Polytopes with screw symmetry, without rotational symmetry

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Polytopes with screw symmetry, without rotational symmetry

Postby hamish_todd » Fri Sep 24, 2021 1:41 pm

Hey folks. Here's two cute things:

https://www.youtube.com/watch?v=ilVS-ZY5QP4 - lecture on a sculpture a very strange symmetry group. Here's the sculpture:
Image

Here's the "laves graph", a 3D structure with screw symmetry but lacking in rotational symmetry:
https://www.shadertoy.com/view/wddBRX

I'd like a polytope which has this kind of symmetry: multiple axes such that, if you do "double rotations" of the polytope, the polytope looks the same. Anyone know of such things?
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Fri Oct 01, 2021 2:49 am

Yes, I'd also like to know about polychora with only double rotation symmetry, not simple rotation symmetry.

I asked quickfur about this here viewtopic.php?f=25&t=2273
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Tue Oct 12, 2021 11:54 pm

Here are some results on the possible symmetry groups. I'll denote a double rotation as (θ,ϕ), for rotating by angle θ in the x,y-plane and ϕ in the z,w-plane.

If either angle is irrational, then applying the double rotation repeatedly, (nθ,nϕ), gives infinitely many different elements of the group, which we don't want (as it would give the polychoron infinitely many vertices). So both θ and ϕ (considered as multiples of 360°) must be rational.

If the symmetry group contains a double rotation that's not isoclinic, then applying it repeatedly gives a non-trivial simple rotation. For example, 5*(72°, 90°) = (360°, 450°) = (0°, 90°). So the rotation must be isoclinic: θ = ±ϕ.

If the group contains both a left-isoclinic rotation L and a right-isoclinic rotation R, then it also contains a non-isoclinic rotation LR (that is their composition), and thus it also contains a simple rotation. Proof: Suppose contrarily that LR = L' is another left-isoclinic rotation. Then we have R = L⁻¹L'. Since the lefts form a group, L⁻¹ is also a left, and the product L⁻¹L' is also a left. But this contradicts R being a right. Similarly, LR = R' being another right leads to a contradiction through L = R'R⁻¹. Therefore, LR is neither left-isoclinic nor right-isoclinic.

If the group contains a left L and a reflection F, or a rotoreflection F, then FLF⁻¹ (which can be thought of as a reflection of the rotation itself) is a right, and thus the group contains a simple rotation.

Therefore, the group must contain only left-isoclinic rotations, or only right-isoclinic rotations, and no other transformations.

Actually there are two isoclinic rotations which are both left and right: The identity (0°, 0°), and (180°, 180°) which simply negates all coordinates. But these are easily accounted for in the above. There is one group containing the latter and a reflection and no simple rotations; its 4 transformations send (x,y,z,w) to (x,y,z,w) itself (the identity), (-x,y,z,w) (a reflection), (x,-y,-z,-w) (a 180° rotoreflection), and (-x,-y,-z,-w) (the 180° double rotation).
Last edited by mr_e_man on Wed Oct 13, 2021 4:02 am, edited 1 time in total.
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Wed Oct 13, 2021 2:30 am

Left-isoclinic rotations in 4D correspond to rotations in 3D.

See some previous posts viewtopic.php?f=27&t=2482 .

In geometric algebra, a 3D rotation by angle θ around unit vector u can be represented by the multivector exp(Iuθ/2) = cos(θ/2) + Iu sin(θ/2), where I is the unit trivector. (Actually there are two unit trivectors, ±I, corresponding to left-handed and right-handed senses of rotation around u; choose one.)

The unit vectors e₁, e₂, and e₃ represent the x-, y-, and z-axes (respectively), so any point or vector can be written as (x,y,z) = xe₁ + ye₂ + ze₃. Vectors can be multiplied according to the rules

e₁² = e₂² = e₃² = 1,
ee₂ = - ee₁, ee₃ = - ee₁, ee₃ = - ee₂,

along with the usual rules for multiplication (such as of matrices): distributivity and associativity. Just as the space of vectors is spanned by 3 basis vectors e₁,e₂,e₃, the space of multivectors is spanned by 2³ = 8 basis multivectors:

1, (grade 0)
e₁, (grade 1)
e₂, (grade 1)
e₃, (grade 1)
i = -Ie₁ = ee₂, (grade 2)
j = -Ie₂ = ee₃, (grade 2)
k = -Ie₃ = ee₁, (grade 2)
I = eee₃ (grade 3).

