Clifford Algebra Paper

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Clifford Algebra Paper

Postby PatrickPowers » Tue Sep 08, 2020 11:04 am

Platonic solids generate their four-dimensional analogues

This is above me. I'm hoping someone can explain it, especially how spinors are the link from 3D to 4D Platonic solids.
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Re: Clifford Algebra Paper

Postby quickfur » Tue Sep 08, 2020 5:53 pm

Wendy is probably the right person to ask for this one. :)

I'm not at all facile with spinors, but in my incomplete biased understanding, the connection here is essentially related to the fact that 3D rotational vectors have a 4-dimensional representation, e.g., via the quaternion representation, or in this case, the spinor representation. Since 3D Platonic solids have rotational symmetries, we can take those symmetries and rewrite them in a 4D representation, then re-interpret the 4D representation as points in 4D Euclidean space. Apparently, the paper is saying that we can derive the 4D regular polytopes from the 3D Platonic solids this way (though I only read the abstract so I'm not clear how exactly this connection is made).

There may be some connection with the Hopf fibration as well (perhaps that's where spinors come in -- since the Hopf fibration can be understood as a particular mapping of a 3D spherical polytope to 2D complex numbers, the latter of which can be interpreted as 4 scalar coordinates of a 4D vector).
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Re: Clifford Algebra Paper

Postby wendy » Thu Sep 10, 2020 8:59 am

The link lies in complex numbers, and algebraic geometry.

The equation of a line is y=ax+b. In complex euclidean geometry, all the dimensions are doubled, so the 2d space becomes 4d, and the 1d space becomes 2. It still means that you can draw only one 'line' through two given points. If we now consider the lines through 0,0 we get y=ax. Consider now the value w = cis(\omega t), ie a rotation. We can put wy = awx, so this shows that one can have in 4 dimensions, a rotation involving every point tracking in a simple circle around the origin. A clifford rotation.

The value a can take complex values as well, and the slope of a line at right angles has a slope -1/a. We can achieve this result with a sphere, of diameter (0,0,0) to (0,0,1), where the plane is the gradients (r, i, 0) [real, imaginary]. The plane is projected onto this sphere, by the intersection of the line drawn from (r, i, 0) to (0, 0, 1), gives a conformal mapping of the plane, and also preserves the feature where lines at right angles to each other are diametrically opposite.

This creates one swirlybob. Representing 4d space as CE2, has the same effect as 2d space as CE1, in that there is only one real line.

The idea here is that having a sphere, one can project the 62 icosahedral points, or 26 octahedral points, or the 20 tetrahedral points onto this sphere, and they represent clifford-rotations that (of varying orders), leave [3,3,5] or [3,4,3] or [3,3,4] invariant. There is a fourth one not discussed in the paper, formed by [[3,4,3]].

In a clifford rotation, not only do all of the points rotate, but for any given direction (arrow = line with direction), it is immediately passed to every point on the sphere so that we end up with every point has a path and direction. Hopf fibulation. But this solution is not unique, If we suppose the point A,A represents a great arrow (a single arrow passing through a point), then A, X represents all paths that belong to the same swirlybob A. A, -A represents the point diametrically opposite A,A here. Reversing the arrow also changes both signs, so A,X and -A,-X follow the same track in different directions.

The paths of the lines in a clifford-rotation is such that if the lines near A,A "rotate" around this line, in that if one were to divide the 4-space into planes 3d spaces that contain a,a, and radiate outwards like a paddle-steamer wheel, the circle of lines around A,A will cross these paddles in succession, spending half above the 4space and half below. If in the direction of travel, this is a clockwise rotation, then it's a clockwise rotation for every paddle-wheel and circle.

There is a whole corresponding set of clifford parallesl that are anticlockwise. In essence, if A,A is perpendicular to A,-A and -A,A. If the clockwise rotation represents the first symbol, the anticlockwise rotation represents the second. So the space of great arrows is in 6D, a prism-product of two spheres. These are still 4D rotations, but it takes 6 (as in 1,4,6,4,1) dimensions to represent all four rotations.

By varying the intensity around the two axies, one can make all points of six dimensions represent a valid rotation in 4d. Most of these would be Lassajur rotations, ie twice around one axis as once around the other.

