crystallomath

80 pie patterns

September 30, 2017

weber1929

Pie patterns are three-dimensional analogues of the much more familiar border patterns, having wallpaper patterns as their cross sections. The term is mine but the concept (‘two-sided plane’) goes back to the late 1920’s, and so does the result that there exist precisely 17+46+17=80 pie patterns: the goal of this post is to revisit this result and concept.

The standard approach to and classification of pie patterns — see section 8.8 (Notes and References to Chapter 8) of Grunbaum & Shephard’s TILINGS AND PATTERNS for details — is a combination of group-theoretic methods and concepts (that I totally avoid, even though one might for example protest that my “cross pattern” is a geometrical disguise of Coxeter’s color-preserving normal subgroup) and simple geometric facts and ideas (that I also use, for example my “level-swapping isometries” are Shubnikov’s antisymmetries). I believe that employing Geometry rather than Group Theory makes this material more accessible and revealing, and so does my notation for pie patterns, directly inspired by that for border patterns and stressing their three-dimensional nature over their two-dimensional ‘representation’. Still, it is a two-dimensional result — the existence of precisely 46 two-colored wallpaper patterns (provable by both Geometry and Group Theory) — that plays the decisive role in my current approach (and in the standard approach as well).

Every cross (‘horizontal’) section of a pie pattern is a wallpaper pattern, inheriting the restriction of the pie pattern’s ‘vertical’ isometries; at each level of the pie pattern the corresponding wallpaper patterns may be distinct, but they have to be of the same type, sharing in fact the same isometries: this type of wallpaper pattern I call “cross pattern” of the pie pattern. Further, and following the basic idea of the preceeding post, I observe that it suffices to consider pie patterns consisting of only two levels; the pie pattern’s ‘full identity’ is now determined by the cross pattern’s isometries and the isometries that swap lower level and upper level.

It is important to point out — remembering that it is isometries rather than motifs that do matter — that a cross pattern is considered as such only in case all its isometries are ‘vertically extended’ across the entire pattern; differently said, the cross pattern is a ‘valid representation’ of the pie pattern’s vertical isometries (that work, at each cross section, as isometries for a horizontal wallpaper pattern).

For example, the two $p112$ border patterns in the right column of the preceeding post do not have $m$ as their cross pattern: indeed the ‘point reflection’ of the one-dimensional pattern $m$ at the top/bottom is not extended to the entire border pattern as a vertical reflection, due to top and bottom being ‘asymetrically placed’ with respect to each other. For a pie pattern analogous situation, consider the pie pattern (right below) having copies of the same $pmm$ wallpaper pattern at both its upper level (blue rectangles) and lower level (black rectangles); the asymetrical placement of the two patterns with respect to each other has killed all the reflections and twofold rotations (‘inherited’ from the $pmm$ and the rectangles), so the resulting pie pattern has nothing but twofold rotoreflection (of axes passing through centers of ‘intersection rectangles’) — an outcome (later to be named $111\bar{2}/p1$ ) also achievable through some ‘ghost’ cross pattern of $p1$ type like the one consisting of the shown black (or blue) right triangles:
pmm-p1 [A slightly more symmetrical placement of rectangles left to the reader … would allow for vertical reflections and vertical twofold rotations, fully exploiting the $pmm$ ‘s rich structure and creating a pie pattern with horizontal glide reflection (later to be named $mab2/pmm$ and roughly corresponding to the border pattern $pma2$ ); this is not to be confused with the pie pattern — later to be named $mmr2/pmm$ and corresponding to the border pattern $pmm2$ — where the upper $pmm$ is directly above the lower $pmm$ .]

The basic symmetry possibilities for border patterns (‘horizontal’ reflection or glide reflection about the pattern’s middle line, ‘vertical’ reflection, and half turn) are now extended to ‘horizontal’ reflection ( $m$ ) or ‘horizontal’ glide reflection ( $a$ ) about the pattern’s middle plane, ‘vertical’ reflection ( $m$ ) or ‘vertical’ glide reflection of vector parallel to the middle plane ( $g$ ), twofold ( $2$ ) or threefold ( $3$ ) or fourfold ( $4$ ) or sixfold ( $6$ ) rotation about an axis perpendicular to the middle plane, AND ‘horizontal’ twofold rotation ( $r$ ) or ‘horizontal’ twofold glide rotation ( $b$ ) about an axis lying on the middle plane. Note that a twofold or a threefold vertical rotation may have a ‘square root’, that is a fourfold ( $\bar{4}$ ) or sixfold ( $\bar{6}$ ) vertical rotoreflection (the square of which yields the respective rotation), and there may also exist twofold rotoreflection ( $\bar{2}$ ), also known as inversion. (Observe that a fourfold or sixfold rotation cannot have such a ‘square root’, as the composition of two nearest parallel eightfold or twelvefold rotoreflections would yield a fourfold or sixfold rotation that would rotate the respective rotoreflection to a nearer one.)

[For background information on three-dimensional isometries I refer the reader to George Martin’s TRANSFORMATION GEOMETRY or to my own ISOMETRIES COME IN CIRCLES. (Note here that I use the term “glide rotation” instead of the better known term “screw rotation”.)]