The even-grade multivectors are exactly isomorphic to the quaternions, which are defined by i² = j² = k² = -1 and ij = k. Indeed,

k² = (ee₁) (ee₁) = (-ee₂) (ee₁) = - e₁(ee₂)e₁ = - e₁(1)e₁ = - ee₁ = -1, (similarly for i² and j²,)

ij = (ee₂) (ee₃) = (-ee₃) (-ee₁) = +e₂(ee₃)e₁ = ee₁ = k.

Moving on to 4D, the even-grade multivectors are isomorphic to two copies of the quaternions. But the identities for the quaternions are not the same scalar 1; instead, with J = eeee₄ being the unit quadvector, the identities are 1L = (1 - J)/2 and 1R = (1 + J)/2. Note that J² = +1, so 1L1R = (1 - J + J - J²)/4 = 0.

There are 2⁴ = 16 basis multivectors for 4D, and half of those, 8, for the even-grade multivectors:

1,
ee₂, ee₃, ee₃,
ee₄, ee₄, ee₄,
eeee₄.

Taking "averages" of dual pairs of these, we get an alternative basis:

1L = (1 - eeee₄)/2, 1R = (1 + eeee₄)/2,
iL = (ee₂ - ee₄)/2, iR = (ee₂ + ee₄)/2,
jL = (ee₃ - ee₄)/2, jR = (ee₃ + ee₄)/2,
kL = (ee₁ - ee₄)/2, kR = (ee₁ + ee₄)/2.

Verify that the 'L's satisfy the equations defining quaternions, and the 'R's do also, and the product of an 'L' and an 'R' is 0.

So, given any 3D rotation, we can represent it as a 3D even multivector, then as a 4D "left" even multivector, then as a 4D rotation.

Specifically, using the same notation as in the beginning of this post, the 4D multivector is 1R + 1Lcos(θ/2) + (-Iu - ue₄)/2 sin(θ/2). For example, if u = e₃ (rotation around the z-axis), then we get

1R + 1Lcos(θ/2) + kLsin(θ/2)
= (1 + cos(θ/2))/2 + eeee₄ (1 - cos(θ/2))/2 + (ee₁ - ee₄)/2 sin(θ/2)
= cos²(θ/4) + eeee₄ sin²(θ/4) + (ee₁ - ee₄) sin(θ/4)cos(θ/4)
= (cos(θ/4) - ee₂sin(θ/4)) (cos(θ/4) - ee₄sin(θ/4))

thus confirming that it represents an isoclinic rotation (-θ/2, -θ/2).

Or, you might just stick with abstract quaternions (not interpreted as multivectors) multiplying other quaternions, and get similar results.
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Wed Oct 13, 2021 3:11 am

I guess I should describe how a multivector M represents a rotation. In short, for rotating a point v around the origin, in any number of dimensions, the formula is M⁻¹vM.

For example, if M = cos(θ/2) + ee₂sin(θ/2), then the basis vectors get sent to

M⁻¹e₁M = (cos(θ/2) - ee₂sin(θ/2)) e₁ (cos(θ/2) + ee₂sin(θ/2))
= (e₁cos(θ/2) - eee₁sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₁ (cos(θ/2) - ee₁sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₁ (cos²(θ/2) - sin²(θ/2) + 2ee₂sin(θ/2)cos(θ/2))
= e₁ (cos(θ) + ee₂sin(θ))
= e₁cos(θ) + e₂sin(θ),

M⁻¹e₂M = (cos(θ/2) - ee₂sin(θ/2)) e₂ (cos(θ/2) + ee₂sin(θ/2))
= (e₂cos(θ/2) - eee₂sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= (e₂cos(θ/2) + eee₂sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₂ (cos(θ/2) + ee₂sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₂ (cos(θ) + ee₂sin(θ))
= e₂cos(θ) - e₁sin(θ),

M⁻¹e₃M = (cos(θ/2) - ee₂sin(θ/2)) e₃ (cos(θ/2) + ee₂sin(θ/2))
= (e₃cos(θ/2) - eee₃sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₃ (cos(θ/2) - ee₂sin(θ/2)) (cos(θ/2) + ee₂sin(θ/2))
= e₃ (cos²(θ/2) + sin²(θ/2) + 0ee₂sin(θ/2)cos(θ/2))
= e₃ (1)
= e₃.

And an arbitrary point v = xe₁ + ye₂ + ze₃ gets sent to

M⁻¹vM = xM⁻¹e₁M + yM⁻¹e₂M + zM⁻¹e₃M
= xe₁cos(θ) + xe₂sin(θ) + ye₂cos(θ) - ye₁sin(θ) + ze
= (x cos(θ) - y sin(θ))e₁ + (x sin(θ) + y cos(θ))e₂ + ze₃;

that is, (x,y,z) gets rotated to (x cos(θ) - y sin(θ), x sin(θ) + y cos(θ), z).


