Apparently the math stuff belongs here, so I am reposting it in case

anyone wants to follow up.

--- In tuning@y..., kalleaho@m... wrote:

> I understand that with microtemperament one can achieve greater

> number of consonant intervals in a scale that is originally tuned

in

> strict Just Intonation.

> This can be a great method for generating new and interesting

scales

> but doesn't it destroy the structure of the original scale?

When you finally get to the point where you can't tell the

difference, the question becomes moot. However, the answer is yes--

introducing approximations increases the flexibility of harmonic

relationships and thereby changes the structure. The question is

really best understood in terms of group theory.

The intervals of any form of just intonation are by definition

positive rational numbers, and so an element of the abelian group of

positive rationals under multiplication (since our hearing, and hence

musical structure, is multiplicative.) We are never really interested

in all rational numbers, but can content ourselves with a finitely

generated subgroup--for one easy example, the group G of all the

numbers of the form 2^a * 3^b * 5^c, where a, b and c are integers,

are the numbers generated by the first three primes, 2, 3 and 5.

An approximate tuning system can very often be seen as a (subset of

a) group homomorphism to a group of smaller rank, and most

significantly to a group of rank 1. For another easy example, the

rank 3 free group G can be sent to a rank 1 free group by a

homomorphism h12(2) = 12, h12(3) = 19, h12(5) = 28. A subset of 88

contiguous elements of this group associated to notes tuned as usual

by setting g12(1) = 440 hz, g12(2) = 2*440 hz, g12(3) = 2^19/12*440

hz, g12(5) = 2^28/12*440 hz is the standard keyboard.

Any such homomorphism is defined by its kernel, which are the

elements sent to the identity. In the case of h12, the kernel is

spanned by 81/80 (the diatonic comma) and 128/125 (the great diesis),

where we have h12(81/80) = h12(128/125) = 0. A tuning system which

does not contain the diatonic comma in its kernel (and this includes

just intonation!) will have a structure quite different that which

musicians normally expect. On the other hand one that does, such as

what we get from the 19 or 31 tone system, will seem more "normal".

Consider the system h72(2) = 72, h72(3) = 114, h72(5) = 167, h72(7) =

202, h72(11) = 249 (the last two values are irrelevant here, but they

do no harm and they cover the range of primes which makes the 72

system interesting.) This 72 system has a 12 system tuning embedded

in it, so we could suppose that structurally they are very similar.

However, intersecting the kernels shows they are remotely related; in

particular h72(81/80) = 1 and h72(128/125) = 3, the second is not so

important but the first shows that the 72 system is fundamentally

different in structure from the 12 system. In contrast, h31(2) = 31,

h31(3) = 49, h31(5) = 72, h31(7) = 87 and h31(11) = 107 *does* have

the property that h31(81/80) = 0; and while h31(128/125) = 1 we still

find h31 is much closer in structre to h12 than is h72.

--- In tuning-math@y..., genewardsmith@j... wrote:

>

> Any such homomorphism is defined by its kernel, which are the

> elements sent to the identity. In the case of h12, the kernel is

> spanned by 81/80 (the diatonic comma) and 128/125 (the great diesis),

> where we have h12(81/80) = h12(128/125) = 0.

What you call the "kernel" is what we call the set

of unison vectors. I have a feeling your conception

will prove to be more useful, since we've been

using language like, "well, you could say that the

two unison vectors of the 5-limit 12-tone

periodicity block are the syntonic comma (81/80)

and the small diesis (128/125), but you could also

say that they're the syntonic comma and the

diaschisma (2025/2048)", for example. We're well

aware that any valid set comes from any other

valid set simply by "adding" and "subtracting" the

unison vectors from one another. But we've run

into some pathological cases -- for example, the

small diesis (128/125) and the schisma (32805/

32768), while they can be derived from the same

two unison vectors, define a periodicity block with

24, instead of 12, notes . . . and not a well-behaved

periodicity block at that. Any insights?

> A tuning system which

> does not contain the diatonic comma in its kernel (and this includes

> just intonation!) will have a structure quite different that which

> musicians normally expect. On the other hand one that does, such as

> what we get from the 19 or 31 tone system, will seem more "normal".

We're all very cognizant of that, and you should be

aware that many of the musicians on this list are

knee-deep in exploring systems which rely

specifically on having other commas in their

kernal.

>

> Consider the system h72(2) = 72, h72(3) = 114, h72(5) = 167, h72(7) =

> 202, h72(11) = 249 (the last two values are irrelevant here, but they

> do no harm and they cover the range of primes which makes the 72

> system interesting.)

You may know that we've been discussing this

system in great detail, especially a few months ago

on the tuning list, where we found some amazing

21-, 31-, and 41-tone subsets of 72-tET.

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

> What you call the "kernel" is what we call the set

> of unison vectors. I have a feeling your conception

> will prove to be more useful, since we've been

> using language like, "well, you could say that the

> two unison vectors of the 5-limit 12-tone

> periodicity block are the syntonic comma (81/80)

> and the small diesis (128/125), but you could also

> say that they're the syntonic comma and the

> diaschisma (2025/2048)", for example.

My way of saying things has the advantage of being standard

mathematical terminology, which allows one to bring relevant concepts

into play. I had noticed that unison vectors seemed to have something

to do with the kernel, but I couldn't tell if it meant generators of

the kernel or any element of the kernel, and I see by your comments

that no one has really decided!

The kernel of some homomorphism h is everything sent to the identity--

if this is the set of unison vectors then for instance 1 is *always*

a unison vector, since h(1) = 0. On the other hand, unison vectors

could be elements of a minimal set of generators for the kernel. In

this case 81/80, 128/125 and 2048/2025 would all belong in the same

kernel generated by any two of them. Depending on which set of

generators you picked, two of them would be unison vectors and the

other one would not be. 1 and 32805/32768 would also both be in the

kernel, but neither would be unison vectors.

Probably the simplest solution at this point would be to drop the

terminology, but if you don't you need to decide what exactly it

means. By the way, I called 128/125 a great diesis, and you call it a

small diesis. Has this been decided? Which is it, and where does one

go to find out?

We're well

> aware that any valid set comes from any other

> valid set simply by "adding" and "subtracting" the

> unison vectors from one another. But we've run

> into some pathological cases -- for example, the

> small diesis (128/125) and the schisma (32805/

> 32768), while they can be derived from the same

> two unison vectors, define a periodicity block with

> 24, instead of 12, notes . . . and not a well-behaved

> periodicity block at that. Any insights?

It's true that anything in the kernel is obtained by adding and

subtracting, since the kernel of an abelian group homomorphism is an

abelian group. It's not true that any linearly independent set of

kernel elements which span the corresponding vector space over the

rationals as a basis is also a minimal set of generators for the

kernel, and that is what you have discovered.

Let's take 81/80 and 128/125 to start with. I may write these

additively, so that 81/80 = 2^-4 * 3^4 * 5^-1 is written [-4, 4, -1]

and 128/125 becomes [7, 0, -3]. We can put these together into a 2x3

matrix, giving us

[-4 4 -1]

[ 7 0 -3]

If we take the absolute value of the determinants of the minors of

this matrix, we recover the homomorphism:

abs(det([[4, -1],[0,-3]]) = 12, abs(det([[-4, -1],[7,-3]]) = 19, and

abs(det([-4,4],[7,0]))= 28, recovering the homomorphism column vector

[12]

[19]

[28]

from the generators of the kernel. This computation shows these

two "unison vectors" do in fact generate the kernel. If we perform a

similar computation for 128/125 and 32805/32768 we first get the

matrix

[7 0 -3]

[-15 8 1]

The column vector we get from the absolute values of the determinants

of the minors of this is:

[24]

[38]

[56]

In other words, these two define the kernel of a homomorphism to the

24 et division of the octave, which in the 5-limit is "pathological"

in the sense that you have two separate 12 divisions a quarter-tone

apart, and we cannot get from one to the other using relationships

taken from 5-limit harmony, as all of the numbers in the above

homomorphism are even.

--- In tuning-math@y..., genewardsmith@j... wrote:

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

>

> > What you call the "kernel" is what we call the set

> > of unison vectors. I have a feeling your conception

> > will prove to be more useful, since we've been

> > using language like, "well, you could say that the

> > two unison vectors of the 5-limit 12-tone

> > periodicity block are the syntonic comma (81/80)

> > and the small diesis (128/125), but you could also

> > say that they're the syntonic comma and the

> > diaschisma (2025/2048)", for example.

>

> My way of saying things has the advantage of being standard

> mathematical terminology, which allows one to bring relevant concepts

> into play.

Absolutely. That's what I'm hoping for.

> I had noticed that unison vectors seemed to have something

> to do with the kernel, but I couldn't tell if it meant generators of

> the kernel or any element of the kernel, and I see by your comments

> that no one has really decided!

Any n linearly dependent elements of the kernel will generate the kernel, in the context where

the unison vectors are completely ignored or tempered out. Any of these elements can be

called a unison vector of the system. Any such set of n can be called "the set of unison vectors

defining" the system [oops, I realized later that this may not be correct]. In other contexts, the

particular choice is relevant, such as in the construction of Fokker periodicity blocks in JI, and

there, I suppose, it makes a difference which elements you call "the generators of the kernel".

Am I understanding you correctly? [not really -- see below]

>

> Probably the simplest solution at this point would be to drop the

> terminology, but if you don't you need to decide what exactly it

> means. By the way, I called 128/125 a great diesis, and you call it a

> small diesis. Has this been decided? Which is it, and where does one

> go to find out?

There are many lists of interval names out there. I've seen "small diesis" more often than "great

diesis" for this interval -- "large diesis" refers to the difference between four minor thirds and an

octave.

>> We're well

> > aware that any valid set comes from any other

> > valid set simply by "adding" and "subtracting" the

> > unison vectors from one another. But we've run

> > into some pathological cases -- for example, the

> > small diesis (128/125) and the schisma (32805/

> > 32768), while they can be derived from the same

> > two unison vectors, define a periodicity block with

> > 24, instead of 12, notes . . . and not a well-behaved

> > periodicity block at that. Any insights?

>

> It's true that anything in the kernel is obtained by adding and

> subtracting, since the kernel of an abelian group homomorphism is an

> abelian group. It's not true that any linearly independent set of

> kernel elements which span the corresponding vector space over the

> rationals as a basis is also a minimal set of generators for the

> kernel, and that is what you have discovered.

Aha -- so I was missing something above. In the case where the unison vectors are tempered

out or completely ignored, there are several valid sets of _generators for the kernel_, but not all

sets of n linearly independent elements of the kernel generate it.

Can we unambiguously classify all elements of the kernel into those that can generate it and

those that can't?

> Let's take 81/80 and 128/125 to start with. I may write these

> additively, so that 81/80 = 2^-4 * 3^4 * 5^-1 is written [-4, 4, -1]

> and 128/125 becomes [7, 0, -3]. We can put these together into a 2x3

> matrix, giving us

>

> [-4 4 -1]

> [ 7 0 -3]

>

> If we take the absolute value of the determinants of the minors of

> this matrix, we recover the homomorphism:

>

> abs(det([[4, -1],[0,-3]]) = 12, abs(det([[-4, -1],[7,-3]]) = 19, and

> abs(det([-4,4],[7,0]))= 28, recovering the homomorphism column vector

>

> [12]

> [19]

> [28]

>

> from the generators of the kernel.

I think this is a computation that Graham Breed has done or was trying to do. Graham?

> This computation shows these

> two "unison vectors" do in fact generate the kernel. If we perform a

> similar computation for 128/125 and 32805/32768 we first get the

> matrix

>

> [7 0 -3]

> [-15 8 1]

>

> The column vector we get from the absolute values of the determinants

> of the minors of this is:

>

> [24]

> [38]

> [56]

>

> In other words, these two define the kernel of a homomorphism to the

> 24 et division of the octave,

I think you're jumping the gun with that intepretation. If you look at the pitches in the Fokker

periodicity block defined by this matrix (without the 2's column), you'll see 12 roughly

equally-spaced pairs of pitches separated by a syntonic comma (81/80). What is a syntonic

comma in this system? Well, _two_ syntonic commas come out to be equal to the product of

128/125 and 32805/32768. But both of these are unison vectors, and must vanish. Therefore

their product must vanish. So two syntonic commas vanishes, thus one syntonic comma must

either vanish or be consided a half-octave. If the syntonic comma vanishes, then you have

12-tone equal temperament, not 24-tone equal temperament. If the syntonic comma is a

half-octave, well that's kinda weird, but it still seems like you get 12-tone equal temperament . . .

in case you're wondering about that half-octave, let's look an an analogous case where a

half-octave is naturally involved.

By the way, I don't think of 24-tET as pathological at all, I just don't think that one can say, in any

useful sense, that it's _generated_ or _defined_ by 128/125 and 32805/32768.

Here we move on to the 7-limit, and the unison vectors are

[Note: I'm switching from "/" to ":" because of our convention to use ":" for intervals and "/" for

pitches]

49:48 = [-4 -1 0 2]

64:63 = [ 6 -2 0 -1]

225:224 = [-5 2 2 -1]

The absolute values of the determinants of the minors should be

[10]

[16]

[23]

[28]

right? Here, the product of 64:63 and 225:224 is equal to _two_ 10:7s.And here, it's clearly a

half-octave we're dealing with, not something that vanishes. So we really have a 10-tone scale

here, and if all three unison vectors are tempered out uniformly, you get 10-tET. (In a case of

particularly great interest to me, the 49:48 is _not_ tempered out, and you get a 10-tone

system embedded in something very close to 22-tone equal temperament).

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

> There are many lists of interval names out there. I've seen "small

diesis" more often than "great

> diesis" for this interval -- "large diesis" refers to the

difference between four minor thirds and an

> octave.

I use the table in Ellis' appendix to Helmholtz mostly. If there is

nothing official, this seems like a good choice of _locus classicus_.

> Can we unambiguously classify all elements of the kernel into those

that can generate it and

> those that can't?

The identity 1 is never a generator, and any element which is a

multiple of another element is never a generator. Beyond that we

cannot go--being a generator is really a property of sets of

elements, not of individual elements.