Focusing for now on the first three possibilities for a pie pattern, that is ‘vertical’ (glide) reflection ( $m/g$ ), ‘horizontal’ (glide) reflection ( $m/a$ ), and ‘horizontal’ (glide) twofold rotation ( $r/b$ ), and denoting their absence by $1$ , we see that there exist 27 possible combinations, of which 10 are ruled out by simple geometrical arguments, yielding thus 17 ‘roots’: mmr, mar, mab, gmb, gar, gab, m1r, m1b, m11, g1r, g1b, g11, 1m1, 1a1, 11r, 11b, 111. (In the cases m1r/g1r and m1b/g1b the vertical (glide) reflection and the horizontal (glide) twofold rotation cannot be parallel to each other as the composition of a vertical (glide) reflection and a parallel to it twofold (glide) rotation yields a horizontal (glide) reflection, the ‘root’ gmr is ruled out by the fact that the composition of a horizontal reflection and a twofold rotation lying on it yields a vertical reflection, etc) Note that a g in the first position (vertical glide reflection) implies the absence of vertical reflection (m), but not vice versa: there could exist some ‘free’ glide reflection, of axis different from that of the reflection, parallel to it or not. Further, an a in the second position (horizontal glide reflection) implies the absence of horizontal reflection. (By rough analogy, the possible ‘roots’ for border patterns are mm, ma, m1, 1m, 1a, 11, allowing the well known $7$ types $pmm2$ , $pma2$ , $pm11$ , $p1m1$ , $p1a1$ , $p112$ , $p111$ .)

[Compositions of isometries can be studied in the traditional manner, that is by expressing them as compositions of reflections. Here are some useful rules (with $m/g$ , $m/a$ , $r/b$ as above and $x$ , $\bar{x}$ denoting vertical rotation and vertical rotoreflection, respectively): $m/g*m/a = r/b$ , $m/g*r/b = \bar{x}$ (when $r/b$ is not parallel to $m/g$ ), $m/g*r/b = m/a$ (when $r/b$ is parallel to $m/g$ ), $r/b*m/a = m/g$ , $m/g*x = m/g$ , $m/a*x = \bar{x}$ , $x*\bar{x} = m/a$ , $r/b*x = r/b$ , $r/b*\bar{x} = m/g$ , $m/g*\bar{x} = r/b$ .]

Note here the kinship between the horizontal (glide) reflection ( $m/a$ ) & half turn ( $2$ ) of border patterns and the horizontal (glide) reflection ( $m/a$ ) & horizontal twofold (glide) rotation ( $r/b$ ) of pie patterns: these are the isometries that swap upper level and low level! Of course in the case of pie patterns vertical rotoreflections are also ‘level-swapping’ isometries; on the contrary, the pie pattern’s vertical rotations are ‘global’ isometries, working across the entire pie patern, yielding two-dimensional rotations at each level.

In the case of two-level pie patterns, do upper level and lower level need to be congruent? In the presence of level-swapping isometries such as horizontal (glide) reflection, horizontal twofold (glide) rotation and rotoreflection, the answer is certainly “yes”; in all other cases the answer is “no”, and the lower level could be just another wallpaper pattern of same type as the upper level or simply empty! (This last observation allows us to understand the absence of a $pm11$ two-level example in the preceeding post … as well as to trivially construct one-level examples simply by deleting the lower level in the $pmm2$ and $pma2$ examples OR by slightly increasing or decreasing the length of bars at the lower level.)

In the following double example, where grey represents upper level and black represents lower level, the pie pattern on the left is of type II (later to be named $mar4/p4m$ ), with vertical reflection planes in four directions, vertical glide reflection planes containing horizontal twofold rotation axes in two (‘vertical’ – ‘horizontal’) directions, horizontal twofold glide rotation axes in two (‘diagonal’) directions, vertical fourfold rotation, vertical fourfold rotoreflection, and vertical twofold rotoreflection, whereas the pattern on the right is of type III (later to be named $m114/p4m$ ), with vertical reflection planes in four directions, vertical glide reflection planes in two (‘vertical’ – ‘horizontal’) directions, vertical fourfold rotation, and ‘free’ vertical twofold rotation. In both patterns the cross pattern is $p4m$ . In the pattern on the left, upper and lower level are congruent to each other; in the pattern on the right, upper and lower level merely share the same isometries.

mar4-114

Our discussion so far suggests that we need three separate investigations: (I) roots mmr, gmb, 1m1 (lower level and upper level mirror images of each other) (II) roots mar, mab, gar, gab, m1r, m1b, g1r, g1b, 1a1, 11r, 11b, 111 (lower level and upper level congruent to each other) (III) roots m11, g11, 111 (lower level and upper level merely sharing the same isometries).

[Perplexed over 111‘s inclusion in (II), too? Just think of rotoreflection!]