In 4D, there is a shortcut for calculating M⁻¹vM when M represents an isoclinic rotation:

M⁻¹vM = v (2⟨M⟩₀ - 1) + 2v•⟨M⟩₂,

where the notation ⟨M⟩n means to take the grade n part of M and ignore the other parts, and v•B is the vector-bivector dot product, which results in a vector orthogonal to v. For example, if M = (1 + cos(θ/2))/2 + eeee₄ (1 - cos(θ/2))/2 + (ee₁ - ee₄)/2 sin(θ/2), representing the isoclinic rotation (-θ/2, -θ/2), then

2⟨M⟩₀ - 1 = cos(θ/2),
2⟨M⟩₂ = (ee₁ - ee₄) sin(θ/2),

and the above formula gives M⁻¹vM = v cos(θ/2) + v•(ee₁ - ee₄) sin(θ/2). In particular, the basis vectors get sent to

M⁻¹e₁M = e₁cos(θ/2) - e₂sin(θ/2),
M⁻¹e₂M = e₂cos(θ/2) + e₁sin(θ/2),
M⁻¹e₃M = e₃cos(θ/2) - e₄sin(θ/2),
M⁻¹e₄M = e₄cos(θ/2) + e₃sin(θ/2),

as expected.
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Thu Oct 14, 2021 2:21 am

If you prefer matrices instead of multivectors, there's another copy of the quaternions we can use:

Code: Select all
      ⌈1 0 0 0⌉
      |0 1 0 0|
1_M = |0 0 1 0|
      ⌊0 0 0 1⌋

      ⌈ 0  0  0  1⌉          ⌈ 0  0 -1  0⌉          ⌈ 0  1  0  0⌉
      | 0  0  1  0|          | 0  0  0  1|          |-1  0  0  0|
I_L = | 0 -1  0  0| ,  J_L = | 1  0  0  0| ,  K_L = | 0  0  0  1|
      ⌊-1  0  0  0⌋          ⌊ 0 -1  0  0⌋          ⌊ 0  0 -1  0⌋

      ⌈ 0  0  0 -1⌉          ⌈ 0  0 -1  0⌉          ⌈ 0  1  0  0⌉
      | 0  0  1  0|          | 0  0  0 -1|          |-1  0  0  0|
I_R = | 0 -1  0  0| ,  J_R = | 1  0  0  0| ,  K_R = | 0  0  0 -1|
      ⌊ 1  0  0  0⌋          ⌊ 0  1  0  0⌋          ⌊ 0  0  1  0⌋

Given a 3D rotation by angle θ around axis u = ae₁ + be₂ + ce₃ with u² = a² + b² + c² = 1, we can convert it to a 4D rotation matrix

cos(θ/2)1M + sin(θ/2)(a IL + b JL + c KL).
Last edited by mr_e_man on Mon Oct 18, 2021 3:46 pm, edited 1 time in total.
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Re: Polytopes with screw symmetry, without rotational symmet

Postby mr_e_man » Thu Oct 14, 2021 3:23 am

Back on the topic of polychoron symmetry groups...

The simplest type of group without simple rotations would be generated by a single isoclinic rotation (θ, θ), where θ = 360°/m for some fixed integer m. The whole group contains m elements, of the form (nθ, nθ) for various integers n. The transformation matrices look like

Code: Select all
⌈ cos(nθ) -sin(nθ)  0        0       ⌉
| sin(nθ)  cos(nθ)  0        0       |
| 0        0        cos(nθ) -sin(nθ) |
⌊ 0        0        sin(nθ)  cos(nθ) ⌋

= cos(nθ)1M - sin(nθ)KL.

So, take some random points (x,y,z,w) (say, 2 to 5 points), apply each of these m matrices to each of those points (to get a total of 2*m to 5*m points, of the form (x cos(nθ) - y sin(nθ), x sin(nθ) + y cos(nθ), z cos(nθ) - w sin(nθ), z sin(nθ) + w cos(nθ)), then take the convex hull, and render! :nod: The result will probably look like a tangled blob of tetrahedra. Perhaps we should take the dual, so that the cells have more distinctive shapes, and only 4 cells meet at each vertex.


