## Cross Section of an n-cube

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

### Cross Section of an n-cube

Take an ordinary cube and balance it on one corner, so that the diagonally opposite corner is up high. Then take a horizontal cross section that goes thorough the center, dividing into two equal parts. That cross section has the shape of a regular hexagon. It's a truncated 3-simplex, or the intersection of two 3-simplexes of the same size and center, one rotated 180 degrees from the other.

So what happens in 4D? I figure the cross section is the intersection of two tetrahedra of the same size and center, one rotated to be the opposite of the other. My question is, is it an icosahedron? What about in higher dimensions?
PatrickPowers
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### Re: Cross Section of an n-cube

Symmetries alone forbid it from being an icosahedron. You are describing the regular compound of two tetrahedra, which Kepler called the stella octangula because it has eight corners in all. The central intersection is a regular octahedron.
I have not investigated whether the octahedron is a diagonal section of a tessaract.
steelpillow
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### Re: Cross Section of an n-cube

What you describe here is a well known sequencing of vertex layers while running through an nD (hyper)cube orthogonally to its (hyper)body diagonal. In fact that sequence of intersections are being described by
• o3o3o3o3...3o3o (top vertex)
• q3o3o3o3...3o3o (simplex, i.e. vertex figure)
• o3q3o3o3...3o3o
• o3o3q3o3...3o3o
• o3o3o3q3...3o3o
• ...
• o3o3o3o3...3q3o
• o3o3o3o3...3o3q (dual simplex, i.e. vertex figure)
• o3o3o3o3...3o3o (bottom vertex)
Here I used the unit-length sided (hyper)cube, i.e. the edge lengths of these sections are the diagonals of a (unit-sided) square, i.e. q=sqrt(2).
Note that the CD diagram of an nD (hyper)cube clearly has n nodes, accordingly all these sections will have n-1 nodes only.

The remainder then is just to splitt n into even and odd values, for if n would be even, then n-1 becomes odd and the above sequence will have a medial entry (which indeed for n=4, the starting figure being a tesseract, happens to be o3q3o, i.e. describes an octahedron).
On the other hand, if n would be odd, then n-1 becomes even and in the above sequence the "medial section" would run midwise between the vertex layers o3o3..3o3q3o3o3...3o3o and o3o3..3o3o3q3o3...3o3o (both ends of the CD having same size here) and the thus obtained midwise cross-section of the nD (hyper)cube becomes o3o3...3o3y3y3o3...3o3o, i.e. (n-1)/2-truncated simplex, where y represents the medial section of half-square triangles, i.e. y=sqrt(2)/2. E.g. for n=3 we'd get (3-1)/2=1, i.e. the truncated triangle y3y (because of (3-1)=2: 2D simplex), which is nothing but the hexagon you mentioned.

--- rk
Klitzing
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### Re: Cross Section of an n-cube

Klitzing wrote:What you describe here is a well known sequencing of vertex layers

Does this sequence have a name I can search for?

I had a quick gander at the Wikipedia pages for tesseract and hypercube, but neither seem to mention this central vertex-first cross-section phenomenon. I'd like to get a layman's overview of this on our wiki.

It doesn't help that I never got the hang of CD symbols other than the basic o and x, either...

Keiji

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### Re: Cross Section of an n-cube

For a 4D cube the cross section begins with a point that is a vertex. Next is a 4-simplex aka tetrahedron, then a truncated tetrahedron. This has four growing triangular faces and four larger faces that are shrinking. These larger faces are irregular or regular hexagons. When we reach the middle the hexagons have shrunken into triangles so there is a octahedron. Then the sequence in a sense reverses.

Evidently mathematicians have declared the truncated tetrahedron with faces that are regular hexagons to be THE truncated tetrahedron. https://www.youtube.com/watch?v=1y0iZuuMLUc

In n dimensions I guess in the interesting region we have something with 2n faces : n (n-1)-simplex faces and n truncated (n-1) simplex faces. In the center the cross section is a Platonic solid with 2n (n-1)-simplex faces.
PatrickPowers
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### Re: Cross Section of an n-cube

In fact the transition from the o3o3o3...3o3q3o3o3...3o3o3o vertex layer section towards the next o3o3o3...3o3o3q3o3...3o3o3o vertex layer section (any numbers of nodes at either side, thence representing any non-polar such vertex layer section) can even be given in a continuous way as o3o3o3...3o3y3z3o3...3o3o3o, where y+z=q=sqrt(2), i.e. y slowly decreases from q to zero, while at the same linear rate z increases from zero to q.

E.g. the non-polar vertex layer sections of a cube (orthogonal to its body diagonal) are a triangle and the dual triangle respectively. The respective transition decreases the original triangle's side lengths while its vertex figures, i.e. tiny new edges at it corners occur and do increase in size (thus providing semiregular hexagons of alternating side lengths) - until the ratio has swapped, i.e. the original sides have decreased to nothing and the new ones have gotten full size instead.

When considering tesseract's sections orthogonal to its hyperbody diagonal (aka vertex-first view) then the vertex layer sections are a tetrahedron, an octahedron, and a dual tetrahedron. Respective transition from tetrahedron towards octahedron clearly are the verious states of a truncated tetrahedron with continuously increasing truncation depth. After the octahedral state it just goes on, but now the decreasing triangles will be where before the hexagons had been - and conversely.

When considering the penteract in its vertex-first view then the vertex layer sections are a pentachoron, a rectified pentachoron, a respectively inverted oriented rectified pentachoron, and finally the dual pentachoron. According transitions would be described as truncated pentachora (of continuously increasing truncation depth) within that first segment, bitruncated pentachora (aka decachora, however again with continuously varying edge ratios), and finaly the tritruncated pentachora (which in here are nothing but truncated pentachora again - within respectively inverted orientation).

This goes on forever: regular simplex (first vertex layer) -> continuity of truncated simplices -> rectified simplex (second vertex layer) -> continuity of bitruncated simplices -> birectified simplex (third vertex layer) -> continuity of tritruncated simplices -> ... -> dual regular simplex (last vertex layer). (For sure, just as above the 2 polar vertices of the hypercube have been ignored within this consideration. According transitions there are easy however: they are just the transition from single vertex to the next true section, i.e. all being regular simplices increasing in size from zero to q=sqrt(2) and, at the opposite end, decreasing dual regular simplices respectively.)

--- rk
Klitzing
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### Re: Cross Section of an n-cube

Very interesting! This seems to correlate with a pattern I've noticed when an n-polytope is progressively truncated (by intersecting it with its dual of decreasing scale). A cube, for example, if progressively truncated by a shrinking dual octahedron, will produce the sequence:

x4o3o
x4x3o
o4x3o
o4x3x
o4o3x

Similarly, a 4-cube produces the sequence:

x4o3o3o
x4x3o3o
o4x3o3o
o4x3x3o
o4o3x3o
o4o3x3x
o4o3o3x

It seems that progressively intersecting an n-cube with a hyperplane perpendicular to <1,1,1,1...> produces a similar sequence in truncated (n-1)-simplices, except without the double-ringed rows? Fascinating.
quickfur
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