Into these bisphere prism we can project the icosahedron, dodecahedron and ID, but if we insert an I * I, D * D and ID * ID we get a series of simple rotations around the {3,3,5} of orders (5, 3, 2 ), For the individual I * pt, D * pt, ID * pt, we have 62 rotations of orders 10, 6, 4.

So we have the 12 great circles of the dodecahedron, representing a set of points A,A, and we can create from this, 5 paths adjacent, 5 paths distant, and one opposite, like the vertices of an icosahedron. We can do it in two different ways.

The regular polytopes {3,3,4}, {3,4,3} and {3,3,5} are formed in this way. The simplex {3,3,3} can not, because opposite a rotation of order 3 is one of order 2. The {4,3,3}, {3,4,3} and {5,3,3} have paths of faces that form Poincare cube, octahedron and dodecahedron, in that if you leave one face you come back on the opposite a half-edge or edge later.

The remaining solid with this property is the octagonny, having 48 faces, each a truncated cube. As with the cube it's an edge, rather than a half-edge, but there is a path of eight faces one covers before returning to the start.
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Re: Clifford Algebra Paper

Postby mr_e_man » Mon Sep 14, 2020 1:48 am

Consider the rotational symmetries of a cube.

Each of the three axes e₁,e₂,e₃ (passing through the centre of a face) has rotations by 90°,180°,270°. A rotation by 90° around the vector e₃, or in the plane e₁e₂, is represented by the rotor

R = cos 45° + e₁e₂ sin 45° = 1/√2 + e₁e₂/√2.

Remember that this acts on vectors by R⁻¹vR, so its opposite -R produces the same rotation. These 9 different rotations thus correspond to 18 different rotors:

± 1/√2 ± e₁e₂/√2,
± 1/√2 ± e₁e₃/√2,
± 1/√2 ± e₂e₃/√2,
± e₁e₂,
± e₁e₃,
± e₂e₃.

(The last few are for 180° rotations.)

Each of the six axes e₁±e₂, e₁±e₃, e₂±e₃ (passing through the centre of an edge) also has a rotation by 180°. The vector e₁+e₂ should be normalized (divided by its magnitude) and dualized (multiplied by the unit trivector) to get a unit bivector:

B = e₁e₂e₃(e₁ + e₂)/√2 = e₂e₃/√2 - e₁e₃/√2,

and then the rotor is

R = cos 90° + B sin 90° = B.

These 6 rotations correspond to 12 rotors:

± e₁e₂/√2 ± e₁e₃/√2,
± e₁e₂/√2 ± e₂e₃/√2,
± e₁e₃/√2 ± e₂e₃/√2.

Finally, each of the four axes e₁±e₂±e₃ (passing through a vertex) has rotations by 120°,240°. The unit bivector dual to e₁+e₂+e₃ is

B = e₁e₂e₃(e₁+e₂+e₃)/√3 = e₂e₃/√3 - e₁e₃/√3 + e₁e₂/√3,

and the rotor is

R = cos 60° + B sin 60° = 1/2 + B√3/2 = 1/2 + e₂e₃/2 - e₁e₃/2 + e₁e₂/2.

These 8 rotations correspond to 16 rotors:

± 1/2 ± e₁e₂/2 ± e₁e₃/2 ± e₂e₃/2.

And we can't forget the identity! That's just R=±1. In total, we have 18+12+16+2=48 rotors. Using the mapping from 3D even multivectors to 4D vectors


we get the coordinates of the vertices of a 24-cell and its dual:

(±1/√2, ±1/√2, 0, 0),
(±1/√2, 0, ±1/√2, 0),
(±1/√2, 0, 0, ±1/√2),
(0, ±1/√2, ±1/√2, 0),
(0, ±1/√2, 0, ±1/√2),
(0, 0, ±1/√2, ±1/√2),
(±1, 0, 0, 0),
(0, ±1, 0, 0),
(0 ,0, ±1, 0),
(0, 0, 0, ±1),
(±1/2, ±1/2, ±1/2, ±1/2).

(Or, interpreting these not as vertices but as normal vectors for facets, we get the octagonny.)

Similarly, the tetrahedron's symmetries produce the coordinates of a single 24-cell.
Last edited by mr_e_man on Tue Sep 15, 2020 6:20 pm, edited 1 time in total.
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Re: Clifford Algebra Paper

Postby PatrickPowers » Mon Sep 14, 2020 10:19 am

Gosh, this is so helpful! Now I feel I will understand it in a reasonable amount of time.
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