Yahoo Tuning Groups Ultimate Backup tuning a pair of unanswered questions

Even if I'm not mathematician, I hope to contribute answering the
unanswered Dan's questions. I shall write my answer in parts on distinct
posts from time to time. (I don't have much time to write).

I would like, first, to thank Paul Erlich who invited me, after "my last
message", to compare periodicity blocks with gammiers. I explored this link
particularly in january and found a great richness permitting to elucidate
questions about complexity dimension. I'm working with that to a theory of
convex bodies in primal lattices.

I would like also to thank Dan Stearns for his questions. I mentioned at
end of "my last message" the need for questions. I needed the Dan's
questions to organize simply, in a flash, a good part of my research. I
don't have time to write a paper now, so it's an opportunity to order
partially a profusion of discoveries.

The first part is like an introduction. Subsequently I shall write, by
example, about generalized Stern-Brocot tree in lattices. At the end of
this part it will be possile to give a better idea of the subjects to come.

Part I

I recall the first question of Dan Stearns in the message 18294 :

<<
(1) Given any arbitrary three-term scale (i.e., a scale made up of "a"
amount of small steps, "b" amount of medium steps, and "c" amount of
large steps), what global rule could govern their arrangement so as to
insure that the resulting scale is trivalent (i.e., has three sizes of
intervals in every unique interval class)?
>>

Let A < B < C be rational numbers corresponding to the steps such that

a log A + b log B + c log C = log 2

In rational systems these scales are subsets in abelian groups <2,p> or
<2,p,q> which contain all integer power combinations of the factors between
the <>. However I will neglect the case <2,p> for the complexity level in
<2,p> has to be too high to be musically pertinent. So the number of steps,
here, equals the base dimension. The numbers p and q are, in theory, simply
independant rational numbers but they are pratically small prime numbers
like 3, 5, 7, 11, 13, ...

We search here global relations independant of the prime numbers used,
therefore we search consistency in modular relations. By isomorphy, all
intervals are represented in the base [log 2 log p log q] by a unique
coordinate vector (x1,x2,x3). (This property results from equality
mentioned about steps and coordinates). I note also the coordinates by (x1
x2 x3)* where (*) indicate, as a transpose sign, that the vector is a
column matrix.

We know soon the number of classes (a + b + c), the degree of octave. How
are determined these classes? The classes are simply the quotient-group
<2,p,q>/<r,s> were r and s are the "commas" of Hellegouarch, or the "unison
vectors" of Fokker, or the "modal srutis" in my terminology. The group
<r,s> is the subgroup of <2,p,q> generated by the linear power combinations
of the vectors r and s.

By equality of dimensions (here) and for we know the order of the steps,
the modal srutis r and s are easily determined. It's simply B - A and C - B
represented by the vectors

(B1-A1 B2-A2 B3-A3)* = (r1 r2 r3)*

(C1-B1 C2-B2 C3-B3)* = (s1 s2 s3)*.

Let X = (x y z)* represent any interval. The image D of X (degree of X) by
the epimorphism applying intervals on Z corresponds to the determinant of
the matrix (X s r).

D = D(X) = det (X s r) =

(x s1 r1)
det (y s2 r2) = x det (s2 r2) + y det (s3 r3) + z det (s1 r1)
(z s3 r3) (s3 r3) (s1 r1) (s2 r2)

So D has the form x D1 + y D2 + z D3.

We could say also that D is the scalar product of X by (D1 D2 D3)* which is
the vectorial product of s and r.

If X = (1 0 0)* then D = D1, the degree of the octave (or the number of
classes modulo 2), for this X represents 2.

If X = (0 1 0)* then D = D2, the degree of p, for this X represents p.
If X = (0 0 1)* then D = D3, the degree of q, for this X represents q.

