Symmetric 5-limit and 7-limit distance measures

Suppose we look at octave equivalence classes in the 5-limit, which
we may represent by vectors [a, b] (representing the class containing
element 3^a 5^b.) Then we define the quadratic form q5 by setting

q5([a, b]) = a^2 + ab + b^2.

The square root of q5 is an ordinary Euclidean distance to the origin
[0, 0], and we may define distance between any two classes v1, v2 by

d(v1, v2) = ||v1 - v2|| = sqrt(q5(v1, v2)).

The 5-limit note classes now defines the regular tessellation of the
plane by equilateral triangles, each triangle representing a triad.
(The dual tessellation by hexagons represents the 5-limit harmonic
relationships between triads.)

We may define interesting, highly symmetric scales by picking a
center point m and taking everything in a sphere of radius r around
m. The simplest example is a triad, where we can take

q5(v - [1/3, 1/3]) <= 1/3,

giving 1 - 5/4 - 3/2.

If we take everything in a sphere of radius 1 around [0, 0] we get

1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2),

while q5(v - [1/3, 1/3]) <= 4/3 gives

1 - 6/5 - 5/4 - 3/2 - 5/3 - 15/8 - (2).

Similarly, we may define

q7([a, b, c]) = a^2 + b^2 + c^2 + ab + ac + bc.

If we take

q7(v - [1/4, 1/4, 1/4]) <= 1/4

we get 1 - 5/4 - 3/2 - 7/4. More interesting is what happens when we
center at [1/2, 1/2, -1/2]:

q7(v - [1/2, 1/2, -1/2]) <= 1/2

gives us

1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2).

This harmonic octahedron is very well supplied with harmony--while it
has only six notes, corresponding to the six verticies of an
octahedron, it has eight chords, corresponding to the eight faces of
an octahedron. (If you've seen D&D dice, the 8-sided dice are
octahedra.)

Concentrating on the octahedra seems to be the best way of
visualizing 7-limit harmony in this symmetrical form. We get figures
like

21/20 ----- 3/2 ----- 15/14
|
| 7/4 5/4
| (6/5) (12/7)
|
7/5 ----- 1 ------ 10/7

Paul or someone could draw something much better, but the 1, 3/2,
15/14, 10/7, 7/5 and 21/20 are all in a plane, the 7/4 and 5/4 are
above, forming the top verticies of two octahedra, and the 6/5 and
12/7 are on the bottom, completing the octahedra. You can also see
the tetrahedron 1 - 5/4 - 3/2 - 7/4 in there if you squint hard. This
may all have something to do with what Monzo is working on.

--- In tuning-math@y..., genewardsmith@j... wrote:
> Suppose we look at octave equivalence classes in the 5-limit, which
> we may represent by vectors [a, b] (representing the class containing
> element 3^a 5^b.) Then we define the quadratic form q5 by setting
>
> q5([a, b]) = a^2 + ab + b^2.
>
> The square root of q5 is an ordinary Euclidean distance to the origin
> [0, 0], and we may define distance between any two classes v1, v2 by
>
> d(v1, v2) = ||v1 - v2|| = sqrt(q5(v1, v2)).
>
> The 5-limit note classes now defines the regular tessellation of the
> plane by equilateral triangles, each triangle representing a triad.
> (The dual tessellation by hexagons represents the 5-limit harmonic
> relationships between triads.)
>
> We may define interesting, highly symmetric scales by picking a
> center point m and taking everything in a sphere of radius r around
> m. The simplest example is a triad, where we can take
>
> q5(v - [1/3, 1/3]) <= 1/3,
>
> giving 1 - 5/4 - 3/2.
>
> If we take everything in a sphere of radius 1 around [0, 0] we get
>
> 1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2),

What happened to 8/5? This _should_ be the 5-limit Tonality Diamond.

> More interesting is what happens when we
> center at [1/2, 1/2, -1/2]:
>
> q7(v - [1/2, 1/2, -1/2]) <= 1/2
>
> gives us
>
> 1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2).

