Last love?

August 2, 2016

EURO 2016 ended with a beautiful Portuguese victory over/in France, but its soccer ball is still seen in other games, and will be remembered for some time, and so will be my previous post, I hope🙂 I thought there would be no other ball in my life, not for a while anyway, but a fateful walk by a sports store proved me wrong:

new (?) Admiral ball on my balcony — Thessaloniki, 7/28/16

Why do I like this ball? Well, the ends of its threefold axes are reminiscent of the fevernova, even though its six ‘feet’ bring us right back to the right ball, that is the teamgeist — hope you share my enthusiasm!

[Note: in the case of the fevernova each threefold axis has one ‘blank’ end and one ‘propeller’ end, with all four propellers ‘turning’ the same way; in this new ball each threefold axis has two ‘propeller’ ends, so there are eight ‘propellers’ in total, with four of them ‘turning’ one way and four the other way — a direct effect of reflection! (Quick question for you: do the ‘propellers’ at the end of the same threefold axis ‘turn’ the same way or opposite ways?)]

EURO 2016 novel (?) spherical tiling

June 12, 2016

Ignoring ‘minor’ details, we see that the EURO 2016 soccer ball is characterized by twelve congruent parallelograms symmetrically placed on the surface of a sphere. These twelve parallelograms may be viewed as equivalent to the twelve ‘oriented’ edges of a cube (corresponding in fact to its four ‘equatorial’ hexagons):

Examining either the ‘oriented’ cube above or the EURO 2016 ball we see that they have the cube’s rotational symmetry: all cubical rotations still work, but no (roto)reflections are present. In other words, the EURO 2016 ball has the symmetry of the World Cup 2014 ball (brazuca). This is further underlined by the presence of brazuca-like stichings on the official EURO 2016 ball (on the right above): each parallelogram is symmetrically placed between two ‘kissing tongues’, so the brazuca‘s symmetries are fully supported (with the emphasis shifted from the six ‘crosses’ to the twenty four ‘tongues’ nonetheless). On the toy EURO 2016 ball (on the left above) there is however a surprise: its stichings, and corresponding spherical tiling (which is new to me), have full cubical symmetry: the cube’s six edges have been replaced by regular octagons, and the cube’s twelve edges have been replaced by ‘bow ties’ (each of them formed by two isosceles pentagons reminiscent of the pentagons present in the Cairo tiling).

[Markus Rissanen observes that the diagonals of the twelve parallelograms split the sphere into six ‘spherical squares’, within each of which it becomes trivial to symmetrically inscribe an octagon, etc]

Finite 3-d symmetry groups

January 25, 2016

In the previous post we determined geometrically all finite three-dimensional rotation groups, that is all finite symmetry groups that contain only three-dimensional rotations and rotoreflections; these are the ‘general’ groups [n]+, [2+, (2n)+], [2, n]+, and the ‘special’ groups [3, 3]+, [4, 3]+, [5, 3]+. Here we complete the job by allowing reflections in: as previous posts, notably this one, have indicated, our task is to determine how reflection planes can coexist with rotation and rotoreflection axes, and the Conjugacy Principle makes that relatively easy.

Starting with the case of a single n-fold rotation, [n]+, we observe that any reflection plane can either contain the rotation axis or be perpendicular to it (otherwise the n-fold rotation would be reflected to another n-fold rotation): in the first case we end up with the group [n+, 2] (single reflection perpendicular to the n-fold rotation with implied ‘trivial’ n-fold rotoreflection along the axis of the n-fold rotation), in the second case we end up with the group [n] (n reflection planes intersecting along the n-fold rotation axis). (Note here that, as a trivial special case of either group at $n = 1$, we obtain the reflection-only group [1].) So, ‘departing’ from [n]+ we end up with the groups [n+, 2], [n], and, of course, [n]+ (with [1]+ being the identity-only, fully asymmetrical group). [The groups [n]+ and [n] are isomorphic to the only finite two-dimensional symmetry groups, the cyclic group Cn and the dihedral group Dn, respectively.]

In the case of a single 2n-fold rotoreflection, [2+, (2n)+],  with implied n-fold rotation — equivalently, single n-fold rotation with a ‘square root’ — we note that no reflection plane can be perpendicular to the rotoreflection axis, for the product of the 2n-fold rotoreflection with the reflection would yield a 2n-fold rotation. (More precisely, a reflection plane perpendicular to the rotoreflection axis leads to the group [(2n)+, 2] obtained above.) Less obviously, no reflection plane could contain the rotoreflection axis: indeed the product of a 2n-fold rotoreflection with a reflection containing it is a twofold rotation perpendicular to the rotoreflection and making an angle of $360^0/4n$ with the reflection plane; this fact has been experimentally verified for $n=2, n=3, n=5$ in Lemmas 5, 9, 14 here and can be rigorously established by expressing the 2n-fold rotoreflection as product of three reflections — the first of them the said reflection that contains the rotoreflection axis (to be canceled out when we take the product), the second a reflection making an angle of $360^0/4n$ with the first one (so that their product will be the rotational part of the given 2n-fold rotoreflection, that is a $360^0/2n$ rotation), and the third one perpendicular to the first two (and being none other than the reflectional part of the rotoreflection). (More precisely, a reflection plane containing the rotoreflection axis leads to the group [2+, 2n] obtained below.) So, in this case we are ‘stuck’ with what we started with, namely the group [2+, (2n)+]. (Note that at $n=1$ we obtain the inversion-only group [2+, 2+]: no problem, inversion is after all a ‘free’ twofold rotoreflection!)

Starting with [2, n]+, that is one ‘central’ n-fold rotation with n twofold rotations perpendicular to it, the Conjugacy Principle tells us that any reflection that may wish to join the party must either contain one of the twofold rotations (reflecting it to itself) or lie half way between two twofold rotations (reflecting them to each other); at the same time, as in the previous two cases, any ‘intruding’ reflection must either contain the n-fold rotation axis or be perpendicular to it. But note here that, by Lemma 1 here, a reflection containing the n-fold axis and one twofold axis implies a reflection perpendicular to the n-fold axis containing all twofold axes, and vice versa; in other words, the only way to avoid a reflection perpendicular to the n-fold axis is to have reflection planes containing the n-fold axis and lying half way between adjacent twofold axes. We finally arrive at only two new groups: the ‘half-way’ group [2+, 2n] (n reflections containing the n-fold rotation axis, each of them half way between two twofold rotation axes and perpendicular to the plane defined by them) and the ‘right-on’ group [2, n] (one reflection perpendicular to the n-fold rotation axis and containing the twofold rotations, plus n reflections containing the n-fold axis and one twofold axis each);  [2+, 2n] has genuine 2n-fold rotoreflection along the n-fold rotation axis, whereas [2, n] has trivial n-fold rotoreflection along the n-fold rotation axis. (Note that $n=1$ is not possible in this case, or rather yields a group already found, [2+, 2], whereas [2, 1] = [2] and [2, 1]+ = [2]+ ‘hold trivially’.)

Putting everything together, we obtain seven general finite symmetry groups: [n]+, [n+, 2], [n][2+, (2n)+], [2, n]+[2+, 2n], [2, n]. Marvelously, these seven three-dimensional groups are in one-to-one correspondence with the seven one-dimensional groups known as border patterns, as indicated here: the reader may refer to that post for visualizations.

Speaking of visualizations and border pattern representation of three-dimensional symmetry groups, the following diagram captures our arguments so far:

(Blank = central n-fold rotation [translation], dotted line = central 2n-fold rotoreflection [glide reflection], dots = twofold axes perpendicular to the n-fold rotation [twofold centers], vertical lines = reflections containing the central axis [vertical reflections], horizontal line = reflection perpendicular to the central axis [horizontal reflection].)

To complete the job, we need to see how reflection might be injected into the special rotation groups [3, 3]+, [4, 3]+, [5, 3]+. This is an easy task: [4, 3]+ and [5, 3]+ allow only the possibilities [4, 3] (cube/octahedron) and [5, 3] (dodecahedron/icosahedron), whereas [3, 3]+ allows the possibilities [3, 3] (regular tetrahedron, jabulani ball) and [3+, 4] (pyritohedron, teamgeist ball). These limitations are implied by the Conjugacy Principle and the “either half-way or right-on” principle for reflections together with the placement of rotation axes in the rotation groups established in the previous post:

— in [5, 3]+ starting with two fivefold axes like AD and BE we get the four remaining axes: AD and BE rotate each other to GL and KI forming two equilateral rotation triples (as shown in the previous post), whereas the product of the fivefold rotations corresponding to AD and BE yields, under the right combination of order and orientation, the two fivefold axes corresponding to FC and JH (by way of (BKJFG)(CLEIH) * (AKCHG)(DIFJL) = (AJEIG)(BKLDH) and (BGFJK)(CHIEL) * (AGHCK)(DLJFI) = (AFELK)(BGIDC), for example);

this placement of fivefold axes makes it clear that every half-way reflection must be right-on (containing two fivefold axes), and from this observation it follows that there exists only one reflection group containing the rotation group [5, 3]+, and that is the full symetry group of the icosahedron/dodecahedron, [5, 3].

— in [4, 3]+ the three orthogonal fourfold axes make it clear that every reflection must be right-on, either containing one fourfold axis and reflecting the other two fourfold axes to each other or containing two fourfold axes being perpendicular to the third one (this is a special case of right-on reflection, with the other two to-be-reflected-to-each-other axes collapsing into one); but it is easy to see, multiplying them by fourfold rotations, that these 4-vertex and 4-edge reflections imply each other, therefore the only reflection group containing the rotation group [4, 3]+ is the full symmetry group of the cube/octahedron, [4, 3].

— in [3, 3]+ there exist four threefold axes, the four diagonals of a cube basically, forming two X’s perpendicular to each other;

more precisely, let the first X consist of AA’, BB’ and the second X consist of CC’, DD’, with AA’, BB’ rotating each other to CC’, DD’ (and vice versa).  It is clear that a half-way reflection with respect to the one X will also be half-way with respect to the other X, whereas any right-on reflection (like one containing AA’) must also contain another axis (like BB’) and reflect the other two axes (CC’, DD’) to each other: we are left with two possibilities, the half-way reflections (containing no threefold axis) of the pyritohedron ([3+, 4]) and the right-on reflections (containing two threefold axes each) of the regular tetrahedron ([3, 3]).

[We point out here — making yet another analogy with infinite two-dimensional symmetry groups (wallpaper patterns) in the spirit of the previous post and Chapter 8 of Isometrica — that p6 and p3 behave like [4, 3]+ and [5, 3]+, allowing only the right-on possibility for reflection (p6m and p3m1, p31m, respectively), whereas p4 behaves like [3, 3]+, allowing both the half-way (p4g) and right-on (p4m) possibilities for reflection. (One may point out here that the case of p31m is more complicated, as some threefold centers lie on no reflection axis … and it is for similar reasons that we leave p2 out of this somewhat eccentric parenthetical discussion altogether!)]

We conclude that there exist fourteen types of finite  three-dimensional symmetry groups, the seven general groups [n]+, [n+, 2], [n][2+, (2n)+], [2, n]+[2+, 2n], [2, n] and the seven special groups [3, 3]+, [3, 3], [3+, 4][4, 3]+[4, 3][5, 3]+[5, 3]. (For a more traditional (algebraic) derivation of this classic Klein-Weyl result and further discussion the reader is referred to Marjorie Senechal’s 1990 paper Finding the Finite Groups of Symmetries of the Sphere.)