Examples:

Let A < B < C be 9/8 < 8/7 < 7/6 represented in [log 2 log 3 log 7] by

(-3 2 0)* < (3 0 -1)* < (-1 -1 1)*

Then r and s correspond to (6 -2 -1)* and (-4 -1 2)* which are the modal
srutis 64/63 and 49/48. The degree function is

(x -4 6)
D(X) = det (y -1 -2) = 5x + 8y + 14 z
(z 2 -1)

So in the slendro family, there are 5 degrees in which the harmonic 3
corresponds to degree 3 (for 8 mod 5 = 3) and the harmonic 7 to degree 4
(for 14 mod 5 = 4). And we can verify that A, B and C have degree 1 :

D(9/8) = D(-3,2,0) = (5)(-3) + (8)(2) + (14)(0) = 1
D(8/7) = D(3,0,-1) = (5)(3) + (8)(0) + (14)(-1) = 1
D(7/6) = D(-1,-1,1) = (5)(-1) + (8)(-1) + (14)(1) = 1

---

Let A < B < C be 16/15 < 10/9 < 9/8 represented in base log <2,3,5> by

(4 -1 -1)* < (1 -2 1)* < (-3 2 0)*

Then r and s correspond to (-3 -1 2)* and (-4 4 -1)*, which are the modal
srutis 25/24 and 81/80. Then D(x,y,z) is

(x -4 -3)
D(X) = det (y 4 -1) = 7x + 11y + 16 z
(z -1 2)

So in the Zarlino family, there are 7 degrees in which the harmonic 3
corresponds to degree 4 (for 11 mod 7 = 4) and the harmonic 5 to degree 2
(for 16 mod 7 = 2). Verifying tonal generator we find :

D(16/15) = D(4,-1,-1) = (7)(4) + (11)(-1) + (16)(-1) = 1
D(10/9) = D(1,-2,1) = (7)(1) + (11)(-2) + (16)(1) = 1
D(9/8) = D(-3,2,0) = (7)(-3) + (11)(2) + (16)(0) = 1

---

Returning to the question, the following relation express the independance
of the numbers 2, p and q. Let (A1 A2 A3)* (B1 B2 B3)* (C1 C2 C3)* be the
coordinates of A, B and C. Then

(A1 B1 C1) (a) (A1*a + B1*b + C1*c) (1)
(A2 B2 C2) (b) = (A2*a + B2*b + C2*c) = (0)
(A3 B3 C3) (c) (A3*a + B3*b + C3*c) (0)

To reach the octave, an equal amount of p (or q) have to be used in
numerator and denominator while an excedent 2 is required. This is the
basic constraint.

----------

So far it's only trite remarks. Now concerning

<< to insure that the resulting scale is trivalent (i.e., has three sizes
of intervals in every unique interval class) >>

I could say that we have just seen there exist an infinity of intervals for
each classes and that the question is to determine the three pertinent
intervals for each classes (except unison). I will show that they are
contained in a convex shape (easily calculated) around the unison.

We could question the necessity to use three values for each classes but
without using the gammier theory, this level is out of reach (I want here
to restrict my comments to Z-module properties).

This new subject is more sophisticated so I will begin with remarks to give
a better contextualization.

---

First we have to note that, independantly of prime numbers used (that we
suppose to be not too high), it subsists in lattice (or Z-module)
representation, a component of the global complexity that I will name, for
the time being, "modular complexity". For each interval represented by
(x,y,z) this modular complexity corresponds to the sum of the absolute
value of the coordinates.

MC = MC(X) = |x| + |y| + |z|

So MC is 0 for unison (0 0 0)* and grows up in any direction. Knowing prime
numbers, we could adjust relative rate of growing in different axis and
calculate complexity or sonance :

complexity = 2^|x| * p^|y| * q^|z|

sonance = |x|*log 2 + |y|*log p + |z|*log q

But independently of primal base used we can understand the importance to
use for each class of intervals those of minimal complexity which are
forcely located around the unison. We are not interested here in metric of
complexity but in topological invariance using MC in place of complexity.