(I) The wallpaper patterns allowed as cross patterns for the pie patterns corresponding to the roots mmr, gmb, 1m1 are — the ‘judges’ here being the first letter (vertical (glide) reflection restricted at each level to two-dimensional (glide) reflection) and the missing fourth letter (vertical rotation restricted at each level to two-dimensional rotation) — { $pm$ , $cm$ , $pmm$ , $cmm$ , $pmg$ , $p4m$ , $p4g$ , $p31m$ , $p3m1$ , $p6m$ }, { $pg$ , $pgg$ }, { $p1$ , $p2$ , $p4$ , $p3$ , $p6$ }, respectively. Note that, in the presence of horizontal reflection, every vertical rotation axis is also a vertical rotoreflection axis (and vice versa), so rotoreflection will never be mentioned in the present case. The resulting $17$ pie patterns are ‘level-by-level’ copies of the $17$ wallpaper patterns and may be listed by root as follows:

mmr (vertical reflection, horizontal reflection, horizontal twofold rotation):

$mmr1/pm$

$mmr1/cm$

$mmr2/pmm$

$mmr2/cmm$

$mmr2/pmg$

$mmr4/p4g$

$mmr4/p4m$

$mmr3/p31m$

$mmr3/p3m1$

$mmr6/p6m$

gmb (vertical glide reflection, horizontal reflection, horizontal twofold glide rotation):

$gmb1/pg$

$gmb2/pgg$

1m1 (horizontal reflection):

$1m11/p1$

$1m12/p2$

$1m14/p4$

$1m13/p3$

$1m16/p6$

(II) The wallpaper patterns possibly allowed as cross patterns for the pie patterns corresponding to the roots mar, mab, gar, gab, m1r, m1b, g1r, g1b, 1a1, 11r, 11b, 111 are, again, { $pm$ , $cm$ , $pmm$ , $cmm$ , $pmg$ , $p4m$ , $p4g$ , $p31m$ , $p3m1$ , $p6m$ } (for the roots mar, mab, m1r, m1b), { $pg$ , $pgg$ } (for the roots gar, gab, g1r, g1b), and { $p1$ , $p2$ , $p4$ , $p3$ , $p6$ } (for the roots 1a1, 11r, 11b, 111). Unlike in case (I), however, where it is obvious that all allowed types do indeed exist (by mere mirroring), here we seem destined to investigate the actual existence of $4\cdot 10 + 4\cdot 2 + 4\cdot 5 = 68$ types. Further, we would need to be careful with rotoreflection: the composition of vertical rotation and horizontal glide reflection yields a vertical rotoreflection of same order about a different axis (roots mar, mab, gar, gab, 1a1), whereas rotoreflection may also be there ‘on its own’ (roots m1r, m1b, g1r, g1b, 111 … but not 11r or 11b, as the composition of vertical rotoreflection with horizontal twofold (glide) rotation would yield vertical (glide) reflection perpendicular to the latter); in the first case only rotation is going to be indicated, with rotoreflection simply ‘implied’.

(III) The wallpaper patterns allowed as cross patterns for the pie patterns corresponding to the roots m11, g11, 111 are, exactly as in (I), { $pm$ , $cm$ , $pmm$ , $cmm$ , $pmg$ , $p4m$ , $p4g$ , $p31m$ , $p3m1$ , $p6m$ }, { $pg$ , $pgg$ }, { $p1$ , $p2$ , $p4$ , $p3$ , $p6$ }, respectively. As in (I), all $17$ predicted types are obviously possible, by placing a wallpaper pattern at the upper level and leaving the lower level empty, for example; and we can do better than that, as our $m114/p4m$ example above shows!

[By comparison, the border patterns corresponding to pie patterns of type I, II, III are { $pmm2$ , $p1m1$ }, { $p1a1$ , $p112$ , $pma2$ }, { $pm11$ , $p111$ }, respectively; again, it is type II border patterns that are the least predictable.]

In view of (II), the classification of pie patterns would promise to be quite complicated. Our two-level representation of pie patterns, however, together with an one-to-one correspondence between two-level pie patterns and two-colored wallpaper patterns established below (implicitly known since 1929), will simplify the situation significantly.

To wit, let us first recall that the 46 two-colored wallpaper patterns, first described as such in 1936 by H. J. Woods, are discussed in detail in Chapter 5 of Dorothy Washburn’s & Donald Crowe’s SYMMETRIES OF CULTURE, as well as in Chapter 6 of my own ISOMETRICA (available both in book form and online); the second source contains a detailed geometrical discussion and classification of two-colored patterns, whereas the first source is more oriented toward examples from anthropology and artwork. (I believe that my classification of the 46 two-colored wallpaper patterns in ISOMETRICA, together with those of H. J. Woods and Arthur Loeb (COLOR AND SYMMETRY), are the only geometrical ones, the others being group-theoretic.)

The promised correspondence between two-level pie patterns and two-colored wallpaper patterns is now established by coloring grey the upper level and black the lower level of a pie pattern. Unless the lower level is ‘straight below’ the upper level (as in (I) and some slightly more complicated cases discussed later), the pie pattern looks, ‘viewed from above’, like a two-colored wallpaper pattern, with the following matching betwen three-dimensional pie pattern isometries and two-dimensional wallpaper pattern isometries:

horizontal glide reflection ( $a$ ) <——> color-reversing translation ( $p'$ )

vertical reflection ( $m$ ) <——> color-preserving reflection ( $m$ )

vertical glide reflection ( $g$ ) <——> color-preserving glide reflection ( $g$ )

horizontal twofold rotation ( $r$ ) <——> color-reversing reflection ( $m'$ )

horizontal twofold glide rotation ( $b$ ) <——> color-reversing glide reflection ( $g'$ )

vertical twofold rotation ( $2$ ) <——> color-preserving twofold rotation ( $2$ )

vertical threefold rotation ( $3$ ) <——> color-preserving threefold rotation ( $3$ )

vertical fourfold rotation ( $4$ ) <——> color-preserving fourfold rotation ( $4$ )

vertical sixfold rotation ( $6$ ) <——> color-preserving sixfold rotation ( $6$ )

vertical twofold rotoreflection ( $\bar{2}$ ) <——> color-reversing twofold rotation ( $2'$ )

vertical fourfold rotoreflection ( $\bar{4}$ ) <——> color-reversing fourfold rotation ( $4'$ )

vertical sixfold rotoreflection ( $\bar{6}$ ) <——> color-reversing sixfold rotation ( $6'$ )

In other words, level-swapping pie pattern isometries are, in effect, color-reversing wallpaper pattern isometries (and vice versa)! An obvious ‘exception’ is made for the ‘invisible’ horizontal reflection, the presence of which simplifies matters greatly anyway (as already noted in (I)).

