5-simplices where each 4-simplex is the topological
join of the achromatic segment and the corresponding
tetrahedron in the triangulation of σ.
The permutations have the structure of a group;
algebraically, they are the elements of the symmetric
group S
5
and, perhaps not an obvious fact, its ele-
ments can be
cyclically sequenced
so that two con-
secutive permutations differ by a transposition of con-
secutive coordinates (Johnson, 1963). From the view-
point of the wavelength domain, two consecutive co-
ordinates of a colour are better related than another
pair of ccordinates. The swapping of two consecu-
tive colour coordinates is called a
mild, hue-family
change
. Also, it so happens that the two tetrahedra of
σ that correspond to such a permutation have exactly
a triangular face in common; also, you can visit each
tetrahedra of σ exactly once going from one tetrahe-
dron to the next through such triangles. Equivalently
stated, there is a Hamiltonian circuit in the Cayley
graph of S
5
(Witte, 1982).
We say that each tetrahedron of σ determines a
family of hues in the sense that in each tetrahedron
the relative contributions of the primaries is fixed.
Also, you have that the 120 hue families are cycli-
cally ordered. It is the families of hue rather than the
hues themsalves that are cyclically ordered, unlike the
trichromatic case where the hue sphere is triangulated
into 6 intervals (Restrepo, 2011), and each interval is
linearly ordered, the points within each tetrahedron of
σ do not have a linear order.
As we said in Section 2, the S
4
boundary Σ of
[0,1]
5
consists of 10 4-cubes; each such 4-cube con-
nects with 8 other 4-cubes via each 3-cube of its
boundary. For example, the 4-cubes {v = 0} and
{z = 1} connect via the 3-cube {v = 0,z = 1}. In the
corresponding connectivity graph, having as nodes
the 4-cubes and as branches the connecting 3-cubes,
there are several Hamiltonian circuits.
Also, the 120 tetrahedra of the S
3
equatorial hue
sphere σ can be grouped into 20 3-cubes; for exam-
ple, the 3-cube {v = 0,w = 0} groups the tetrahedra
{v = 0,w = 0,x ≤ y ≤ z}, {v = 0,w = 0,y ≤ x ≤ z},
{v = 0,w = 0,y ≤ z ≤ x}, {v = 0,w = 0,z ≤ y ≤ x},
{v = 0,w = 0,z ≤ x ≤ y} and {v = 0,w = 0,x ≤ z ≤ y}.
That is, σ can also be
celled
into 20 3-cubes, on
each of which, as in the trichromatic case, the colour
points can be given the trichromatic attributes of hue
saturation and value. Each 3-cube in σ connects with
each of 6 other 3-cubes via a 2-square. For example,
the 3-cubes {v = 1, x = 0} and {x = 0,z = 1} connect
via the square {v = 1, x = 0,z = 1}; there are several
Hamiltonian circuits in the graph corresponding to
this cellular decomposition of σ into 3-cubes. in such
a way that each 3-cube (the union of six tetrahedra)
is visited exactly once and every pair of consecutive
3-cubes have a square (the union of two triangles)
face in common.
We indicate an iterative procedure of obtaining
Hamiltonian circuits in the graph of S
n
. Let e
n
denote
the identity element of S
n
. In Figure 1 the permuta-
tions of two elements are used to get the permutations
of a set of three elements in such a way that you get
a factorisation, into transpositions of consecutive el-
ements, of e
3
from that of e
2
. The symmetric group
S
3
with three elements is generated by the transposi-
tions t
0
:= (0,1) and t
1
:= (1,2). Denote the action
of a transposition t acting on an ordered set, or triple
p = [p
0
, p
1
, p
2
] as t(p), or simply tp; thus, for exam-
ple, (1, 2)[0, 1, 2] = [0, 2, 1]. From Figure 1, the
permutations of [0,1,2] are cyclically ordered as
Figure 1: The lines mustard and red represent the symmet-
ric group S
2
. A new element is positioned at each possible
position for after each transposition in the previous permu-
tation to get S
3
. The iteration of this gives a Hamilton circuit
on each S
n
.
[0,1,2]
[1,0,2] = (0,1).[0,1,2]
[1,2,0] = (1,2)(0,1).[1,2,0]
[2,1,0] = (0,1)(1,2)(0,1).[0,1,2]
[2,0,1] = (1, 2)(0, 1)(1, 2)(0, 1).[0,1,2]
[0,2,1] = (0,1)(1,2)(0,1)(1,2)(0,1).[0,1,2]
[0,1,2] = (1,2)(0,1)(1,2)(0,1)(1,2)(0,1).[0,1,2]
The last ordering being the initial one. Thus, at
the last line you have a factorization of the identity e
3
of the symmetric group S
3
as
e
3
= Π
3!−1
i=0
t
i
= (1,2)(0,1)(1,2)(0,1)(1,2)(0,1)
into 6 transpositions of consecutive elements, either
(0,1) or (1,2). In fact, the 6 permutations
π
j
:= Π
j
i=0
t
i
, j ∈ /0,5/
of S
3
are thus cyclically ordered. We simplify further
the notation by writing only the first element of each
transposition in the factorisations, for example e
3
=
1.0.1.0.1.0.
The 24 permutations in S
4
are likewise cyclically
ordered. By interleaving the sequences of transposi-
tions 0:=(0, 1), 1:=(1,2), 2:=(2,3), 2:=(2, 3), and its re-
verse, alternatively in between each of the 6 transpo-
sitions in the factorisation of e
3
∈ S
3
and taking care
of adding a one in the notation of the transpositions in
e
3
, after the end of an inserted sequence 2, 1, 0, but
not after an inserted sequence 0, 1, 2, (extending the
idea in Figure 1) you get
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