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I'm sorry if this question is too simple but I'm unable to find a good answer. I searched through the site and could not find a similar question.

A vertex in a 3D polyhedron is where 3 edges meet. As an extension a vertex in N dimensions polyhedron is the intersection of N hyperplanes.

How to define the vertex on a cone? More importantly how to reconcile it with the definition of a polygon vertex.

In practice we can make a cone from paper which is a 2D polyhedron with vertices. Is there is a way to formalize this into proving the vertex of cone is indeed similar to the vertex of a rectangle? If so can it be extended to N dimensions?

Or am I thinking about it all wrong and is there a much simpler answer.

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    There must be a mathematical way to define the pointy top. – rambalachandran Aug 18 '21 at 13:14
  • There are several ways you might define the pointy top. But there is a significant category mistake you are making in this question: the definition of a vertex of a polytope relies on the fact that a polytope is the intersection of finitely many half spaces. This may not be the case for a cone (take the conical hull of $S^1 \subset \mathbb{R}^3$, embedded in $x_3 = 1$), so you can't extend the definition in any natural way. – While I Am Aug 18 '21 at 14:54

2 Answers2

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There is a general concept which incorporates both polyhedra and cones, namely the concept of a convex body. And the concept of a vertex can be defined in that general setting.

By definition, a subset $C \subset \mathbb R^3$ is a convex body if there exists a set of half-spaces $\{H_i\}_{i \in I}$ such that $$C = \bigcap_{i \in I} H_i $$ And, to be complete, a (closed) half space $H$ is the solution set of an equation of the form $ax + by + cz + d \ge 0$, for some constants $a,b,c,d \in \mathbb R$ such that at least one of $a,b,c$ is nonzero: $$H = \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d \ge 0\} $$ Some extra terminology: the boundary plane of $H$ is $$\partial H = \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d =0\} $$ If $H$ is a half-space, if $C \subset H$, and if $C \cap \partial H \ne \emptyset$, then we say that $H$ is a supporting half-space of $C$. In this terminology, $C$ is equal to the intersection of the set of all supporting half-spaces of $C$.

So, for example, a finite sided (convex) 3-D polyhedron can be defined as a convex body $C \subset \mathbb R^3$ which is the intersection of a finite set of closed half spaces (this allows for finite sided polyhedra to be unbounded; one could also restrict ones attention just to finite sided polyhedra that are bounded).

Let me also define a cone in this setting, more specifically a right circular cone (this is just to keep things simple; one could study more general cones). First, one is given a 2-dimensional plane $P \subset \mathbb R^3$, a line $L \subset \mathbb R^3$ perpendicular to $P$, a point $Q = P \cap L$, another point $R \in L - \{Q\}$, and a radius $s > 0$. One takes $C$ to be the circle in $P$ of radius $s$ centered on $Q$. For each $x \in C$ one takes $H_x$ to be the half space such that its boundary plane $\partial H_x$ passes through $R$ and intersects the plane $P$ in a line that is tangent to $C$ at the point $x$, and such that the half space $H_x$ entirely contains the circle $C$.

With that general definition and these two special examples, here's the definition of a vertex. First, given a cone $C \subset \mathbb R^3$ and a point $p \in c$, to say that $p$ is an extreme point of $C$ means that there exists a closed half-space $H \subset \mathbb R^3$ such that $H \cap C = \{p\}$ (this implies that $p$ is contained in the boundary plane $\partial H$; this also implies that the "opposing" half space of $H$ is a supporting half-space of $C$). Let's also say that any half-space $H$ satisfying the above property is a witness to the extremity of $p$.

A half space that witnesses the extremity of $p \in C$ might not be unique, and we need to quantify the non-uniqueness. For each half-space $H$ that witnesses the extremity of $p$, let $v^\perp_H$ be the unique unit vector based at $p$ which is orthogonal to $\partial H$ and points into $H$. The half-space $H$ determines and is determined by the vector $v^\perp_H$, so I'll extend the terminology to say that the vector $v^\perp_H$ witnesses the extremity of $p$.

To say that an extreme point $p$ is a vertex means that there exist three non-coplanar vectors $v^\perp_1,v^\perp_2,v^\perp_3$ each based at $p$, and each witnessing the extremity of $p$.

The unique vertex of the right circular cone defined above is the point $R$. Also, in a finite sided polyhedron, a point is a vertex if and only if it is a common endpoint of a set of three edges. (This begs the definition of edges. First, let's say that an extreme point $x \in C$ is an edge point if the set of vectors $v_H$ which witness the extremity of $x$ traces out a circular arc in the unit sphere centered on $x$; and then let's say that an edge is a set of edge points all of which share the same set of witnessing vectors.)

One other interesting example is the 3-dimensional ball $B^3 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^3 \le 1\}$, which is a convex body. It's set of extreme points is the unit sphere $S^2 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 = 1\}$, but $B^3$ has no vertices and no edges, because the witnessing half-space of each extreme point is unique.

Lee Mosher
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  • Thank you for the detailed answer. This makes sense. Any references you would recommend to study these concepts in more detail. – rambalachandran Aug 19 '21 at 08:30
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    I do believe there is a literature on convex bodies, but unfortunately I do not know that literature well. My answer is more based on what bits and pieces of that topic I happened to ingest over the years. That wikipedia link is brief but following some of its sublinks might be enlightening, for example here you'll more-or-less find my definition of supporting half-space. – Lee Mosher Aug 19 '21 at 23:02
  • This doesn't sound quite right: Your definition for 'extreme point' seems to actually be that of an 'exposed point'. But while every exposed point is extreme, not every extreme point is exposed. (See for instance https://en.wikipedia.org/wiki/Exposed_point.) That said, for polytopes the two notions are equivalent. – Semiclassical Aug 20 '21 at 01:13
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You start by talking about the vertices of a polyhedron and then switch to a cone. But by any definition I know, a cone is not a polyhedron. Exactly what is your definition of "cone"?

user247327
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  • My question is how do you define a vertex for a cone and how to reconcile it with the definition of a vertex of a polyhedron since in practice it's possible to create a cone from a paper which is a 2D polyhedron. – rambalachandran Aug 18 '21 at 13:17
  • You "create a cone from a paper" by bending the paper so it is no longer a 2D polyhedron! Again, what is your definition of a "cone"? – user247327 Aug 18 '21 at 13:27
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    @rambalachandran Thanks to your question, I've learnt that the formal definition of vertex only exists for polyhedra or more general polytopes; which have "flat sides" (I assume Wikipedia is being deliberately loose with their language there). In other words, I gather that a cone is not an object with a definition of a vertex; the pointy bit is perhaps a cusp instead - I'm not sure – FShrike Aug 18 '21 at 13:30
  • It's a function or curve $f(\theta,h) = (h \cos\theta, h \sin\theta, h )$. How to define a vertex for this curve. Maybe you can say the point that whose value is dependent only on h and not on theta. But I'm unable to reconcile it with the standard definition. – rambalachandran Aug 18 '21 at 13:33
  • Yes, when h= 0, that is (0, 0, 0) which is independent of $\theta$. What "standard definition" are you referring to? – user247327 Aug 20 '21 at 11:52