Are there other ‘exceptions’? It seems at first that we might run into problems when a vertical glide reflection plane ( $g$ ) contains a horizontal twofold rotation axis ( $r$ ), or, ‘conversely’, a vertical reflection plane ( $m$ ) contains a horizontal twofold glide rotation axis ( $b$ ): in these two cases the corresponding lines in the planar ‘symmetry plan’ would coincide, seemingly requiring a separate investigation! (Note here that the other two cases are impossible: a horizontal twofold rotation axis ( $r$ ) contained in a vertical reflection plane ( $m$ ) creates horizontal reflection ( $m*r$ ); and a horizontal twofold glide rotation axis ( $b$ ) contained in a vertical glide reflection plane ( $g$ ) is also impossible … as an equality of minimal gliding vectors would again imply horizontal reflection ( $g*b^{-1}$ ), whereas unequal minimal gliding vectors would violate the minimality of either $b$ ‘s gliding vector (via $g*b^{-1}*g$ and $|2x-y|<y$ for $x<y$ ) or $g$ ‘s gliding vector (via $b*g^{-1}*b$ and $|2y-x|<x$ for $y<x$ ).)

The ‘exceptions’ discussed above pose no problems for our perspective (pun intended), as the following example illustrates:
bem-reg
Indeed, in this example — involving separate views of the upper level (grey triangles) and of the lower level (black triangles) — the lines passing through columns of triangles represent both vertical glide reflection planes and horizontal twofold rotation axes lying on them ( $r\in g$ ), whereas the lines passing between columns of triangles represent both vertical reflection planes and horizontal twofold glide rotation axes lying on them ( $b\in m$ ). It seems at first that this example may not correspond to any two-colored types, being impossible to represent through a single planar two-colored pattern; but the following ‘unified’ pattern (later to be known as a $mar1/cm$ ) does just that, sharing the same isometries with the initial example:

In other words, the ‘exceptional’ cases $r\in g$ and $b\in m$ simply correspond, in the context of two-colored wallpaper patterns, to the cases of a color-reversing reflection axis also being a color-preserving glide reflection axis and of a color-preserving reflection axis also being a color-reversing glide reflection axis, respectively; in both cases, color-reversing translation in the direction of (glide) reflection plays the decisive role … and implies, in the context of pie patterns, that the cases $r\in g$ and/or $b\in m$ may only occur in the presence of horizontal glide reflection ( $a$ ) and only with the roots mar, mab, gar, gab.

Below I provide examples of the $10$ two-colored wallpapern patterns that have (glide) reflection but no rotation, corresponding to the $10$ type II pie-patterns that have no vertical rotation or rotoreflection but do have vertical (glide) reflection or horizontal twofold (glide) rotation (or both); we use the pie pattern name (to be provided further below) rather than the two-colored wallpaper pattern name. The pie pattern is ‘viewed from above’, so both (glide) reflection planes and twofold (glide) rotation axes are shown in blue and red lines, respectively, whereas the cases $b\in m$ and $r\in g$ are represented by purple lines:

ten-types2

[The existence of only $13$ possible types for rotationless pie patterns — the $10$ ‘two-colored’ types shown above (g-b, b-b, m/b-g/r, r-r, m-r, m/b-m/b, g/r-g/r, r-b, g-r, m-b) together with the $3$ ‘one-colored’ types (m-m, g-g, m-g) — is not that hard to establish, and it should be a first step for a new geometrical classification of pie patterns. (Further, the only way to set the two $11r1/p1$ pie patterns above apart from each other is to observe that one is of r-r type and the other is of r-b type!) Indeed this post’s approach relies heavily on the geometrical classification of the $46$ two-colored wallpaper patterns, as well as on the geometrical classification of the $17$ one-colored wallpaper patterns, presented in Chapters 6 and 8 of ISOMETRICA; my future goal is a ‘combined’ approach that would depart from the $13$ rotationless pie patterns and move on to richer types, examining which ‘roads’ of mutually parallel vertical (glide) reflections and horizontal twofold (glide) rotations could pass through ‘forrests’ of vertical rotations and rotoreflections — much in the spirit of Chapter 8 of ISOMETRICA, where I examine how (glide) reflection axes of the three rotationless wallpaper patterns ( $pm$ , $pg$ , $cm$ ) could pass through lattices of rotation centers, giving rise to richer types.]

