<p>Yes, at least if you define your terms correctly.</p>

<p>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 <em>empty</em> set of closed hemispheres containing $S^n$!) Your definition of convexity allows some non-standard convex sets, like the complement of a single point.</p>

<p>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 <a href="http://en.wikipedia.org/wiki/Convex_cone">convex cone</a> 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 <a href="http://en.wikipedia.org/wiki/Dual_cone">dual cone</a> 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 $S^n$.</p>

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2011-11-07T12:47:54.687Z
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.
Amir
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2011-11-07T15:00:56.027Z
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
Sasho Nikolov
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2011-11-07T11:17:06.663Z
2011-11-07T11:17:06.663Z
Jeffε
<p>I hope I am not misunderstanding your definitions, but I feel like the following is a counterexample.</p>

<p>Let's work in <a href="http://en.wikipedia.org/wiki/Spherical_coordinate_system" rel="nofollow">spherical coordinates</a> 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 $X$?</p>

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2011-11-06T18:16:19.303Z
Oops, missed your answer before posting my comment! Sorry.
Piyush
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2011-11-07T02:24:16.770Z
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.
Amir
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2011-11-07T02:38:11.543Z
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.
Amir
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2011-11-07T03:46:36.303Z
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
Sasho Nikolov
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2011-11-07T04:15:08.753Z
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.
Amir
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2011-11-07T13:03:08.860Z
Sasho, thank you for helping me get the basic intuition.
Amir
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2011-11-06T17:51:08.683Z
2011-11-06T17:51:08.683Z
Sasho Nikolov
<p>While pondering a problem on separating objects in $R^n$ by hyperplanes, I came across the following puzzle:</p>

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

<p>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 $v(p)$).</p>

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

<p>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 $M$. </p>

<p>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?</p>

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2011-11-06T09:55:52.530Z
$S^n$ 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?
Piyush
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2011-11-06T15:46:43.607Z
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.
Amir
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2011-11-06T18:15:29.897Z
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 $X$.
Piyush
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2011-11-07T02:09:23.637Z
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.
Amir
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2011-11-07T10:18:34.507Z
I corrected the title; convexity is geometry, not topology.
Jeffε
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2011-11-07T12:49:59.957Z
Thank you for the correction, Jeff and for the rigorous answer. Piyush, thank you for making me clarify (some) definitions, and attempting a solution.
Amir
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2011-11-08T04:39:57.077Z
Amir, sorry, I meant the unit vectors in the directions of $x_1$ and $x_2$ actually. With that change my comment was roughly equivalent to Sasha Nikolov's answer below.  The idea was just that if you take any two vectors $x_1$ and $x_2$ in $S_n$, and look at one of the two 2D "cones" generated by these with vertex at the origin, the intersection of this cone with the sphere is one of the "convex hulls" of these two points.  I think Jeffe's answer expresses this much more clearly.
Piyush
<cg.comp-geom>
Determining the convexity of a set on $S^n$
Amir
# Determining the convexity of a set on $S^n$

(Asked by: username_0 on 2011-11-06)

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?

**Tags:** cg.comp-geom
> **Comments:**
> **username_1**: $S^n$ 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?  
> **username_0**: 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.  
> **username_1**: 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 $X$.  
> **username_0**: 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.  
> **username_2**: I corrected the title; convexity is geometry, not topology.  
> **username_0**: Thank you for the correction, Jeff and for the rigorous answer. Piyush, thank you for making me clarify (some) definitions, and attempting a solution.  
> **username_1**: Amir, sorry, I meant the unit vectors in the directions of $x_1$ and $x_2$ actually. With that change my comment was roughly equivalent to Sasha Nikolov's answer below.  The idea was just that if you take any two vectors $x_1$ and $x_2$ in $S_n$, and look at one of the two 2D "cones" generated by these with vertex at the origin, the intersection of this cone with the sphere is one of the "convex hulls" of these two points.  I think Jeffe's answer expresses this much more clearly.  

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## 2 answers

### Answer by username_2 on 2011-11-07. Score: 5 | Accepted answer

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](http://en.wikipedia.org/wiki/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](http://en.wikipedia.org/wiki/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 $S^n$.

> **Comments:**
> **username_0**: 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.  
> **username_3**: 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  

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### Answer by username_3 on 2011-11-06. Score: 4

I hope I am not misunderstanding your definitions, but I feel like the following is a counterexample.

Let's work in [spherical coordinates](http://en.wikipedia.org/wiki/Spherical_coordinate_system) 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 $X$?

> **Comments:**
> **username_1**: Oops, missed your answer before posting my comment! Sorry.  
> **username_0**: 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.  
> **username_0**: 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.  
> **username_3**: 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  
> **username_0**: 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.  
> **username_0**: Sasho, thank you for helping me get the basic intuition.  

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