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From Wiki Dual polyhedron

For example the smplices:

3-simplex (tetrahedron) is self-dual

2-simplex (triangle) is self-dual

1-simplex (line segment) ??

hypercubes:

3-cube (cube) dual is octahedron

2-cube (square) dual is square, self-dual

1-cube (line segment) ??

In the above link you can see it says all regular simplices are self-dual, but I can't see how a line segment is self-dual. Shouldn't its dual have 1 vertex and 2 edges? How would you draw that? Is there a rigorous definition of duality in this geometric sense?

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    $\begingroup$ Isn't this more a conceptual question? ie. for a line segment, would you call the outer points vertices? In my opinion the line segment is just one edge. Therefore its dual would be one vertex, namely a point. I'm not sure about this, never heard of the conceps fo dual polyhedra. If someone could confirm, that would be nice. $\endgroup$ – Tom Ultramelonman May 8 '16 at 16:07
  • $\begingroup$ the person who wrote that wiki article also said "The self-dual regular polytopes are: ... In general, all regular n-simplexes", you can find it at the bottom. So I wonder how is a line self dual? All I can think of is one point and to infinite edges <----(*)----> $\endgroup$ – nik May 8 '16 at 16:20
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    $\begingroup$ In 3D the boundary of a polyhedron has vertices, edges, and faces; the dual interchanges vertices and faces. In 2D the boundary of a polygon has only vertices and edges; the dual interchanges vertices and edges. In 1D the boundary of an interval has only vertices. $\endgroup$ – Rahul May 8 '16 at 16:30
  • $\begingroup$ that explains it then, thanks $\endgroup$ – nik May 8 '16 at 16:34
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To understand why a line segment (which is both the cube and the simplex in dimension 1) is "its own dual" you have to think a little more carefully about duality and dimensions.

The dual of an $n$-dimensional polytope $P$ has a face of dimension $n-d-1$ for each $d$-dimensional face of $P$. You can see that clearly for the cube and octahedron. The eight vertices of the cube (dimension $0$) give the eight faces of the octahedron (dimension $3-0-1=2$). The twelve edges of each naturally correspond.

The dual of a square shouldn't be thought of as a square, but rather as a diamond whose edges come from the vertices of the square. They are perpendicular to the edges of the original square and form a square, but that's a coincidence.

On the line, with dimension $1$, each of the two endpoints of a segment, with dimension $0$, gives a face of the dual of dimension $1-0-1 = 0$, which is just a point. The dual has two points, so is a line segment.

You can get all this from the algebra of polarities as described on the wikipedia page.

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