Basic higher-dimensional shapes

This is a non-extensive list of basic geometric shapes, mainly in 4 spacial dimensions, but sometimes extended to higher dimensions.

## Extending Platonic solids

In 3D, there are 5 platonic solids (but at least 48 regular polyhedra).
Those are the tetrahedron (triangular pyramid), the cube, the octahedron (two square pyramids stuck together), the dodecahedron, and the icosahedron.

The first thing of note is, that the faces of those regular polyhedra are themselves regular polygons.
Without further explanation, this generalizes to all dimensions, in that the hyperfaces of n-dimensional regular hyper-polyhedra are themselves n-1 dimensional hyper-polyhedra.
To construct the regular polyhedra of some dimension, the regular polyhedra of the next lower dimension are used.
For regular 2D polyhedra, i.e. regular polgons, the building blocks consist of the only 1D regular polyhedron, a line.

The method of construction is to first attach two building blocks at one of their faces, then attach another at an adjacent face, another at an adjacent cell of that new building block, until the reach back onto themselves in their own lower dimension. Prior to that point, if at least 3 blocks are connected, it is possible to lift the chain into the new dimension and connect it in a rounded shape. In this way the total angle of the chain is less than 360°, but it can still connect up on itself. If the resulting part can be completed, it is a regular polyhedron of the next higher dimension.

As an example of this proccess, we can construct the regular 3D polyhedra:

### Constructing the 3D regular polyhedra

Chaining equilateral triangles, they can connect by 3, 4, or 5 elements, before reaching up on themselves. We can see this by noting that the angle of 60° between faces of a triangle (called the dihedral angle) dividing 360° 6 times. Any number lower than that but at least 3 will work.

Continuing with squares, their dihedral angle of 90° divides 360° 4 times, allowing only a body with 3 faces per chain.

For pentagons, their dihedral angle of 108° divide 360° somewhat over 3 times, allowing again only 3 faces per chain.

For hexagons or higher many-gons, their dihedral angles are 120° or more, allowing only chains shorter than 3, which cannot exist. Therefore they produce no 3D polyhedra.

This leaves 5 chains, which all form a respective platonic solid:

* 3 triangles -> tetrahedron
* 4 triangles -> octahedron
* 5 triangles -> icosahedron
* 3 squares -> cube
* 3 pentagons -> dodecahedron

### Finding Dihedral angles of regular polyhedra

As we have seen, using just the dihedral angle, which is tha angle between the normal direction of two adjacent hyper-faces, is the sole factor on what regular polyhedra exist in the next higher dimension. Since the process for finding that dihedral angle of even 3D shapes is a lot more involved, this chapter provides a basic equation for working out that angle from the dihedral angle of the regular polyhedra making up the hyper-face itself.

Arange the haper-faces in a chain of length

n

{\displaystyle n}

. Call the dihedral angle of the faces

a

{\displaystyle a}

. The faces will cover an angle of

a
⋅
n

{\displaystyle a\cdot n}

, which through folding them up into the new dimension, is brought up to

360

∘

{\displaystyle 360^{\circ}}

.
Looking top down from the new dimesnion, folding the faces up is equivalent to flatening or squashing them towards the center. If a point is moved inwards by a factor

s

{\textstyle s}

, it is moved up to

1
−

s
2

{\textstyle \sqrt{1-s^2}}

.

The apparent dihedral angles from the top down perspective shift with that squash, which through a simple geometric argument can be shown to transform them to

s
⋅
tan
⁡
(
b

/

2
# )
tan
⁡
(
a

/

2
)

{\displaystyle s\cdot\tan(b/2) = \tan(a/2)}

 which solves for the apparant dihedral angle

# b
2
⋅
arctan
⁡
(
tan
⁡
(
a

/

2
)

/

s
)

{\displaystyle b = 2\cdot\arctan(\tan(a/2)/s)}

. We can suppose the right apparent squash will lead to b\*n equaling

360

∘

{\displaystyle 360^{\circ}}

, since, when closed, the shapes will connect to each other. This can be solved as

360
∘

=
n
⋅
# b
n
⋅
2
⋅
arctan
⁡
(
tan
⁡
(
a

/

2
)

/

s
)

{\displaystyle 360^\circ = n\cdot b = n\cdot 2\cdot\arctan(\tan(a/2)/s)}

 giving

# s
tan
⁡
(
a

/

2
)

/

tan
⁡
(

180
∘

/

n
)

{\displaystyle s = \tan(a/2)/\tan(180^\circ/n)}

.