( Beside, I suggest than N-limit is more pertinent with
complexity than with odds or primes. Convexity has really
sense with complexity in constrast with the two others.
Convexity means constraint to use all possibilities
within the limit. It would be easy to show, by some
examples, like slendro scales, the non-pertinence of
convexity for odds and primes, and I found that the
pertinent tonal structures are convex bodies.)

As alternative to MC, we can use MC2 = sqrt (xx + yy + zz), the usual
euclidian norm, in plunging our Z-module in R^3, to analyse ball containers
of intervals behind a limit of complexity. It's particularly helping in
dimension > 3 to figure a parallelotope having his vertex belonging to an
hyperquadric.

Knowing primal base we could represent by a spherical body (in R^3) the
domain containing the set of discrete intervals (of Z^3) having a maximal
distance limit from unison. This distance (MC2) is a represention of the
"sonance". (We have to use the log of complexity, the sonance, to be linear
with MC2).

This container spherical shape is not preserved by the representation in
Z^3 but we expect to obtain a container ellipsoidal shape with

1/log 2 : 1/log p : 1/log q

axis ratios and we expect that this shape be easily described by an
equation using a "diagonal" matrix M such that

(x y z) M (x y z)* = K

---

Another important remark concerns the effect of octave modularity on MC and
MC2. Independently of primal base, octave modularity means using <p,q>
rather than <2,p,q>. Thus we have to use only "reduced modular complexity"
with RMC = |y| + |z| and RMC2 = sqrt (yy + zz). So we can easily understand
that RMC and RMC2 are globaly more reduced relatively to MC and MC2 if

yz > 1 rather than yz < 1.

For example the intervals 15/8 and 5/3 have the same RMC = 2, but they have
respectively 5 and 2 for the MC ; the intervals 45/32 and 9/5 have also the
same RMC = 3, but respectively 8 and 3 for the MC.

By principle, if y and z in (y,z) have the same sign we can expect a great
value of |x| to obtain an interval of the first octave.

So a cube domain (having vertex on axis) with norm MC < K in <2,p,q> would
appear as a compressed hexagonal shape rotated counterclockwise in <p,q>.
We expect, with usual euclidian norm, that our ellipsoid container would
appear also like a counterclockwise rotated ellipse responding to the equation

(1 y z) M (1 y z)* = K

But the matrix M would be no more diagonal and would have the form

(1 0 0)
M = (0 a d)
(0 d c)

(1 0 0)
(1 y z) (0 a d) (1 y z)* = K
(0 d c)

axx + bxy + cyy = K (where b = 2d)

---

I will anticipate on arguments here to give now examples insuring that this
abstracted matter has sense in tonal structures.

By periodicity, the vectors s, r+s, r, -s, -r-s and -s determine 6 limit
points of complexity. (We will develop that later, and we will use the
convex hexagonal outline to study the algebraic properties). But we can
exhibit now the elliptic container of complexity determined by A, B and C
which satisfied to the six points. By symmetry, the s, s+r and r values are
sufficient to calculate this ellipse.

In the first example we had r = (6 -2 -1)* and s = (-4 -1 2)*. So the
points are (-1,2) (-3,1) (-2,-1) and we have to solve :

a - 2b + 4c = K
9a - 3b + c = K
4a + 2b + c = K

We find easily the quadratic form 3xx + 3xy + 7yy = K and we may verify
that the six points give K = 25 the square of classes.

In the second example we had r = (-3 -1 2)* and s = (-4 4 -1)*. So the
points are (4,-1) (3,1) (-1,2) and we have to solve :

16a - 4b + c = K
9a + 3b + c = K
a - 2b + 4c = K

We find easily the ellipses family 3xx + 3xy + 13yy = K and we may verify
that the six points give K = 49 the square of classes.

