## Equidistant rotation axes and Platonic Equations

The preceding posts may have indicated that it is hard for an n-fold rotation to coexist (within a finite group) with rotations other than twofold rotations perpendicular to it, perhaps already impossible at n = 6: the purpose of this post was to confirm this, for n > 6 as well, but along the way we will also have a look at the exceptional cases (2 < n < 6) and the ‘genesis’ of the Platonic solids (or their rotation groups at least).

So, let us first prove that a sixfold or sevenfold etc rotation may only coexist (within a finite group of isometries) with a twofold rotation perpendicular to it (rotating the twofold rotation to other perpendicular twofold rotations, and rotated by the twofold rotation to itself — Conjugacy Principle). Our strategy: assuming a smallest possible angle between two n-fold rotations, where n > 5, we will produce another pair of n-fold rotations at a still smaller angle from each other (by rotating each n-fold rotation axis to a new one by the other n-fold rotation); note here that, assuming the coexistence of an n-fold rotation with a non-twofold rotation (or even a twofold rotation not perpendicular to it), we may as well assume the existence of at least two (and finitely many) n-fold rotations because that other rotation rotates the n-fold rotation to another n-fold rotation.

The strategy outlined above is very similar to the strategy employed in Planar Crystallography and the standard proof of the impossibility of the coexistence of two fivefold rotations in a finite planar symmetry group: we assume two nearest fivefold rotation centers and we rotate each one of them by the other to get two new fivefold centers at a still smaller distance from each other:

In our three-dimensional situation involving two rotations by $\gamma\leq60^0$ at a minimal angular distance $\phi$ from each other … we simply place rotation axes OK, OL in such a way that triangle KOL lies on the xz-plane and the bisector of angle KOL is the z-axis, assuming for convenience $|OK|=|OL|=1$; axis OK is rotated by axis OL to a new axis OK’ (shown below), with $K'=(x, y, z)$, and axis OL is rotated by axis OK to a new axis OL’ (not shown below), with $L'=(-x, y, z)$:

Our goal is to arrive at a contradiction by establishing $|K'L'|<|KL|$, that is $x; to achieve this we need to compute $x$.

From $|PK'|=|PK|=sin\phi$ we obtain $(sin\phi /2)x+(cos\phi /2)z=cos\phi$, and from the Law of Cosines in PKK’ we obtain $(sin\phi /2)x-(cos\phi /2)z=sin^2\phi -1-sin^2\phi cos\gamma$; solving the system we get

$x=\displaystyle\frac{cos\phi +sin^2\phi -1-sin^2\phi cos\gamma }{2sin\phi /2}$.

At this point we notice that the desired inequality $x holds rather trivially for $\gamma \leq90^0$, via $cos\phi +sin^2\phi <1+2sin^2\phi/ 2$: what is the relevance of the assumption $\gamma \leq60^0$?

The answer turns out to be quite deep and wonderful: our arguments above have missed the possibility $x=0$ and $K'\equiv L'$ (when, as we are going to see further below, there are three rotation axes at equal angular distances from each other)! It is easy to see that $x=0$ is equivalent to $(1-cos\gamma )cos^2\phi -cos\phi +cos\gamma =0$; solving the quadratic and rejecting the trivial solution $\phi =0^0$ we arrive at

$cos\phi =\displaystyle\frac{cos\gamma }{1-cos\gamma }$      (I)

… an equation that holds only trivially at $\gamma =60^0$ and is outright impossible for $\gamma <60^0$.

Putting everything together, we see that we have indeed proved that it is not possible for two rotations by an angle $\gamma\leq60^0$ to coexist within a finite group.

On the other hand, equation (I) captures the ‘genesis’ of both the cube/octahedron (at $\gamma=90^0$) and the dodecahedron/icosahedron (at $\gamma=72^0$), that is four of the five Platonic solids! [Note here that the equation in question is impossible at either $\gamma=180^0$ or $\gamma=120^0$ , leading to an obtuse, hence non-minimal, $\phi$. (By the way, in case it is not ‘obvious’ why only angles of the form $360^0/n$ matter, an argument similar to that employed in section 4.0.6 (proof of the Crystallographic Restriction) of Isometrica applies.)]

To wit, at $\gamma=90^0$ equation (I) yields $\phi=90^0$, capturing the fact that any two of the three fourfold axes of the cube (perpendicular to each other) rotate each other to the third fourfold axis. Likewise, but less obviously, at $\gamma=72^0$ (I) yields, with $cos\gamma=(\sqrt{5}-1)/4$, $\phi=arccos(1\sqrt{5})\approx63.43^0$, the angle between any two adjacent fivefold axes in the icosahedron (as one could verify with the help of a standard pentagon-hexagon soccer ball); again, each one of these two fivefold axes rotates each other to a third, ‘equidistant’, fivefold axis. [In both cases, cube and icosahedron, the endpoinds of the axes form an equilateral triangle: this could follow from our formulas, but it is geometrically obvious as well — think of the construction of the equilateral triangle for example!]

