_1$ and $x_2$ (by the definition you gave) is not contained in $X$. $\endgroup$ – Piyush Nov 6 '11 at 18:15
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    $\begingroup$ Piyush, $S^n$ generally denotes the surface of the unit spehre. Over here,neither of the points $x_1$ or $x_2$ lie in $S_n$, and hence cannot be in $X$ by the definition I gave. $\endgroup$ – Amir Nov 7 '11 at 2:09
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    $\begingroup$ I corrected the title; convexity is geometry, not topology. $\endgroup$ – Jeffε Nov 7 '11 at 10:18
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Yes, at least if you define your terms correctly.

Following standard usage in Riemannian geometry, a subset $M \subseteq S^n$ is convex if and only if, for every pair of points $x$ and $y$ in $M$, some shortest path from $x$ to $y$ lies entirely within $M$. With this standard definition, any convex subset $M \subseteq S^n$ is both the convex closure of the points in $M$ and the intersection of all closed hemispheres that contain $M$. In particular, the only convex subset of $S^n$ that does not lie entirely in a closed hemisphere is the entire sphere $S^n$. (And yes, $S^n$ is the intersection of the empty set of closed hemispheres containing $S^n$!) Your definition of convexity allows some non-standard convex sets, like the complement of a single point.

Let's assume that the sphere $S^n$ is centered at the origin of $\mathbb{R}^{n+1}$. Following the standard definitions, a subset $M\subseteq S^n$ is convex if and only if $M$ is the intersection of $S^n$ with a convex cone in $\mathbb{R}^{n+1}$. So let $C$ be a convex cone in $\mathbb{R}^{n+1}$, and let $M = C \cap S^n$. Let $C^*$ denote the dual cone of $C$; this is also a convex cone, so the intersection $S^n\cap C^*$ is convex. Your set $X$ is $S^n \cap (C^* \cup -C^*)$ = $(S^n\cap C^*) \cup -(S^n\cap C^*)$. Thus, $X$ is the union of two antipodal convex subsets of $S^n$.

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  • $\begingroup$ Thank you Jeff. Yes, this seems clear to me now, to the extent that I feel somewhat dumb for not understanding the problem clearly initially. $\endgroup$ – Amir Nov 7 '11 at 12:47
  • $\begingroup$ I was thinking about this dual cone way to think about his definition of $X$, too. I think his $X$ is actually (the closure of) $S^n \setminus (C^* \cup - C^*)$. Could $X$ still be the union of two convex sets? I am not so sure $\endgroup$ – Sasho Nikolov Nov 7 '11 at 15:00
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I hope I am not misunderstanding your definitions, but I feel like the following is a counterexample.

Let's work in spherical coordinates in 3 dimensions. Then $S^3 = \{(1, \theta, \phi)\}$. Let $M = \{(1, \theta, 0): -\theta_0 \leq \theta \leq \theta_0\}$. In that case I think $X$ is the the part of $S^3$ that lies between the planes to which the vectors $(1, -\theta_0, 0)$ and $(1, \theta_0, 0)$ are normal. Look at two points that lie on the boundary of $X$ but do not lie on the same plane. I don't think the major or minor arc between them will be in $X$?

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  • $\begingroup$ Oops, missed your answer before posting my comment! Sorry. $\endgroup$ – Piyush Nov 6 '11 at 18:16
  • $\begingroup$ Sasho, $X$ is indeed the part of $S^3$ that lies between the planes normal to vectors $(1, -\theta_0, 0)$ and $(1, \theta_0, 0)$ , and which pass through the origin. Yes, your counterexample is correct. I am pondering though whether we can say something weaker about the topology of $X$ instead. For example, here we can decompose $X$ into two convex components. $\endgroup$ – Amir Nov 7 '11 at 2:24
  • $\begingroup$ Of course, this may not be true for more complex examples. If $M$ was a thin ellipse containing the north pole of $S^3$, then $X$ would be more weirdly shaped. I still think it would yield us two convex components, but I am unsure. $\endgroup$ – Amir Nov 7 '11 at 2:38
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    $\begingroup$ I was thinking that too, but I also could not imagine what more complex $M$ would do. My intuition is that if you start with the line I described and start "fattening" it until you have a disk, then $X$ would be less and less concave until it becomes a "ring" between two circles that are parallel to a great circle $\endgroup$ – Sasho Nikolov Nov 7 '11 at 3:46
  • $\begingroup$ Sasho, the only constructive comment I can add right now is that your intuition is correct for the example you just gave. I will spend some time thinking on this, and whether in general $X$ will always be decomposable into two convex components. $\endgroup$ – Amir Nov 7 '11 at 4:15

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