This post is dedicated to the gradually disappearing kiosks of Greek streets known as periptera (wings-all-around, a word known from peripteros naos = column-surrounded temple that plausibly entered Greek through certain flames of love in Song of Songs 8.6 (περίπτερα αὐτῆς περίπτερα πυρός, φλόγες αὐτῆς) and the Septuagint rendering of Hebrew רִשְׁפֵּי = spark): initially conceived as phone-booth-sized roofy stores to be owned by amputated soldiers, they grew in size in recent decades … to the point of selling various symmetry-minded balls that partially inspired this series of posts … and made my Thessaloniki strolls — talk the walk — more interesting🙂

Equidistant rotation axes and Platonic Equations

December 6, 2015

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.

Cubical/icosahedral subgroups AND crystallographic point groups: border pattern representation

October 29, 2015

The observation presented below is definitely not new, as the ubiquitous Marjorie Senechal came up with it 40 years ago in Point Groups and Color Symmetry; I know this thanks to Alain Bossavit’s Point Groups (1995). Still, I think it is worth discussing it here, in some detail and color, due to its great didactic potential.

Unaware of Senechal’s 1975 paper, and having not noticed it on first reading of Bossavit’s 1995 paper, I rediscovered the connection between ‘monoaxial’ cubical/icosahedral subgroups and border patterns through my observation of various commercial balls; indeed most ideas presented here are also present in Cube on the ball. Further, this border pattern connection extends to crystallographic point groups (briefly discussed below) and beyond, that is to all finite subgroups of the sphere (not discussed below); note here that point groups are finite spherical subgroups where only rotations of orders two, three, four, six and rotoreflections of orders two, four, six are allowed. (This extension may of course remind my readers of Cubically yours, icosahedron🙂 )

Our presentation is going to be succint and visual: for each type of border pattern, finite versions (bands) are shown, and a brief explanation is made on how the two-dimensional isometries of the border pattern turn into three-dimensional isometries of some cubical/icosahedral group or point group … simply by wrapping the band! (Note here that every cubical subgroup happens to be a point group, but not vice versa, whereas some icosahedral subgroups (specifically, those consisting of five or ten ‘units’) are not point groups: details will be provided along as needed.) Please excuse various drawing imprecisions!

p111: [2]+, [3]+, [4]+, [5]+, [6]+

The band’s finite/circular n-unit translation turns into n-fold rotation about the axis of the cylinder created by wrapping the band (i.e., by joining its right and left bands). Note that [6]+ is a point group but not a subgroup of either the cube or the icosahedron (which possess no sixfold rotation); conversely, [5]+ is an icosahedral subgroup, but not a point group.

p1a1: [2+, 2+], [2+, 4+], [2+, 6+], [2+, 10+]

The band’s glide reflection (along the band’s middle line, featuring 2n units) turns into 2n-fold rotoreflection (with implied n-fold rotation). Of special is interest is the first case above ([2+, 2+]), where the two units (one upward, one downward) yield twofold rotoreflection, that is inversion, without rotation. Inversion is absent from the next case, [2+, 4+], which is now equipped with twofold rotation, but it reappears in the last two cases. Since eightfold and twelvefold rotoreflection are not allowed in either cubical/icosahedral subgroups or point groups, the spherical subgroups [2+, 8+] and [2+, 12+] are not listed.

p1m1: [2+, 2], [3+, 2], [4+, 2], [6+, 2]

The band’s horizontal reflection turns now into reflection perpendicular to the n-fold rotation. (Observe here that, in the same way horizontal reflection (together with translation) yields ‘trivial’ glide reflection in a border pattern, reflection and n-fold rotation perpendicular to each other combine into ‘trivial’ rotoreflection.) Notice that [3+, 2] is a point group but not a cubical/icosahedral subgroup: as we have seen in previous posts, no threefold rotation can be perpendicular to a reflection in either the cube or the icosahedron! (For a very similar reason the spherical subgroup [5+, 2] is not a cubical/icosahedral subgroup, and, since it is not a point group, either, it is not listed.)

pm11: [2], [3], [4], [5], [6]

The band’s vertical reflection turns into reflection containing the axis of the n-fold rotation. There are two kinds of vertical reflection in the band, one running through units and one running half way between units, which leads to two kinds of reflection planes when n is even and to reflection planes of non-equivalend ‘ends’ when n is odd; either way, we obtain n reflection planes intersecting each other along the n-fold rotation axis.

p112: [2, 2]+, [2, 3]+, [2, 4]+, [2, 5]+, [2, 6]+

The band’s half turn centers turn into endpoints of axes of twofold rotation, all n of them perpendicular to the axis of the n-fold rotation. As in the previous case, the two kinds of half turn centers lead to two kinds of twofold axes when n is even and to twofold axes of non-equivalent ‘ends’ when n is odd.

pma2: [2+, 2], [2+, 4], [2+, 6], [2+, 10]

The band’s glide reflection, n vertical reflections (of one kind only), and n half turns (of one kind only) yield 2n-fold rotoreflection (and implied n-fold rotation along the same axis), n reflections (of one kind only) containing the rotoreflection axis, and n twofold rotations (of one kind only) perpendicular to the rotoreflection axis, respectively. Note that [2+, 2] has been obtained already out of p1m1!  (Again, there is inversion at 2n = 6 and 2n = 10, but not at 2n = 4 (tennis ball).)

pmm2: [2, 2], [2, 3], [2, 4], [2, 6]

The band’s horizontal reflection, vertical reflection (of two kinds), and half turn (of two kinds) yield reflection perpendicular to the n-fold rotation (and ‘trivial’ rotoreflection along the same axis), n reflections (of two kinds when n is even) containing the n-fold rotation axis, and n twofold rotations (of two kinds when n is even) perpendicular to the n-fold rotation axis, respectively. Again, [2, 3], the only group above without inversion, is a point group, but not a cubical/icosahedral subgroup; and [2, 5], being neither of the two, is not listed.

… There exist 32 crystallographic point groups, how many have we listed above? Subtracting 1 for the one and only repetition ([2+, 2]), and not including the five icosahedral, non-point-groups listed ([5]+, [2+, 10+], [5], [2, 5]+, [2+, 10]), we count (4 + 3 + 4 + 4 + 4 + 3 + 4) – 1 = 25 groups. To these we must obviously add [1]+ (the trivial identity-only group that needs no representation) and [1] (the reflection-only group represented by the symmetry group of the isosceles triangle — especially when on a wrapped band or sphere rather than on a flat surface!), so we move from 25 to 25 + 2 = 27 groups. The remaining five groups are the so-called cubical point groups, that is the ‘polyaxial’ symmetry groups of the cube and … the 2002-2014 World Cup soccer balls!

Cubical/icosahedral subgroups: diagnostic diagrams

October 12, 2015

[M = reflection, P = inversion, R = rotation, S = rotoreflection]

Cubical/icosahedral subgroups: lists of elements

September 30, 2015

Building on the previous three posts, I offer here cubical and icosahedral examples for most of the proper subgroups, specifically for all but the ‘polyaxial’ ones. (Recall here that cubical ‘polyaxial’ subgroups ([3, 3]+, [3, 3], [3+, 4], [4, 3]+) have been presented in the context of 2002-2014 World Cup soccer balls, whereas the only ‘polyaxial’ icosahedral, non-cubical subgroup is the icosahedron’s rotation subgroup, [5, 3]+.) Specifically, I list the isometries in each example — as presented in my Concluding Remarks here — employing the die and soccer ball notations introduced in previous posts, using I for the identity, red color for rotations based on the ‘main axis’ of the ‘monoaxial’ subgroup, green for other (‘free’) rotations, orange for inversion, purple for rotoreflections (by necessity along the ‘main axis’), and blue for reflections. In the spirit of the preceding post, cubical and icosahedral examples for each subgroup type common to the cube and the icosahedron are presented together. There exists precisely one example for each icosahedral subgroup, and up to three algebraically but not geometrically equivalent examples for each cubical subgroup.

[1]+ = {Ι}

[1] = {I, (16)} = {I, (23)(45)} = {Ι, (AG)(DL)(HK)(JI)}

[2+, 2+] = {I, (16)(25)(34)} = {I(AD)(BE)(CF)(GL)(HJ)(IK)}

[2]+ = {I(25)(34)} = {I, (16)(23)(45)} = {I, (AG)(BF)(CE)(DL)(HJ)(IK)}

[2] = {I(25)(34), (25), (34)} = {I, (25)(34)(23)(45), (24)(35)} = {I, (16)(23)(45), (16), (23)(45)} = {I, (AG)(BF)(CE)(DL)(HJ)(IK), (AG)(DL)(HK)(JI), (BF)(CE)(HI)(JK)

[2+, 2] = {I, (25)(34), (16)(25)(34), (16)} = {I, (16)(23)(45), (16)(25)(34), (24)(35)} = {I, (AG)(BF)(CE)(DL)(HJ)(IK), (AD)(BE)(CF)(GL)(HJ)(IK), (AL)(BC)(DG)(EF)}

[2, 2]+ = {I, (25)(34)(16)(23)(45), (16)(24)(35)} = {I, (25)(34)(16)(25), (16)(34)} = {I, (AG)(BF)(CE)(DL)(HJ)(IK), (AD)(BC)(EF)(GL)(HK)(IJ), (AL)(BE)(CF)(DG)(HI)(JK)}

[2, 2] = {I, (25)(34)(16)(23)(45), (16)(24)(35), (16)(25)(34), (16), (24)(35), (23)(45)} = {I, (25)(34)(16)(25), (16)(34), (16)(25)(34), (16), (34), (25)} = {I, (AG)(BF)(CE)(DL)(HJ)(IK), (AD)(BC)(EF)(GL)(HK)(IJ), (AL)(BE)((CF)(DG)(HI)(JK), (AD)(BE)(CF)(GL)(HJ)(IK), (AL)(BC)(DG)(EF), (BF)(CE)(HI)(JK), (AG)(DL)(HK)(IJ)}

[2+, 4+] = {I, (25)(34), (16)(2354), (16)(2453)}

[2+, 4] = {I, (25)(34), (16)(2354), (16)(2453), (16)(23)(45), (16)(24)(35), (25), (34)} = {I, (25)(34), (16)(2354), (16)(2453), (16)(25), (16)(34), (23)(45), (24)(35)}

[3]+ = {I, (123)(465), (132)(456)} = {I(ABG)(CIJ)(DEL)(FKH)(AGB)(CJI)(DLE)(FHK)}

[3] = {I, (123)(465), (132)(456), (12)(56), (13)(46), (23)(45)} = {I, (ABG)(CIJ)(DEL)(FKH), (AGB)(CJI)(DLE)(FHK), (AB)(CJ)(DE)(FH), (AG)(DL)(HK)(IJ), (BG)(CI)(EL)(FK)}

[2, 3]+ = {I(123)(465), (132)(456), (15)(26)(34), (16)(24)(35), (14)(25)(36)} = {I, (ABG)(CIJ)(DEL)(FKH)(AGB)(CJI)(DLE)(FHK), (AD)(BL)(CK)(EG)(FI)(HJ), (AE)(BD)(CH)(FJ)(GL)(IK), (AL)(BE)(CF)(DG)(HI)(JK)}

[2+, 6+] = {I, (123)(465), (132)(456), (142635), (153624),  (16)(25)(34)} = {I(ABG)(CIJ)(DEL)(FKH)(AGB)(CJI)(DLE)(FHK), (ALBDGE)(CHIFJK), (AEGDBL)(CKJFIH)(AD)(BE)(CF)(GL)(HJ)(IK)}