That's the hexany. You should familiarize yourself with CPS (Combination Product Set) scales . . . the hexany is the 2)4
(1,3,5,7) hexany, meaning it's the set of numbers you get when you take products of 2 numbers at a time out of {1,3,5,7}. The
3)6 (1,3,5,7,9,11) and 3)6 (1,3,7,9,11,15) are called Eikosanies, and are similarly symmetrical in the equilateral-triangular
lattice. So is the 2)5 (1,3,5,7,9) dekany, when 9 gets its own axis -- Dave Keenan created a splendid rotating dekany that
actually plays the notes as the dekany rotates in 4-dimensional space. We went through a lot of interesting mathematics
on the tuning list -- for example, we got in touch with George Olshevsky who has a vast knowledge of polychora in 4-, 5-,
and 6-dimensional space.
>
> Concentrating on the octahedra seems to be the best way of
> visualizing 7-limit harmony in this symmetrical form. We get figures
> like
>
> 21/20 ----- 3/2 ----- 15/14
> |
> | 7/4 5/4
> | (6/5) (12/7)
> |
> 7/5 ----- 1 ------ 10/7
>
> Paul or someone could draw something much better, but the 1, 3/2,
> 15/14, 10/7, 7/5 and 21/20 are all in a plane, the 7/4 and 5/4 are
> above, forming the top verticies of two octahedra, and the 6/5 and
> 12/7 are on the bottom, completing the octahedra. You can also see
> the tetrahedron 1 - 5/4 - 3/2 - 7/4 in there if you squint hard.

As well as the tetrahedron 1 - 6/5 - 3/2 - 12/7, the so-called "minor tetrad".

> This
> may all have something to do with what Monzo is working on.

Hmm . . . the triangular lattice is the way _I_ do things . . . Monz uses a different lattice definition (with different angles),
and only shows the connections corresponding to _primes_ -- so intervals like 5:3 don't get a connector.

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> > If we take everything in a sphere of radius 1 around [0, 0] we get
> >
> > 1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2),

> What happened to 8/5? This _should_ be the 5-limit Tonality Diamond.

What happened is that I forgot to put it in. Maybe I should have run
a computer program. :)

> > More interesting is what happens when we
> > center at [1/2, 1/2, -1/2]:
> >
> > q7(v - [1/2, 1/2, -1/2]) <= 1/2
> >
> > gives us
> >
> > 1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2).
>
> That's the hexany. You should familiarize yourself with CPS
(Combination Product Set) scales . . . the hexany is the 2)4
> (1,3,5,7) hexany, meaning it's the set of numbers you get when you
take products of 2 numbers at a time out of {1,3,5,7}.

I've been thinking of it as an octahedron for the last quarter-
century, but by whatever name it is a fundamental doodad in the 7-
limit. Does anyone but me think of the hexany as a regular
octahedron, and the 7-limit note-classes as forming the (unique)
semiregular 3D honeycomb, with cells consisting of tetrahedra and
octahedra, and vertex figure the cubeoctahedron? Or is the hexany a
purely combinatorial idea, s your definition suggests?

The
> 3)6 (1,3,5,7,9,11) and 3)6 (1,3,7,9,11,15) are called Eikosanies,
and are similarly symmetrical in the equilateral-triangular
> lattice.

Hmmm. In the first place, the hexany is not in an equilateral-
triangle lattice, but a 3D semiregular lattice, at least in my
metric. In the second place, you have both 3 and 9 in the above list,
so it can't be entirely symmetrical.

So is the 2)5 (1,3,5,7,9) dekany, when 9 gets its own axis -- Dave
Keenan created a splendid rotating dekany that
> actually plays the notes as the dekany rotates in 4-dimensional
space.

3 and 9 have separate generators, you figure?

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > > If we take everything in a sphere of radius 1 around [0, 0] we get
> > >
> > > 1 - 6/5 - 5/4 - 4/3 - 3/2 - 5/3 - (2),
>
> > What happened to 8/5? This _should_ be the 5-limit Tonality Diamond.
>
> What happened is that I forgot to put it in. Maybe I should have run
> a computer program. :)
>
> > > More interesting is what happens when we
> > > center at [1/2, 1/2, -1/2]:
> > >
> > > q7(v - [1/2, 1/2, -1/2]) <= 1/2
> > >
> > > gives us
> > >
> > > 1 - 15/14 - 5/4 - 10/7 - 3/2 - 12/7 - (2).
> >
> > That's the hexany. You should familiarize yourself with CPS
> (Combination Product Set) scales . . . the hexany is the 2)4
> > (1,3,5,7) hexany, meaning it's the set of numbers you get when you
> take products of 2 numbers at a time out of {1,3,5,7}.
>
> I've been thinking of it as an octahedron for the last quarter-
> century, but by whatever name it is a fundamental doodad in the 7-
> limit. Does anyone but me think of the hexany as a regular
> octahedron,

Everyone does. You should have been on the tuning list for the last year. Erv Wilson drew this octahedron in the 60's, I
believe.

> and the 7-limit note-classes as forming the (unique)
> semiregular 3D honeycomb, with cells consisting of tetrahedra and
> octahedra,

I've been drawing those for at least 10 years now. Didn't you see, for example, the lattices I just posted for Rami Vitale's
scale (ASCII) and Justin White's scale (.gif) on the tuning list?