What happens in the cube with fourfold rotations and in the icosahedron with fivefold rotations is actually very similar to a planar situation: any two (nearest or not) sixfold centers rotate each other to the same sixfold center.

Further, in both the cube and the icosahedron, the product of any two distinct members of such an ‘equilateral triple’ may yield, depending on order and orientation, the threefold rotation that rotates all three axes to each other (exactly as it happens in the standard hexagonal lattice p6, where sixfold centers are rotated to each other by threefold centers and vice versa): the product of either any two distinct fourfold rotations in the cube or any two fivefold rotations in the icosahedron may be, depending on the order and orientation of the factors, a threefold rotation rotating them — or at least their axes — to each other.

[In the cube, given the fourfold rotations (1364), (2354), (1265), and their inverses, notice that (1364) * (2354) = (135)(264), (1364) * (2453) = (132)(456), (1463) * (2354) = (142)(356), (1463) * (2453) = (145)(263), with all resulting threefold rotations/axes rotating each one of the three fourfold axes/rotations to another one. In the icosahedron, given the fivefold rotations (BKJFG)(CLEIH), (AGHCK)(DLJFI), (ABHIF)(CDEJK) and their inverses, observe that (BKJFG)(CLEIH) * (AGHCK)(DLJFI) = (ABK)(CJG)(DEI)(FHL) and (BGFJK)(CHIEL) * (AKCHG)(DIFJL) = (ABG)(CIJ)(DEL)(FKH), two threefold rotations/axes rotating each one of the three fivefold axes/rotations to another one; on the other hand, (BKJFG)(CLEIH) * (AKCHG)(DIFJL) = (AJEIG)(BKLDH) and (BGFJK)(CHIEL) * (AGHCK)(DLJFI) = (AFELK)(BGIDC). (This difference between the cube and the icosahedron will be explained further below.)]

The emergence of threefold rotations as products of fourfold rotations (cube) or fivefold rotations (icosahedron) discussed above is just the tip of the iceberg, hiding a much deeper fact: the composition of any two distinct members of any equilateral rotation triple is a threefold rotation; that is, given any three rotations by an angle $\gamma$, with their axes forming an angle $\phi$ with each other, and $\gamma$, $\phi$ satisfying (I) (so that any two members of the triple rotate each other to the third member), the product of any two of them may yield, depending on order and orientation, a threefold rotation rotating one to the other.

The phenomenon described above follows easily (by way of half-angle formulas, expressing everything in terms of either $\gamma$ or $\phi$) from the following result (which I first posted here, with no serious originality claims): the rotation angle of the composition of two rotations, each of them by an angle $\gamma$, with intersecting axes forming an angle $\phi$, is given by

$2\eta =4\cdot arcsin\sqrt{sin^2\phi /2+(cos^2\phi /2)(cos^2\gamma /2)}$

This rotation formula follows from the following picture (again first posted here), where each of the two rotation axes (OK, OL) is represented as intersection/product of one ‘base’ reflection plane (OKL)  and one ‘lateral’ reflection plane (MOK, MOL) intersecting each other at an angle of $\gamma/2$:

The picture above also helps prove that the dihedral angle between the product rotation (OM) and the plane defined by the two initial rotations (OK, OL) is equal to

$\theta=arcsin\displaystyle\left(\frac{(sin\phi /2)(sin\gamma /2)}{\sqrt{sin^2\phi /2+(cos^2\phi /2)(cos^2\gamma /2)}}\right)$

These two formulas help us re-invent the cube (or at least its rotation subgroup, [4, 3]+, symmetry group of the brazuca ball): starting with two perpendicular fourfold rotations ($\gamma =\phi =90^0$) we get a third one perpendicular to both (as we have seen already), and any two of them combined will produce two ‘symmetrical’ rotations by

$4\cdot arcsin\sqrt{1/2+(1/2)(1/2)}=4\cdot 60^0=240^0$

intersecting at an angle of

$2\theta=2\cdot arcsin\displaystyle\left(\frac{(1/\sqrt{2})(1/\sqrt{2})}{\sqrt{1/2+(1/2)(1/2)}}\right)=arccos(1/3)$;

and multiplying these two threefold rotations (now with $\gamma =120^0$, $\phi =arccos(1/3)$), we obtain a rotation by an angle of

$4\cdot arcsin\displaystyle\left(\sqrt{(1/3)+(2/3)(1/4)}\right)=4\cdot 45^0=180^0$,

and about an axis making with the plane defined by the two threefold rotations an angle of

$arcsin\displaystyle\left(\frac{(1/\sqrt{3})(\sqrt{3}/2)}{\sqrt{(1/3)+(2/3)(1/4)}}\right)=45^0$,

that is … a twofold rotation along a fourfold axis, a so-called 2-side twofold rotation! (There also exist 2-edge twofold rotations emerging as products of one threefold rotation and one fourfold rotation.)