[2+, 6] = {I(123)(465), (132)(456), (142635), (153624)(15)(26)(34), (16)(24)(35), (14)(25)(36), (16)(25)(34), (12)(56), (13)(46), (23)(45)} = {I(ABG)(CIJ)(DEL)(FKH)(AGB)(CJI)(DLE)(FHK), (ALBDGE)(CHIFJK), (AEGDBL)(CKJFIH)(AD)(BL)(CK)(EG)(FI)(HJ), (AE)(BD)(CH)(FJ)(GL)(IK), (AL)(BE)(CF)(DG)(HI)(JK)(AD)(BE)(CF)(GL)(HJ)(IK), (AB)(CJ)(DE)(FH), (AG)(DL)(HK)(IJ), (BG)(CI)(EL)(FK)}

[4]+ = {I, (2354), (2453), (25)(34)}

[4] = {I(2354), (2453), (25)(34), (25), (34), (23)(45), (24)(35)}

[4+, 2] = {I(2354), (2453)(25)(34), (16)(2354), (16)(2453), (16)(25)(34), (16)}

[2, 4]+ = {I(2354), (2453)(25)(34), (16)(25), (16)(34), (16)(23)(45), (16)(24)(35)}

[2, 4] = {I(2354), (2453)(25)(34), (16)(2354), (16)(2453), (16)(25)(34)(16)(25), (16)(34), (16)(23)(45), (16)(24)(35), (16), (25), (34)(23)(45), (24)(35)}

[5]+ = {I, (BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI)}

[5] = {I(BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI), (CH)(GK)(FJ)(IL), (BF)(CE)(HI)(JK), (BK)(EI)(HL)(GJ), (BG)(CI)(EL)(FK). (BJ)(CL)(EH)(FG)}

Cubical/icosahedral subgroups: geometrical derivation

September 6, 2015

After the ‘instinctive’ approach of the previous two posts, I offer here a geometrical, quite possibly novel, derivation of the subgroups of the cube/octahedron and the icosahedron/dodecahedron. This derivation is based on

(i) fifteen lemmas of broader interest stated in the Appendix with quick verification based on ‘experimental’ composition of isometries by way of permutation multiplication – rather than complete geometrical proofs that I might offer in a later post — as illustrated in the previous two posts (Cube on the ball and Cubically yours, icosahedron).

(ii) repeated use of the Conjugacy Principle (as illustrated – with geometrical proofs of several two-dimensional cases – in Isometrica, 4.0.4, 4.0.5, 6.4.4, 8.0.3, etc): if two isometries f, g belong to a group of isometries (by leaving a certain set invariant, for example), then the image f[g] of g under f, equal in fact to the composition/product f*g*f’ (where f’ is the inverse of f), also belongs to the group; the Conjugacy Principle is accepted here as an intuitive axiom (that the reader may easily verify in the cube or the icosahedron, reflecting for example rotation (axes) or reflection (planes) and rotating reflection (planes) or rotation (axes) to valid isometries).

[For examples and background information on three-dimensional isometries I refer to the previous two posts – where you may also find explicit descriptions of the subgroups obtained here, complete with the die (cube) and the soccer ball (truncated icosahedron) notations – and my 2001 paper Isometries come in circles.]

Since, as we saw in the previous two posts, the cube and the icosahedron share many subgroups, my approach to their subgroups is a ‘unified’ one, based on each considered subgroup’s smallest rotation: this smallest rotation could be by 180, 120, 90, or 72 degrees, whereas the two rotation-free subgroups will be treated separately. [This approach echoes the classification of wallpaper patterns in chapter 8 of Isometrica, with the crucial difference that the rotation-free subgroups will play a very marginal role here.]

Let’s start with rotation-free subgroups: as the product of two reflections is a rotation by twice their intersection angle (direct (‘level-by-level’) generalization of an easy two-dimensional theorem) and the square of a 2n-fold rotoreflection is, more or less by definition, an n-fold rotation about the 2n-fold rotoreflection axis, such a rotation-free subgroup may only contain – in addition to the identity, always – either precisely one cubical/icosahedral reflection or the cube/icosahedron’s inversion (twofold rotoreflection); it cannot contain both inversion and reflection, because in that case their product would generate a twofold rotation (of axis lying on the reflection plane), for example (16)*(16)(25)(34) = (25)(34) or (16)(25)(34)*(12)(56) = (15)(26)(34) (cube), (AG)(DL)(HK)(JI)*(AD)(BE)(CF)(GL)(HJ)(IK) = (AL)(BE)(CF)(DG)(HI)(JK) (icosahedron). We conclude that the cube and the icosahedron share only two rotation-free subgroups, the inversion-only subgroup [2+, 2+] (parallelepiped) and the reflection-only subgroup [1] (isosceles triangle). (The wikipedia subgroup hierarchy diagrams cited at the previous two posts also list the trivial identity-only subgroup [1]+.)

We proceed now into the derivation of the subgroups in four sections as promised: in the first two sections (smallest rotation 180 or 120 degrees) cubical and icosahedral subgroups are treated together, whereas the third section (smallest rotation 90 degrees) is devoted to cubical subgroups and the fourth section (smallest rotation 72 degrees) is devoted to icosahedral subgroups.

(I) Cubical/icosahedral subgroups of smallest rotation 180 degrees

Let us first observe that such a subgroup has twofold rotation either in one direction or in three, mutually perpendicular, directions: there cannot be more than three directions because in that case there must exist twofold rotations of axes intersecting at an angle smaller than 90 degrees, therefore producing, by Lemma 1, rotations by less than 180 degrees; and there cannot be precisely two directions because the product of two perpendicular twofold rotations is a twofold rotation perpendicular to both (Lemma 2).

Next, let us observe that, in the cube, a twofold rotation R2 may or may not have a ‘square root’, that is a fourfold rotoreflection S4 such that S4*S4 = R2; more specifically, 2-side twofold rotations do have a square root, for example (25)(34) = (16)(2354)*(16)(2354), whereas 2-edge twofold rotations do not.

We conclude that, rotationwise, an icosahedral subgroup may look like R2 or R2 x R2 x R2 and a cubical subgroup may look like R2 or S4 or R2 x R2 x R2 or R2 x R2 x S4 – the cubical possibilities R2 x S4 x S4 and S4 x S4 x S4 are ruled out because the product of two perpendicular fourfold rotoreflections is a threefold rotation (Lemma 3).

It is interesting that the fourth cubical case above, R2 x R2 x S4, must include reflection, as the product of a twofold rotation and a fourfold rotoreflection perpendicular to each other is a reflection … containing the rotoreflection axis and making a 45-degree angle with the rotation axis (Lemma 4).

(Ia) Single twofold rotation not generated by fourfold rotoreflection (R2)

How can a single twofold rotation coexist with reflection? By the Conjucacy Principle, the reflection plane must either contain the rotation axis or be perpendicular to it, otherwise the twofold rotation would be reflected to a second twofold rotation. There do indeed exist examples of both cases, [2] and [2+, 2], respectively: the first subgroup is of order 4, the symmetry group of the right isosceles triangular prism, consisting of one twofold rotation, two perpendicular reflections (with their product being equal to the twofold rotation), and the identity; and the second subgroup – already discussed indirectly near the beginning of this post – is also of order 4 and contains, in addition to the identity and twofold rotation, one reflection and implied inversion. (Notice, echoing earlier remarks, ‘factorizations’ of the inversion like (16)(25)(34) = (16)*(25)(34) = (12)(56)*(15)(26)(34) and (AD)(BE)(CF)(GL)(HJ)(IK) = (AG)(DL)(HK)(JI)*(AL)(BE)(CF)(DG)(HI)(JK).)

Without reflection we end up with the twofold-rotation-only 2-element subgroup [2]+ (the symmetry group of the parallelogram, a subgroup of both [2] and [2+, 2]).

(Ib) Single fourfold rotoreflection (S4) [cube only]

A single fourfold rotoreflection cannot coexist with reflection: indeed the reflection plane must, by the Conjugacy Principle again, either contain the rotoreflection axis or be perpendicular to it; in the first case the product would be a twofold rotation perpendicular to the rotoreflection (Lemma 5), in the second case the product would be a fourfold rotation (Lemma 6). But it is of course possible to have a single fourfold rotoreflection – with implied twofold rotation – all alone (subgroup [2+, 4+], of order 4).

Vincent Callebaut’s Agora Tower in Taipei, Taiwan

(Ic) Three mutually perpendicular twofold rotations, no fourfold rotoreflection (R2 x R2 x R2)

In this case the Conjugacy Principle tells us that a reflection plane must either contain (and be defined by) two rotation axes or contain one rotation axis and reflect the other two axes to each other. The first case yields the rectangular parallelepiped subgroup (of order 8, with three mutually perpendicular reflections), [2, 2]: the product of a reflection and a twofold rotation contained in it is a new reflection perpendicular to the original reflection and containing the twofold rotation (Lemma 7). The second case is theoretically possible only in the cube, but even there impossible because … the product of a reflection with a twofold rotation making a 45-degree with it would be a fourfold rotoreflection (Lemma 8).

Without reflection we end up with the brazuca II subgroup, [2, 2]+ (a subgroup of [2, 2] of order 4).

(Id) Two twofold rotations and one fourfold rotoreflection (R2 x R2 x S4) [cube only]

Recall that this combination must include reflection, in fact Lemma 8 indicates – as discussed right above in (Ic) – that a reflection plane could contain the fourfold rotoreflection axis (Rz) and reflect the two twofold rotation axes (Rx, Ry) to each other: indeed that works, and there are certainly precisely two reflection planes with that property (M+, M-), and the outcome is the famous tennis ball, subgroup [2+, 4].

Of course we need to rule out a couple of other possibilities: there cannot be a third reflection plane, either perpendicular to the fourfold rotoreflection (by Lemma 6) or containing one twofold rotation axis and the fourfold rotoreflection axis (by Lemma 7 and previous case). And it needs to be stressed that, whereas there exists only one possibility for the fourfold rotoreflection (along one of the cube’s three fourfold axes), there exist two ‘isomorphic’ possibilities for the pairs of the reflections and the twofold rotations: the latter can be either two perpendicular 2-edge rotations perpendicular to the fourfold rotoreflection or two perpendicular 2-side rotations perpendicular to the fourfold rotoreflection, whereas the former can respectively be either two perpendicular 4-edge reflections containing the fourfold rotoreflection or two perpendicular 4-vertex reflections containing the fourfold rotoreflection; the two possibilities have been discussed in detail toward the end of Cube on the ball.

(II) Cubical/icosahedral subgroups of smallest rotation 120 degrees

The existence of more than one threefold rotations in a cubical subgroup implies the existence of all threefold rotations: this follows from the Conjugacy Principle, as, for example, (123)(465) rotates (154)(236) to (153)(246) to (124)(365) to (154)(236); notice also how (154)(236)*(123)(465) = (135)(264) and (145)(263)*(132)(456) = (124)(365), etc (Note here that the term “threefold rotation” includes any given threefold rotation plus its inverse along the same axis.) And here comes an important difference between the cube and the icosahedron: no icosahedral subgroup of smallest rotation 120 degrees may contain ‘adjacent’ threefold rotations, as the product of them would yield a fivefold rotation, for example (ABG)(CIJ)(DEL)(FKH)*(ABK)(CJG)(DEI)(FHL) = (AGIEJ)(BHDLK). Still, the icosahedron has room for threefold rotations just a little bit further apart from each other – at ‘distance two’ rather than ‘distance one’ from each other – making in fact the same angle (about 70.53 degrees) with each other as the cubical threefold rotations: as in the case of the cube, there exist quadruples of such icosahedral threefold rotations with each one of them rotating the other three to each other; for example, (ABG)(CIJ)(DEL)(FKH) rotates (AJK)(BFL)(CGE)(DHI) to (ALI)(BCH)(DGK)(EFJ) to (AHE)(BDJ)(CLK)(FGI) to (AJK)(BFL)(CGE)(DHI). (Note here that the product of any two of these threefold rotations is – depending on order and orientation, exactly as in the case of the cube – either one of the other two threefold rotations or a twofold rotation.)