The angle between the faces can be calculated from their normal vectors, which initially while laying flat where pointing upwards. Suppose the upward direction where categorized by a unit vector

e
→

2

{\textstyle \vec e\_2}

, a face was laying with its center facing in the e1 direction, and the chain rotated around the

e
→

1

{\textstyle \vec e\_1}

e
→

3

{\textstyle \vec e\_3}

 axis. The normal vector of the face initially in e1 direction is rotated to

n
→

1

=
s
⋅

e
→

1

1
−

s
2

⋅

e
→

2

{\textstyle \vec n\_1 = s\cdot\vec e\_1 + \sqrt{1-s^2}\cdot\vec e\_2}

.
The normal vector of its neighbouring face is rotated by the squashed dihedral angle

# b

360
∘

/

n

{\textstyle b = 360^\circ/n}

 to

cos
⁡
(

360
∘

/

n
)
⋅

e
→

1

sin
⁡
(

360
∘

/

n
)
⋅

e
→

3

{\textstyle \cos(360^\circ/n)\cdot\vec e\_1 + \sin(360^\circ/n)\cdot\vec e\_3}

 and then folded up to

n
→

2

=
s
⋅
cos
⁡
(

360
∘

/

n
)
⋅

e
→

1

s
⋅
sin
⁡
(

360
∘

/

n
)
⋅

e
→

3

1
−

s
2

⋅

e
→

2

{\textstyle \vec n\_2 = s\cdot\cos(360^\circ/n)\cdot\vec e\_1 + s\cdot\sin(360^\circ/n)\cdot\vec e\_3 + \sqrt{1-s^2}\cdot\vec e\_2}

. The angle between them can then be calculated as

arccos
⁡
(

n
→

1

⋅

n
→

2

/

(

n
→

1

⋅

n
→

2

)

)

, though since the dihedral angle is measured the opposite way to the angle of the normal vectors, it will here be

180
∘

−
arccos
⁡
(

n
→

1

⋅

n
→

2

/

(

n
→

1

⋅

n
→

2

)

)

. This can be simplified to

180
∘

−
arccos
⁡
(
cos
⁡
(

360
∘

/

n
)

s
2

⋅
(
1
−
cos
⁡
(

360
∘

/

n
)
)
)

{\displaystyle 180^\circ-\arccos(\cos(360^\circ/n)+s^2\cdot(1-\cos(360^\circ/n)))}

, where we can replace

s

{\displaystyle s}

 with

tan
⁡
(
a

/

2
)

/

tan
⁡
(

180
∘

/

n
)

{\textstyle \tan(a/2)/\tan(180^\circ/n)}

 to further get

arccos
⁡

(

1
−
2

cos
2

⁡
(

180
∘

/

n
)

cos
2

⁡
(
a

/

2
)

)

{\displaystyle \arccos \left(1-2{ \cos^2(180^\circ/n)\over \cos^2(a/2) }\right)}

There is an interactive recreation of this process in desmos here.

### Finding all regular polytopes

Polytopes are hyper-polyhedra. The regular polytopes are the extension of regular polygons and platonic solids into higher dimensions.
Using the function for dihedral angles in the last chapter, we can calculate the dihedral angles of the 5 platonic solids:

* tetrahedron: arccos(1-2\*cos^2(180°/3)/cos^2(60°/2)) = arccos(1/3) ≈ 70.529°
* octahedron: arccos(1-2\*cos^2(180°/4)/cos^2(60°/2)) = arccos(-1/3) ≈ 109.471°
* icosahedron: arccos(1-2\*cos^2(180°/5)/cos^2(60°/2)) = arccos(-sqrt(5)/3) ≈ 138.190°
* cube: arccos(1-2\*cos^2(180°/3)/cos^2(90°/2)) = arccos(0) = 90°
* dodecahedron: arccos(1-2\*cos^2(180°/3)/cos^2(108°/2)) = arccos(-1/sqrt(5)) ≈ 116.565°

Remembering we need chains of at least three faces, we can only construct regular 4-polytypes from regular polyhedra with dihedral angles below 120°. This is the case for all platonic solids except for the icosahedron, with the tetrahedron being able to form chains of four and five too, meaning we can construct a total of 6 4-polytypes

#### The 6 regular 4-polytopes

* 3 tetrahedra -> 5-cell (4-simplex)
* 4 tetrahedra -> 16-cell (4-orthoplex)
* 5 tetrahedra -> 600-cell
* 3 octahedra -> 24-cell
* 3 cubes -> 8-cell (Tesseract, 4D hypercube, or just 4-cube)
* 3 dodecahedra -> 120-cell

The names are derived from the number of hyperfaces (i.e. platonic solids) making up the shape. The tesseract or 8-cell is made from 8 cubes, the 120-cell from 120 dodecahedra.
This is analogous to the 3-polytypes (an octahedron is made from 8 triangles), and the 2-polytypes (an octagon is made from 8 line segments).