> > In other words, these two define the kernel of a homomorphism to

the

> > 24 et division of the octave,

> I think you're jumping the gun with that intepretation.

Well, I just showed it was true, though I haven't explained why the

method works. However, I was unclear about one thing--I wasn't

talking about periodicity blocks at all, simply about homomorphisms

and ets. Any homomorphism uniquely determines a kernel, and vice-

versa. If we take everything generated by the above two elements, we

get a subgroup of our abelian group, and regarded as a kernel it will

uniquely be associated to a homomorphism--which in this case has an

image which happens to be the 24-et in the 5-limit.

It is clear however that there is a very close relationship between

sets of generators for a kernel and periodic blocks. In fact, we can

uniquely associate a periodic block with a set of generators. Let

{g1, g2, ... gk} be a set of generators. If we take equivalence

classes by octaves, we get another set {h1, h2, ... hk} of generators

for the group of equivalence classes (this is just another sort of

homomorphism, actually.) If we take the absolute value of the

determinant of {h1, h2, ... hk}, which is the same as the first minor

with the 2-column removed we considered before, we get an integer N

which is the content ("hypervolume") of the parallelepiped defined by

{h1, h2, ... hk}--which fact if you think about its implications

should make clear why the method for finding the homomorphism works.

Now take the set S of all group elements a1*h1+a2*h2+ ... + ak*hk,

where the ai are rational numbers 0 <= ai < 1. These are lattice

elements in the parallelepiped, and because its content is N there

are N of them. These N elements S define a periodic block uniquely

associated to the set {h1, h2, ... hk} and hence to {g1, g2, ... gk}.

We can make it a little more palatable by moving the identity to

somewhere in the middle instead of at a corner, of course. :)

Hence in the example you gave, you do get a periodic block with 24

elements, and I don't see anything pathological about it beyond the

fact that some elements will be a comma apart.

By the way, a periodic block is really a special case of a JT scale,

and I don't know why it should be the most interesting case.

> By the way, I don't think of 24-tET as pathological at all, I just

don't think that one can say, in any

> useful sense, that it's _generated_ or _defined_ by 128/125 and

32805/32768.

Well, it is however--these two elements generate a kernel, and the

kernel is *uniquely* associated to a homomorphism.

> Here we move on to the 7-limit, and the unison vectors are

>

> [Note: I'm switching from "/" to ":" because of our convention to

use ":" for intervals and "/" for

> pitches]

>

> 49:48 = [-4 -1 0 2]

> 64:63 = [ 6 -2 0 -1]

> 225:224 = [-5 2 2 -1]

>

> The absolute values of the determinants of the minors should be

>

> [10]

> [16]

> [23]

> [28]

>

> right?

I don't know if they should be, but they are. :) Right!

Here, the product of 64:63 and 225:224 is equal to _two_ 10:7s.

... minus an octave.

(In a case of

> particularly great interest to me, the 49:48 is _not_ tempered out,

and you get a 10-tone

> system embedded in something very close to 22-tone equal

temperament).

This is a tempered 10-tone scale, in other words.

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

> I think you're jumping the gun with that intepretation.

You were right--my method doesn't work for finding the homomorphism

in these cases.

I'll try to sort this out tomorrow if I have time, but the image

under the homomorphism has to have a 2-torsion part. I think the deal

is h([a, b, c]) gets sent to [12*a+19*b+28*c, a+b+c (mod 2)].

Paul wrote:

> > Let's take 81/80 and 128/125 to start with. I may write these

> > additively, so that 81/80 = 2^-4 * 3^4 * 5^-1 is written [-4, 4, -1]

> > and 128/125 becomes [7, 0, -3]. We can put these together into a 2x3

> > matrix, giving us

> >

> > [-4 4 -1]

> > [ 7 0 -3]

> >

> > If we take the absolute value of the determinants of the minors of

> > this matrix, we recover the homomorphism:

> >

> > abs(det([[4, -1],[0,-3]]) = 12, abs(det([[-4, -1],[7,-3]]) = 19, and

> > abs(det([-4,4],[7,0]))= 28, recovering the homomorphism column vector

> >

> > [12]

> > [19]

> > [28]

> >

> > from the generators of the kernel.

>

> I think this is a computation that Graham Breed has done or was trying

> to do. Graham?

Hello!

Octave-specific vectors for an octave-invariant system are a fudge. So I

add the octave to the kernel:

[ 1 0 0]

[-4 4 -1]

[ 7 0 -3]

Invert that and multiply by the determinant and you get:

[-12 0 0]

[-19 -3 1]

[-28 0 4]

The left hand column is the "homomorphism column vector" above (sign isn't

important as long as it's consistent). It's identical to Gene's formula

by the definition of matrix inversion. The 12 is Fokker's determinant.

The other columns happen to be the generator mappings for the equivalent

column being a chromatic unison vector. I don't think there's a proof for

this always working yet, but it does. Note that this seems to be a

different meaning of "generator" from above.

Lots of background is at <http://x31eq.com/tuning.htm>. See

the "temperaments from unison vectors" program. You'll need Numeric

Python which I couldn't find on the FTP site last time I looked.

Apparently, the ActiveState download includes it, so try that if you're on

Windows.

An aside: the operation "invert and multiply by determinant" could be

made primary. It's actually simpler to calculate than a regular inverse,

because the last thing you do is to divide by the determinant. This would

give an algebra containing only integer matrices. I'd be interested to

know if this exists in the mathematical literature anywhere.

> > This computation shows these

> > two "unison vectors" do in fact generate the kernel. If we perform a

> > similar computation for 128/125 and 32805/32768 we first get the

> > matrix

> >

> > [7 0 -3]

> > [-15 8 1]

> >

> > The column vector we get from the absolute values of the determinants

> > of the minors of this is:

> >

> > [24]

> > [38]

> > [56]

> >

> > In other words, these two define the kernel of a homomorphism to the

> > 24 et division of the octave,

>

> I think you're jumping the gun with that intepretation.

This has been acknowledged now. The easy way you know it isn't 24-equal

is that 24, 38 and 56 are all even numbers. So you're only defining every

other note from 24-equal, which is identical to 12-equal.

> Here we move on to the 7-limit, and the unison vectors are

>

> [Note: I'm switching from "/" to ":" because of our convention to use

> ":" for intervals and "/" for pitches]

>

> 49:48 = [-4 -1 0 2]

> 64:63 = [ 6 -2 0 -1]

> 225:224 = [-5 2 2 -1]

>

> The absolute values of the determinants of the minors should be

>

> [10]

> [16]

> [23]

> [28]

>

> right? Here, the product of 64:63 and 225:224 is equal to _two_

> 10:7s.And here, it's clearly a half-octave we're dealing with, not

> something that vanishes. So we really have a 10-tone scale here, and if

> all three unison vectors are tempered out uniformly, you get 10-tET.

> (In a case of particularly great interest to me, the 49:48 is _not_

> tempered out, and you get a 10-tone system embedded in something very

> close to 22-tone equal temperament).

In this case one entry is and odd number, so it really is 10 notes and

not 5.

Graham

--- In tuning-math@y..., graham@m... wrote:

> Octave-specific vectors for an octave-invariant system are a

fudge. So I

> add the octave to the kernel:

>

> [ 1 0 0]

> [-4 4 -1]

> [ 7 0 -3]

Adding the octave to the kernel changes the image from a rank one

free group (something isomorphic to Z) to a cyclic group of order 12

(isomorphic to Z/12Z.) The homomorphism is simply

h([a, b, c]) = 12*a+19*b+28*c (mod 12) = 7*b+4*c (mod 12). In other

words, a fifth is 7 semitones, a major third is 4 semitones, and

octaves we are ignoring.

> The other columns happen to be the generator mappings for the

equivalent

> column being a chromatic unison vector. I don't think there's a

proof for

> this always working yet, but it does. Note that this seems to be a

> different meaning of "generator" from above.

As I've remarked already, one can't prove something without a

statement of the conjectured result, and one can't understand the

statement without definitions. So--what's the claim?

> An aside: the operation "invert and multiply by determinant" could

be

> made primary. It's actually simpler to calculate than a regular

inverse,

> because the last thing you do is to divide by the determinant.

This would

> give an algebra containing only integer matrices. I'd be

interested to

> know if this exists in the mathematical literature anywhere.

It's called the adjoint matrix. If M is a square matrix, you would

notate it as adj(M).

If R is any ring, we can define a ring M_n(R) of nxn matricies with

entries in R, for any positive integer n. M_n(R) is actually an R-

algebra, since scalar mulitplication by R is defined. If R is a

commutative ring, we can define the determinant and hence the adjoint

matrix in M_n(R). In particular if R is the integers Z, then we have

an adjoint to any matrix with elements in Z, as you noted. We always

have

M*adj(M) = det(M)I = adj(M)*M,

where "det" is the determinant mapping the matrix M to an element of

R, "*" denotes matrix multiplication, and I is the identity matrix.

genewardsmith@juno.com () wrote:

> --- In tuning-math@y..., graham@m... wrote:

>

> > Octave-specific vectors for an octave-invariant system are a

> fudge. So I

> > add the octave to the kernel:

> >

> > [ 1 0 0]

> > [-4 4 -1]

> > [ 7 0 -3]

>

> Adding the octave to the kernel changes the image from a rank one

> free group (something isomorphic to Z) to a cyclic group of order 12

> (isomorphic to Z/12Z.) The homomorphism is simply

> h([a, b, c]) = 12*a+19*b+28*c (mod 12) = 7*b+4*c (mod 12). In other

> words, a fifth is 7 semitones, a major third is 4 semitones, and

> octaves we are ignoring.

Well, this goes comfortably beyond what I know about group theory. Is Z

the set of integers? Making the rank one like the (1) in SU(1)? A

"cyclic group of order 12" would make sense if it means the same 12 notes

repeat each octave. The rest I think I understand. So you weren't

assuming octave invariance to start with?

> > The other columns happen to be the generator mappings for the

> equivalent

> > column being a chromatic unison vector. I don't think there's a

> proof for

> > this always working yet, but it does. Note that this seems to be a

> > different meaning of "generator" from above.

>

> As I've remarked already, one can't prove something without a

> statement of the conjectured result, and one can't understand the

> statement without definitions. So--what's the claim?

The concept of generator is defined in the Carey/Clampitt paper that

Paul's already pointed you towards. Or maybe it's only referred to, but

you'll get the idea. I'm claiming I can uniquely define a generated scale

from a set of unison vectors. The full process is defined by a Python

script. It's something like:

Put the octave at the top of the matrix and the chromatic unison vector

next. Invert the matrix and multiply by the lowest common denominator.

The left hand column is the number of scale steps to each prime interval,

maybe not in its lowest terms as we know. The second column is the

generator mapping. The highest common factor gives you the number of

periods (or intervals of periodicity) to an octave. Divide through by

that and you have the numbers of generators that period-reduced give the

prime intervals. Multiply through by -1 and it still works, but no other

mapping will.

> > An aside: the operation "invert and multiply by determinant" could

> be

> > made primary. It's actually simpler to calculate than a regular

> inverse,

> > because the last thing you do is to divide by the determinant.

> This would

> > give an algebra containing only integer matrices. I'd be

> interested to

> > know if this exists in the mathematical literature anywhere.

>

> It's called the adjoint matrix. If M is a square matrix, you would

> notate it as adj(M).

Thanks. How could it be defined for a non-square matrix?

> If R is any ring, we can define a ring M_n(R) of nxn matricies with

> entries in R, for any positive integer n. M_n(R) is actually an R-

> algebra, since scalar mulitplication by R is defined. If R is a

> commutative ring, we can define the determinant and hence the adjoint

> matrix in M_n(R). In particular if R is the integers Z, then we have

> an adjoint to any matrix with elements in Z, as you noted. We always

> have

What's a ring?

> M*adj(M) = det(M)I = adj(M)*M,

>

> where "det" is the determinant mapping the matrix M to an element of

> R, "*" denotes matrix multiplication, and I is the identity matrix.

Yes, that follows.

Graham

> From: <genewardsmith@juno.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Saturday, August 18, 2001 10:26 PM

> Subject: [tuning-math] Re: Microtemperament and scale structure

>

>

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

>

> > There are many lists of interval names out there. I've seen

> > "small diesis" more often than "great diesis" for this interval

> > -- "large diesis" refers to the difference between four minor

> > thirds and an octave.

>

> I use the table in Ellis' appendix to Helmholtz mostly. If there is

> nothing official, this seems like a good choice of _locus classicus_.

Hi Gene, and welcome to the tuning lists.

I've come up against this confusion of terminology before, and

tried to grapple with a solution for it in this post:

http://www.ixpres.com/interval/td/monzo/o483-26new5limitnames.htm

Hope you find that helpful. This actually is in the archives of

</tuning>, but there was only very

limited response to it. I still think my ideas present a useful

classification.

love / peace / harmony ...

-monz

http://www.monz.org

"All roads lead to n^0"

_________________________________________________________

Do You Yahoo!?

Get your free @yahoo.com address at http://mail.yahoo.com

--- In tuning-math@y..., graham@m... wrote:

> Well, this goes comfortably beyond what I know about group theory.

Is Z

> the set of integers?

Z is the ring of integers (from German, "Zahlen" = numbers), but

since you asked what a ring is let's make some definitions:

(1) An abelian group (G, +, 0) is a set of elements G closed under an

addition operation "+", with an identity element 0 and an inverse -a

for any a in G, such that -a+a=0. The "+" operation is associative,

a+(b+c)=(a+b)+c, and since the group is abelian, commutative also, so

that a+b=b+a. We can also write the group multiplicatively as (G, *,

1) when that is convenient.

(2) A ring (R, +, *, 0, 1) is an abelian group under "+", and closed

under the associative (mulitplication) operation "*", with a

multiplicative identity 1. It also satisfies the distributive laws:

a*(b+c) = a*b+a*c

(b+c)*a = b*a+c*a

If the multiplication is commutative, so that a*b=b*a, we have a

commutative ring.

(3) A free abelian group of rank n, Z^n, is n copies of Z considered

as an additive group: ZxZxZ ... xZ n times. Concretely, it consists

of vectors of length n with integer values, under addition.

