Breed tempered cubic lattice of 7-limit tetrads

I've often mentioned the cubic lattice of 7-limit tetrads:

http://66.98.148.43/~xenharmo/sevlat.htm

If we temper, we introduce equivalences between tetrads. In
particular, the otonal tetrad with root 2401/2400 is represented by (2
3 -3), and
for breed tempering we want a unique representative modulo this
equivalence. One way to do this is to use [a b] = (a a b) as the
unique representative; there is one and only one such chord in each
breed equivlance class. We then have that (1 0 0) ~ [3 -3], (0 1 0) ~
(-2 3) and of course (0 0 1) ~ [0 1], from which we may find the
unique representative.

The cubic lattice has the property that two tetrads share an interval
iff they are at a distance of one apart. If we define a norm by

|| [a b] || = sqrt(12a^2 + 20ab + 9b^2)

then two tetrads share an interval iff they are at a distance of three
apart. However, three is not the mininal lattice distance. The
situation is very much reminiscent of the breed plane, where the
haexany projects to a hexagon. Here the set of chords sharing an
interval with a central chord, at the verticies of an octahedron in
the cubic lattice, projects to a hexagon. We have two chords at a
distance of one from [0 0], namely [-1 1] and [1 -1], and two chords
at a distance of 2sqrt(2), [-1 2] and [1 -2], before be get to the six
chords at a distance of three, [0 -1], [- 1], [-2 3], [2 -3], [-3 3],
[3 -3]. It should be noted that [a b] represents an otonal tetrad if b
is even, and a utonal tetrad if it is odd, so we have two utonal
tetrads at a distance of one from each otonal tetrad, and vice-verse,
two of the same kind at a distance of 2sqrt(2), and the six at a
distance of three must be of opposite kind, sharing an interval.

The otonal chord with root 225/224 is represented by (1 1 4), which is
already of the form (a a b), so it is breed represented by [1 4]. If
we reduce [a b] modulo [1 4], we can reduce to [0 b-a], which maps
everything down to a single line of chords; this is the miracle line
of chords, since of course putting together 2401/2400 and 225/224
gives miracle.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
We have two chords at a
> distance of one from [0 0], namely [-1 1] and [1 -1], and two chords
> at a distance of 2sqrt(2), [-1 2] and [1 -2], before be get to the six
> chords at a distance of three, [0 -1], [- 1], [-2 3], [2 -3], [-3 3],
> [3 -3].

I left off two of the same kind at a distance of two, [-2 2] and
[2 -2].

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> One way to do this is to use [a b] = (a a b) as the
> unique representative; there is one and only one such chord in each
> breed equivlance class.

There are several advantages to making a further coordinate
transformation, so that (a b c) is transformed to <b+c 3a-b+c>. If we
do this, we can use (b-a b-a 2a-b) to serve as a representitive of
<a b>. Then <a b> is otonal when b is even, and utonal when it is odd.
Moreover, the otonal tetrad <1 0> corresponds to (-1 -1 2), which is
the otonal chord with root 60/49, the neutral third. <0 1> is a utonal
tetrad, but <0 2> corresponds to (2 2 -1), which is the otonal tetrad
with root 343/240, which modulo 2401/2400 is the same as 10/7.
Similarly, 60/49 is the same as 49/40. Hence <1 0> corresponds to root
49/40, and <0 2> to root 10/7, and we have coordinates corresponding
to the {49/40,10/7} Hermite reduced coordinates I've been using for
the breed plane.

Moreover, the geometry is no longer skewed. The utonal tetrads sharing
a common interval with <0 0> are the four <+-1 +-1> tetrads, plus
<0 +-3>, we can consider this to lie on a hexagon or an ellipse, as we
prefer.

I think I'll nominate these as the canonical breed tetrad coordinates.