( We can note, in this case, that the elliptic shape
contains erroneously the points (4,0) and (-4,0)
with K = 48 which are 81/64 and 128/81. So the
approximative euclidian norm fail here. Strictly,
only the hexagonal shape inside, as we will see,
is truly periodical -- while ellipses may overlap --
and contains exclusively the three intervals of
minimal complexity for each classes. But this
hexagonal shape has his vertex located on the
calculated ellipse. And all that is generalizable
at N dimensions. )

By anticipation we presumed here that the set of intervals included in the
hexagonal shape corresponds strictly to the three periodical blocks of less
complexity. These blocks of one interval by class surround unison and have
only unison as intersection, so we have with N = a + b + c classes only the
3N-2 intervals searched. With gammier theory we could go deeper and
consider, with algebraic arguments, lesser numbers N of intervals included
in the first shape and see that these subsets are also convex bodies.

The two precedent examples permit to see how the ideal N-limit ball of
sonance (which has, by definition of sonance, a cubic form in R^3 with his
vertex on primal axis) is approximately approched by systems with
epimorphism to Z (i.e. having classes). The hexagonal shape correspond to
the sonance parallepiped defined by sonance limit of the modal srutis.

The following figures show an image of our slendro and zarlino examples. We
see the hexagonal shape defined by r and s vectors and the three periodical
blocks of 5 and 7 distinct classes surrounding unison. The intervals of
degree 1 and -1 are colored. In the second example, we distinct the 13
intervals of the Zarlino gammier among the complete 19 intervals gammier
(using harmonic 27 and 45). So we see how modulation at dominant or
sub-dominant keeps the confinement.

http://www.aei.ca/~plamothe/pix/r8.gif

http://www.aei.ca/~plamothe/pix/r7.gif

----------

I stop here the introduction. In the following parts I shall show how a
universal matrix transform successively a given periodical block in the
other surrounding blocks. In dimension N, this matrix appears as a root N+1
of the identity matrix. If we change negative signs in this matrix by
positive signs, the successive powers exibit N-bonacci series and we have
the analogy of Fibonacci in 2D. We can also construct an analog of Stern-
Brocot tree with N branches.

Beside, I found a simple method to represent 3D and 4D lattices using
Tribonacci series. It's so simple I imagine it's already known. If not
I shall show that.

I don't give other details. The matter grows each day and I did'nt take
time to organize.

Pierre Lamothe

Hi Pierre,

Your post looks very interesting, and I've gone through the first part of it fairly
slowly.

Here is how I understand the main points - maybe it will help others
on the TL who are relatively new to the idea of a periodicity block, and maybe will help
give another perspective:

Notation:
lowercase for numbers of steps, upper case for step sizes.
a steps of size A, b of size B, and c of size C adding up to the octave.

<2,p,q> = all numbers of form 2^x1*p^x2*q^x3 where x1, x2 and x3 can be positive or negative
exponents.

Ex. For j.i. diatonic, p=3, q = 5, and <2,p,q> = all numbers of form 2^x1*3^x2*5^x3.

(for those completely new to this idea, note that this definition automatically givies
you only one representative for each ratio. E.g. 6/5 and 12/10 would both be
expressed as 2^1*3^1*5^-1.)

We can use vector notation:
j.i. diatonic
Notes:
1/1 3^2/2^3 5/2^2 2^2/3 3/2 5/3 3*5/2^3 (2/1)
=
(0,0,0), (-3, 2, 0), (-2, 0, 1), (2, -1, 0), (-1, 1, 0), (0, -1, 1), (-3, 1, 1), (1, 0, 0)

e.g. (0,-1,1) short for (3^-1)*(5^1), i.e. 5/3.

Now, let r = (r1,r2,r3) and s = (s1,s2,s3) be two fixed vectors, corresponding to the
Fokker unison vectors.

We need a third vector to make a parallelepiped in 3D, call this X = (x,y,z)

Then the volume of the parallelepiped with vertices (0, r, s, X, r*s, s*X , r*X, r*s*X)
is given by the determinant (this is a well known result)
> (x s1 r1)
> det (y s2 r2) = x det (s2 r2) + y det (s3 r3) + z det (s1 r1)
> (z s3 r3) (s3 r3) (s1 r1) (s2 r2)

since it is a unit lattice, each point occupies unit volume, so the volume of the parallelopiped
gives the number of lattice points in it, wherever it is located
in the lattice (with certain conventions on how one treats points on the boundaries
of it). (also a well known result, but not quite so much so as the volume of the parallelopiped).