(4) The cyclic group of order n, C(n) is the set of elements of Z

modulo n, considered as an additive group; as a ring it is denoted

Z_n or Z/nZ. Z considered as an abelian group is sometimes called the

cyclic group of infinite order as well as the free abelian group of

rank one.

(5) The units of Z/nZ, represented by the elements relatively prime

to n, form the unit group U(n) (not to be confused with the unitary

group!) under multiplication. (I add this since you mentioned SU(1)

below--that's pretty boring, by the way!--and that made me think of

the two meanings of U(n).)

> > It's called the adjoint matrix. If M is a square matrix, you

would

> > notate it as adj(M).

> Thanks. How could it be defined for a non-square matrix?

You could define something using wedge products but I wouldn't

recommend bothing with that now.

--- In tuning-math@y..., genewardsmith@j... wrote:

> > > In other words, these two define the kernel of a homomorphism to

> the

> > > 24 et division of the octave,

>

> > I think you're jumping the gun with that intepretation.

>

> Well, I just showed it was true, though I haven't explained why the

> method works. However, I was unclear about one thing--I wasn't

> talking about periodicity blocks at all, simply about homomorphisms

> and ets.

Well there may be an important difference then.

>

> Hence in the example you gave, you do get a periodic block with 24

> elements, and I don't see anything pathological about it beyond the

> fact that some elements will be a comma apart.

Right -- but here's what's pathological.

Normally, if you temper out the defining unison

vectors of the PB, you get an ET, where the

number of notes is the determinant of the matrix

of unison vectors.

But, in this case, if you temper out the schisma and

the diesis, you're tempering out their sum, which

means you're tempering out _two_ syntonic

commas . . . which means that you're either

tempering out the syntonic comma, or setting it to

half an octave. If you're tempering out the syntonic

comma, then the number of notes in the ET is not

the determinant (24), but only half that (12). If

you're setting it to half an octave . . . whatever that

means . . . it seems you still get only 12-tET.

>

> By the way, a periodic block is really a special case of a JT scale,

> and I don't know why it should be the most interesting case.

Tell me what JT means. And, as you can see in the

_Gentle Introduction_, one could choose hexagons

instead of parallelograms. In 3D, one could choose

hexagonal prisms or rhombic dodecahedra instead

of paralellepipeds. In fact, one can choose any

weird shape that has one and only one element

from the group (to the extent that group theory

actually works here . . . ).

But if my hypothesis is correct, then

parallelograms will be the only way to construct

hyper-MOS scales, I think -- since otherwise the

period boundaries will not always be straight lines,

and thus MOSs won't always result from

tempering out all but one of the unison vectors.

>

> > By the way, I don't think of 24-tET as pathological at all, I just

> don't think that one can say, in any

> > useful sense, that it's _generated_ or _defined_ by 128/125 and

> 32805/32768.

>

> Well, it is however--these two elements generate a kernel, and the

> kernel is *uniquely* associated to a homomorphism.

Try understanding my explanation above as to

why, in a _musical_ context, I have a problem with

seeing 24-tET as the result of this. Your math may

be fine but the association with an ET with the full

number of notes, I'm arguing, may not always be

appropriate.

How, in actual practice, could one take the infinite

5-limit just lattice, alter all the intervals slightly so

that both 128/125 and 32805/32768 vanish, and

end up with 24-tone equal temperament?

> Here, the product of 64:63 and 225:224 is equal to _two_ 10:7s.

>

> ... minus an octave.

Right . . . I guess I'm used to being cavalier about

octaves in this context (though not in the context

of lattice metrics) . . . I mean, each rung in the

lattice is defined by an octave-invariant interval, for

if it weren't, you wouldn't get anything like unisons

between notes the edges of the periodicity block . .

. right?

> (In a case of

> > particularly great interest to me, the 49:48 is _not_ tempered out,

> and you get a 10-tone

> > system embedded in something very close to 22-tone equal

> temperament).

>

> This is a tempered 10-tone scale, in other words.

Yes -- I call it the decatonic scale. It comes in two

main forms -- the symmetrical decatonic (rotations

of LssssLssss) and pentachordal decatonic

(rotations of LsssssLsss). You'll see that it's

especially rich in 4:5:6:7 and 1/7:1/6:1/5:1/4 tetrads.

--- In tuning-math@y..., genewardsmith@j... wrote:

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

>

> > I think you're jumping the gun with that intepretation.

>

> You were right--my method doesn't work for finding the homomorphism

> in these cases.

Whew! I thought I was going crazy :)

>

> I'll try to sort this out tomorrow if I have time, but the image

> under the homomorphism has to have a 2-torsion part. I think the deal

> is h([a, b, c]) gets sent to [12*a+19*b+28*c, a+b+c (mod 2)].

Well, I look forward to the explanation. Let me

know if there's a reference I should study to make

this all more comprehensible.

--- In tuning-math@y..., graham@m... wrote:

>

> [-12 0 0]

> [-19 -3 1]

> [-28 0 4]

>

> The left hand column is the "homomorphism column vector" above (sign isn't

> important as long as it's consistent). It's identical to Gene's formula

> by the definition of matrix inversion. The 12 is Fokker's determinant.

>

> The other columns happen to be the generator mappings for the equivalent

> column being a chromatic unison vector.

> I don't think there's a proof for

> this always working yet, but it does.

Can you show with examples?

--- In tuning-math@y..., graham@m... wrote:

> > So--what's the claim?

>

> The concept of generator is defined in the Carey/Clampitt paper that

> Paul's already pointed you towards. Or maybe it's only referred to, but

> you'll get the idea. I'm claiming I can uniquely define a generated scale

> from a set of unison vectors. The full process is defined by a Python

> script. It's something like:

>

> Put the octave at the top of the matrix and the chromatic unison vector

> next.

You still haven't told Gene what the claim is

Gene -- first of all, start with a set of n unison

vectors. The unison vectors that are tempered out

or completely ignored are called "commatic unison

vectors". The unison vectors that amount to a

musically significant difference, but not (often)

large enough to move you from one scale step to

the next, are called "chromatic unison vectors".

The weak form of the hypothesis simply says that

if there is 1 chromatic unison vector, and n-1

commatic unison vectors, then what you have is a

linear temperament, with some generator and

interval of repetition (which is usually equal to the

interval of equivalence, but sometimes turns out to

be half, a third, a quarter . . . of it).

The strong form says that if you construct the

Fokker (hyperparallelepiped) periodicity block

from the n unison vectors, and again 1 is

chromatic and n-1 are commatic, then the notes in

the PB form an MOS scale.

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

> Gene -- first of all, start with a set of n unison

> vectors. The unison vectors that are tempered out

> or completely ignored are called "commatic unison

> vectors". The unison vectors that amount to a

> musically significant difference, but not (often)

> large enough to move you from one scale step to

> the next, are called "chromatic unison vectors".

Thanks! I'd guessed that was what it meant. I think you are adding to

the confusion by calling both of them "unison vectors", though--why

not unison and step vectors instead?

> The weak form of the hypothesis simply says that

> if there is 1 chromatic unison vector, and n-1

> commatic unison vectors, then what you have is a

> linear temperament, with some generator and

> interval of repetition (which is usually equal to the

> interval of equivalence, but sometimes turns out to

> be half, a third, a quarter . . . of it).

At last we are making progress! I don't see much role for

the "chromatic" element here, though. If the n-1 unison vectors are

linearly independent, we've already seen recently how to tell if they

generate a kernel of something mapping to Z: compute the gcd of the

determinant minors, and see if it is 1 or not. If they have no common

factor, then they define such a mapping, and the "chromatic vector"

will go to a certain number of steps in this mapping--hopefully 1,

but perhaps 2, 3, 4 ... etc.

As for temperment, that has to do with tuning and you cannot draw any

conclusions about tuning unless you introduce it into your statement

somewhere--nothing in, nothing out.

> The strong form says that if you construct the

> Fokker (hyperparallelepiped) periodicity block

> from the n unison vectors, and again 1 is

> chromatic and n-1 are commatic, then the notes in

> the PB form an MOS scale.

PB I presume means periodicity block, and MOS is some kind of jumped-

up well-formed scale, I understand. Could you similarly define MOS

(and WF while you are at it?)

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

> But, in this case, if you temper out the schisma and

> the diesis, you're tempering out their sum, which

> means you're tempering out _two_ syntonic

> commas . . . which means that you're either

> tempering out the syntonic comma, or setting it to

> half an octave.

I'm afraid that is where the "torsion" I was talking about comes in.

Suppose you color all 5-limit notes either green or red, by making

[a,b,c] green if a+b+c is even, and red if it is odd. Then two reds

add up to a green, a green and a red to a red, and two greens a green.

Your two generators are green, but the comma is red. The generators

generate only greens, but you need two reds to get a green. Hence the

image under the homomorphism goes to a 12 et note, but there is a red

keyboard and a green keyboard!

> Tell me what JT means.

To me, something defined in terms of rational numbers. What does it

mean to you?

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

> Whew! I thought I was going crazy :)

I knew my method didn't always work, but I had concluded it worked in

the "interesting" cases. You came up with an "uninteresting" case of

a kind I hadn't thought about, which turned out to be interesting.

> Well, I look forward to the explanation. Let me

> know if there's a reference I should study to make

> this all more comprehensible.

I hope the red-green show made some kind of sense. An introductory

textbook on abstract algebra would be the place to start if you want

to learn this stuff.

I must say I am surprised and pleased with the attitude around here.

The one time I tried to publish about music, the Computer Music

Journal turned it down as "too mathematical", so I thought people

were a little allergic. I would like a copy of that paper now, and I

could put it up on a web page--I think I sent a copy to some just

intonation library in San Francisco--does that ring any bells?

Forgive me for stepping in here guys, but I'm online and

figure that sooner is better...

> I must say I am surprised and pleased with the attitude around

> here. The one time I tried to publish about music, the Computer

> Music Journal turned it down as "too mathematical", so I thought

> people were a little allergic. I would like a copy of that paper

> now, and I could put it up on a web page--I think I sent a copy

> to some just intonation library in San Francisco--does that ring

> any bells?

The Just Intonation Network is here in SF:

>PB I presume means periodicity block, and MOS is some kind of

>jumped-up well-formed scale, I understand. Could you similarly

>define MOS (and WF while you are at it?)

MOS, WF, and Myhill's property are all equivalent. They are

usually given as something like:

MOS or WF- any pythagorean-type scale in which the generating

interval always spans the same number of scale degrees.

While strict pythagorean scales are usually generated with

3:2's against 2:1's, MOS and WF allow any generator, and

sometimes the interval of equivalence is allowed to be non-2:1.

Myhill's property- all generic scale intervals have exactly

two specific sizes.

-Carl

--- In tuning-math@y..., carl@l... wrote:

> The Just Intonation Network is here in SF:

>

> http://www.dnai.com/~jinetwk/

Thanks. Do you know if it has a library and if it would still have a

paper I sent to it back in the mid-80's? People have been getting

copies somehow, I've heard, and I suspect it comes from there.

>> The Just Intonation Network is here in SF:

>>

>> http://www.dnai.com/~jinetwk/

>

> Thanks. Do you know if it has a library and if it would still

> have a paper I sent to it back in the mid-80's? People have been

> getting copies somehow, I've heard, and I suspect it comes from

> there.

They do in fact have a tremendous library, mostly of stuff from

the 80's, when the Network was at its peak. Unfortunately it

is very disorganized, to the point where the chance they'll know

if they have thing x is less than 50%, and it would take hours,

even days to say for sure. Xeroxes, dot-matrix printouts abound,

in boxes in Henry Rosenthal's basement.

-Carl

In-Reply-To: <9lq0ik+d44m@eGroups.com>

In article <9lq0ik+d44m@eGroups.com>, genewardsmith@juno.com () wrote:

> At last we are making progress! I don't see much role for

> the "chromatic" element here, though. If the n-1 unison vectors are

> linearly independent, we've already seen recently how to tell if they

> generate a kernel of something mapping to Z: compute the gcd of the

> determinant minors, and see if it is 1 or not. If they have no common

> factor, then they define such a mapping, and the "chromatic vector"

> will go to a certain number of steps in this mapping--hopefully 1,

> but perhaps 2, 3, 4 ... etc.

Yes, the chromatic UV should be redundant. That reminds me of two

hypotheses I didn't get round to implementing in Python code:

1) If you use an octave-invariant matrix, with the chromatic UV at the

top, the left hand column of the adjoint matrix is the mapping by

generator, like the second column was before. (May need to be divided

through by the GCD.) That makes the adjoint matrix a list of generator

mappings (the octave being a special case where it's specific).

2) Any simple interval will do for the chromatic UV, so long as all the

UVs are linearly independent. I don't have strict criteria for

"simple" here, any more than criteria for what work as unison vectors in

the first place. But trying [1 0 0 ...], [0 1 0 ...], etc until something

works (non-zero determinant) should do the trick. I don't think such

things can be described as unison vectors, but I believe they do work in

this context.

So are you saying you have an equivalent to (2)? I don't think the GCD

has to be 1.

> As for temperment, that has to do with tuning and you cannot draw any

> conclusions about tuning unless you introduce it into your statement

> somewhere--nothing in, nothing out.

I suppose it depends on how you define "temperament". Is "meantone" a

temperament or a class of temperaments? The chromatic UV is used to

define the tuning. If you want to push the definition and make a third a

unison vector, you can define quarter comma meantone by setting it just.

So the commatic UVs define the temperament class and the chromatic UV is

used to define the specific tuning. I make the octave explicit for the

same reason.

> > The strong form says that if you construct the

> > Fokker (hyperparallelepiped) periodicity block

> > from the n unison vectors, and again 1 is

> > chromatic and n-1 are commatic, then the notes in

> > the PB form an MOS scale.

>

> PB I presume means periodicity block, and MOS is some kind of jumped-

> up well-formed scale, I understand. Could you similarly define MOS

> (and WF while you are at it?)

Whatever they mean, MOS and WF are the same thing: a generated scale with

only two step sizes. It may be a useful property for alternative systems

of tonality, but we don't have enough examples to pronounce on that yet.