Verifying the computational predictions above, we see that (124)(365) * (132)(456) = (16)(34), a twofold rotation along the axis of (1364), and (132)(456) * (124)(365) = (25)(34), a twofold rotation along the axis of (2453). At the same time, we observe that (124)(365) * (123)(465) = (145)(263) and (132)(456) * (142)(356) = (154)(236): depending on order and orientation, our two threefold rotations/axes may produce the other two threefold rotations/axes instead of twofold rotations!

Such ‘discrepancies’ as the ‘double’ nature of products of threefold rotations (noted right above) or products of fivefold rotations (noted further above) are of course due, as hinted already, to order of factors and orientation of rotations, captured in the following picture:

This picture simply complements the previous one: if the two rotations do not have the same orientation … then the reflection planes into which they are analysed intersect each other not straight above the internal bisector of the angle between the two axes, but straight above the external bisector; once this is understood, the formulas involved in the ‘old’ internal bisector/product case are derived in exactly the same way as in the ‘new’ external bisector/product case (with $\phi /2$ replaced by $\pi /2-\phi /2$).

The examples included at the bottom of the picture explain the product ‘discrepancies’ mentioned earlier: the internal product of two threefold rotations in the cube is a twofold rotation, whereas their external product is a threefold rotation; and the internal product of two fivefold rotations in the icosahedron is a threefold rotation, whereas their external product is a fivefold rotation.

Note that in the case $\phi =90^0$ the two formulas produce the same results for both the product-rotation-angle and the angle $\theta$ between the product-rotation-axis and the internal/external bisector of the angle between the two rotation axes;  this ‘simplicity’ is the reason I have not included any examples involving products of two fourfold rotations (by necessity perpendicular to each other). And it is for the opposite reason that I am not including any more examples from the icosahedron, where threefold axes may meet at two different angles, $arccos(1/3)$ or $arccos(1/\sqrt{5}$), and fivefold rotations may be short ($72^0$) or long ($144^0$); for such details and examples the reader is referred either here (in Greek) or, hopefully, to a future post in this blog devoted entirely to the icosahedron’s rotation group, [5, 3]+ (featured right below on a most unexpected find purchased from a peripteron at the corner of Venizelou and Tsimiski streets here in Thessaloniki).

Having derived and discussed the rotations of the cube/octahedron and the icosahedron/dodecahedron, we may wonder: where is the fifth Platonic solid, that is the regular tetrahedron? That perfect solid (and its rotation group) corresponds to another equation and one more case we missed in our earlier discussion, namely the case $x=sin\phi /2$ … when two rotation axes at an angular distance $\phi$ from each other rotate each other to two new axes at an angular distance $\phi$ from each other! Indeed $x=sin\phi /2$ is equivalent to the quadratic $(1-cos\gamma )cos^2\phi -2cos\phi +(1+cos\gamma )=0$, the only non-trivial solution of which is

$cos\phi =\displaystyle\frac{1+cos\gamma }{1-cos\gamma }$      (II)

… an equation that breaks down into two possibilities: $\gamma =180^0$, $\phi = 90^0$ (two perpendicular twofold axes), and $\gamma =120^0$, $\phi =arccos(1/3)$ (four threefold axes intersecting each other at about $70.53^0$).

In the former case above the ‘image’ axes coincide with the initial axes, but a third twofold axis emerges as a product of them … and that’s it: three mutually perpendicular twofold rotations that create no new rotations of any kind; it’s the symmetry group of the brazuca II ball, known as [2, 2]+ (special case of the group [2, n]+ discussed at the end).

In the latter case above we get [3, 3]+, the rotation group of the regular tetrahedron, that is the symmetry group of the fevernova ball, consisting of four threefold axes (and eight threefold rotations), as well as three twofold rotations. We have already studied products of threefold rotations in the context of the cube/die (containing the regular tetrahedron in a very natural way, as illustrated for example here). As for ‘genesis’, it is easy to see, starting with (123)(465) and (124)(365), that (123)(465) and its inverse, (132)(456), rotate (124)(365) to (154)(236) and (153)(246), whereas (124)(365) and its inverse, (142)(356), rotate (123)(465) to (153)(246) and (154)(236).

[Note here that substituting (II) into the internal product-rotation-angle formula results into a twofold rotation: instead of getting one threefold rotation rotating all three rotations by $\phi$ to each other, as in the case of (I), we get three twofold rotations each of which swaps all four rotations by $\phi$ in two pairs!]