As in (I), where some twofold rotations were generated by a fourfold rotoreflection and some didn’t, a threefold rotation R3 may or may not have a ‘square root’, that is a sixfold rotoreflection S6 such that S6*S6 = R3. This time, however, there is no obvious reason to rule out the coexistence of more than one sixfold rotoreflections. On the other hand, since no threefold rotation may leave the axis of another threefold rotation invariant, and every threefold rotation would rotate a sixfold rotoreflection to another sixfold rotoreflection, either all four threefold rotations in our cubical/icosahedral subgroup have a ‘square root’ or none of them does. (By contrast, in (Id) the tennis ball’s fourfold rotoreflection rotates the two twofold rotations to each other, and is rotated by each one of them back to itself.)

Putting everything together, we conclude that, rotationwise, our subgroup looks like R3 or S6 or R3 x R3 x R3 x R3 or S6 x S6 x S6 x S6: in each of these cases we need to address the existence and location not only of reflection, as in (I), but also of twofold rotation.

(IIa) Single threefold rotation not generated by sixfold rotoreflection (R3)

In the case of R3 (threefold rotation without sixfold rotoreflection in only one direction) let us first observe that, by the Conjugacy Principle, twofold rotation may coexist with the threefold rotation only if the former rotates the axis of the latter back to itself: so, in both the cube and the icosahedron, for every given threefold rotation there is a unique triplet of twofold rotations – all perpendicular to it – rotating its axis to itself (and rotated in turn, consistently with the Conjugacy Principle again, to each other by it). For example, the triplet corresponding to the cube’s threefold rotation (123)(465) is, denoting each 2-edge twofold rotation’s axis’ ends by the end-defining sides, {1|4-3|6, 2|6-1|5, 3|5-2|4}, whereas the triplet corresponding to the icosahedron’s threefold rotation (ABG)(CIJ)(DEL)(FKH) is, denoting each twofold rotation’s axis’ ends by the end-defining pentagons, {C|H-F|J, I|F-K|C, J|K-H|I}. This ‘triplet uniqueness’ is corroborated by way of isometry composition: the product of members of the other possible triplets corresponding to (123)(465), that is {1|2-5|6, 2|3-4|5, 3|1-6|4} and {1|6, 2|5, 3|4}, with (123)(465) … would be equal to a fourfold rotation and to another threefold rotation, respectively; and the product of members of the other possible triplets corresponding to (ABG)(CIJ)(DEL)(FKH), that is {A|B-D|E, B|G-L|E, G|A-L|D}, {A|J-D|H, B|C-E|F, G|I-L|K}, {A|F-D|C, B|K-E|I, G|H-L|J}, with (ABG)(CIJ)(DEL)(FKH) … would be equal to a short fivefold rotation, a long fivefold rotation, and another threefold rotation, respectively.

Further, as in (I), the Conjugacy Principle dictates that a possible reflection plane must either be perpendicular to the threefold rotation axis or contain the threefold rotation axis: the first case is geometrically impossible simply by cubical/icosahedral structure; and in the second case there cannot be twofold rotation as, by the Conjugacy Principle again, the reflection plane would have to contain one twofold axis and reflect the other two twofold axes to each other … and then the product of the reflection with any one of the two twofold rotations would be a sixfold rotoreflection about the threefold axis. (Demonstration: in the cube, where (23)(45) contains the axis of (123)(465) and reflects 2|4-3|5 = (16)(24)(35) to itself and 1|4-3|6 = (14)(25)(36) & 1|5-2|6 = (15)(26)(34) to each other, (23)(45)*(14)(25)(36) = (153624) and (23)(45)*(15)(26)(34) = (142635), two sixfold rotoreflections that are inverses of each other; in the icosahedron, where A|G-D|L = (AG)(DL)(HK)(IJ) contains the axis of (ABG)(CIJ)(DEL)(FKH) and reflects H|I-J|K = (AL)(BE)(CF)(DG)(HI)(JK) to itself and C|H-F|J = (AE)(BD)(CH)(FJ)(GL)(IK) & C|K-F|I = (AD)(BL)(CK)(EG)(FI)(HJ) to each other, (AG)(DL)(HK)(IJ)*(AE)(BD)(CH)(FJ)(GL)(IK) = (AEGDBL)(CKJFIH) and (AG)(DL)(HK)(IJ)*(AD)(BL)(CK)(EG)(FI)(HJ) = (ALBDGE)(CHIFJK), two sixfold rotoreflections that are inverses of each other.)

We conclude that a unique threefold rotation without a sixfold rotoreflection generating it may only coexist with either reflection only (subgroup [3], with three reflection planes intersecting each other along the threefold rotation axis) or twofold rotation only (subgroup [2, 3]+, with three twofold rotation axes perpendicular to the threefold rotation axis); and it is possible of course to have the unique threefold rotation all alone (subgroup [3]+).

(IIb) Single sixfold rotoreflection (S6)

In this case we note that, since the square of the sixfold rotoreflection is a threefold rotation, all the restrictive results obtained in (IIa) are still valid; in particular, the possible locations for reflection and twofold rotation remain the same. We also note that it is possible now to have both twofold rotation (three axes) perpendicular to the sixfold rotoreflection and reflection (three planes) containing the sixfold rotoreflection axis: in fact this has already been demonstrated in (IIb) above, where we proved that this combination results into sixfold rotoreflection along the threefold reflection axis (subgroup [2+, 6]). But notice at the same time that it is no longer possible to have reflection planes containing the sixfold rotoreflection axis without ‘bonus’ twofold rotation (as in the case of subgroup [3] above): the product of a reflection and a sixfold rotoreflection contained in it is a twofold rotation (Lemma 9). Nor is it possible anymore to have sixfold rotoreflection perpendicular to the twofold rotation axes without ‘bonus’ reflection (as in the case of subgroup [2, 3]+ above): the product of a sixfold rotoreflection and a twofold rotation perpendicular to each other is a reflection containing the sixfold rotoreflection axis (Lemma 10). The only remaining possibility is to have the unique sixfold rotoreflection all alone (subgroup [2+, 6+]).

(IIc) Threefold rotation in four directions without sixfold rotoreflection (R3 x R3 x R3 x R3)

A cubical subgroup of smallest rotation 120 degrees that contains four, that is all, threefold rotations must also contain all 2-side twofold rotations, as seen through products like (132)(456)*(153)(246) = (16)(25): sometimes the product of two cubical threefold rotations is a threefold rotation – as in (132)(456)*(135)(264) = (124)(365) – and sometimes it is a twofold rotation! On the other hand, the said subgroup may not contain any 2-edge twofold rotations, as the product of any of those with any threefold rotation is always equal to a fourfold rotation, for example (123)(465)*(12)(34)(56) = (1364) and (123)(465)*(13)(25)(46) = (2453).

To a large extent, these cubical observations may be extended to the icosahedron. As we have seen at the beginning of (II), the cube’s ‘global’ quadruple of threefold rotations is replaced in the icosahedron by ‘local’ quadruples of threefold rotations (the ends of axes of which form a ‘cuboid’):

dodecahedral cube (by Euclid et al)

such quadruples and ‘cuboids’ — five of them inside every icosahedron — come again complete with twofold rotations, those twofold rotations that correspond to the ‘cuboid”s equivalent of 2-side twofold rotations; for example, the three twofold rotations corresponding to the quadruple {(ABG)(CIJ)(DEL)(FKH), (AJK)(BFL)(CGE)(DHI), (ALI)(BCH)(DGK)(EFJ), (AHE)(BDJ)(CLK)(FIG)} – and the ‘cuboid’ {(ABG), (DEL), (AJK), (DHI), (BCH), (EFJ), (CLK), (FIG)} – are A|F-C|D, B|K-E|I, G|H-J|L. (The reader may check how any one of these twofold rotations, like for example G|H-J|L = (AD)(BI)(CF)(GH)(EK)(JL), rotates the four threefold rotations to each other – modulo inverses – in pairs, like (ABG)(CIJ)(DEL)(FKH) to (AKJ)(BLF)(CEG)(DIH) and vice versa, and is rotated by them to A|F-C|D, B|K-E|I, B|K-E|I, B|K-E|I, respectively, and by their inverses to B|K-E|I, A|F-C|D, A|F-C|D, A|F-C|D, respectively, etc)

How about reflection? The Conjugacy Principle tells us that if our cubical subgroup contains one 4-edge reflection then it contains all three 4-edge reflections, as for example (123)(465) rotates (16) to (25) to (34) to (16); likewise, one 4-vertex reflection implies them all, as for example (123)(465) rotates (12)(56) to (23)(45) to (13)(46) to (12)(56), and (25)(34) rotates (12)(56) to (15)(26) and (13)(46) to (14)(36), etc On the other hand, our cubical subgroup cannot contain both 4-edge reflections and 4-vertex reflections, as the product of a 4-edge reflection and a 4-vertex reflection (intersecting each other at 45 degrees) is a fourfold rotation (special case of a well known result), for example (16)*(12)(56) = (1265). So, if our cubical subgroup has reflection, then it contains either all reflection planes defined by any two threefold axes or all reflection planes containing no threefold axis; but the second case needs to be ruled out – or rather be moved to (IId) further below – as the product of a cubical reflection containing no threefold rotation with a threefold rotation is a sixfold rotoreflection, as for example in (16)*(123)(465) = (123654).

Again, these observations largely apply to the icosahedron, too – where, mind you, there are no reflection planes containing no threefold rotation axes … but there are reflections not containing our subgroup’s threefold rotations! For example, in the ‘cuboid’ defined by the quadruple of threefold rotations {(ABG)(CIJ)(DEL)(FKH), (AJK)(BFL)(CGE)(DHI), (ALI)(BCH)(DGK)(EFJ), (AHE)(BDJ)(CLK)(FGI)}, there exist three ‘working’ reflections (reflecting our threefold rotations to each other in pairs, and rotated by them to each other, as they ought to): A|F-C|D, B|K-E|I, G|H-J|L. (This is clear from the picture of the dodecahedral/icosahedral ‘cuboid’ above, but the reader may also check algebraically how, for example, the reflection G|H-J|L = (AF)(BI)(CD)(EK) reflects (ABG)(CIJ)(DEL)(FKH) to (AEH)(BJD)(CKL)(FIG) and vice versa, whereas it is rotated by (ABG)(CIJ)(DEL)(FKH) to A|F-C|D = (BK)(EI)(GJ)(HL); and how indeed (AF)(BI)(CD)(EK)*(ABG)(CIJ)(DEL)(FKH) = (AIJDKH)(BGFELC), etc: as in the case of the cube, the product of a reflection containing no threefold rotation with a threefold rotation is a sixfold rotoreflection.) Now, how about reflection planes containing our ‘cuboid”s threefold rotations? Well, here we are out of luck: limiting ourselves to our ‘cuboid’ always, there is no working reflection containing a threefold rotation! (For the same reason some other cubical isometries – notably fourfold rotations and fourfold rotoreflections – are not icosahedral: they work in the cube but not in its circumscribed icosahedron.)

Putting everything together, we end up with only two possibilities and subgroups: [3, 3]+ (fevernova – all threefold rotations and all ‘2-side’ twofold rotations (as discussed above for both the cube and the icosahedron) with no reflection) and [3, 3] (jabulani – all threefold rotations and all ‘2-side’ twofold rotations plus reflection with planes containing the threefold axes); note here how the observation at the very end of the preceding paragraph rules out [3, 3] as an icosahedral subgroup! ([3, 3] is also the symmetry group of the fifth Platonic solid, the regular tetrahedron; in contrast to our observation about the cube and its circumscribed icosahedron, all tetrahedral isometries are still valid in the circumscribed cube: the tetrahedron symmetry group is a subgroup of the cube/octahedron symmetry group, but the cube/octahedron symmetry group is not a subgroup of the icosahedron/dodecahedron symmetry group.)