It's certainly useful to think about when designing generalized keyboards

or alternative notation systems. (See

<http://www.anaphoria.com/xen3b.PDF> if you haven't already.)

Graham

In-Reply-To: <9lpte7+27ap@eGroups.com>

Paul wrote:

> > The other columns happen to be the generator mappings for the

> > equivalent column being a chromatic unison vector.

> > I don't think there's a proof for

> > this always working yet, but it does.

>

> Can you show with examples?

It's what <http://x31eq.com/vectors.html> is all about.

<http://x31eq.com/vectors.out> is a list of examples.

Graham

--- In tuning-math@y..., genewardsmith@j... wrote:

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

>

> > Gene -- first of all, start with a set of n unison

> > vectors. The unison vectors that are tempered out

> > or completely ignored are called "commatic unison

> > vectors". The unison vectors that amount to a

> > musically significant difference, but not (often)

> > large enough to move you from one scale step to

> > the next, are called "chromatic unison vectors".

>

> Thanks! I'd guessed that was what it meant. I think you are adding

to

> the confusion by calling both of them "unison vectors", though--why

> not unison and step vectors instead?

Three reasons:

1) The number of notes in the scale should be (normally) the

determinant of the matrix of unison vectors. One has to include both

the chromatic and the commatic unison vectors in this calculation.

2) In the "prototypical" case, the commatic unison vector is "the

comma", 81:80; and the chromatic unison vector is "the chromatic

unison" or "augmented unison", 25:24. These define a 7-tone

periodicity block: the diatonic scale. You see how the terminology is

just a generalization of this case.

3) "Step vectors" would refer to something else. In the prototypical

example, the step vectors would be 16:15, 10:9, and 9:8.

>

> > The weak form of the hypothesis simply says that

> > if there is 1 chromatic unison vector, and n-1

> > commatic unison vectors, then what you have is a

> > linear temperament, with some generator and

> > interval of repetition (which is usually equal to the

> > interval of equivalence, but sometimes turns out to

> > be half, a third, a quarter . . . of it).

>

> At last we are making progress! I don't see much role for

> the "chromatic" element here, though.

You're right . . . it plays no role here.

> If the n-1 unison vectors are

> linearly independent, we've already seen recently how to tell if

they

> generate a kernel of something mapping to Z:

No -- you did that with n unison vectors -- I'm not counting the 2

axis as a "dimension" here.

>

> > The strong form says that if you construct the

> > Fokker (hyperparallelepiped) periodicity block

> > from the n unison vectors, and again 1 is

> > chromatic and n-1 are commatic, then the notes in

> > the PB form an MOS scale.

>

> PB I presume means periodicity block, and MOS is some kind of

jumped-

> up well-formed scale, I understand. Could you similarly define MOS

> (and WF while you are at it?)

MOS means that there is an interval of repetition (normally equal to

the interval of equivalence [usually octave], but sometimes it comes

out as half, third, quarter . . . of the IE). The scale repeats

itself exactly within each interval of repetition. Within each, there

is a generating interval, which is iterated some number of times such

that the scale has two step sizes.

Examples:

The diatonic scale (LsssLss) is MOS: the IoR is an octave, and the

generator is L+s+s.

The melodic minor scale (LssssLs) is not MOS: there is no generator

that produces all the notes and no others.

--- In tuning-math@y..., genewardsmith@j... wrote:

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

>

> > But, in this case, if you temper out the schisma and

> > the diesis, you're tempering out their sum, which

> > means you're tempering out _two_ syntonic

> > commas . . . which means that you're either

> > tempering out the syntonic comma, or setting it to

> > half an octave.

>

> I'm afraid that is where the "torsion" I was talking about comes

in.

> Suppose you color all 5-limit notes either green or red, by making

> [a,b,c] green if a+b+c is even, and red if it is odd. Then two reds

> add up to a green, a green and a red to a red, and two greens a

green.

>

> Your two generators are green, but the comma is red. The generators

> generate only greens, but you need two reds to get a green. Hence

the

> image under the homomorphism goes to a 12 et note, but there is a

red

> keyboard and a green keyboard!

Are you saying that both keyboards are tuned identically, or that

there may be an offset?

>

> > Tell me what JT means.

>

> To me, something defined in terms of rational numbers. What does it

> mean to you?

I was just asking what it stood for. "Just Tuning"?

--- In tuning-math@y..., carl@l... wrote:

>

> MOS, WF, and Myhill's property are all equivalent.

This is not quite true -- for example, LssssLssss is MOS but not WF

and doesn't have Myhill's property.

--- In tuning-math@y..., graham@m... wrote:

>

> I suppose it depends on how you define "temperament".

Is "meantone" a

> temperament or a class of temperaments? The chromatic UV is used

to

> define the tuning.

You mean the commatic UVs (81:80 in the case of meantone)?

> If you want to push the definition and make a third a

> unison vector, you can define quarter comma meantone by setting it

just.

Now I think you're pushing definitions too far. Let's not forget the

strong form of the hypothesis!

> So the commatic UVs define the temperament class and the chromatic

UV is

> used to define the specific tuning.

Hmm . . . perhaps one _can_ define things this way, but it's by no

means universal. How would one define LucyTuning in this way??

>

> Whatever they mean, MOS and WF are the same thing: a generated

scale with

> only two step sizes.

Not the same thing. Clampitt lists all the WFs in 12-tET, and there

is no sign of the diminished (octatonic) scale, or any other scale

with an interval of repetition that is a fraction of an octave. These

are all MOS scales, though.

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9lpte7+27ap@e...>

> Paul wrote:

>

> > > The other columns happen to be the generator mappings for the

> > > equivalent column being a chromatic unison vector.

> > > I don't think there's a proof for

> > > this always working yet, but it does.

> >

> > Can you show with examples?

>

> It's what <http://x31eq.com/vectors.html> is all about.

> <http://x31eq.com/vectors.out> is a list of examples.

>

> Graham

I meant for the particular case which you erased above.

>> MOS, WF, and Myhill's property are all equivalent.

>

> This is not quite true -- for example, LssssLssss is MOS but not WF

> and doesn't have Myhill's property.

What single generator produces the scale?

-Carl

--- In tuning-math@y..., carl@l... wrote:

> >> MOS, WF, and Myhill's property are all equivalent.

> >

> > This is not quite true -- for example, LssssLssss is MOS but not

WF

> > and doesn't have Myhill's property.

>

> What single generator produces the scale?

>

> -Carl

One possibility is s -- here the interval of repetition is the half-

octave.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Hi Carl,

>

> It's a 12-tet scale with a 5/12 generator.

I'm not seeing the 12-tET-ness or the 5/12-ness of this at all:

> > > This is not quite true -- for example, LssssLssss is MOS but not

> WF

> > > and doesn't have Myhill's property.

>>>> MOS, WF, and Myhill's property are all equivalent.

>>>

>>> This is not quite true -- for example, LssssLssss is MOS but not

>>> WF and doesn't have Myhill's property.

>>

>> What single generator produces the scale?

>>

>> -Carl

>

> One possibility is s -- here the interval of repetition is the half-

> octave.

Then my reply is that the scale is MOS/WF at the half-octave.

-Carl

--- In tuning-math@y..., carl@l... wrote:

> Then my reply is that the scale is MOS/WF at the half-octave.

Well there's no point in going into a big debate on terminology here,

but note that Clampitt' list of WFs in 12-tET is sorely incomplete if

you allow this kind of construction.

Let's just say "MOS" and forget about it.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Paul Erlich wrote,

>

> <<I'm not seeing the 12-tET-ness or the 5/12-ness of this at all>>

>

> Hmm, why not?

Sorry, I hadn't seen your subsequent message where you said you were

interpreting this scale as a 12-tET scale. As I'm sure you know, the

discussion of this scale started with them in 22-tET or something

close to it. There one could say the generator is 1/11 of an octave.

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

> --- In tuning-math@y..., carl@l... wrote:

> > MOS, WF, and Myhill's property are all equivalent.

> This is not quite true -- for example, LssssLssss is MOS but not WF

> and doesn't have Myhill's property.

If "L" is a large scale step and "s" is a small scale step, then this

has two sizes of steps. If that is Myhill's property then it should

have it, so why doesn't it?

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

> Are you saying that both keyboards are tuned identically, or that

> there may be an offset?

It certainly would be more interesting musically with an offset, but

it doesn't matter in the sense that this is a tuning question, not a

structure question. Either way, a comma interval is represented by

jumping from the green keyboard to the red keyboard or vice-versa--

therefore, there is no distinction between a comma up and a comma

down, and two commas are a unison.

> I was just asking what it stood for. "Just Tuning"?

Sorry, just trying to be one of the boys in this alphabet soup of

acronyms around here.

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

> Are you saying that both keyboards are tuned identically, or that

> there may be an offset?

It certainly would be more interesting musically with an offset, but

it doesn't matter in the sense that this is a tuning question, not a

structure question. Either way, a comma interval is represented by

jumping from the green keyboard to the red keyboard or vice-versa--

therefore, there is no distinction between a comma up and a comma

down, and two commas are a unison.

> I was just asking what it stood for. "Just Tuning"?

Sorry, just trying to be one of the boys in this alphabet soup of

acronyms around here.

--- In tuning-math@y..., genewardsmith@j... wrote:

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

> > --- In tuning-math@y..., carl@l... wrote:

>

> > > MOS, WF, and Myhill's property are all equivalent.

>

> > This is not quite true -- for example, LssssLssss is MOS but not

WF

> > and doesn't have Myhill's property.

>

> If "L" is a large scale step and "s" is a small scale step, then

this

> has two sizes of steps. If that is Myhill's property then it should

> have it, so why doesn't it?

Myhill's property isn't just about the step sizes. Recall the melodic

minor scale, which has two step sizes but isn't WF. Myhill's property

says it has two sizes of _every_ generic interval size. But in the

case of LssssLssss, all "sixths" are the same size: L+4*s. There's

only one size of "sixth" -- so Myhill fails.

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

> The diatonic scale (LsssLss) is MOS: the IoR is an octave, and the

> generator is L+s+s.

>

> The melodic minor scale (LssssLs) is not MOS: there is no generator

> that produces all the notes and no others.

Shouldn't all those "L"s be "s"s and vice versa?

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

> 1) The number of notes in the scale should be (normally) the

> determinant of the matrix of unison vectors. One has to include

both

> the chromatic and the commatic unison vectors in this calculation.

What you are calling the determinant is just the determinant of the

minor you get by setting 2 aside--and there is 2 on the brain again.

From the point of view of approximations and real life, your comment

is true. From the point of view of pure algebra, it isn't. From an

algebraic point of view, the 7-et might be [7, 11, 17] and not

[7, 11, 16]--they are different homomorphisms. To recover the whole

homomorphism, and not just the number of steps in an octave, we need

all three minor determinants.

> 2) In the "prototypical" case, the commatic unison vector is "the

> comma", 81:80; and the chromatic unison vector is "the chromatic

> unison" or "augmented unison", 25:24. These define a 7-tone

> periodicity block: the diatonic scale. You see how the terminology

is

> just a generalization of this case.

Both of these are elements of the kernel of the 7-homomorphism

[7, 11, 16] and together they generate it. There really is no

distinction to be drawn beyond the obvious fact that 25/24 is bigger

than 81/80. You could always call the biggest element in your

generating set the chromatic unison and the rest commatic unison

vectors, but I don't see the point. Anyway, what is chromatic for one

set will end up being commatic for another!

> > > The weak form of the hypothesis simply says that

> > > if there is 1 chromatic unison vector, and n-1

> > > commatic unison vectors, then what you have is a

> > > linear temperament, with some generator and

> > > interval of repetition (which is usually equal to the

> > > interval of equivalence, but sometimes turns out to

> > > be half, a third, a quarter . . . of it).

Now I translate this to saying that if the rank of the kernel is n,

then we get a linear temperament. Since the rank of the set of notes

is n+1, this means the codimension is 1 and hence the rank of the

homomorphic image is 1, meaning we have an et--which is precisely

what we did get in the case where we had the 7-et. Why do you say

linear temperament, which we've just determined means rank 2?

> > At last we are making progress! I don't see much role for

> > the "chromatic" element here, though.

>

> You're right . . . it plays no role here.

Aha! So perhaps what you are saying is if the codimension is 2, then

the rank of the homomorphic image is 2, and we have a linear

temperament.

> No -- you did that with n unison vectors -- I'm not counting the 2

> axis as a "dimension" here.

Not a good idea in this context--you should.

> MOS means that there is an interval of repetition

What do the letters of the acronym stand for?

--- In tuning-math@y..., graham@m... wrote:

> It's what <http://x31eq.com/vectors.html> is all about.

> <http://x31eq.com/vectors.out> is a list of examples.

It was pretty hard to figure out what they were examples of.

Let me give an example matrix computation, and see if it looks

familiar. Let's take three et's in the 5-limit, for 12, 19, and 34.

If we make a matrix out of them, we have

[12 19 34]

S = [19 30 54]

[28 44 79]

Since this consists of three column vectors pointing in more or less

the same direction, the determinant is likely to be small; however

none of these three is a linear combination of the other two (as

often will happen--ets tend to be sums of other ets) the determinant

is nonzero--in this case, 1. If we invert it, we get

[-6 -5 6]

S^(-1) = [11 -4 -2]

[-4 4 -1]

The row vectors of S^(-1) are now 15625/15552, 2048/2025, and 81/80.

Taken in pairs, these give generators for the kernel of each of the

above systems, and hence good unison vectors for a PB. Each is a step

vector in one system, and a unison vector in the other two, in the

obvious way (given how matrix multiplication works.)

In the same way, we could start with three linearly independent

unison vector candidates, and get a matrix of three ets by inverting.

The single vectors generate the intersection of the kernels of a pair

of ets, and so define a linear temperament which factors through to

each of the ets. That is, 81/80 generates the intersection of the

kernel of the 12-system and the 19-system, and produces the mean tone

temperaments. Both 12 and 19 belong to this system--we can send it to

first the mean tone, then to either 12 or 19 (then to tuning as the

last step!) Similarly, 2048/2025 defines a temperament which is

common to both the 12 and the 34 system. It essentially defines what

they have in common.