There exist 63 two-colored wallpaper patterns, 17 of which may be viewed as one-colored … in the sense that, although the two colors are in perfect symmetry — or rather ‘balance’ — with each other, there are no color-reversing isometries: such wallpaper patterns correspond to pie patterns having no level-swapping isometries (roots m11, g11, 111), yielding therefore excellent (‘quasi-symmetric’) examples for the $17$ types of (III). These $17$ pie patterns I list, in direct analogy to (I), right below (with a corresponding example from ISOMETRICA in brackets):

m11 (vertical reflection):

$m111/pm$ [Fig. 6.15]

$m111/cm$ [Fig. 6.25]

$m112/pmm$ [Fig. 6.70]

$m112/cmm$ [Fig. 6.80]

$m112/pmg$ [Fig. 6.61]

$m114/p4g$ [Fig. 6.109]

$m114/p4m$ [Fig. 6.117]

$m113/p31m$ [Fig. 6.122]

$m113/p3m1$ [Fig. 6.125]

$m116/p6m$ [Fig. 6.138]

g11 (vertical glide reflection):

$g111/pg$ [Fig. 6.4]

$g112/pgg$ [Fig. 6.48]

111:

$1111/p1$ [Fig. 6.8]

$1112/p2$ [Fig. 6.5]

$1114/p4$ [Fig. 6.100]

$1113/p3$ [Fig. 6.121]

$1116/p6$ [Fig. 6.129]

Unfortunately the situation is not that simple in (II): not only because several of the ‘plausible’ $68$ types may not be possible (as already hinted on), but also because of situations like the one described below (where the same 4-root (mar2) and the same cross pattern ( $pmg$ ) may generate distinct pie patterns):

pb'gm-pc'mg

[The pattern on the left, to be named $mar_g2/pmg$ right below, has vertical reflection (m) in one (‘horizontal’) direction and horizontal twofold rotation (r) in two directions; there also exists vertical glide reflection (g) along both kinds of ‘vertical’ horizontal twofold rotations (r), indicated by g/r. The pattern on the right, to be named $mar2/pmg$ right below, has vertical reflection (m) with ‘in-between’ horizontal twofold glide rotation (b) in one (‘horizontal’) direction, and horizontal twofold rotation (r) with ‘in-between’ vertical glide reflection (g) in the other (‘vertical’) direction. (Global isometries are shown in blue lines (solid for vertical reflection (m), dotted for vertical glide reflection (g)), level-swapping isometries are shown in red lines (solid for horizontal twofold rotation (r), dotted for horizontal twofold glide rotation (b)), and the ‘mixed’ case g/r is shown in solid purple lines. Both patterns share vertical twofold rotation and vertical twofold rotoreflection, not shown above; their axes pass through color-preserving centers and color-reversing centers, respectively, found at the intersections of perpendicular lines.)]

It is therefore prudent at this point to save the $17$ roots mainly for notational purposes, and simply follow the route opened by the one-to-one correspondence between pie patterns and two-colored wallpaper patterns. Below I provide a list of the $46$ type II pie patterns by root & cross pattern and under both the ‘root name’ and the corresponding two-colored wallpaper pattern type — as recorded in both SYMMETRIES OF CULTURE and ISOMETRICA — with the ‘broader’ (one-colored) wallpaper type in brackets:

mar (vertical reflection, horizontal glide reflection, horizontal twofold rotation):

$mar1/cm$ = $c'm$ [ $pm$ ]

$mar1/pm$ = $p_b'1m$ [ $pm$ ]

$mar2/pmm$ = $p_b'mm$ [ $pmm$ ]

$mar2/cmm$ = $c'mm$ [ $pmm$ ]

$mar2/pmg$ = $p_c'mg$ [ $cmm$ ]

$mar_g2/pmg$ = $p_b'gm$ [ $pmm$ ]

$mar4/p4g$ = $p_c'4gm$ [ $p4m$ ]

$mar4/p4m$ = $p_c'4mm$ [ $p4m$ ]

mab (vertical reflection, horizontal glide reflection, horizontal twofold glide rotation):

$mab1/pm$ = $p_c'm$ [ $cm$ ]

$mab_m1/pm$ = $p_b'm$ [ $pm$ ]

$mab2/pmm$ = $p_c'mm$ [ $cmm$ ]

$mab2/pmg$ = $p_b'mg$ [ $pmg$ ]

gar (vertical glide reflection, horizontal glide reflection, horizontal twofold rotation):

$gar1/pg$ = $p_c'g$ [ $cm$ ]

$gar_g1/pg$ = $p_b'g$ [ $pm$ ]

$gar2/pgg$ = $p_c'gg$ [ $cmm$ ]

$gar_g2/pgg$ = $p_b'gg$ [ $pmg$ ]

gab (vertical glide reflection, horizontal glide reflection, horizontal twofold glide rotation):

$gab1/pg$ = $p_b'1g$ [ $pg$ ]

m1r (vertical reflection, horizontal twofold rotation):

$m1r\bar{2}/pm$ = $pmm'$ [ $pmm$ ]

$m1r\bar{2}/cm$ = $cmm'$ [ $cmm$ ]

$m1r\bar{4}/pmm$ = $p4'mm'$ [ $p4m$ ]

$m1r\bar{4}/cmm$ = $p4'm'm$ [ $p4m$ ]

$m1r\bar{6}/p31m$ = $p6'm'm$ [ $p6m$ ]

$m1r\bar{6}/p3m1$ = $p6'mm'$ [ $p6m$ ]

m1b (vertical reflection, horizontal twofold glide rotation):

$m1b\bar{2}/pm$ = $pmg'$ [ $pmg$ ]