(IId) Sixfold rotoreflection in four directions (S6 x S6 x S6 x S6)

Observe first that the presence of four sixfold rotoreflections (of axes as in (IIc)) implies reflection: for example, (142635)*(142635)*(132645) = (25) (cube), (ALBDGE)(CHIFJK)*(ALBDGE)(CHIFJK)*(AGIDLK)(BFHECJ) = (BK)(EI)(GJ)(HL) (icosahedron/’cuboid’). Together with our discussion in (IIc), this observation yields precisely one subgroup for this case: [3+, 4] (teamgeist – all sixfold rotoreflections and all ‘2-side’ twofold rotations plus reflection containing twofold but no sixfold/threefold axes).

(III) Cubical subgroups of smallest rotation 90 degrees

By the Conjugacy Principle, there exists either one or all three fourfold rotations: every fourfold rotation rotates another fourfold rotation to the third fourfold rotation. A fourfold rotation cannot have a ‘square root’ as there is no eightfold rotoreflection in the cube, but its axis may trivially work as a fourfold rotoreflection axis, too (if and only if our subgroup contains reflection perpendicular to the fourfold rotation). So this case is best split into just two cases: R4 and R4 x R4 x R4.

(IIIa) Single fourfold rotation (R4)

By the Conjugacy Principle, there can be neither 2-edge twofold rotation non-perpendicular to the fourfold rotation axis nor threefold rotation: such rotations would rotate the fourfold rotation to another fourfold rotation. On the other hand, there exists 2-side twofold rotation (along an axis perpendicular to the fourfold axis corresponding to a ‘dormant’ fourfold rotation) if and only if there exists 2-edge twofold rotation (along an axis perpendicular to the fourfold axis): the product of a fourfold rotation and a twofold rotation perpendicular to each other is a twofold rotation perpendicular to the fourfold rotation and making a 45-degree angle with the twofold rotation (Lemma 11); this is also related to Lemma 1. Likewise, there exists 4-vertex reflection containing the fourfold rotation axis if and only if there exists 4-edge reflection containing the fourfold reflection axis: the product of a fourfold rotation and a reflection containing it is another reflection making a 45-degree angle with the initial reflection and containing the fourfold rotation (Lemma 12). Observe here that, as in (I) and, only in theory, (II), the Conjugacy Principle implies that the only possible reflection not containing the fourfold rotation axis is the one perpendicular to it (in which case there exists, as pointed out already, ‘trivial’ fourfold rotoreflection along the fourfold rotation axis); this would also follow from Lemma 8.

Considering all eight possible combinations of answers Y (yes) or N (no) to the three questions concerning the existence of twofold rotation (four axes) perpendicular to the fourfold rotation, the existence of reflection (four planes) containing the fourfold rotation, and the existence of reflection perpendicular to the fourfold rotation, respectively, we end up – in view of the discussion above – with five subgroups and three impossibilities:

YYY: subgroup [2, 4] (an enriched version of [2+, 4])

YYN: impossible

YNY: impossible

YNN: subgroup [2, 4]+ (analogous to [2, 2]+ and [2, 3]+)

NYY: impossible

NYN: subgroup [4] (analogous to [2] and [3])

NNY: subgroup [4+, 2] (a ‘doubling’ of [2+, 2])

NNN: subgroup [4]+ (analogous to [2]+ and [3]+)

[The case NYY is trivially impossible (the intersection of two reflection planes, one containing the fourfold axis and the other perpendicular to it, yields a twofold rotation perpendicular to the fourfold rotation), whereas YNY and YYN are ruled out by Lemma 7.]

(IIIb) Three mutually perpendicular fourfold rotations (R4 x R4 x R4)

Since the product of two perpendicular fourfold rotations is a threefold rotation (Lemma 13), our subgroup is easily seen to contain all threefold rotations; in fact each fourfold rotation rotates all four threefold rotations to each other (like (2354) rotating (123)(465) to (135)(264) to (154)(236) to (142)(356) to (123)(465)), and each threefold rotation rotates all fourfold rotations to each other (like (123)(465) rotating (2354) to (1463) to (1265) to (2354)). Further, our subgroup contains all 2-side twofold rotations, which may be viewed as either squares of fourfold rotations or products of two threefold rotations; but it also contains all 2-edge rotations, as products of one threefold rotation and one fourfold rotation, for example (13)(25)(46) = (2354)*(123)(465). We conclude that our subgroup contains all cubical rotations, so it is equal either to the (roto)reflection-free subgroup [4, 3]+ (brazuca), that is the rotation subgroup of the cube, or to the symmetry group of the cube itself ([4, 3]). (The cube’s symmetry group is of order 48, so every subgroup containing its rotation subgroup, which is already of order 24, must be the cube itself; and there are more geometrical ways to check this, as the product of any (roto)reflection with the rotations would yield all the other (roto)reflections.)

(IV) Icosahedral subgroups of smallest rotation 72 degrees

Unlike the case of icosahedral threefold rotations, which we studied in (II), any two distinct icosahedral fivefold rotations are ‘adjacent’ to each other. Further, the existence of more than one fivefold rotation implies, by the Conjugacy Principle, the existence of all six fivefold rotations; and exactly the same holds true for tenfold rotoreflections. So this case splits naturally into four cases, R5, S10, R5 x R5 x R5 x R5 x R5 x R5, and S10 x S10 x S10 x S10 x S10 x S10.

(IVa) Single fivefold rotation (R5)

Obviously there cannot be fourfold rotation (cube only). A bit less obviously, the Conjugacy Principle implies that there cannot be threefold rotation. As for twofold rotation, it is possible only if its axis is perpendicular to the fivefold rotation: this follows again from the Conjugacy Principle, as in the cases of single threefold rotation and single fourfold rotation in (IIa) and (IIIa). Likewise, any reflection plane must either contain the fivefold rotation axis or be perpendicular to it: like in (II) and unlike in (III), the latter case is ruled out by the very structure of the icosahedron.

We conclude, exactly as in (IIa), that there exist three options associated with a single fivefold rotation not generated by tenfold rotoreflection: the fivefold rotation left alone (subgroup [5]+), the fivefold rotation combined with five twofold rotations perpendicular to it (subgroup [2, 5]+), and the fivefold rotation combined with five reflections containing it (subgroup [2, 5]).

(IVb) Single tenfold rotoreflection (S10)

This case parallels (IIb): in view of the observations in (IVa), and also with the help of Lemmas 14 & 15 (which parallel Lemmas 9 & 10), we conclude that the only possibilities here are single tenfold rotoreflection all alone (subgroup [2+, 10+]) and single tenfold rotoreflection combined with both twofold rotation (five axes) perpendicular to it and reflection (five planes) containing it (subgroup [2+, 10]).

(IVc) Six fivefold rotations (R5 x R5 x R5 x R5 x R5 x R5)

A subgroup containing all six fivefold rotations must also contain all threefold rotations: indeed any two ‘adjacent’ fivefold rotations yield the two threefold rotations ‘between’ them, for example (BGFJK)(CHIEL)*(AKCHG)(DIFJL) = (ABG)(CIJ)(DEL)(FKH) and (AKCHG)(DIFJL)*(BGFJK)(CHIEL) = (AKB)(CGJ)(DIE)(FLH). It follows, in view of observations made in (II), that our subgroup also contains all twofold rotations, so it is the (roto)reflection-free subgroup [5, 3]+, the rotation subgroup (of order 60) of the icosahedron’s full symmetry group (of order 120). (Beyond the order argument employed, observe that our subgroup cannot contain any reflections, for in that case it would have to contain, by the observations at the end of (IVa) for example, tenfold rotoreflections, etc)

(IVd) Six tenfold rotoreflections (S10 x S10 x S10 x S10 x S10 x S10)

Since the square of every tenfold rotoreflection is a fivefold rotation about the same axis, we conclude, by the observations made right above, that the subgroup in question is the icosahedron’s full symmetry group ([5, 3]).

Concluding remarks

Cubical & icosahedral subgroups are not as unpredictable as one might at first think (by looking at the wikipedia subgoup hierarchy diagrams, for example), and as I certainly thought only four months ago (while preparing my 5/20/15 lecture at Aristotle University); indeed the said lecture makes it clear that most cubical subgroups may be obtained by way of group tables from [2, 4] and [2+, 6], and after that has been realized it is even easier to obtain most icosahedral subgroups as subgroups of [2, 2], [2+, 6], and [2+, 10]. These ‘easy’ subgroups are the ‘monoaxial’ ones, that is subgroups where there is only one axis of smallest rotation; notice here that if non-strict inequalities are allowed, even the subgroups [2+, 4] (tennis ball) and [2, 2] (rectangular parallelepiped) may be viewed as ‘monoaxial’: one twofold axis perpendicular to two ‘different’ twofold axes. (In the case of the tennis ball, one of the three mutually perpendicular twofold axes is deeply different, as it happens to hide fourfold rotoreflection underneath; in fact, as the Coxeter notation strongly suggests, [2+, 4] is very similar to [2+, 6]* and [2+, 10], with single fourfold rotoreflection instead of single sixfold or tenfold rotoreflection.) As for ‘polyaxial’ subgroups, we have seen how restrictive and unforgiving symmetry is: beyond the four subgroups featured on the 2002-2014 World Cup soccer balls, only the icosahedral subgroup [5, 3]+ comes forward to intrigue (those unfamiliar with the concept of rotation subgroup, that is)! And, among World Cup subgroups, only the teamgeist ([3+, 4]) comes as a surprise of sorts: indeed the jabulani ([3, 3]), besides being isomorphic to the symmetry group of the regular tetrahedron, may be viewed as [2+, 4] ‘tripled’ (‘created’ in fact by perpendicular fourfold rotoreflections and Lemma 3); as for the fevernova ([3, 3]+) and the brazuca ([4, 3]+), these happen to be the rotation subgroups of the regular tetrahedron and of the cube, respectively.

*the similarity between [2+, 6] and [2+, 4] is best illustrated through a precious ball found in a toys & gifts store near Thessaloniki’s Vardar Square:

**the teamgeist stands between the rectangular parallelepiped and the cube the same way the fevernova stands between the brazuca II and the brazuca: throwing in the threefold rotations we jump from [2, 2] and [2, 2]+ to [3+, 4] and [3, 3]+, respectively; and then throwing in the fourfold rotations we move from [3+, 4] and [3, 3]+ to [4, 3] and [4, 3]+, respectively.

I hope to provide additional viewpoints on cubical & icosahedral subgroups in future posts. For the time being, here is a brief summary of our findings and strategy:

APPENDIX: Some product lemmas of broader interest

[In all lemmas and cube/icosahedron illustrations below, all rotation and rotoreflection axes and all reflection planes do pass through the same point, in our case the cube/icosahedron’s center.]

Lemma 1: The product of two twofold rotations is a rotation by an angle twice the angle between the two axes. (This is quite easy to prove geometrically – along the fact that the product rotation is perpendicular to both factors — by writing each twofold rotation as a product of two perpendicular reflections, the ‘middle’ one being the one defined by the plane of their axes.)

Cube: (15)(26)(34)*(16)(24)(35) = (123)(465) [threefold rotation]

(16)(25)*(16)(24)(35) = (2453) [fourfold rotation]

Icosahedron: (AK)(BJ)(CF)(DI)(EH)(GL)*(AF)(BE)(CD)(GJ)(HL)(IK) = (ACI)(BHG)(DFK)(EJL) [threefold rotation]

(AK)(BJ)(CF)(DI)(EH)(GL)*(AJ)(BE)(CI)(DH)(FK)(GL) = (ABHIF)(CDEJK) [short fivefold rotation]

(AK)(BJ)(CF)(DI)(EH)(GL)*(AH)(BG)(CF)(DJ)(EL)(IK) = (AEGJI)(BLHKD) [long fivefold rotation]

Lemma 2: The product of two perpendicular twofold rotations is a twofold rotation perpendicular to them. (This is a special case of Lemma 1.)