There are other types of matrix computations we could make, but I'm

wondering if this seems familiar?

In-Reply-To: <9lt510+dcmg@eGroups.com>

In article <9lt510+dcmg@eGroups.com>, genewardsmith@juno.com () wrote:

> --- In tuning-math@y..., graham@m... wrote:

>

> > It's what <http://x31eq.com/vectors.html> is all about.

> > <http://x31eq.com/vectors.out> is a list of examples.

>

> It was pretty hard to figure out what they were examples of.

Um, yes, it would be. You'll find the discussion in the archives.

> Let me give an example matrix computation, and see if it looks

> familiar. Let's take three et's in the 5-limit, for 12, 19, and 34.

> If we make a matrix out of them, we have

>

> [12 19 34]

> S = [19 30 54]

> [28 44 79]

>

> Since this consists of three column vectors pointing in more or less

> the same direction, the determinant is likely to be small; however

> none of these three is a linear combination of the other two (as

> often will happen--ets tend to be sums of other ets) the determinant

> is nonzero--in this case, 1. If we invert it, we get

>

> [-6 -5 6]

> S^(-1) = [11 -4 -2]

> [-4 4 -1]

>

> The row vectors of S^(-1) are now 15625/15552, 2048/2025, and 81/80.

> Taken in pairs, these give generators for the kernel of each of the

> above systems, and hence good unison vectors for a PB. Each is a step

> vector in one system, and a unison vector in the other two, in the

> obvious way (given how matrix multiplication works.)

Aaaaaah! So they are! I hadn't thought of doing that. Suddenly

everything is a lot clearer.

The main difference with what I do is that I consider two ETs and perfect

octaves instead of three ETs. Presumably, chromatizing a unison vector is

the same as junking one of the three ETs you get from the inverse.

> In the same way, we could start with three linearly independent

> unison vector candidates, and get a matrix of three ets by inverting.

So is that what I'm doing? Hmm. Actually, I'd take two UVs along with

the octave. Hmm.

> The single vectors generate the intersection of the kernels of a pair

> of ets, and so define a linear temperament which factors through to

> each of the ets. That is, 81/80 generates the intersection of the

> kernel of the 12-system and the 19-system, and produces the mean tone

> temperaments. Both 12 and 19 belong to this system--we can send it to

> first the mean tone, then to either 12 or 19 (then to tuning as the

> last step!) Similarly, 2048/2025 defines a temperament which is

> common to both the 12 and the 34 system. It essentially defines what

> they have in common.

Yes, that figures. Let's go back to the example.

> [12 19 34]

> S = [19 30 54]

> [28 44 79]

> [-6 -5 6]

> S^(-1) = [11 -4 -2]

> [-4 4 -1]

Rows and columns are interchanged when you take the inverse. So the [-4 4

-1] corresponds to 34. This is the unison vector that results from taking

34 *out* of the system. Take out the bottom two, and you should end up

with 12=

|[ 1 0 0]|

|[11 -4 -2]| = 12

|[-4 4 -1]|

so that works.

Generalizing to more dimensions, presumably considering n unison vectors,

and taking out n-2 of them, will give you the linear temperament.

So, as I already have a program for generating consistent ETs, I could use

it to generate a list of candidate unison vectors. And then use them to

go back to ETs. All of it without assuming octave equivalence anywhere.

> There are other types of matrix computations we could make, but I'm

> wondering if this seems familiar?

Okay, I think I see the connection. What you're doing with octave

*specific* matrices is analogous to what I did with octave *invariant*

matrices. So you invert the matrix of unison vectors to get sets of

generators, where the generator is a step size in an ET. I invert a

matrix of one less unison vector to get sets of generators, where the

generator is the interval so called in WF or MOS theory. Algebraically,

it's exactly the same operation.

What I do with octave specific matrices is a bit more complicated.

Because I'm considering one fewer ET, and setting the octave just, that

means the inverse contains a mixture of step-size and MOS generators. I

really should be *reducing* the number of unison vectors, but it is very

interesting to see your method that works with one more.

Graham

In-Reply-To: <9lrmea+klq4@eGroups.com>

Paul) wrote:

> > I suppose it depends on how you define "temperament".

> Is "meantone" a

> > temperament or a class of temperaments? The chromatic UV is used

> to

> > define the tuning.

>

> You mean the commatic UVs (81:80 in the case of meantone)?

No, the commatic UVs define the approximation, the chromatic UVs are used

to define the tuning.

> > If you want to push the definition and make a third a

> > unison vector, you can define quarter comma meantone by setting it

> just.

>

> Now I think you're pushing definitions too far. Let's not forget the

> strong form of the hypothesis!

I'd be quite happy to forget the strong form of the hypothesis.

> > So the commatic UVs define the temperament class and the chromatic

> UV is

> > used to define the specific tuning.

>

> Hmm . . . perhaps one _can_ define things this way, but it's by no

> means universal. How would one define LucyTuning in this way??

( 1 0 0) (1 )

(-2 0 1)H' =~ (1/pi) Oct

(-4 4 -1) (0 )

or

(0 1)h' = (1/pi) Oct

(4 -1) (0 )

> > Whatever they mean, MOS and WF are the same thing: a generated

> scale with

> > only two step sizes.

>

> Not the same thing. Clampitt lists all the WFs in 12-tET, and there

> is no sign of the diminished (octatonic) scale, or any other scale

> with an interval of repetition that is a fraction of an octave. These

> are all MOS scales, though.

But Carey & Clampitt also say the two concepts are identical. Don't they?

Yes, note 1 says they are "equivalent".

They list as Example 2 of CLAMPITT.PDF "the 21 nondegenerate well-formed

sets in the twelve-note universe" but say nothing about those 12 notes

defining an *octave*. The octatonic scale would belong to the 2 tone

universe. Also, on page 2, "By /interval of periodicity/ we mean an

interval whose two boundary pitches are functionally equivalent.

Normally, the octave is the interval of periodicity." They don't define

Well Formedness, but refer to another paper. I'm guessing it depends on

the interval of periodicity, not the octave. The same seems to be true of

Myhill's Property.

Graham

In-Reply-To: <009701c12a17$911e9700$5340d63f@stearns>

Dan Stearns wrote:

> Rather than thinking of Myhill's property, MOS and WF as one big

> group, perhaps it's better to pair WF and Myhill's property on one

> side and MOS and maximal evenness on the other -- here's the idea:

>

> all MOS scales will have a ME rotation

>

> all WF scales will have Myhill's property

So WF/Myhill assumes an octave of 2:1 but MOS/ME can be any period? That

would make sense, and I agree with you that MOS/ME is better. But I

wonder if WF/Myhill was ever intended to be so restricted.

Graham

In-Reply-To: <9lt3q7+gtig@eGroups.com>

In article <9lt3q7+gtig@eGroups.com>, genewardsmith@juno.com () wrote:

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

>

> > 1) The number of notes in the scale should be (normally) the

> > determinant of the matrix of unison vectors. One has to include

> both

> > the chromatic and the commatic unison vectors in this calculation.

>

> What you are calling the determinant is just the determinant of the

> minor you get by setting 2 aside--and there is 2 on the brain again.

> From the point of view of approximations and real life, your comment

> is true. From the point of view of pure algebra, it isn't. From an

> algebraic point of view, the 7-et might be [7, 11, 17] and not

> [7, 11, 16]--they are different homomorphisms. To recover the whole

> homomorphism, and not just the number of steps in an octave, we need

> all three minor determinants.

If the octave counts as a "unison vector" then the determinant *does* come

out right. If not, there's only a determinant for the octave-invariant

matrix anyway! You seem to have a low view of algebra.

Paul doesn't mention equal temperament. Only "the number of notes in the

scale". This is correct.

For example, here's that pair of octave-invariant unison vectors:

[ 4 -1]

[ 0 -3]

Invert it to get

[-3 1]

[ 0 4]

Each column is a mapping of generators.

The left hand column is what you get by making [0 -3] the chromatic UV.

As there's a common factor of 3, it means the octave is divided into three

equal parts. A fifth is (-)1 generators, and a third is an exact octave

division.

The right hand column is the usual meantone mapping. The fifth is the

generator, and a third is four generators. This is the full

octave-invariant homomorphism.

Equal temperaments are tricky when they're octave invariant. Effectively,

it means the new period is a scale step. Hence all notes are identical.

That would make equal temperaments zero-dimensional in octave invariant

terms, which happens to agree with the Hausdorff dimension. To get the

rest of the information out, I agree it makes sense to consider

octave-specific matrices, and I have argued that before. (Check the

archives.) However, as Carey and Clampitt show, you can convert between

the MOS generator and scale steps, so the homomorphism is still there.

> > 2) In the "prototypical" case, the commatic unison vector is "the

> > comma", 81:80; and the chromatic unison vector is "the chromatic

> > unison" or "augmented unison", 25:24. These define a 7-tone

> > periodicity block: the diatonic scale. You see how the terminology

> is

> > just a generalization of this case.

>

> Both of these are elements of the kernel of the 7-homomorphism

> [7, 11, 16] and together they generate it. There really is no

> distinction to be drawn beyond the obvious fact that 25/24 is bigger

> than 81/80. You could always call the biggest element in your

> generating set the chromatic unison and the rest commatic unison

> vectors, but I don't see the point. Anyway, what is chromatic for one

> set will end up being commatic for another!

That's all true, but missing the point. Commas and chromatic semitones

are certainly treated differently in diatonic music.

> Now I translate this to saying that if the rank of the kernel is n,

> then we get a linear temperament. Since the rank of the set of notes

> is n+1, this means the codimension is 1 and hence the rank of the

> homomorphic image is 1, meaning we have an et--which is precisely

> what we did get in the case where we had the 7-et. Why do you say

> linear temperament, which we've just determined means rank 2?

Because the rest of the world says "linear temperament" and has done so

for longer than we've been alive.

> > No -- you did that with n unison vectors -- I'm not counting the 2

> > axis as a "dimension" here.

>

> Not a good idea in this context--you should.

How about you stop telling us what we should be doing, and start listening

to what we're trying to say?

> > MOS means that there is an interval of repetition

>

> What do the letters of the acronym stand for?

Moments of Symmetry.

Graham

--- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:

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

> > The diatonic scale (LsssLss) is MOS: the IoR is an octave, and

the

> > generator is L+s+s.

> >

> > The melodic minor scale (LssssLs) is not MOS: there is no

generator

> > that produces all the notes and no others.

>

> Shouldn't all those "L"s be "s"s and vice versa?

Whoops! Of course.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> <<Myhill's property isn't just about the step sizes. Recall the

> melodic minor scale, which has two step sizes but isn't WF. Myhill's

> property says it has two sizes of _every_ generic interval size. But

> in the case of LssssLssss, all "sixths" are the same size: L+4*s.

> There's only one size of "sixth" -- so Myhill fails.>>

>

>

> Hi Paul (and everyone),

>

> Rather than thinking of Myhill's property, MOS and WF as one big

> group, perhaps it's better to pair WF and Myhill's property on one

> side and MOS and maximal evenness on the other -- here's the idea:

>

> all MOS scales will have a ME rotation

What do you mean, an ME rotation? Surely you don't mean maximal

evenness -- which is independent of rotation. Many MOS scales are not

ME in any reasonable embedding -- the Blackjack scale, for instance.

But basically, you're right -- there is very little difference

between MOS and WF and Myhill's property.

--- In tuning-math@y..., genewardsmith@j... wrote:

> > 2) In the "prototypical" case, the commatic unison vector is "the

> > comma", 81:80; and the chromatic unison vector is "the chromatic

> > unison" or "augmented unison", 25:24. These define a 7-tone

> > periodicity block: the diatonic scale. You see how the

terminology

> is

> > just a generalization of this case.

>

> Both of these are elements of the kernel of the 7-homomorphism

> [7, 11, 16] and together they generate it. There really is no

> distinction to be drawn beyond the obvious fact that 25/24 is

bigger

> than 81/80.

There is an additional, formal distinction drawn, which you're not

taking into account in your formalism.

> You could always call the biggest element in your

> generating set the chromatic unison and the rest commatic unison

> vectors,

It's not always done that way.

> but I don't see the point.

The point is that, for example, if only one of the unison vectors is

chromatic and the rest are commatic, you end up with an MOS scale,

which has up to two specific sizes for each generic interval. 7-tET

has only one specific size for each generic interval. Did you read

my "proof" of the Hypothesis?

> Anyway, what is chromatic for one

> set will end up being commatic for another!

Sure -- but within a given complete set of unison vectors, choosing

one of the unison vectors to be chromatic and the rest to be commatic

leads uniquely to an MOS scale -- that's what we're trying to prove

here.

> > > > The weak form of the hypothesis simply says that

> > > > if there is 1 chromatic unison vector, and n-1

> > > > commatic unison vectors, then what you have is a

> > > > linear temperament, with some generator and

> > > > interval of repetition (which is usually equal to the

> > > > interval of equivalence, but sometimes turns out to

> > > > be half, a third, a quarter . . . of it).

>

> Now I translate this to saying that if the rank of the kernel is n,

> then we get a linear temperament. Since the rank of the set of

notes

> is n+1, this means the codimension is 1 and hence the rank of the

> homomorphic image is 1, meaning we have an et--

We most definitely do not have an ET. There are two step sizes, and

in general, up to two sizes for each generic interval.

> which is precisely

> what we did get in the case where we had the 7-et.

In 5-limit space, if 81/80 and 25/24 are both commatic unison

vectors, then you essentially have 7-et. But if 81/80 is commatic

while 25/24 is chromatic, you get a diatonic scale, where the cycle

of steps is sLLLsLL, the cycle of thirds is sLssLsL, the cycle of

fourths is sssssLs, etc.

> Why do you say

> linear temperament, which we've just determined means rank 2?

Because the diatonic scale described above has a single generator,

namely s+2*L (or any octave-equivalent, such as s+3*L).

> > > At last we are making progress! I don't see much role for

> > > the "chromatic" element here, though.

> >

> > You're right . . . it plays no role here.

>

> Aha! So perhaps what you are saying is if the codimension is 2,

then

> the rank of the homomorphic image is 2, and we have a linear

> temperament.