$m1b\bar{4}/cmm$ = $p4'g'm$ [ $p4g$ ]

g1r (vertical glide reflection, horizontal twofold rotation):

$g1r\bar{2}/pg$ = $pm'g$ [ $pmg$ ]

$g1r\bar{4}/pgg$ = $p4'gm'$ [ $p4g$ ]

g1b (vertical glide reflection, horizontal twofold glide rotation):

$g1b\bar{2}/pg$ = $pgg'$ [ $pgg$ ]

1a1 (horizontal glide reflection):

$1a11/p1$ = $p_b'1$ [ $p1$ ]

$1a12/p2$ = $p_b'2$ [ $p2$ ]

$1a14/p4$ = $p_c'4$ [ $p4$ ]

11r (horizontal twofold rotation):

$11r1/p1$ [r-r]= $pm'$ [ $pm$ ]

$11r1/p1$ [r-b]= $cm'$ [ $cm$ ]

$11r2_b/p2$ = $pm'g'$ [ $pmg$ ]

$11r2_r/p2$ = $pm'm'$ [ $pmm$ ]

$11r2_{rb}/p2$ = $cm'm'$ [ $cmm$ ]

$11r4/p4$ = $p4g'm'$ [ $p4g$ ]

$11r4_r/p4$ = $p4m'm'$ [ $p4m$ ]

$11r3/p3$ = $p31m'$ [ $p31m$ ]

$11r3_r/p3$ = $p3m'$ [ $p3m1$ ]

$11r6/p6$ = $p6m'm'$ [ $p6m$ ]

11b (horizontal twofold glide rotation):

$11b1/p1$ = $pg'$ [ $pg$ ]

$11b2/p2$ = $pg'g'$ [ $pgg$ ]

111:

$111\bar{2}/p1$ = $p2'$ [ $p2$ ]

$111\bar{4}/p2$ = $p4'$ [ $p4$ ]

$111\bar{6}/p3$ = $p6'$ [ $p6$ ]

The subscripts in our list indicate the isometry on which the subscripted isometry lies (or passes through); such ‘subscript’ information is provided only for disambiguation purposes, specifically to distinguish $p_b'gm$ from $p_c'mg$ , $p_b'm$ from $p_c'm$ , $p_b'g$ from $p_c'g$ , $p_b'gg$ from $p_c'gg$ , $p4m'm'$ from $p4g'm'$ , $p3m'$ from $p31m'$ , and, last but not least, $pm'g'$ AND $pm'm'$ from $cm'm'$ : the last three types share $p2$ as their cross pattern, so a double subscript for $cm'm'$ becomes necessary exactly as the diagram below indicates.

p2111
Although there is no need for disambiguation in this case, the ‘double’ (two-level) diagrams below show the three types sharing $p3$ as their cross pattern:

p3m11 Three kinds of vertical threefold rotation axes, all of them intersecting horizontal twofold rotation axes, one of them passing through centers of ‘tripods’.

p31m Two kinds of vertical threefold rotation axes, one of them intersecting horizontal twofold rotation axes and passing through centers of ‘tripods’.

p6'

Vertical sixfold rotoreflection axes passing through the centers of ‘tripods’, ‘free’ vertical threefold rotation axes.

Next come the three types that, although of distinct cross patterns, belong to the $p4g$ family (when viewed as two-colored wallpaper patterns, that is):

p4g3

$g1r\bar{4}/pgg$ = $p4'gm'$ : vertical glide reflection in two (‘diagonal’) directions, horizontal twofold rotation and horizontal twofold glide rotation in the other two directions, vertical fourfold rotoreflection, ‘free’ vertical twofold rotation.

$11r4/p4$ = $p4g'm'$ : horizontal twofold rotation and horizontal twofold glide rotation in two directions, horizontal twofold glide rotation in the other two (‘diagonal’) directions, vertical fourfold rotation, ‘free’ vertical twofold rotation.

$m1b\bar{4}/cmm$ = $p4'g'm$ : vertical reflection and vertical glide reflection in two directions, horizontal twofold glide rotation in the other two (‘diagonal’) directions, vertical fourfold rotoreflection, ‘free’ vertical twofold rotation.

We conclude with the two type II pie patterns that have $p4g$ and $p4m$ as their cross pattern, shown in both ‘double’ and ‘unified’ form:

pc4gm-gammas Vertical reflection with horizontal twofold glide rotation along ( $b\in m$ ) and in-between horizontal twofold rotation in two (‘vertical’ – ‘horizontal’) directions; horizontal twofold rotation and in-between vertical glide reflection in the other two (‘diagonal’) directions; vertical fourfold rotation, vertical fourfold rotoreflection, vertical twofold rotoreflection.

mar4

Vertical reflection with horizontal twofold glide rotation along ( $b\in m$ ) in two (‘diagonal’) directions; in-between vertical glide reflection with horizontal twofold rotation along ( $r\in g$ ) in the same two (‘diagonal’) directions; vertical reflection with in-between horizontal twofold glide rotation in the other two (‘vertical’-‘horizontal’) directions; vertical fourfold rotation, vertical fourfold rotoreflection, vertical twofold rotoreflection.

[A pie pattern of same type appears early in this post, but what is ‘diagonal’ here (m/b–g/r) is ‘vertical’-‘horizontal’ there, and what is ‘diagonal’ there (m–b) is ‘vertical’-‘horizontal’ here.]

