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Let $S$ be a set of points in $\mathbf{R}^n$:

$$S=\{ (x_1,\dots,x_m) \ | \ x_i \in \mathbf{R}^n \ ; i=1,\dots, m \}$$

How do we define and find the dual space or dual polytope of $S$?

0x90
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    In your question $S=\mathbf R^n.$ – Rgkpdx Feb 23 '18 at 12:04
  • @Ton what do you mean? S is a subset of $R^n$ – 0x90 Feb 23 '18 at 12:09
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    But then you say $S= {\text{every point in }\mathbf R^n}$, right? – Rgkpdx Feb 23 '18 at 12:51
  • @Ton why is it every point? I am not sure? It's a set of n vectors in R^n. – 0x90 Feb 23 '18 at 15:32
  • If you meant $ S={ (x_1,\dots,x_m) \ | \ x_i \in \mathbf{R}^n \ ; i=1,\dots, m }$, then $S$ is the set of all subsets of $\mathbb R^n$ of size $m$. I would define a subset of $m$ vectors of $\mathbb R^n$ as $S={x_i}_{1\le i\le m,i\in\mathbb N}$ for some $x_i\in \mathbb R^n$. – Rgkpdx Feb 23 '18 at 15:43
  • @Ton I really don't get why there is a confusion. For 2D with 2 points S could be: {(1.5,6), (8,3.14)} – 0x90 Feb 23 '18 at 15:52
  • @0x90 You have used n to denote both the dimensionality of R and the number of points in S. This common use of n creates an identity between the numbers. Rgkpdx is suggesting you use m for one of them. But they are wrong to suggest that you have used n to denote the number of points in R. – Guy Inchbald Sep 23 '20 at 09:07
  • Duality is a structural concept unrelated to point coordinates. There are many solutions, each with different coordinates. Do you have any particular one in mind? Also, can you be explicit whether you are restricting yourself to convex polytopes? Are all the points in S vertices of it? – Guy Inchbald Sep 23 '20 at 09:11
  • @GuyInchbald, I revised the question. If it makes it's easier to assume convex polytope I am fine with that. – 0x90 Sep 23 '20 at 12:47

1 Answers1

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Take an arbitrary point for center and any sphere around that, and then invert your points wrt. to that sphere along the ray from that center to those. Next errect at those inverted points the hyperplanes. These then are the mutual dual facets to those points.

Esp. when you had for the polytope the convex hull of that set of points, then you'd get for the dual polytope the intersection kernel of those halfspaces, defined by the hyperplanes and containing the center point.

--- rk

Edit:
The main issue here is to derive a hyperplane as the dual of a point. This is being done by means of spherical reciprocation. Let's consider the point $P=(p,0,...,0)$. Next consider e.g. the unit sphere around the origin. Then the reciprocal point $P'$ would be located at $(\frac{1}{p},0,...,0)$ and accordingly the hyperplane through that point and orthogonal to that very ray would by given by $x_1=\frac{1}{p}$.

--- rk

Dr. Richard Klitzing
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