Perhaps -- if I count your way.

>

> > No -- you did that with n unison vectors -- I'm not counting the

2

> > axis as a "dimension" here.

>

> Not a good idea in this context--you should.

OK -- I'll try to do it your way from now on.

>

> > MOS means that there is an interval of repetition

>

> What do the letters of the acronym stand for?

You don't want to know. It's Moment of Symmetry -- I know, that means

something completely different in other contexts. But I think the

person who came up with it was a non-mathematician looking at the

results of iterating a fixed generator. Imagine that the generator

keeps building on top of itself, around and around the cycle that is

the octave. At stage N in this process, we have a scale with N notes.

If the scale has only two step sizes, then this "moment" in the

process has a special "symmetry" -- thus the name.

--- In tuning-math@y..., genewardsmith@j... wrote:

> --- In tuning-math@y..., graham@m... wrote:

>

> > It's what <http://x31eq.com/vectors.html> is all

about.

> > <http://x31eq.com/vectors.out> is a list of examples.

>

> It was pretty hard to figure out what they were examples of.

>

> Let me give an example matrix computation, and see if it looks

> familiar. Let's take three et's in the 5-limit, for 12, 19, and 34.

> If we make a matrix out of them, we have

>

> [12 19 34]

> S = [19 30 54]

> [28 44 79]

>

> Since this consists of three column vectors pointing in more or

less

> the same direction, the determinant is likely to be small; however

> none of these three is a linear combination of the other two (as

> often will happen--ets tend to be sums of other ets) the

determinant

> is nonzero--in this case, 1. If we invert it, we get

>

> [-6 -5 6]

> S^(-1) = [11 -4 -2]

> [-4 4 -1]

>

> The row vectors of S^(-1) are now 15625/15552, 2048/2025, and

81/80.

> Taken in pairs, these give generators for the kernel of each of the

> above systems, and hence good unison vectors for a PB. Each is a

step

> vector in one system, and a unison vector in the other two, in the

> obvious way (given how matrix multiplication works.)

>

> In the same way, we could start with three linearly independent

> unison vector candidates, and get a matrix of three ets by

inverting.

>

> The single vectors generate the intersection of the kernels of a

pair

> of ets, and so define a linear temperament which factors through to

> each of the ets. That is, 81/80 generates the intersection of the

> kernel of the 12-system and the 19-system, and produces the mean

tone

> temperaments. Both 12 and 19 belong to this system--we can send it

to

> first the mean tone, then to either 12 or 19 (then to tuning as the

> last step!) Similarly, 2048/2025 defines a temperament which is

> common to both the 12 and the 34 system. It essentially defines

what

> they have in common.

The latter are called diaschismic temperaments.

>

> There are other types of matrix computations we could make, but I'm

> wondering if this seems familiar?

Yes it does -- see Herman Miller's posts to this forum -- but thanks

for this formalism, it's sure to be very valuable.

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9lrmea+klq4@e...>

> Paul) wrote:

>

> > > I suppose it depends on how you define "temperament".

> > Is "meantone" a

> > > temperament or a class of temperaments? The chromatic UV is

used

> > to

> > > define the tuning.

> >

> > You mean the commatic UVs (81:80 in the case of meantone)?

>

> No, the commatic UVs define the approximation, the chromatic UVs

are used

> to define the tuning.

I see what you mean by that now. By varying the interval that the

chromatic unison vector is mapped to, you can get different

variations on the tuning system which obeys the given "approximation"

you refer to. But you're missing the point of the chromatic unison

vector. You can take _any_ non-commatic vector, vary the interval

that that vector is mapped to, and likewise get different variations

on the tuning system which obeys the "approximation".

The chromatic unison vector comes into play by _delimiting_ the

number of notes in the scale. As you know, if there is only one

chromatic unison vector, then ignoring it and just looking at the

commatic unison vectors defines an "approximation" that implies a

particular linear temperament. This linear temperament is potentially

infinite in extent. The chromatic unison vector allows for a sort of

imperfect closure of the system because, after a certain number of

iterations of the generator of the linear temperament, you'll

generate the chromatic unison vector (this should be easy to prove).

And that number of iterations of the generator is where you stop

adding notes to your scale.

> I'd be quite happy to forget the strong form of the hypothesis.

That's too bad. But OK, let's address the weak form first. Then we

throw out the one chromatic unison vector, as it doesn't come into

play here. We already have many arguments as to why it works. Let's

get it on firm mathematical footing with Gene. Maybe Gene can prove

the rule about how to calculate the generator of the linear

temperament.

> >

> > Not the same thing. Clampitt lists all the WFs in 12-tET, and

there

> > is no sign of the diminished (octatonic) scale, or any other

scale

> > with an interval of repetition that is a fraction of an octave.

These

> > are all MOS scales, though.

>

> But Carey & Clampitt also say the two concepts are identical.

Don't they?

> Yes, note 1 says they are "equivalent".

Well obviously they made a mistake either here or in their list!

>

> They list as Example 2 of CLAMPITT.PDF "the 21 nondegenerate well-

formed

> sets in the twelve-note universe" but say nothing about those 12

notes

> defining an *octave*. The octatonic scale would belong to the 2

tone

> universe.

You mean the 3 tone universe.

> Also, on page 2, "By /interval of periodicity/ we mean an

> interval whose two boundary pitches are functionally equivalent.

> Normally, the octave is the interval of periodicity." They don't

define

> Well Formedness, but refer to another paper. I'm guessing it

depends on

> the interval of periodicity, not the octave. The same seems to be

true of

> Myhill's Property.

But in the symmetrical decatonic scale, the boundary pitches are

_not_ functionally equivalent. They're a half-octave apart, and one

only assumes that _octaves_ are functionally equivalent when

constructing the periodicity block. The half-octave periodicity comes

in as a new feature, not as a previously assumed equivalence relation.

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <009701c12a17$911e9700$5340d63f@stearns>

> Dan Stearns wrote:

>

> > Rather than thinking of Myhill's property, MOS and WF as one big

> > group, perhaps it's better to pair WF and Myhill's property on one

> > side and MOS and maximal evenness on the other -- here's the idea:

> >

> > all MOS scales will have a ME rotation

> >

> > all WF scales will have Myhill's property

>

> So WF/Myhill assumes an octave of 2:1 but MOS/ME can be any period?

No, WF/Myhill only assumes that the interval of repetition is equal

to the interval of equivalence. ME shouldn't be in this discussion --

it assumes an embedding universe and Blackjack isn't an ME set in 72-

tET.

--- In tuning-math@y..., graham@m... wrote:

> For example, here's that pair of octave-invariant unison vectors:

>

> [ 4 -1]

> [ 0 -3]

>

> Invert it to get

>

> [-3 1]

> [ 0 4]

>

The inverse times the determinant.

> Equal temperaments are tricky when they're octave invariant.

Effectively,

> it means the new period is a scale step. Hence all notes are

identical.

Here you're going with the WF-related view that the interval of

repetition must be equal to the interval of equivalence. I suggest

instead the alternate view that they be allowed to differ, but that

the interval of equivalence is always an integer number of intervals

of repetition.

> How about you stop telling us what we should be doing, and start

listening

> to what we're trying to say?

Graham, I agree with you but maybe we should be willing to meet Gene

halfway? Dave Keenan expressed an opinion on this but I'm not sure

what it was.

--- In tuning-math@y..., "D.Stearns" <STEARNS@C...> wrote:

> Paul Erlich wrote,

>

> <<What do you mean, an ME rotation?>>

>

> In the 12-tet diatonic, the II, or dorian rotation, is ME.

That is not the accepted definition of ME, according to any of the

papers defining it by Clough, et al. It's a rotationally invariant

property of scales. It does assume an embedding universe of notes --

normally a 12-note universe, but this can vary.

>

> <<Surely you don't mean maximal evenness -- which is independent of

> rotation. Many MOS scales are not ME in any reasonable embedding --

> the Blackjack scale, for instance.>>

>

> Yes, maximal evenness -- show me any single generator MOS scale and

> I'll show you its ME rotation!

The blackjack scale is not ME in 72. The diatonic scale is not ME in

31. We've been over this before.

--- In tuning-math@y..., graham@m... wrote:

> > In the same way, we could start with three linearly independent

> > unison vector candidates, and get a matrix of three ets by

inverting.

> So is that what I'm doing? Hmm. Actually, I'd take two UVs along

with

> the octave. Hmm.

This is probably going to annoy you, but I think it is true so I'll

say it anyway--the octave is merely adding to the confusion. Let's

take an example consisting of 128/125, 25/24 and 81/80. We get the

matrix

[ 7 0 -3]

M = [-3 -1 2]

[-4 4 -1]

Then

[ 7 12 3]

M^(-1) = [11 19 5]

[16 28 7]

The columns of which are the 7, 12 and 3 divisions. (Little known but

in some contexts actually useful fact--in the 3 et, a comma equals a

third.)

You can recover the same information by replacing one row of M in

succession by [1 0 0] and taking adjoints or inverses, which you did

before in the case of the top row. When you take the three adjoint

matricies in succession, you get:

[ 7 0 0]

-[11 1 2]

[16 4 1],

[ 0 12 0]

-[-1 19 3]

[-4 28 0],

[ 0 0 3]

-[-2 -3 5]

[-1 0 7].

You indicated you had an interpretation of the other columns of these

matricies, but I don't see anything interesting about them, and hence

my comment. I'd like to know if I am wrong!

--- In tuning-math@y..., graham@m... wrote:

>You seem to have a low view of algebra.

Actually, I'm an algebraist and number theorist, and algebraists are

notorious for wanting to solve a problem using only algebraic methods

when possible, which is what I was doing. If you calculate the "7"

using determinants, you can get the rest of the homomorphism by using

floating point calculations: 7 log_2(3) = 11.09... and 7 log_2(5) =

16.25... As an algebraist, I naturally want to do the computation

using only algebra. As an algebraist, I also want to say "Now hold

on! What if I want to say a third has 3 steps, instead of two?"

Therefore I distinguish the systems [7, 11, 16] and [7, 11, 17].

> For example, here's that pair of octave-invariant unison vectors:

>

> [ 4 -1]

> [ 0 -3]

>

> Invert it to get

>

> [-3 1]

> [ 0 4]

>

> Each column is a mapping of generators.

Each column is a mapping of everything in the 5-limit modulo octaves,

but I don't see how you draw the conclusions you do about them.

> That's all true, but missing the point. Commas and chromatic

semitones

> are certainly treated differently in diatonic music.

If you are looking *only* at diatonic music and not allowing any

accidentals, then neither one is a scale step; they seem therefore to

be treated in the same way. The comma, being smaller, lends itself

better to approximations, of course.

> > Now I translate this to saying that if the rank of the kernel is

n,

> > then we get a linear temperament. Since the rank of the set of

notes

> > is n+1, this means the codimension is 1 and hence the rank of the

> > homomorphic image is 1, meaning we have an et--which is precisely

> > what we did get in the case where we had the 7-et. Why do you say

> > linear temperament, which we've just determined means rank 2?

> Because the rest of the world says "linear temperament" and has

done so

> for longer than we've been alive.

You're missing my point--the world does not call something we get

from codimension 1 a linear temperament, it calls codimension 2 (with

2 generators) a linear temperament.

> How about you stop telling us what we should be doing, and start

listening

> to what we're trying to say?

In a mathematical subject, mathematics is your friend; which means

potentially so is a friendly mathematician. Why not view me as an

ally in this endeavor?

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9lt3q7+gtig@e...>

> In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:

> > Why do you say

> > linear temperament, which we've just determined means rank 2?

>

> Because the rest of the world says "linear temperament" and has done

so

> for longer than we've been alive.

Sorry if I'm responsible for confusing you, Gene.

Meantone remains a 5-limit linear temperament whether you are

including 2s or ignoring them. When including 2s it appears 2

dimensional, when ignoring 2s it appears 1 dimensional.

I really like your approach of treating 2s just like the other primes.

We then need to switch our thinking to integer-limit rather than

odd-limit when deciding which errors we care about. Or how about an

a*b complexity limit for ratios a:b?

-- Dave Keenan

In-Reply-To: <9lufbl+8f7m@eGroups.com>

Paul wrote:

> --- In tuning-math@y..., graham@m... wrote:

>

> > For example, here's that pair of octave-invariant unison vectors:

> >

> > [ 4 -1]

> > [ 0 -3]

> >

> > Invert it to get

> >

> > [-3 1]

> > [ 0 4]

> >

>

> The inverse times the determinant.

Yes.

> > Equal temperaments are tricky when they're octave invariant.

> Effectively,

> > it means the new period is a scale step. Hence all notes are

> identical.

>

> Here you're going with the WF-related view that the interval of

> repetition must be equal to the interval of equivalence. I suggest

> instead the alternate view that they be allowed to differ, but that

> the interval of equivalence is always an integer number of intervals

> of repetition.

I don't think MOS is any more clearly defined than WF in this respect. So

let's assume they are identical. In which case for the things my program

spits out to be MOSes, and therefore your hypothesis to be true, they must

be the same. Or, at least, we need to redefine both to use the "period"

instead of "interval of equivalence".

The octave-equivalent algebra doesn't distinguish the different repeating

blocks. The determinant tells you the number of steps to the octave, but

to get the number of steps to any given interval you need to go

octave-specific. The linear temperament result tells you the number of

times the period goes into the interval of equivalence (octave) and the

mapping in terms of generators. Again, you need the octave-specific

algebra, or the metric, to get the mapping by generators *and* periods.

Effectively, the octave-equivalent algebra collapses so that the interval

of equivalence becomes the period in any given context.

> > How about you stop telling us what we should be doing, and start

> listening

> > to what we're trying to say?

>

> Graham, I agree with you but maybe we should be willing to meet Gene

> halfway? Dave Keenan expressed an opinion on this but I'm not sure

> what it was.

Gene's a better mathematician than I am, so he should be able to

understand the octave-equivalent case. From his comments, it looks like

he doesn't, so I'll keep trying to explain it until he does. If I can't

manage that, what hope will I have with people who don't know their

non-Abelian ring from their anomalous symmetric suspension?