Cube: (25)(34)*(16)(34) = (16)(25), (25)(34)*(16)(23)(45) = (16)(24)(35), (16)(23)(45)*(16)(24)(35) = (25)(34)

Lemma 3: The product of two perpendicular fourfold rotoreflections is a threefold rotation.

Cube [only]: (25)(1463)*(16)(2354) = (132)(456), (25)(1364)*(16)(2354) = (145)(263), (16)(2354)*(25)(1364) = (153)(246), etc

Lemma 4: The product of a twofold rotation and a fourfold rotoreflection perpendicular to each other is a reflection containing the rotoreflection axis and making a 45-degree angle with the rotation axis.

Cube [only]: (16)(2354)*(16)(34) = (23)(45), (16)(2354)*(16)(23)(45) = (25)

Lemma 5: The product of a reflection and a fourfold rotoreflection of axis lying on the reflection plane is a twofold rotation perpendicular to the fourfold rotoreflection and making a 45-degree angle with the reflection plane. (This is obviously related to Lemma 4.)

Cube [only]: (16)(2354)*(25) = (16)(24)(35), (23)(45)*(16)(2354) = (16)(34)

Lemma 6: The product of a reflection and a fourfold rotoreflection of axis perpendicular to the reflection plane is a fourfold rotation about the fourfold rotoreflection axis. (This almost follows from the definition, taking into account that, in the cube, the two reflection planes involved are one and the same.)

Cube [only]: (16)(2354)*(16) = (2354)

Lemma 7: The product of a reflection and a twofold rotation of axis lying on it is a new reflection perpendicular to the original reflection and containing the twofold rotation axis. (This is an easy consequence of the product of two perpendicular reflections being a twofold rotation along the intersection.)

Cube: (25)*(25)(34) = (34), (23)(45)*(25)(34) = (24)(35), (23)(45)*(16)(23)(45) = (16)

Icosahedron: (AK)(BJ)(CF)(DI)(EH)(GL)*(AK)(CG)(DI)(FL) = (BJ)(CL)(EH)(FG), (AK)(BJ)(CF)(DI)(EH)(GL)*(BJ)(CL)(EH)(FG) = (AK)(CG)(DI)(FL)

Lemma 8: The product of a reflection with a twofold rotation making a 45-degree with it is a fourfold rotoreflection of axis perpendicular to the rotation axis and lying on the reflection plane. (This is closely related to Lemmas 4 and 5.)

Cube [only]: (25)(34)*(13)(46) = (1463)(25), (16)(23)(45)*(34) = (16)(2354)

Lemma 9: The product of a reflection and a sixfold rotoreflection of axis lying on the reflection plane is a twofold rotation perpendicular to the sixfold rotoreflection and making a 30-degree angle with the reflection.

Cube: (23)(45)*(153624) = (14)(25)(36)

Icosahedron: (AG)(DL)(HK)(IJ)*(AEGDBL)(CKJFIH) = (AE)(BD)(CH)(FJ)(GL)(IK)

Lemma 10: The product of a sixfold rotoreflection and a twofold rotation perpendicular to each other is a reflection containing the sixfold rotoreflection and making a 30-degree angle with the twofold rotation. (This is closely related to Lemma 9.)

Cube: (14)(25)(36)*(153624) = (12)(56)

Icosahedron: (AE)(BD)(CH)(FJ)(GL)(IK)*(AEGDBL)(CKJFIH) = (BG)(CI)(EL)(FK)

Lemma 11: The product of a fourfold rotation and a twofold rotation perpendicular to each other is a twofold rotation perpendicular to the fourfold rotation and making a 45-degree angle with the twofold rotation.

Cube [only]: (2354)*(16)(23)(45) = (16)(25), (2354)*(16)(25) = (16)(24)(35)

Lemma 12: The product of a fourfold rotation and a reflection containing it is another reflection making a 45-degree angle with the initial reflection and containing the fourfold rotation.

Cube [only]: (2354)*(24)(35) = (34), (2354)*(25) = (24)(35)

Lemma 13: The product of two perpendicular fourfold rotations is a threefold rotation.

Cube [only]: (2354)*(1463) = (123)(465)

Lemma 14: The product of a reflection and a tenfold rotoreflection of axis lying on the reflection plane is a twofold rotation perpendicular to the tenfold rotoreflection and making an 18-degree angle with the reflection.

Lemma 15: The product of a tenfold rotoreflection and a twofold rotation perpendicular to each other is a reflection containing the tenfold rotoreflection axis and making an 18-degree angle with the twofold rotation. (This is closely related to Lemma 14.)

[The reader may have already noticed the interplay among several of these lemmas. For example, Lemma 8, related to Lemma 4 and Lemma 5, can be generalized in a way encompassing lemmas not included above — yet implicitly present in (IIa) and (IVa) — but analogous to Lemmas 10 & 9 and Lemmas 15 & 14, respectively:

Theorem: The product of a reflection M with a twofold rotation R is a rotoreflection of axis perpendicular to R and lying on M and of angle twice the angle between M and R. (Proof: it suffices to write R as M’*M”, where M’ contains R being perpendicular to M and M” contains R being perpendicular to M’, so that R*M = (M’*M”)*M = M’*(M”*M), where M”*M is a rotation of axis perpendicular to R and lying on M, hence perpendicular to M’, and of angle twice the angle between M and R.)]

… This post is dedicated to the memory of my mother Eudoxia (“Good Thinker”), who passed away three years ago this day … and also to the memory of the victims of Istanbul’s anti-Greek pogrom sixty years ago this day.

Cubically yours, icosahedron

July 4, 2015

In the previous article in this blog, Cube on the ball, we saw how a small collection of commercial balls, notably World Cup soccer balls, may help us understand and verify the cube’s subgroup hierarchy. Here we will see how the icosahedron/dodecahedron’s subgroups are related to the cube/octahedron’s subgroups; we will see in particular that, despite the higher order of its symmetry group (120 vs 48), the icosahedron has essentially nothing new to add to the cube in terms of subgroups!

First we need of course to understand the isometries of the icosahedron. We will do that by way of an old fashion soccer ball consisting of 20 white hexagons and 12 black or whatever pentagons, widely available in street kiosks (periptera) and beyond in toy size … and by way of labeling its pentagons exactly as below:

Rather ironically in view of what we saw in World Cup symmetries and Cube on the ball, back in Fall 2001 I had seen it fit to ‘identify’ the soccer ball with the icosahedron — after all, the 2002 ball had not surfaced yet, and I am not old enough to remember the pre-1970 balls🙂 At any rate, above you see a classification of the icosahedron’s isometries into ten kinds of isometries — exactly as in the case of the cube, but with no ‘grouped group table’ this time — with the new element being fivefold & tenfold isometries that need to be understood before going further. (And a small detail that also needs to be pointed out: D is missing from the diagram above … simply because it is A’s antipodal!)

Following the examples given above, and considering always rotations (and rotoreflections) about the rotation axis A-D (and the ‘equator’ reflection plane perpendicular to it): a short fivefold rotation, like (BGFJK)(CHIEL), is what one might indeed ‘expect’, rotating ten out of twelve pentagons in an ‘upper’ circle (B, G, F, J, K) and in a ‘lower’ circle (C, H, I, E, L), moving each pentagon to the pentagon ‘right next’ to it (by 72 degrees); a long fivefold rotation, like (BFKGJ)(CILHE), ‘jumps’ every other pentagon in both circles, progressing each time by 144 degrees, returning every pentagon to its initial position after two, rather than just one, full circlings of the ball; a short fivefold rotoreflection, like (AD)(BHGIFEJLKC), swaps A & D and ‘rotates’ the other ten pentagons up and down around the ball, moving back and forth from the upper circle to the lower circle, progressing each time by 36 degrees; and a long fivefold rotoreflection, like (AD)(BIJCGEKHFL), swaps A & D and moves every pentagon of the upper circle to a pentagon of the lower circle 108 degrees away, and vice versa, returning every pentagon to its initial position after three circlings of the ball. [Observe that the square of (AD)(BHGIFEJLKC) equals (BGFJK)(CHIEL), whereas the square of (AD)(BIJCGEKHFL) equals the inverse of (BFKGJ)(CILHE), that is (BJGKF)(CEHLI).]

Considering the subgroup diagram above, taken from wikipedia, and comparing it to the analogous wikipedia diagram for the cube, we see that the icosahedron’s ‘large’ subgroups (and their own subgroups) were, in essence, also present in the cube; indeed we see above the teamgeist, [3+, 4], the fevernova, [3, 3]+, the rotation subgroup [5, 3]+, completely analogous to the brazuca‘s [4, 3]+, the 6-T ball, [2+, 6], and a ‘new’ subgroup, [2+, 10], ‘analogous’ to the 6-T ball.

That the teamgeist is a subgroup of the icosahedron becomes clear if we view as its six ‘feet’ the ‘units’ CL, FG, BH, EJ, AK, DI. These ‘units’ give away instantly the three reflections and three twofold rotations, whereas the axes for the threefold rotations are ABG-DEL, AFJ-CHD, BCK-EFI, GHI-JKL; the same axes work for sixfold rotoreflections like  (ALBDGE)(CHIFJK) (axis ABG-DEL). Notice that all the isometries discussed here either preserve or swap/rotate the six ‘units’, the sixfold rotoreflection (ALBDGE)(CHIFJK) for example has the effect AK —> CL —> BH —> DI —> FG —> EJ —> AK.

It is easy now to see how the fevernova is a subgroup of the teamgeist: the regular tetrahedron with vertices at the ‘centroids’ of the ‘triangles’ ABG, CHD, EFI, JKL has reflections (and fourfold rotoreflections) that do not work for the icosahedron, but its threefold rotations are the teamgeist‘s threefold rotations discussed above, and its twofold rotations, of axes AK-DI, BH-EJ, CL-FG, work for the teamgeist, too.

Due to their fourfold symmetries, there is no room in the icosahedron for either the jabulani or the brazuca, but the latter has a close relative inside the icosahedron, consisting of all the icosahedral rotations; it is not easy to ‘represent’ this subgroup on our labeled soccer ball, but there are, thankfylly, other intriguing examples:

The example on the left is an origami-like creation by Krystyna Burczyk, hosted by George Hart at the Museum of Mathematics. The example on the right is a pendant light by Tom Dixon which I first saw during a late evening walk in a Thessaloniki store last summer. And right in the middle you see a precious ball the internet location of which I unfortunately no longer recall: there is fivefold rotation at the middle of every regular pentagon, threefold rotation wherever three irregular pentagons come together symmetrically, and twofold rotation in the middle of every two-pentagon unit. (There are not that many examples of commercial balls with partial icosahedral, yet not subcubical, symmetry!)

The 6-T ball subgroup, [2+, 6], is easily recreated on our labeled soccer ball either by keeping only the ABG and DEL triangles or by keeping only the remaining triangles, FKH and CIJ: in both cases there are three reflection planes (each of them bisecting precisely two pentagons, antipodal of each other), one threefold rotation axis (defined by the triangles above) that also acts as a sixfold rotoreflection axis (of effect either (ALBDGE) or (CHIFJK)), three twofold rotations (with axes that may be written either as AE-DB, AL-DG, BL-EG or as FJ-CH, JK-HI, FI-CK), and, last but not least, inversion (either in the form (AD)(BE)(GL) or in the form (CF)(IK)(HJ)). [Alternatively, we may also represent [2+, 6] as either AEL-LAB-BLD-DBG-GDE-EGA or FIJ-JFK-KJC-CKH-HCI-IHF, with the six-up-and-down-isosceles-‘triangles’ set remaining invariant under the isometries already discussed, respectively.]