Calculating the angles of the 6 regular 4-polytypes

* 4-simplex: arccos(1-2\*cos^2(180°/3)/cos^2(arccos(1/3)/2)) = arccos(1/4) ≈ 75.522°
* 16-cell: arccos(1-2\*cos^2(180°/4)/cos^2(arccos(1/3)/2)) = arccos(-1/2)) = 120°
* 600-cell: arccos(1-2\*cos^2(180°/5)/cos^2(arccos(1/3)/2)) = arccos(-(1+3\*sqrt(5))/8) ≈ 164.478°
* 24-cell: arccos(1-2\*cos^2(180°/3)/cos^2(arccos(-1/3)/2)) = arccos(-1/2) = 120°
* Tesseract: arccos(1-2\*cos^2(180°/3)/cos^2(90°/2)) = arccos(0) = 90°
* 120-cell: arccos(1-2\*cos^2(180°/3)/cos^2(arccos(-1/sqrt(5))/2)) = arccos(-1/(sqrt5-1)) = 144°

We notice only the 4-simplex and tesseract can from regular 5-polytopes, with the 4D simplex being able to form two.
More importantly though, notice that the angle of the cube-series remains at 90° for each iteration of the dihedral angle formula. Since the formula remains the same, the same input will always yield the same output, the cube series will always have dihedral angles of 90°. With that realization, we can look into generalizing for the n-polytypes

#### The 3 regular n-polytopes (for dimensions 5+)

In the previous chapter we noticed how the n-cubes always have dihedral angle 90°. This is because the angle fromula will produce the same output for the same input, and 90° produces 90° in the next higher shape. With this realization, we found that the n-cube will always be one of the n-polytopes for a given dimension n. At 90° it can always form chains of 3 and no more in the next higher dimension.

Looking at the other remaining 4-polytope, the 4D simplex or 5-cell, we notice it has an angle of arccos(1/4), while the tetrahedron had an angle of arccos(1/3) and the triangle 60° = arccos(1/2).
After having noticed this pattern, we can confirm by inputing arccos(1/n) into the angle function, yielding arccos(1-2\*cos^2(180°/3)/cos^2(arccos(1/n)/2)) = arccos(1/(n+1)), confirming our hunch. Since arccos(1/n) < arccos(0) = 90°, we can not only always construct another 3 chain from an nth dimensional simplex, we can even construct a 4-chain. Since the angle of the shape of the 4 long chain in 4D was already over 120°, we know for larger n it will only increase further, preventing further shapes from it.

With that we have found the 3 series of shapes the two remaining 4D candidates for 5-polytopes create:

* 3 (n-1)-simplices -> n-simplex
* 4 (n-1)-simplices -> n-orthoplex
* 3 (n-1)-cubes -> n-cube

The angles of those are

* n-simplex: arccos(1/n)
* n-orthoplex: arccos(1-2\*cos^2(180°/4)/cos^2(arccos(1/(n-1))/2)) = arccos(2/n-1)
* n-cube: 90°

### Further terminology

The construction described above used chains of n hyperfaces, that got rolled up into regular polytopes. The numbers involved in that process show up in proper mathematical notation too, but under the name "Schläfli Symbol". A Schläfli symbol is an angled bracket

{
}

{\displaystyle \{\}}

, containing the numbers involved in constructing a shape from the beginning. For example, a cube is constructed by taking

4

{\displaystyle 4}

 lines and rolling them into a square, then taking

3

{\displaystyle 3}

 squares and rolling them into a cube. Thus the Schläfli symbol of a cube is

{
4
,
3
}

{\displaystyle \{4,3\}}

. A 24-cell is

{
3
,
4
,
3
}

{\displaystyle \{3,4,3\}}

, since it is made from triangles (

3

{\displaystyle 3}

 lines), made into octahedra (

4

{\displaystyle 4}

 triangles), made into the 24-cell (

3

{\displaystyle 3}

 octahedra). The first dimension is skipped by definition, as it doesn not align well with this concept, thus a Schläfli symbol for n-polytopes has n-1 numbers within it. If too many numbers occur in a single Schläfli symbol, they can be abbreviated as

a
k

{\displaystyle a^k}

, where

a

{\displaystyle a}

 is the number and

k

{\displaystyle k}

 is how many times it should be repeated. A n-cube therefore has the Schläfli symbol

{
4
,

3

n
−
2

}

{\displaystyle \{4, 3^{n-2}\}}

, and an n-simplex

{

3

n
−
1

}

{\displaystyle \{3^{n-1}\}}

.