> and vertex figure the cubeoctahedron?

I think George Olshevsky explained "vertex figure" to us. And the cuboctahedron is the shape of the 7-limit Tonality
Diamond, etc, etc. . . .

> Or is the hexany a
> purely combinatorial idea, s your definition suggests?

Combinatorial _and_ geometrical.
>
> The
> > 3)6 (1,3,5,7,9,11) and 3)6 (1,3,7,9,11,15) are called Eikosanies,
> and are similarly symmetrical in the equilateral-triangular
> > lattice.
>
> Hmmm. In the first place, the hexany is not in an equilateral-
> triangle lattice, but a 3D semiregular lattice, at least in my
> metric.

What I mean is that the lattice has a lot of equilateral triangles in it. Of course it has a lot of squares in it too, but I'm just
trying to distinguish it, in layman's terms, from the Cartesian square lattice, which is common too (Euler and Fokker come
to mind).

> In the second place, you have both 3 and 9 in the above list,
> so it can't be entirely symmetrical.

9 gets its own axis, and is _not_ assumed to equal 3*3.
>
> So is the 2)5 (1,3,5,7,9) dekany, when 9 gets its own axis -- Dave
> Keenan created a splendid rotating dekany that
> > actually plays the notes as the dekany rotates in 4-dimensional
> space.
>
> 3 and 9 have separate generators, you figure?

Well, one can think of it either way, but the figure is more symmetrical when one puts 3 and 9 on different axes. The
problem is that there are a couple of 9-limit consonances that don't show up as direct connections. But with the
alternative, having only axes for prime numbers, one doesn't have direct connections for _any_ of the ratios of 9. One
solution is to use a taxicab metric, along with "wormholes" (search the tuning list archives).

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> I've been drawing those for at least 10 years now. Didn't you see,
for example, the lattices I just posted for Rami Vitale's
> scale (ASCII) and Justin White's scale (.gif) on the tuning list?

I did, and it looked suspiciously regular (and hence familiar) to me;
that's why I suggested you could do a better job of drawing it.

> What I mean is that the lattice has a lot of equilateral triangles
in it. Of course it has a lot of squares in it too, but I'm just
> trying to distinguish it, in layman's terms, from the Cartesian
square lattice, which is common too (Euler and Fokker come
> to mind).

Do people ever look at the reciprocal lattices--the hexagons in the 5-
limit, and in the 7-limit the bee honeycomb of rhombic dodecahedra,
which is what you get by squashing the spheres in a regular close-
packing together. These are lattices of chords.

> Well, one can think of it either way, but the figure is more
symmetrical when one puts 3 and 9 on different axes.

I've never found an ideal solution to this either, though you can
certainly make the 3 half as big as 5,7,9,11 and 13.

--- In tuning-math@y..., genewardsmith@j... wrote:
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
>
> > I've been drawing those for at least 10 years now. Didn't you
see,
> for example, the lattices I just posted for Rami Vitale's
> > scale (ASCII) and Justin White's scale (.gif) on the tuning list?
>
> I did, and it looked suspiciously regular (and hence familiar) to
me;
> that's why I suggested you could do a better job of drawing it.
>
> > What I mean is that the lattice has a lot of equilateral
triangles
> in it. Of course it has a lot of squares in it too, but I'm just
> > trying to distinguish it, in layman's terms, from the Cartesian
> square lattice, which is common too (Euler and Fokker come
> > to mind).
>
> Do people ever look at the reciprocal lattices--the hexagons in the
5-
> limit, and in the 7-limit the bee honeycomb of rhombic dodecahedra,
> which is what you get by squashing the spheres in a regular close-
> packing together. These are lattices of chords.

So far, plotting notes has been more useful -- it encapsulates all
the information in terms of the simplest elements musicians must deal
with: individual tones.
>
> > Well, one can think of it either way, but the figure is more
> symmetrical when one puts 3 and 9 on different axes.
>
> I've never found an ideal solution to this either, though you can
> certainly make the 3 half as big as 5,7,9,11 and 13.

Let's take a step back. If we don't assume octave equivalence, then
the Tenney lattice, with a taxicab metric, is quite wonderful.
There's an axis for each prime, and distance between rungs along each
is log(p). Then the Tenney "harmonic distance" (HD) between two notes
turns out to be log(n*d) where n/d is the ratio between the notes.
That's quite wonderful -- the harmonic entropy of the simpler ratios
n/d turns out to behave very much like log(n*d).

Now when we assume octave equivalence . . . and use 9-limit or
higher . . . things get hairy.