Graham

In-Reply-To: <9lul29+57gl@eGroups.com>

Gene wrote:

> Actually, I'm an algebraist and number theorist, and algebraists are

> notorious for wanting to solve a problem using only algebraic methods

> when possible, which is what I was doing. If you calculate the "7"

> using determinants, you can get the rest of the homomorphism by using

> floating point calculations: 7 log_2(3) = 11.09... and 7 log_2(5) =

> 16.25... As an algebraist, I naturally want to do the computation

> using only algebra. As an algebraist, I also want to say "Now hold

> on! What if I want to say a third has 3 steps, instead of two?"

> Therefore I distinguish the systems [7, 11, 16] and [7, 11, 17].

This is because you're expecting to get an octave-specific result. The

octave-equivalent equivalent of an equal temperament is an MOS. (Or

family of MOS scales, there's nothing special about the particular numbers

of generators that give two interval sizes.) Hence the system gets

defined a different way, but *can* be defined by pure algebra.

For example, you could have 5-limit scales generated by fifths, with the

octave as the period. The most common of these is meantone, where a third

is four fifths. You can write that as [1 4]. An alternative is schismic,

explained by either Ellis or Helmholtz, where a third is eight fourths.

That's written [1 -8]. Both of these are fifth scales, but where a third

is mapped differently. That's the same as the examples of 7 scales you

gave, where a third is also mapped differently. Both cases submit to

exactly the same algebra. The only difference is that there's a

complication when you convert octave-equivalent vectors into pitch sizes.

Another example of a fifth scale is the diaschismic family, which has

already come up. There, a third is 2 half-octave reduced fourths.

Crazily enough, that's written [2 -4]. It means a homomorphism of [1 -2]

where the interval of equivalence is a half-octave.

> > For example, here's that pair of octave-invariant unison vectors:

> >

> > [ 4 -1]

> > [ 0 -3]

> >

> > Invert it to get

> >

> > [-3 1]

> > [ 0 4]

> >

> > Each column is a mapping of generators.

>

> Each column is a mapping of everything in the 5-limit modulo octaves,

> but I don't see how you draw the conclusions you do about them.

Ah, well, that'll be why you think algebra doesn't apply.

Usually, you think of 12-equal as being

... Bb B C C# D Eb E F F# G G# A Bb B C C# D ...

Where it continues infinitely in either direction. In which case 5 steps

is always a fourth, 7 is a fifth, 4 is a major third, etc. Truncate it to

within an octave and you have

C C# D Eb E F F# G G# A Bb B

Now, the interval from B to C isn't a scale step any more. And the

interval from F to C isn't a fifth. This is a problem. It means equal

temperaments can't be expressed in terms of scale steps in an octave

equivalent system, unless you explicitly define the system to be cyclic.

That is, say "the note after B is C again".

However, those notes can also be written like this:

Eb Bb F C G D A E B F# C# G#

That's the circle of fifths. You'll have seen it before. Now, the

interval from F to C is a fifth, and C to F is a fourth. B to C is the

same as E to F, and so that can be called a "semitone". However, the

interval from G# to Eb should be a fifth, but isn't. So the system still

needs to be made cyclic:

... Db Ab Eb Bb F C G D A E B F# C# G# D# ...

By making G#=Ab, C#=Db, and so on, we have another definition of 12-equal.

However, by taking the sequence at face value, turning the circle into a

spiral we've defined meantone. All the notes are in a line, so it's a

linear temperament. It's this temperament that setting 81:80 to be a

commatic unison vector gives. Its homomorphism is given by the right hand

column of

[-3 1]

[ 0 4]

Or [1 4] as a row vector. It tells you that a fifth is 1 step and a major

third is four steps. Check with the line above, you'll see it's correct.

All the algebra doesn't tell us is how to tune that "generator" step ...

> If you are looking *only* at diatonic music and not allowing any

> accidentals, then neither one is a scale step; they seem therefore to

> be treated in the same way. The comma, being smaller, lends itself

> better to approximations, of course.

Thanks! I think I understand the term "chromatic unison vector" better

now!

> You're missing my point--the world does not call something we get

> from codimension 1 a linear temperament, it calls codimension 2 (with

> 2 generators) a linear temperament.

I'm fully aware that the world has this wrong, but I go along with the

accepted usage. Still, as I show above, you can describe a linear

temperament using a codimension of 1.

> In a mathematical subject, mathematics is your friend; which means

> potentially so is a friendly mathematician. Why not view me as an

> ally in this endeavor?

But I do! Why else would I spend so much time with you?

Graham

--- In tuning-math@y..., genewardsmith@j... wrote:

>

> If you are looking *only* at diatonic music and not allowing any

> accidentals,

We _are_ allowing accidentals, and they play a very specific role in

diatonic music (as opposed to, say, atonal serial music).

--- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:

> --- In tuning-math@y..., graham@m... wrote:

> > In-Reply-To: <9lt3q7+gtig@e...>

> > In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:

> > > Why do you say

> > > linear temperament, which we've just determined means rank 2?

> >

> > Because the rest of the world says "linear temperament" and has

done

> so

> > for longer than we've been alive.

>

> Sorry if I'm responsible for confusing you, Gene.

>

> Meantone remains a 5-limit linear temperament whether you are

> including 2s or ignoring them. When including 2s it appears 2

> dimensional, when ignoring 2s it appears 1 dimensional.

>

> I really like your approach of treating 2s just like the other

primes.

> We then need to switch our thinking to integer-limit rather than

> odd-limit when deciding which errors we care about. Or how about an

> a*b complexity limit for ratios a:b?

When discussing _tuning_, then these can all be useful

considerations. But when discussing _scales which use octave-

equivalence_, then a*b turns into odd-limit, as I've demonstrated on

harmonic_entropy.

--- In tuning-math@y..., graham@m... wrote:

> I don't think MOS is any more clearly defined than WF in this

respect. So

> let's assume they are identical. In which case for the things my

program

> spits out to be MOSes, and therefore your hypothesis to be true,

they must

> be the same. Or, at least, we need to redefine both to use

the "period"

> instead of "interval of equivalence".

I must be on Saturn today. None of this makes any sense to me.

> The octave-equivalent algebra doesn't distinguish the different

repeating

> blocks.

Sure it does. The symmetrical decatonic scale (LssssLssss) has 10

notes -- two repeating blocks per octave.

> The linear temperament result tells you the number of

> times the period goes into the interval of equivalence (octave)

Exactly! So the WF definition, where the period _is_ the interval of

equivalence, won't do.

> and the

> mapping in terms of generators. Again, you need the octave-

specific

> algebra, or the metric, to get the mapping by generators *and*

periods.

> Effectively, the octave-equivalent algebra collapses so that the

interval

> of equivalence becomes the period in any given context.

What do you mean? For the symmetrical decatonic, the interval of

equivalence remains _two_ periods whether you're looking octave-

equivalently or not.

--- In tuning-math@y..., graham@m... wrote:

> This is because you're expecting to get an octave-specific result.

The

> octave-equivalent equivalent of an equal temperament is an MOS.

In what way? What do you mean by that? That certainly doesn't seem to

be true.

> (Or

> family of MOS scales, there's nothing special about the particular

numbers

> of generators that give two interval sizes.)

As opposed to the ones that don't??

In-Reply-To: <9m1338+rgjv@eGroups.com>

Paul wrote:

> > This is because you're expecting to get an octave-specific result.

> The

> > octave-equivalent equivalent of an equal temperament is an MOS.

>

> In what way? What do you mean by that? That certainly doesn't seem to

> be true.

There both one dimensional sets of points.

> > (Or

> > family of MOS scales, there's nothing special about the particular

> numbers

> > of generators that give two interval sizes.)

>

> As opposed to the ones that don't??

Not for these purposes.

Graham

In-Reply-To: <9m12rc+9cqs@eGroups.com>

Paul wrote:

> > The octave-equivalent algebra doesn't distinguish the different

> repeating

> > blocks.

>

> Sure it does. The symmetrical decatonic scale (LssssLssss) has 10

> notes -- two repeating blocks per octave.

But how can you say "per octave" in a system that doesn't recognize

octaves?

> > The linear temperament result tells you the number of

> > times the period goes into the interval of equivalence (octave)

>

> Exactly! So the WF definition, where the period _is_ the interval of

> equivalence, won't do.

Only if their definition of "interval of equivalence" is the same as

yours. Whatever the definition, I don't expect they intended to exclude

this case.

> > and the

> > mapping in terms of generators. Again, you need the octave-

> specific

> > algebra, or the metric, to get the mapping by generators *and*

> periods.

> > Effectively, the octave-equivalent algebra collapses so that the

> interval

> > of equivalence becomes the period in any given context.

>

> What do you mean? For the symmetrical decatonic, the interval of

> equivalence remains _two_ periods whether you're looking octave-

> equivalently or not.

It doesn't look that way to me. That scale can be defined as a

periodicity block by

[-1 2 0]

[ 0 2 -2]

[-2 0 -1]

The adjoint is

[-2 2 -4]

[ 4 1 -2]

[ 4 -4 -2]

The left hand column defines the generator mapping, which we can write as

[ 1]

-2*[-2]

[-2]

The minus sign isn't important. The 2 tells us the new interval of

equivalence is half the old one. The

[ 1]

[-2]

[-2]

gives us the mapping by the generator modulo this new interval of

equivalence. 16:15 is [-1 -1 0] times

[ 1]

[-2]

[-2]

or 1 generator. 3:2 is [1 0 0] or 1 generator. Where does the algebra

tell us these two intervals are not equivalent? You could define it by

giving each vector a "spin", but I still don't see how to derive that from

the octave equivalent algebra.

Graham

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9m1338+rgjv@e...>

> Paul wrote:

>

> > > This is because you're expecting to get an octave-specific

result.

> > The

> > > octave-equivalent equivalent of an equal temperament is an MOS.

> >

> > In what way? What do you mean by that? That certainly doesn't

seem to

> > be true.

>

> There both one dimensional sets of points.

Yes, but it seemed you were talking about a more "specific" type

of "equivalence".

>

> > > (Or

> > > family of MOS scales, there's nothing special about the

particular

> > numbers

> > > of generators that give two interval sizes.)

> >

> > As opposed to the ones that don't??

>

> Not for these purposes.

Well I disagree. The hypothesis, in fact, is all about showing that

there _is_ something special about the particular numbers of

generators that give two interval sizes, as opposed to those that

don't.

--- In tuning-math@y..., graham@m... wrote:

> In-Reply-To: <9m12rc+9cqs@e...>

> Paul wrote:

>

> > > The octave-equivalent algebra doesn't distinguish the different

> > repeating

> > > blocks.

> >

> > Sure it does. The symmetrical decatonic scale (LssssLssss) has 10

> > notes -- two repeating blocks per octave.

>

> But how can you say "per octave" in a system that doesn't recognize

> octaves?

What do you mean, doesn't recognize octaves? Octave-equivalence is

assumed right from the beginning, in the construction of the 7-limit

periodicity block.

>

> > > The linear temperament result tells you the number of

> > > times the period goes into the interval of equivalence (octave)

> >

> > Exactly! So the WF definition, where the period _is_ the interval

of

> > equivalence, won't do.

>

> Only if their definition of "interval of equivalence" is the same

as

> yours. Whatever the definition, I don't expect they intended to

exclude

> this case.

Then why on earth didn't they list the Messaien scale and the

diminished (octatonic) scale and the augmented (hexatonic) scale,

etc. in their list?? Certainly the list was geared toward those

thinking about a tuning of 12 notes per _octave_.

>

>

> > > and the

> > > mapping in terms of generators. Again, you need the octave-

> > specific

> > > algebra, or the metric, to get the mapping by generators *and*

> > periods.

> > > Effectively, the octave-equivalent algebra collapses so that

the

> > interval

> > > of equivalence becomes the period in any given context.

> >

> > What do you mean? For the symmetrical decatonic, the interval of

> > equivalence remains _two_ periods whether you're looking octave-

> > equivalently or not.

>

> It doesn't look that way to me. That scale can be defined as a

> periodicity block by

>

> [-1 2 0]

> [ 0 2 -2]

> [-2 0 -1]

>

> The adjoint is

>

> [-2 2 -4]

> [ 4 1 -2]

> [ 4 -4 -2]

>

> The left hand column defines the generator mapping, which we can

write as

> [ 1]

> -2*[-2]

> [-2]

>

> The minus sign isn't important. The 2 tells us the new interval of

> equivalence is half the old one.

That's your interpretation. But it isn't correct! We have 10

equivalence classes in this group, and this half-octave takes you

from one equivalence class to another -- not to the same one.

> The

>

> [ 1]

> [-2]

> [-2]

>

> gives us the mapping by the generator modulo this new interval of

> equivalence. 16:15 is [-1 -1 0] times

>

> [ 1]

> [-2]

> [-2]

>

> or 1 generator. 3:2 is [1 0 0] or 1 generator. Where does the

algebra

> tell us these two intervals are not equivalent?

Where does it tell us they are? Yes, they're _both_ generators.

Because of the symmetry, there is more than one possible generator.

Think of a prime-numbered ET -- how many generators does that have?

Paul wrote:

> > The left hand column defines the generator mapping, which we can

> write as

> > [ 1]

> > -2*[-2]

> > [-2]

> >

> > The minus sign isn't important. The 2 tells us the new interval

of

> > equivalence is half the old one.

>

> That's your interpretation. But it isn't correct! We have 10

> equivalence classes in this group, and this half-octave takes you

> from one equivalence class to another -- not to the same one.

We have 10 notes in the periodicity block, but only 5 equivalence

classes. We also have a 24 note periodicity block with only 12

equivalence classes. Or would you prefer to inflate the equivalence

interval to 2 octaves in that case? Here, it's simple to redefine

the unison vectors to give a 5 note periodicity block with half-

octave equivalence:

[-1 2 0]

[ 0 1 -1]

[-2 0 -1]

Which means, for my interpretation to be correct, we do need to

define "small" for unison vectors.

> gives us the mapping by the generator modulo this new interval of

> > equivalence. 16:15 is [-1 -1 0] times

> >

> > [ 1]

> > [-2]

> > [-2]

> >

> > or 1 generator. 3:2 is [1 0 0] or 1 generator. Where does the

> algebra

> > tell us these two intervals are not equivalent?