If instead of removing the ‘north pole’ (ABG) and the ‘south pole’ (DEL) corresponding to a threefold axis, as we did in the case of the 6-T ball subgroup above (second possibility), we remove the ‘north pole’ and the ‘south pole’ corresponding to a fivefold axis like A-D … we end up with a subgroup where all threefold rotations and sixfold rotoreflections, as well as all fivefold rotations and rotoreflections, save for those about A-D, are gone; as above, there are five — rather than three — reflection planes (each of them bisecting two pentagons) and five — rather than three — twofold rotations (with axes BH-EJ, HG-JL, GI-LK, IF-KC, FE-CB).  All together — don’t forget inversion and, goes without saying, the identity — we end up with a 20-element subgroup known as [2+, 10]: its similarities to the 12-element subgroup [2+. 6] are clear! (And both groups are of course very similar to the tennis ball ([2+, 4]) — a subgroup of the cube but not of the icosahedron — with the notable difference that the latter has no inversion.) We may also represent [2+, 10] as ACH-DBG-AHI-DGF-AIE-DFJ-AEL-DJK-ALC-DKB (with the ten-up-and-down isosceles-‘triangles’ set remaining invariant under the short and long tenfold rotoreflections (AD)(BHGIFEJLKC) & (AD)(BCKLJEFIGH) and (AD)(BIJCGEKHFL) & (AD)(BLFHKEGCJI) — and the short and long fivefold rotations they ‘create’ — and under the reflections bisecting each isosceles ‘triangle’, as well as under the five twofold rotations listed already).

Recall at this point (Cube on the ball) that the three subgroups of order 6 of the 6-T ball ([2+, 6]) were the equilateral triangle or triangular pyramid ([3]), the three-‘wings’ ball ([2, 3]+), and the six-up-and-down-slanted-isosceles-‘triangles’ ball ([2+, 6+]). On the icosahedron these groups are represented by ABG (with the three-pentagons set remaining invariant under the threefold rotations about ABG-DEL and under the reflections that bisect the sides of the equilateral ‘triangle’ ABG), CH-JK-IF (with the three-‘wings’ set remaining invariant under the threefold rotations about ABG-DEL and under the twofold rotations about CH-FJ, JK-HI, IF-KC), and AHI-LIF-BFJ-DJK-GKC-ECH (with the six-up-and-down-slanted-isosceles-‘triangles’ set remaing invariant under under the threefold rotations and sixfold rotoreflections about ABG-DEL, as well as under inversion), respectively.

The corresponding subgroups of order 10 of [2+, 10] are, in complete analogy to the subgroups of order 6 of [2+, 6] (or the subgroups of order 4 of the tennis ball for that matter), the regular pentagon or pentagonal pyramid ([5]), the five-‘wings’ ball ([2, 5]+), and the ten-up-and-down-slanted-isosceles-‘triangles’ ball ([2+, 10+]). Replacing the threefold axis ABG-DEL by the fivefold axis A-D we get analogous representation of these ‘new’ groups by BGFJK (with the five-pentagons set remaining invariant under the short and long fivefold rotations about A-D and under the reflections that bisect the sides of the regular ‘pentagon’ BGFJK), BC-GH-FI-JE-KL (with the five-‘wings’ set remaining invariant under the short and long fivefold rotations about A-D and under the twofold rotations about BC-EF, GH-LJ, FI-CK, JE-HB, KL-IG), and BJL-HLK-GKC-ICB-FBH-EHG-JGI-LIF-KFE-CEJ (with the ten-up-and-down-slanted-isosceles-‘triangles’ set remaing invariant under the short and long fivefold rotations and rotoreflections about A-D, as well as under inversion), respectively.

[2, 5]+ … in Thessaloniki … and in Lionel Messi’s tender hand

Sitting inside all three subgroups of [2+, 10] discussed above is the group [5]+, that is the rotation subgroup of the regular pentagon, consisting of two short fivefold rotations (inverses of each other), two long fivefold rotations (inverses of each other), and, of course, the identity. If one insists on a representation on our labeled soccer ball, a beautiful way to do that is BCL-GHC-FIH-JEI-KLE: it is easy to check that the only non-trivial isometries leaving this five-slanted-isosceles-‘triangles’ set invariant are (BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI).

An analogous representation for the group [3]+, the rotation subgroup of the equilateral triangle, sitting inside the subgroups [3], [2, 3]+, and [2+, 6+], is AIE-BJL-GCD: the only non-trivial isometries leaving this three-slanted-isosceles-‘triangles’ set invariant are (ABG)(FKH)(CIJ)(DEL) and (AGB)(FHK)(CJI)(DLE).

To verify now that [2, 2] (‘four-feet’ ball, parallelepiped) is indeed a subgroup of [3+, 4] (teamgeist), one simply needs to ‘shrink’ the latter’s representation from CL-FG-BH-EJ-AK-DI to CL-FG-BH-EJ: the threefold rotations (and associated sixfold rotoreflections) no longer work, and we are left with three reflections, three twofold rotations, the inversion and the identity. A further reduction to CL-FG — or merely to C-L or C-G — yields the subgroup [2] (rectangle), whereas C-F yields [2+, 2], a subgroup corresponding to [2+. 6] and [2+, 10] and consisting of one twofold (rather than threefold or fivefold) rotation, one reflection, the inversion, and the identity. (But … which reflection, and which twofold rotation? There are five choices for each, and they must be paired in such a way that the product of each pair is the inversion; in other words, the reflection and the twofold rotation must be perpendicular to each other, like for example (AG)(DL)(HK)(IJ) and (AL)(BE)(CF)(DG)(HI)(JK).)

It should be clear by now that, due to the very rich symmetry of our soccer ball, it gets increasingly difficult to represent on it the smaller subgroups, like [2]+ (one twofold rotation — parallelogram), which may also be viewed as the rotation subgroup of the rectangle ([2]), analogously to [3]+ (rotation group of the fevernova‘s ‘unit’, also rotation subgroup of the equilateral triangle, [3]), [4]+ (rotation group of the brazuca‘s cross-like ‘unit’, also rotation subgroup of the square ([4]), a subgroup of the cube but not of the icosahedron), [5]+ (rotation subgroup of the regular pentagon ([5])), and other cyclic groups beyond the cube and the icosahedron. We have already obtained representaions for [3]+ and [5]+, a representation for [2]+ could be a ‘parallelogram’ like ACDG, invariant under only the twofold rotation (AD)(BH)(CG)(EJ)(FL)(IK). Likewise, the ‘isosceles triangle’ ABL represents [1], the smallest dihedral group, remaining invariant only under the reflection (AB)(CJ)(DE)(FH).

How could we represent the group [2+, 2+] (inversion only)? We need two antipodal, totally asymmetrical ([1]+) sets on the 12-pentagon ball: one working possibility is ABLE-DEGB. (It seems at first that [2+,2+] may also be represented on a ball by two antipodal slanted isosceles triangles: this is not quite right, as the isosceles triangle’s inner reflection line turns into a reflection — about a ‘new’ equator plane defined by the two antipodal (hence parallel) reflection lines — for the entire pair of antipodal isosceles triangles; we end up, once again, with [2+, 2].)

[At the risk of going slightly off topic, let us mention here that [2+, 4+] — a subgroup of order 4 of the tennis ball and the cube but not of the icosahedron — may be represented on a ball by four up-and-down slanted isosceles triangles, in full analogy to [2+, 6+], [2+, 10+] and even [2+, 2+] right above, but with a curious twist, that is the lack of inversion: it consists of fourfold rotoreflection and twofold rotation about a single axis. (This explains the lack of ‘inclusion segment’ from [2+, 2+] to [2+, 4+] in the cube’s subgroup diagram, but not from [2+, 2+] to either [2+, 6+] or [2+, 10+] in the icosahedron’s subgroup diagram above: the corresponding ‘inclusion segments’ have apparently been omitted by accident (just like the [2]+ to [2+, 6+] ‘connection’, in both subgroup diagrams this time).)]

And — finally — how to represent [2, 2]+, that is that good old brazuca II, geometrically equivalent to a ‘two-wings’ ball, which features only three mutually perpendicular twofold rotations? Just as in the case of [5, 3]+, the icosahedron’s full rotation subgroup, I do not see a solution on the 12-pentagon ball — when you start with so many reflections … it may not be easy to have a rotations-only party!

I would like to dedicate this post to the manager of Jo-Ann Fabric and Craft Stores, Oswego (summer 2001): when she understood that I needed more than the 4 or 5 black-and-white toy soccer balls available at her store … she suggested that she could probably get more balls from other regional Jo-Ann stores she had to visit anyway … and surely she ended up providing me with no fewer than 28 — a perfect number for MAT 103’s last lab!

Cube on the ball

June 13, 2015

Expanding on a 4/21/15 workshop with 10th grade students at the Experimental High School of the University of Macedonia (at the invitation of their teacher, Nikos Terpsiadis), I delivered a 5/20/15 lecture at Aristotle University (at the invitation of Mathematics Department chairperson, Nikos Karampetakis) on Cubical Symmetries on World Cup soccer balls; both activities were of course closely related to my World Cup symmetries article (2/14/14) in this blog (and to the introduction of cubical and icosahedral symmetries to MAT 103 following my 2000-2001 sabbatical at SUNY Oswego).

During the workshop students were exposed to the isometries of the cube, and were subsequently asked to determine which isometries were present on the World Cup balls of 2002, 2006, 2010, 2014, whereas during the lecture I showed how answering the hardest of these questions may naturally lead to the concept of group, and how these soccer balls, and other balls as well, correspond to various subgroups of the cube. The goal of this article is to present these activities and ideas in some detail, translating from Greek where necessary and elaborating on parts of both the ppt and the video (part 1 & part 2).

Following a brief overview of World Cup soccer balls and cubical vs icosahedral stichings and designs (slides #4 through #8), the introduction to the isometries of the cube is illustrated by way of a die and permutation notation (slide #9). A new idea is the introduction of fourfold rotoreflection (S4) by way of a two-colored tennis ball (part 1, 21:30-23:20 — downward black ‘tongue’ to upward green ‘tongue’ to downward black ‘tongue’ and so on, four times up-and-down around the tennis ball, and about the axis defined by the middles of the two ‘tongues’) and likewise the introduction of sixfold rotoreflection (S6) by way of a certain 6-T ball (part 1, 26:30-28:15 — downward T to upward T to downward T and so on, six times up-and-down around the 6-T ball): this is done not only because the cube’s magnificent  ‘equator’ (sixfold rotoreflection plane corresponding/perpendicular to diagonal threefold rotation axis)

is so much harder to understand than the corresponding sixfold rotoreflection plane of the 6-T ball, but also because fourfold rotereflection is ‘too easy to see’ on the cube — too much symmetry may at times obscure the symmetries!

[Previously, and much in the same spirit, ‘simple’ balls (with minimal symmetry) were used (part 1, 14:10-16:30) to illustrate the difference between inversion and twofold rotation (two concepts that coincide in dimension two); in particular, the ball with the ‘back-to-back-thinking couple’ has no inversion (P)  … because, although at the antipodes of the couple we get the couple again, the man’s head is antipodal of the woman’s head and vice versa🙂 (Note that inversion and twofold rotation may easily coexist, in the case of two antipodal isosceles triangles along the ball’s equator, for example — with reflection along the equator being a ‘bonus’, mind you🙂 ) ]

In their worksheet (slide #10) the students had to provide a “yes” (NAI) or “no” (OXI) answer to seven questions, regarding the existence, in each one of the four World Cup soccer balls, of inversion, reflection, twofold rotation, fourfold rotation, threefold rotation, fourfold rotoreflection, and sixfold rotoreflection; further, and in the form of two ‘footnote’ challenging questions, they had to distinguish between 4-edge reflection (ME) and 4-vertex reflection (MV), and likewise between 2-side twofold rotation (RS) and 2-edge twofold rotation (RE) — if there to begin with, of course.