The reader may find many additional examples of type II patterns, in the form of two-colored wallpaper patterns, in both Chapter 5 of SYMMETRIES OF CULTURE (patterns classified by way of flow charts) and Chapter 6 of ISOMETRICA (patterns clasified by way of symmetry plans). (Please note that in my book I mistakenly use $p'm$ instead of $p_b'm$ (as I discovered only on 9/17/17) 🙂 )

[A good project left to the reader is to verify, following our approach, that there exist precisely $7+17+7=31$ “stick patterns” (three-dimensional analogues of border patterns with border patterns as their cross sections, also known since the late 1920’s); the needed material on two-colored border patterns may be found in both Chapter 4 of SYMMETRIES OF CULTURE and Chapter 5 of ISOMETRICA.]

I would like to dedicate this post to Dorothy Washburn and Donald Crowe: their SYMMETRIES OF CULTURE inspired me to introduce two-colored wallpaper patterns to SUNY Oswego’s MAT 103 (from the very first time I taught it in Fall 1995) and to write Chapter 6 of ISOMETRICA (during three unforgettable weeks in January 2000).

ADDENDUM (8/25/2018): the following clarification concerns the writing in red above!

Sadly, and noticed only after posting this article, two pie patterns ended up having the same name, $11r1/p1$ ! I have already commented on this accident (in red), but would like to further elaborate below.

The cross patterns of the two pie patterns, $11r1/p1$ [r-r] = $pm'$ (horizontal twofold rotation) and $11r1/p1$ [r-b] = $cm'$ (horizontal twofold rotation and horizontal twofold glide rotation), are of $p1$ type. But these are not ‘arbitrary’ $p1$ patterns: they both have translation in two perpendicular directions (something not always true in a $p1$ ). Further, in the $p1$ cross pattern of the r-b pie pattern there is a ‘diagonal’ translation that is not composition of two valid translations in the said two perpendicular directions, whereas this is not the case for the $p1$ cross pattern of the r-r pie pattern! One way to remove the said ambiguity is to set $11r1/p1$ [r-b] = $11r1/p1$ * and $11r1/p1$ [r-r] = $11r1/p1$ ** … recognizing thus two special subcategories of $p1$ , that is $p1$ * (perpendicular translations not generating all translations) and $p1$ ** (perpendicular translations generating all translations).

[Although we do not need this for disambiguation purposes, let us mention here that we could in the same way consider/define a subcategory of $p2$ , denoted by $p2$ ** and referring to those $p2$ wallpaper patterns where the half turn centers form rectangles rather than merely parallelograms. This ‘ruling’ of half turn centers may certainly happen even in the absence of glide reflection, see for example 8.2.1 and 8.2.2 in ISOMETRICA. As for the ‘ruling’ of ‘units’ in $11r1/p1$ * and $11r1/p1$ **, and the said existence of perpendicular translations in their cross pattern, that is a result of the existence of horizontal twofold rotation in them … in about the same way that reflection in a $pm$ yields translation not only perpendicular to it (by mere composition), but also — see for example 6.3.4 in ISOMETRICA — parallel to it! And the more subtle difference between $p1$ * and $p1$ ** could be explained in a manner entirely analogous to the difference between $cm$ and $pm$ , having to do with the analysis or not of ‘diagonal’ translations into valid ‘vertical’ and ‘horizontal’ translations — see 8.1.9 in ISOMETRICA. (We note here — and in relation to ISOMETRICA 8.1.9 again (and stressing the ‘affinity’ between $pm$ and $pg$ ) — that the cross pattern of $11b1/p1$ = $pg'$ is always a $p1$ **, whereas the cross pattern of $1a11/p1$ = $p_b'1$ or $111\bar{2}/p1$ = $p2'$ may well be an ‘arbitrary’ $p1$ … in about the same way the cross pattern of $1a12/p2$ = $p_b'2$ may well be an ‘arbitrary’ $p2$ .)]

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Border patterns … revisited

June 20, 2017

Border patterns, also known as freeze patterns, have been known to humanity for thousands of years, typically decorating the frame of a painting, mosaic, or some other two-dimensional design/pattern. They are straight symmetrical strips, of length finite by necessity but infinite in principle, and of constant, finite width. There exist seven types of them, discussed in several books, including my own ISOMETRICA (Chapter 2): p111, pm11, p1m1, p1a1, p112, pmm2, pma2. Below I am briefly discussing them from another point of view … with something else in mind (that may materialize in the future) 🙂

Border patterns are often considered to be one dimensional: this is quite imprecise, as there exist only two truly one-dimensional patterns, one with translation only (p) and one with reflection or rotation (m); below I show how these two patterns, appropriately placed at an upper and a lower level, can create six of the seven border patterns (the sad exception being pm11 — the only type besides p111 that has no level-swapping isometries).

border-one-two-ConvertImage

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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: adidas072016

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?)]

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EURO 2016 novel (?) spherical tiling

June 12, 2016

euro2016-octagon-brazuca

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): four-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 (Markus’ ‘proto-ball’ would look a bit like this ancient ball, with twelve ‘spherical pentagons’ instead.)]

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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!)

[2+, 4]

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:

border-3d

(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);

six-fivefold-axes

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;

cubetetra-2X

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]).

fevernova-to-jabulani-and-teamgeist

[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 🙂

peripterompales

http://greece.greekreporter.com/2015/10/14/the-uniquely-greek-kiosks-soon-to-become-history/

peripteron

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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:

fivefolds

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 $x<sin\phi/2$ ; 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 $x<sin\phi/2$ 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.

sixfolds

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).