Again, Planar Crystallography offers an analogue of the rotation of two threefold axes by each other to two new threefold axes: right below we see how two initial fourfold centers (1) rotate each other to four new fourfold centers (2), and, even if not directly relevant here, how the centers (1) and (2) rotate each other to a new generation of centers (3). (As in the case of the cube — but also of the icosahedron (not discussed right above) — the product of two fourfold rotations may be a twofold rotation, but twofold rotation centers are not shown in the p4 lattice below.)

Equations (I) and (II) capture the genesis of the Platonic solids, so let us call them, even if not found by Plato — and following a very old Greek literary tradition of spurious attribution 🙂 —  first Platonic equation and second Platonic equation, respectively.

The curious reader may at long last ask: how about the case $x>sin\phi /2$, that is two rotation axes rotated by each other to two new axes further apart from each other? We have seen already, near the beginning of this post, that this inequality is impossible for $\gamma \leq90^0$, how about $\gamma = 120^0$ or $\gamma = 180^0$? Could the two-dimensional analogues — presented below in the spirit of the previous $90^0$ planar example on mutually rotated centers– offer any insights?

lattice of threefold centers (p3)

lattice of twofold centers (p2)

[In the case of $120^0$ centers, the entire (infinite) lattice of centers will be reached via mutual rotation of centers (first three stages shown above) and product of rotations (that will get the ‘missed’ centers). In the case of $180^0$ centers, both mutual rotations and products will only get the line of centers defined by the two initial nearest centers. (In more crystallographic terms, instead of getting the full p2 lattice we get a p112 lattice.)]

Well, $x>sin\phi /2$ turns out to be impossible at $\gamma = 120^0$. Indeed, at $\gamma = 120^0$$x>sin\phi /2$ would be equivalent to $(3cos\phi -1)(cos\phi -1)<0$, that is to $1/3, so that $1/4: the product-rotation-angle formula would then yield (for the product rotation of two initial rotations by $\gamma = 120^0$ at an angle $\phi$ from each other) an impossible rotation angle strictly between $4\cdot 30^0=120^0$ and $4\cdot 45^0=180^0$.

[Similar arguments can be employed to show that the only way for more than four threefold rotations to coexist is to coexist with fivefold rotations in the icosahedron, that no more than three fourfold rotations can coexist (with implied fivefold rotations), and that no more than six fivefold rotations can coexist (icosahedron).]

On the other hand, there is no problem at $\gamma = 180^0$, where $x>sin\phi /2$ is equivalent to $cos\phi (1-cos\phi )>0$, which certainly holds for $0<\phi <90^0$: it is possible, in other words, for two intersecting twofold rotations/axes to rotate each other to two new twofold axes further apart from each other! For example, two twofold axes intersecting at $30^0$ would rotate each other to two new axes intersecting at $90^0$, and in one more ‘step’ we would get the full picture, that is six coplanar twofold axes meeting in perfect symmetry. Likewise, two twofold axes intersecting at $60^0$ would eventually create a perfectly symmetric trio of coplanar twofold axes, two twofold axes intersecting at $20^0$ would eventually be part of a perfectly symmetric set of nine twofold axes, and so on for $\phi =180^0/n$. And there is no reason to worry about the product of the initial two twofold rotations: with or without the formulas employed in this post, it is easy to see that this product is a ‘central’ rotation by $\phi =90^0/n$ perpendicular to the initial rotations. And likewise (same axis, different angles) for the products of the ‘new’ twofold rotations. And the product of the ‘central’ rotation with any one of the twofold rotations perpendicular to it is easily seen to be the ‘next’ perpendicular twofold rotation. This is simply the story of [2, n]+, the rotation-only group consisting of one n-fold rotation and $n$ twofold rotations perpendicular to it, in effect the rotation group of the isosceles n-gonal bipyramid or of the right n-prism (exhibited for example here)!

Putting everything together, we see that the following finite rotation groups are possible in our three-dimensional space: [n]+ (single n-fold rotation without a ‘square root’), [2+, 2n+] (single n-fold rotation with a 2n-fold rotoreflection as ‘square root’), [2, n]+ (discussed right above), [3, 3]+ (rotation group of regular tetrahedron), [4, 3]+ (rotation group of cube/octahedron), [5, 3]+ (rotation group of icosahedron/dodecahedron). [After this derivation of the finite rotation groups, the derivation of all finite symmetry groups is relatively easy but deferred to the next post; note that both derivations are achieved group theoretically in the Appentices of Hermann Weyl’s Symmetry (1952).]

… On this St. Nicholas Day, I would like to dedicate this post to three Nikolaoi, three mathematicians who retriggered my interest in three-dimensional symmetry: these friends are, in chronological order of ‘intervention’, Nikos Kastanis, Nikos Terpsiadis, and Nikos Karampetakis.