Now let's look at the more interesting 4D groups, which come from 3D rotation groups. There are 2 infinite families: One type of group has m rotations around a single axis; but that just gives us 2m isoclinic rotations generated by a single isoclinic rotation, which we've already seen. The other type of group has m rotations around one axis, and m 180° rotations around perpendicular axes, as the rotational symmetries of an m-gon prism. Let θ = 360°/(2m). The quaternions representing these rotations have the form

cos(nθ) + k sin(nθ) ,
j cos(nθ) - i sin(nθ) ,

for various integers n. This gives us two types of matrix

Code: Select all
⌈ cos(nθ) -sin(nθ)  0        0       ⌉    ⌈ 0        0       -cos(nθ) -sin(nθ) ⌉
| sin(nθ)  cos(nθ)  0        0       |    | 0        0       -sin(nθ)  cos(nθ) |
| 0        0        cos(nθ) -sin(nθ) |    | cos(nθ)  sin(nθ)  0        0       |
⌊ 0        0        sin(nθ)  cos(nθ) ⌋ ,  ⌊ sin(nθ) -cos(nθ)  0        0       ⌋

in the m-gonal prismatic swirl group. (There may appear to be some inconsistent signs (±), but it doesn't matter because n may be replaced with -n.)


In a linked post, I've shown the cube's symmetries and the corresponding multivectors, which can instead be written as quaternions, and then re-interpreted as matrices, which will form the cubic swirl group:

±1,
±i , ±j , ±k ,
(±1 ± i ± j ± k)/2 ,
(±1 ± i)/√2 , (±1 ± j)/√2 , (±1 ± k)/√2 ,
(±i ± j)/√2 , (±i ± k)/√2 , (±j ± k)/√2 .

(All combinations of signs are taken.) For tetrahedral symmetry, simply omit the ones with √2. I'll show just a few of the resulting matrices:

(1 + i + j + k)/2 , (1 + i - j + k)/2 , (1 + k)/√2

Code: Select all
⌈ 1/2  1/2 -1/2  1/2⌉    ⌈ 1/2  1/2  1/2  1/2⌉    ⌈ 1/√2  1/√2   0     0  ⌉
|-1/2  1/2  1/2  1/2|    |-1/2  1/2  1/2 -1/2|    |-1/√2  1/√2   0     0  |
| 1/2 -1/2  1/2  1/2|    |-1/2 -1/2  1/2  1/2|    |  0     0    1/√2  1/√2|
⌊-1/2 -1/2 -1/2  1/2⌋ ,  ⌊-1/2  1/2 -1/2  1/2⌋ ,  ⌊  0     0   -1/√2  1/√2⌋

In fact the tetrahedral swirl group is generated by the first two of these matrices, and the cubic swirl group is generated by the third and either one of the first two.


The dodecahedron's rotational symmetries are represented by the quaternions

±1 ,
±i , ±j , ±k ,
(±1 ± i ± j ± k)/2 ,
(±φ ± φ⁻¹i ± j)/2 , (±φ ± φ⁻¹j ± k)/2 , (±φ ± φ⁻¹k ± i)/2 ,
(±φ⁻¹ ± i ± φj)/2 , (±φ⁻¹ ± j ± φk)/2 , (±φ⁻¹ ± k ± φi)/2 ,
(±1 ± φi ± φ⁻¹j)/2 , (±1 ± φj ± φ⁻¹k)/2 , (±1 ± φk ± φ⁻¹i)/2 ,
(±φi ± j ± φ⁻¹k)/2 , (±φj ± k ± φ⁻¹i)/2 , (±φk ± i ± φ⁻¹j)/2 .

Converting a few of these to matrices:

(1 + i + j + k)/2 , (1 + i - j + k)/2 , (φ + φ⁻¹k + i)/2

Code: Select all
⌈ 1/2  1/2 -1/2  1/2⌉    ⌈ 1/2  1/2  1/2  1/2⌉    ⌈  φ/2   φ⁻¹/2   0     1/2  ⌉
|-1/2  1/2  1/2  1/2|    |-1/2  1/2  1/2 -1/2|    |-φ⁻¹/2   φ/2   1/2     0   |
| 1/2 -1/2  1/2  1/2|    |-1/2 -1/2  1/2  1/2|    |  0    -1/2     φ/2   φ⁻¹/2|
⌊-1/2 -1/2 -1/2  1/2⌋ ,  ⌊-1/2  1/2 -1/2  1/2⌋ ,  ⌊-1/2     0    -φ⁻¹/2   φ/2 ⌋

Again the tetrahedral swirl group is generated by the first two, and the dodecahedral swirl group is generated by the third and either one of the first two.
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