Since the vectors that define the parallelopiped are also vectors joining lattice points, then in fact,
the determinant gives the number of distinct points in the lattice up to equivalence
by its three defining vectors.

So far, this is just a brief summary of the notion of a periodicity block.

Now, the interesting move, you want to make a scale with three step sizes
A, B and C, where A, B and C are expressible in terms of (2,p,q).

So you choose r = B-A, s = C-B.

where here we are talking about vector subtraction,

r= (B1-A1 B2-A2 B3-A3)* = (r1 r2 r3)*

s= (C1-B1 C2-B2 C3-B3)* = (s1 s2 s3)*.

As ratios, r=B/A, and s = C/B

In other words, you are treating lattice points as
equivalent if you can get from one to another
via B/A, or via C/B.

So for example if the steps are 9/8, 10/9 and 16/15
and if you start the scale with 10/9, then you have to leave
out 9/8 and 16/15, because they are equivalent to
10/9 under the unison vectors r and s.

Continuing with the diatonic j.i. steps
9/8, 10/9 and 16/15, and these values
of r, s, and using (x,y,z) = X as a third unison
vector, one can find the determinant from these.

Your determinant equation gives

D(X) = 7x + 11y + 16 z

so we can immediately read off from this the
number of notes you get using any other
type of equivalence.

So for octave equivalence, x=1, y=0, z=0,
so D(X)=7, so one has 7 notes.

For equivalence of 1/1 and 3/1
as in BP type scales, you will have
X = (0,1,0) (i.e. 2^0*3^1*5^0)
and so x=0, y=1, z=0.

So this time, we have 11 notes.

Similarly, equivalence under 5/1 gives a scale of 16 notes.

Here, there is one point I don't quite get.
You deduced that the 3/1 will be the fourth
degree of the diatonic scale, and I don't quite
see where you get that.

> So in the Zarlino family, there are 7 degrees in which the harmonic 3
> corresponds to degree 4 (for 11 mod 7 = 4) and the harmonic 5 to degree 2
> (for 16 mod 7 = 2). Verifying tonal generator we find :

I can see that 11 mod 7 = 4, but not the connection of this to
the fourth degree of the scale.

Another example:

for 9/8, 8/7 and 7/6,
treating 9/8
D(X) = 5x + 8y + 14 z
so one can read out that there are 5 notes up to octave equivalence.
(and 8 up to equivalence under 3/1, and 14 up to equivalence under
7/1)

One could also put x=-1, y=1, D(x)=3, so there are 3 notes up to
equivalence by 3/2, and so forth.

Actually finding what these three notes seems a bit tricky.

(this will sometimes give negative numbers for D(X), but that is
okay, as the volume is defined as abs(D(x)), and is always positive).

Are there any nice easy ways to find these three notes, or any of the scales,
without doing a lot of 3D geometry?

That's about as far as I've got, though I've looked ahead, and
see you haven't yet given the trivalence result, so look forward to see
how it works out.

I expect some on the TL may have immediately understood what you
are saying, but for others, this area may also be relatively new, or your
approach to it is new, or some of the notation and terminology,
so maybe what I've said here will help them to ask the right questions
to understand the rest of what you have said.

I look forward to what you have to say about trivalence. I can't at
present see how trivalence is going to emerge from this, and if you
can do that, I'm sure it will be fascinating!

Robert

a pair of unanswered questions

🔗D.Stearns <STEARNS@CAPECOD.NET>

🔗Pierre Lamothe <plamothe@aei.ca>

🔗Paul H. Erlich <PERLICH@ACADIAN-ASSET.COM>

🔗Robert Walker <robert_walker@rcwalker.freeserve.co.uk>

🔗D.Stearns <STEARNS@CAPECOD.NET>