[5, 3]+

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.)

fourfolds

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)

threefolds

lattice of twofold centers (p2)

twofolds

[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 $1/3<cos\phi <1$ , so that $1/4<sin^2\phi /2+(cos^2\phi /2)(cos^2\gamma /2)<1/2$ : 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)!

[2, 4]+

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.

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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!

Tags: Alain Bossavit, border patterns, cubical subgroups, finite spherical subgroups, icosahedral subgroups, Marjorie Senechal, point crystallographic groups
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Cubical/icosahedral subgroups: diagnostic diagrams

October 12, 2015

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

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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)}

[2, 5]+ = {I, (BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI), (AD)(BC)(EF)(GL)(HK)(IJ), (AD)(BH)(CG)(EJ)(FL)(IK), (AD)(BI)(CF)(EK)(GH)(JL), (AD)(BE)(CJ)(FH)(GI)(KL), (AD)(BL)(CK)(EG)(FI)(HJ)}

[2+, 10+] = {I, (BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI), (AD)(BHGIFEJLKC), (AD)(BCKLJEFIGH), (AD)(BLFHKEGCJI), (AD)(BIJCGEKHFL), (AD)(BE)(CF)(GL)(HJ)(IK)}

[2+. 10] = {I, (BGFJK)(CHIEL), (BKJFG)(CLEIH), (BFKGJ)(CILHE), (BJGKF)(CEHLI), (AD)(BHGIFEJLKC), (AD)(BCKLJEFIGH), (AD)(BLFHKEGCJI), (AD)(BIJCGEKHFL), (AD)(BC)(EF)(GL)(HK)(IJ), (AD)(BH)(CG)(EJ)(FL)(IK), (AD)(BI)(CF)(EK)(GH)(JL), (AD)(BE)(CJ)(FH)(GI)(KL), (AD)(BL)(CK)(EG)(FI)(HJ), (AD)(BE)(CF)(GL)(HJ)(IK), (CH)(GK)(FJ)(IL), (BF)(CE)(HI)(JK), (BK)(EI)(HL)(GJ), (BG)(CI)(EL)(FK), (BJ)(CL)(EH)(FG)}

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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).

[2+, 4+]-taipei

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.

Not quite exactly as in (IIa), we see that, in the presence of single fivefold rotation, the coexistence of twofold rotation (perpendicular to the fivefold rotation) and reflection (containing the fivefold rotation) implies a tenfold rotoreflection underneath the fivefold rotation. Indeed, by the Conjugacy Principle, there must exist five twofold axes, all perpendicular to the fivefold rotation and rotated to each other by it, and five reflection planes, all containing the fivefold rotation and rotated to each other by it; and the product of any one of the five reflections with any one of the four twofold rotations not perpendicular to it yields a tenfold rotoreflection about the fivefold rotation’s axis. (Demonstration: given the fivefold rotation A-D = (BGFJK)(CHIEL), contained in the reflection A-B-D-E = (CH)(FJ)(GK)(IL) that reflects G|I-K|L = (AD)(BE)(CJ)(FH)((GI)(KL) to itself, and B|C-E|F = (AD)(BC)(EF)(GL)(HK)(IJ) & B|H-E|J = (AD)(BH)(CG)(EJ)(FL)(IK) to each other, and C|K-F|I = (AD)(BL)(CK)(EG)(FI)(HJ) & G|H-J|L = (AD)(BI)(CF)(EK)(GH)(JL) to each other, we obtain (CH)(FJ)(GK)(IL)*(AD)(BC)(EF)(GL)(HK)(IJ) = (AD)(BHGIFEJLKC), (CH)(FJ)(GK)(IL)*(AD)(BH)(CG)(EJ)(FL)(IK) = (AD)(BCKLJEFIGH), (CH)(FJ)(GK)(IL)*(AD)(BL)(CK)(EG)(FI)(HJ) = (AD)(BIJCGEKHFL), (CH)(FJ)(GK)(IL)*(AD)(BI)(CF)(EK)(GH)(JL) = (AD)(BLFHKEGCJI); note that the first two outcomes are short tenfold rotoreflections of axis A-D, inverses of each other, whereas the last two outcomes are long tenfold rotoreflections of axis A-D, inverses of each other.)

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)

Icosahedron: (AK)(BJ)(CF)(DI)(EH)(GL)*(AD)(BH)(CG)(EJ)(FL)(IK) = (AI)(BE)(CL)(DK)(FG)(HJ)

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.

Icosahedron [only]: (CH)(GK)(FJ)(IL)*(AD)(BHGIFEJLKC) = (AD)(BC)(EF)(GL)(HK)(IJ)

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.)

Icosahedron [only]: (AD)(BC)(EF)(GL)(HK)(IJ)*(AD)(BHGIFEJLKC) = (BK)(EI)(GJ)(HL)

[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.

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crystallomath

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Border patterns … revisited

Last love?

EURO 2016 novel (?) spherical tiling

Finite 3-d symmetry groups

Equidistant rotation axes and Platonic Equations

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

Cubical/icosahedral subgroups: diagnostic diagrams

Cubical/icosahedral subgroups: lists of elements

Cubical/icosahedral subgroups: geometrical derivation

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