The answers appear in slides #11 through #14. Looking here at the ‘footnote’ questions, we see that the easier ones to answer are the ones about reflection, thanks to the presence of threefold rotation (R3): since the corresponding axes are contained in reflection planes in the case of the jabulani (2010) but not in the case of the teamgeist (2006), we conclude that the jabulani has 4-vertex reflection (MV) whereas the teamgeist has 4-edge reflection (ME).

At a higher level of difficulty, unless one resorts to counting axes — for the cube has three axes of 2-side twofold rotation (RS) and six axes of 2-edge twofold rotation (RE) — we decide that, whereas the brazuca (2014) has obviously both kinds of twofold rotation, both the teamgeist and the jabulani have only 2-side twofold rotation (RS): this follows from the rather subtle observation (slide #15) that their twofold rotation axes are intersections of the same kind of reflection planes (ME in the case of teamgeist, MV in the case of jabulani); indeed the axes of the cube’s 2-side twofold rotation may be obtained either as intersections of two 4-vertex reflections or as intersections of two 4-edge reflections, whereas the axes of the cube’s 2-edge twofold rotations require one 4-vertex reflection and one 4-edge reflection. These thoughts lead naturally to the concept of composition of perpendicular reflections, either in a geometrical context (slide #16, not easy to extend to compositions of other types of isometries) or in an algebraic context, by way of permutation multiplication (slide #17, easy to extend to other isometries).

The last workshop question concerns the nature of the twofold rotation in the fevernova (2002): this is again a 2-side twofold rotation (RS), either because there are three twofold rotation axes in the fevernova or because the fevernova‘s twofold rotations are, unlike 2-edge twofold rotations (RE), closed under multiplication/composition (slide #18, right below); indeed always RS * RS = RS (or perhaps RS * RS = I, where I stands for the identity tansformation), whereas RE * RE= RS is possible, for example (15)(26)(34) (12)(34)(56) = (16)(25). (A third way of answering this question would be an ‘external’ one, requiring the observation that the product of two threefold cubical rotations (R3) may be a 2-side twofold rotation (RS), but not a 2-edge twofold rotation (RE).)

The above notion of closedness under isometry multiplication leads naturally to the concepts of (sub)group and group table. The closedness of 2-side twofold rotations is in particular celebrated in the brazuca II ball (slide #19, complete with full group table, right below), which has no other isometries (except of course for the identity (I)), and whose subgroup-of-cube status is illustrated in slide #20, by way of grouped group table (employed for the same purpose for the four balls in slides #21 through #24, as well as in World Cup symmetries — where, of course, all answers to the workshop questions may also be found🙂 ). [Note here that the brazuca II (part 2, 3:32-5:20) may be viewed as a tennis ball with asymmetrical ‘tongues’ — much in the same way the tennis ball ‘tongues’ may be viewed as less symmetrical versions of the die’s 3-1-4 and 2-6-5 ‘sections’.]

Despite this affinity, the brazuca II appears at first not to be a subgroup of the tennis ball, as the latter contains only two 2-side twofold rotations (slide #25): this is not correct, and the subgroup status of the brazuca II within the tennis ball becomes clear if one observes that the latter contains three mutually perpendicular twofold rotations that are closed under isometry composition. (The distinction between 2-edge twofold rotations and 2-side twofold rotations — and likewise between 4-edge reflections and 4-vertex reflections — becomes … counterproductive in the absence of threefold rotation!)

The tennis ball’s full 8 x 8 group table is exhibited in slide #26, whereas the 16 x 16 group table of a ‘four-ovals’ ball (part 2, 13:15-14:02, slides #28 and #29), geometrically/symmetrically equivalent to a square prism, that does contain the tennis ball as a subgroup (slide #30) is shown in full in slide #27; and two more subgroups of order 8, the ‘four-feet’ ball (part 2, 14:50-16:02), equivalent to the rectangular parallelepiped, and the ‘four-wings’ ball (part 2, 16:05-16:56), each of them containing the brazuca II as a subgroup, are shown in slides #31 through #35. (An interesting observation here is that the brazuca II has the same isometries as a ‘two-wings’ ball, where the two ‘wings’ (rectangles) are antipodal of each other.)

Slide #36 (right above), taken from wikipedia and employing the Coxeter notation, shows the cube’s full subgroup structure (which we are merely reproducing here ‘experimentally’). We see that the soccer balls of the last three World Cups correspond, remarkably, to the three subgroups of the cube of order 24 (teamgeist = [3+, 4], jabulani = [3, 3], brazuca = [4, 3]+), each of them containing the fevernova = [3, 3]+ as a subgroup of order 12 (slide #37).

Slides #38 through #43 are devoted to the 6-T ball ([2+, 6], the only other subgroup of the cube of order 12) and its three subgroups of order 6: these are the only subgroups of the cube of order 6 and are ‘orphan’ … in the sense that I have not (yet) found commercial balls corresponding to them, so they are provided only by way of group tables — but check toward the end of this post, too! The 6-T ball is not a subgroup of any World Cup soccer ball: it is not a subgroup of either the jabulani or the brazuca because it has, as we have already seen, sixfold rotoreflection (S6); and it is not a subgroup of the teamgeist because it has four-vertex reflection (MV) — or, in broader terms, its reflection planes contain a threefold rotation axis🙂 [A non-convex solid isomorhic to the 6-T ball is obtained by raising two regular tetrahedra off opposite sides of a Star of David, each one having one of the equilateral triangles as its base.]

Slide #44 — shown below with a couple of corrections involving the tennis ball — conludes the show with a partial cube diagram (subgroup relations), showing all the balls discussed here … in pictures rather than scientific notation. Two subgroups of order 8 and all subgroups of order lower than 6 but one (brazuca II = [2, 2]+) have not been included, but they are easily found by way of group tables etc The ‘four-ovals’ ball = [2, 4] is the only subgroup of order 16 and contains all subgroups of order 8, including the tennis ball = [2+, 4] (also a subgroup of the jabulani), the ‘four-feet’ ball = [2, 2] (also a subgroup of the teamgeist), and the ‘four-wings’ ball = [2, 4]+ (also a subgroup of the brazuca). [The die at the top may of course be replaced by any ball with full cubical symmetry, like this one for example.]

The three ~ signs above would push the lecture into a next level and concept, that of group isomorpism: the two ‘orphan’ groups of order 6 on the right (slides #42 and #43) are isomorphic by way of mapping every 4-vertex reflection of #42 to the unique 2-edge twofold rotation perpendicular to it, for example (13)(46) to (14)(25)(36), and every threefold rotation to itself; the tennis ball (slide #29) is isomorphic to the ‘four-wing’ ball (slide #34) by multiplying every reflection and rotoreflection by (16) and leaving everything else invariant; and the jabulani is isomorphic to the brazuca by mapping 4-vertex reflections to 2-edge twofold rotations as above (extending the isomorphism between the two subgroups of order 6) and every fourfold rotoreflection to the inverse of the fourfold rotation naturally associated with it, like (16)(2453) to (2354), and leaving everything else invariant.

An important lesson here is that two subgroups may be geometrically distinct yet algebraically identical. It follows in particular, by way of the ‘four-wings’ ball and the brazuca, that the tennis ball is an algebraic subgroup of the jabulani:

This isomorphism, mapping reflections to twofold rotations and vice versa, may create the impression that, just as the brazuca is isomorphic to the jabulani algebraically but not geometrically, the tennis ball is an algebraic but not geometrical subgroup of the jabulani: this is quite easily seen not to be correct, especially in case one views the jabulani as a symmetrical version of the fevernova as seen further above, that is consisting of four symmetrically placed curvy equilateral triangles A, B, C, D, and subsequently views the A-B, C-D pairs as equivalent to the ‘tongues’ of the tennis ball; the tennis ball subgroup consists then of those jabulani isometries that happen to either swap or preserve the two pairs (OR the 3-1-4 and 2-6-5 ‘sections’ previously mentioned🙂 ):

(AB) = (24)(35), (CD) = (23)(45), (AB)(CD) = (25)(34), (AC)(BD) = (16)(34), (AD)(BC) = (16)(25), (ADBC) = (16)(2354), (ACBD) = (16)(2453)

It is clear here that, consistently with the previously stated principle against distinguishing — in the absence of threefold rotation — between 2-edge and 2-side twofold rotations, and likewise between 4-edge and 4-vertex reflections, we have replaced the tennis ball’s twofold rotations (16)(23)(45), (16)(24)(35) by the twofold rotations (16)(25), (16)(34), and likewise we have replaced the tennis ball’s reflections (25), (34) by the reflections (24)(35), (23)(45):

It follows from the observations above and the well known full isomorphism between the jabulani and the regular tetrahedron (just join the centroids of the jabulani‘s four curvy equilateral triangles) that the tennis ball is a subgroup of the tetrahedron (a well known fact mentioned for example here or here in passing). (That the tennis ball is a subgroup of the jabulani also becomes clear if one removes either two antipodal ‘units’ or the other four ‘units’ from the fully isomorphic to the jabulani six-‘units’ ball exhibited at the end of the lecture (part 2, 33:40-37:15), receiving a ball fully isomorphic to the tennis ball.)

It should be easy now to verify that one of the two subroups of order 8 that has not been discussed so far, namely the square pyramid (or merely … the square), {I, (25), (34), (23)(45), (24)(35), (2453), (2354), (25)(34)} = [4], is algebraically isomorphic to the tennis ball (therefore to the ‘four-wings’ ball as well); it can be represented on a ball by four equally spaced congruent isosceles triangles, all pointing toward the north pole … whereas inverting every other triangle leads, once again, to the tennis ball — but geometrically, too, this time — and renders the jabulani a special case of it (with isosceles triangles turning equilateral)!  (This last idea, applied to six isosceles triangles, leads once again to the 6-T ball, [2+, 6], whereas … slanting the six up-and-down isosceles triangles — in such a way that reflections and twofold rotations are gone but inversion, threefold rotation and sixfold rotoreflection survive — leads to the ‘orphan’ subgroup of slide #41, [2+, 6+]; as for slides #42 and #43, those correspond to the three-straight-isosceles-triangles ball, that is the triangular pyramid (or merely the equilateral triangle), [3], and to a ‘three-wings’  ball, [2, 3]+, analogous to the ‘four-wings’ ball, respectively.)

[As for the last subgroup of order 8, {I, (16)(25)(34), (16), (2453), (2354), (16)(2453), (16)(2354), (25)(34)} = [4+, 2], that one may be represented on a ball by four equally spaced congruent isosceles triangles along the equator, all pointing in the same direction. (Note here that if “four” is replaced by “three” … we simply depart from the cube, ending up with a ball, which — just like a ball already existing out there as a basketball ball, basically a ball with the equator and three meridians, an ‘aligned’ version of the 6-T ball also known as triangular bipyramid🙂  — has reflection and threefold rotation perpendicular to each other!)]

… Only a few days before their fourth birthday, I would like to dedicate this post to my partner’s twin nieces, Sophia and Sandy: their two-colored tennis ball toy, complete with ‘magnetic’ racket, inspired me to delve into its symmetries — and realize that a rotoreflection is what ‘mirrors’ one half of the tennis ball to the other half — in July 2013!