>

> Where does it tell us they are? Yes, they're _both_ generators.

> Because of the symmetry, there is more than one possible generator.

> Think of a prime-numbered ET -- how many generators does that have?

Any ET has one generator. Although in octave equivalent terms it has

none, because all notes are equivalent. "Octave equivalence" means

the octave is considered equivalent to a unison. How can a unison be

divided into equal steps?

An octave equivalent linear temperament also has one generator. If

two different intervals span the same number of generators, they are

equivalent in that temperament. That's the case here with the

approximations to 3:2 and 16:15.

Graham

--- In tuning-math@y..., "Graham Breed" <graham@m...> wrote:

> Paul wrote:

>

> > > The left hand column defines the generator mapping, which we

can

> > write as

> > > [ 1]

> > > -2*[-2]

> > > [-2]

> > >

> > > The minus sign isn't important. The 2 tells us the new

interval

> of

> > > equivalence is half the old one.

> >

> > That's your interpretation. But it isn't correct! We have 10

> > equivalence classes in this group, and this half-octave takes you

> > from one equivalence class to another -- not to the same one.

>

> We have 10 notes in the periodicity block, but only 5 equivalence

> classes.

I'm afraid that's an incorrect interpretation.

> We also have a 24 note periodicity block with only 12

> equivalence classes.

That's a very different case! It's pathological. We didn't use the

generators of the kernel. The decatonic case is not pathological --

the generators of the kernel clearly define 10 equivalence classes,

not 5. Gene?

> Or would you prefer to inflate the equivalence

> interval to 2 octaves in that case? Here, it's simple to redefine

> the unison vectors to give a 5 note periodicity block with half-

> octave equivalence:

That's obvious. But why do that?

>

> [-1 2 0]

> [ 0 1 -1]

> [-2 0 -1]

>

> Which means, for my interpretation to be correct, we do need to

> define "small" for unison vectors.

I agree that unison vectors should be "small", but I don't think your

interpretation (here) is correct!

>

> > gives us the mapping by the generator modulo this new interval of

> > > equivalence. 16:15 is [-1 -1 0] times

> > >

> > > [ 1]

> > > [-2]

> > > [-2]

> > >

> > > or 1 generator. 3:2 is [1 0 0] or 1 generator. Where does the

> > algebra

> > > tell us these two intervals are not equivalent?

> >

> > Where does it tell us they are? Yes, they're _both_ generators.

> > Because of the symmetry, there is more than one possible

generator.

> > Think of a prime-numbered ET -- how many generators does that

have?

>

> Any ET has one generator.

Incorrect. 7-tET, for example, can be generated by 1/7 octave, 2/7

octave, 3/7 octave . . .

> Although in octave equivalent terms it has

> none, because all notes are equivalent.

Ridiculous. :)

> "Octave equivalence" means

> the octave is considered equivalent to a unison. How can a unison

be

> divided into equal steps?

By going around in a circle.

Hi Gene!

I'm only just now getting around to studying what's been

discussed on tuning-math from August and September, shortly

after you came on board (and when I was planning a trip and

then traveling in Europe). I've only recently gotten back

into the groove here, so much of the last 5 months has

slipped past me. I have some comments for you.

Please, in reading any of my posts to this list, realize that

I am one of the most math-challenged posters here. I somehow

passed Algebra 1 in 9th grade without learning a thing, and spent

most of my time in Geometry class in 10th grade sitting in the

back row studying my score to Neilsen's 4th Symphony (the

terrific "Inextinguishable"... my, how I'd like to do a

microtonal computer realization of *that* some day!...).

Then, after moving and getting the chair of the new school's

math department for my Algebra 2 teacher, and being forced to

pay attention and learn, and struggling heroically to earn a

final "D", I chose no more math classes during the rest of

my formal education. This is a lack which I sorely feel now,

being so caught up in tuning math.

> From: <genewardsmith@juno.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Saturday, August 18, 2001 2:18 PM

> Subject: [tuning-math] Re: Microtemperament and scale structure

>

>

> My way of saying things has the advantage of being standard

> mathematical terminology, which allows one to bring relevant concepts

> into play. I had noticed that unison vectors seemed to have something

> to do with the kernel, but I couldn't tell if it meant generators of

> the kernel or any element of the kernel, and I see by your comments

> that no one has really decided!

>

> The kernel of some homomorphism h is everything sent to the identity--

> if this is the set of unison vectors then for instance 1 is *always*

> a unison vector, since h(1) = 0. On the other hand, unison vectors

> could be elements of a minimal set of generators for the kernel. In

> this case 81/80, 128/125 and 2048/2025 would all belong in the same

> kernel generated by any two of them. Depending on which set of

> generators you picked, two of them would be unison vectors and the

> other one would not be. 1 and 32805/32768 would also both be in the

> kernel, but neither would be unison vectors.

>

> Probably the simplest solution at this point would be to drop the

> terminology, but if you don't you need to decide what exactly it

> means.

This is an important suggestion for me, since I'm the guy who

created and maintains the Tuning Dictionary. What have the rest

of you heavyweights decided about this? I've been slugging around

the regular periodicity-block and unison-vector terminology a lot

here lately, and no-one has complained. Do we go with "kernel"?

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

>

> > We're well aware that any valid set comes from any other

> > valid set simply by "adding" and "subtracting" the

> > unison vectors from one another. But we've run

> > into some pathological cases -- for example, the

> > small diesis (128/125) and the schisma (32805/

> > 32768), while they can be derived from the same

> > two unison vectors, define a periodicity block with

> > 24, instead of 12, notes . . . and not a well-behaved

> > periodicity block at that. Any insights?

>

> It's true that anything in the kernel is obtained by adding and

> subtracting, since the kernel of an abelian group homomorphism is an

> abelian group. It's not true that any linearly independent set of

> kernel elements which span the corresponding vector space over the

> rationals as a basis is also a minimal set of generators for the

> kernel, and that is what you have discovered.

>

> Let's take 81/80 and 128/125 to start with. I may write these

> additively, so that 81/80 = 2^-4 * 3^4 * 5^-1 is written [-4, 4, -1]

> and 128/125 becomes [7, 0, -3]. We can put these together into a 2x3

> matrix, giving us

>

> [-4 4 -1]

> [ 7 0 -3]

I used to use [ ] around a matrix because I thought that was the correct

notation. Then switched to | |, then Graham told me that that is wrong

because it's for the determinant, and he uses ( ) so I switched to that.

Now I see you using [ ], and my sense is that your methodology on this

list is more highly respected than anyone else's, so I'm inclined to

follow (and go back to what I felt was right in the first place!).

What's the deal on this?

> If we take the absolute value of the determinants of the minors of

> this matrix, we recover the homomorphism:

>

> abs(det([[4, -1],[0,-3]]) = 12, abs(det([[-4, -1],[7,-3]]) = 19, and

> abs(det([-4,4],[7,0]))= 28, recovering the homomorphism column vector

>

> [12]

> [19]

> [28]

>

> from the generators of the kernel. This computation shows these

> two "unison vectors" do in fact generate the kernel. If we perform a

> similar computation for 128/125 and 32805/32768 we first get the

> matrix

>

> [7 0 -3]

> [-15 8 1]

>

> The column vector we get from the absolute values of the determinants

> of the minors of this is:

>

> [24]

> [38]

> [56]

>

> In other words, these two define the kernel of a homomorphism to the

> 24 et division of the octave, which in the 5-limit is "pathological"

> in the sense that you have two separate 12 divisions a quarter-tone

> apart, and we cannot get from one to the other using relationships

> taken from 5-limit harmony, as all of the numbers in the above

> homomorphism are even.

Thanks for using easy examples to explain your methods here.

I'm finally beginning to understand what you write.

This is an interesting result for 24-EDO, but doesn't really describe

accurately the case Paul used as his illustration. It appears here:

<http://www.ixpres.com/interval/td/erlich/srutipblock.htm>, about

2/3 of the way down the page. I've added graphs of the pitch-height

of the notes in the scale, and of the step-sizes between each degree.

The periodicity-block I derived there is a JI system where the

unison-vectors which are not tempered out are not quarter-tones,

but rather the syntonic comma 81:80 = (-4 4 -1) = ~21.5 cents,

and the diaschisma = (11 -4 -2) = ~19.55 cents.

The numbers in the homomorphism still work to count the scale-steps

which represent the prime-factors, but the scale-steps are not

equal. Arranged in descending pitch-height:

ratio ~cents

256/135 1107.821284

15/8 1088.268715

9/5 1017.596288

16/9 996.0899983

27/16 905.8650026

5/3 884.358713

8/5 813.6862861

405/256 794.1337173

3/2 701.9550009

40/27 680.4487113

64/45 609.7762844

45/32 590.2237156

27/20 519.5512887

4/3 498.0449991

81/64 407.8200035

5/4 386.3137139

6/5 315.641287

32/27 294.1349974

9/8 203.9100017

10/9 182.4037121

16/15 111.7312853

135/128 92.17871646

81/40 21.5062896

1/1 0

So the four step-sizes between degrees of the scale,

in ~cents, are:

90.225 _purana_ sruti

70.672 _nyuna_ sruti

21.506 } both of these are conflated

19.553 } into the _pramana_ sruti

and with an anomalous interval of ~92.179 cents between the

highest degree and the "8ve".

If you didn't before, I think now you can see why Paul calls

this "not a well-behaved periodicity-block"... at any rate,

now *I* finally understand!

The homomorphism column vector resulting from your calculations

is also off for this scale; here, it should be:

[24]

[39]

[56]

where the middle value is [39] instead of your [38].

What does this mean? I can see that it's because in this

scale the 38th degree is much farther away from 3/2 than

is the 39th degree, whereas in 24-EDO it's the other

way around. Is there a way to adjust your formula so

that it works for this JI case?

-monz

_________________________________________________________

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

[snipped old stuff from Gene]

> Thanks for using easy examples to explain your methods here.

> I'm finally beginning to understand what you write.

>

> This is an interesting result for 24-EDO, but doesn't really

describe

> accurately the case Paul used as his illustration.

Why not?

> It appears here:

> <http://www.ixpres.com/interval/td/erlich/srutipblock.htm>, about

> 2/3 of the way down the page. I've added graphs of the pitch-height

> of the notes in the scale, and of the step-sizes between each

degree.

>

> The periodicity-block I derived there is a JI system where the

> unison-vectors which are not tempered out are not quarter-tones,

Nor were they in the (snipped) example above.

> but rather the syntonic comma 81:80 = (-4 4 -1) = ~21.5 cents,

> and the diaschisma = (11 -4 -2) = ~19.55 cents.

Monz, you're equivocating! On the page you reference, you write,

"I had determined the finity of this

pitch-set by the unison vectors which form the intervals of a

skhisma [= 3^8 * 5^1 = ~2.0 cents] and a diesis [= 5^-3 = ~41.1

cents]." These are exactly the unison vectors of the (snipped)

example above!

> The numbers in the homomorphism still work to count the scale-steps

> which represent the prime-factors,

No they don't -- for example, look at all the 3:2s in the scale and

count how many scale steps they subtend.

> but the scale-steps are not

> equal.

Why should they be? They wouldn't be equal for any JI periodicity

block.

Arranged in descending pitch-height:

>

> ratio ~cents

>

> 256/135 1107.821284

> 15/8 1088.268715

> 9/5 1017.596288

> 16/9 996.0899983

> 27/16 905.8650026

> 5/3 884.358713

> 8/5 813.6862861

> 405/256 794.1337173

> 3/2 701.9550009

> 40/27 680.4487113

> 64/45 609.7762844

> 45/32 590.2237156

> 27/20 519.5512887

> 4/3 498.0449991

> 81/64 407.8200035

> 5/4 386.3137139

> 6/5 315.641287

> 32/27 294.1349974

> 9/8 203.9100017

> 10/9 182.4037121

> 16/15 111.7312853

> 135/128 92.17871646

> 81/40 21.5062896

> 1/1 0

>

>

> So the four step-sizes between degrees of the scale,

> in ~cents, are:

>

> 90.225 _purana_ sruti

> 70.672 _nyuna_ sruti

> 21.506 } both of these are conflated

> 19.553 } into the _pramana_ sruti

>

> and with an anomalous interval of ~92.179 cents between the

> highest degree and the "8ve".

I'd call that five step-sizes, then. Why is one "anomalous"? It also

occurs between 81/64 and 4/3, for example.

> If you didn't before, I think now you can see why Paul calls

> this "not a well-behaved periodicity-block"... at any rate,

> now *I* finally understand!

You do?

> The homomorphism column vector resulting from your calculations

> is also off for this scale; here, it should be:

>

> [24]

> [39]

> [56]

>

> where the middle value is [39] instead of your [38].

> What does this mean? I can see that it's because in this

> scale the 38th degree is much farther away from 3/2 than

> is the 39th degree, whereas in 24-EDO it's the other

> way around. Is there a way to adjust your formula so

> that it works for this JI case?

I think you're *not* understanding why this is "not a well-behaved

periodicity-block". Sorry!

> From: paulerlich <paul@stretch-music.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Wednesday, December 26, 2001 12:39 PM

> Subject: [tuning-math] Re: Microtemperament and scale structure

>

>

> > The numbers in the homomorphism still work to count the scale-steps

> > which represent the prime-factors,

>

> No they don't -- for example, look at all the 3:2s in the scale and

> count how many scale steps they subtend.

> >

> > So the four step-sizes between degrees of the scale,

> > in ~cents, are:

> >

> > 90.225 _purana_ sruti

> > 70.672 _nyuna_ sruti

> > 21.506 } both of these are conflated

> > 19.553 } into the _pramana_ sruti

> >

> > and with an anomalous interval of ~92.179 cents between the

> > highest degree and the "8ve".

>

> I'd call that five step-sizes, then. Why is one "anomalous"? It also

> occurs between 81/64 and 4/3, for example.

Oh... when I looked I saw it ocurring only in the one place

I described.

> > If you didn't before, I think now you can see why Paul calls

> > this "not a well-behaved periodicity-block"... at any rate,

> > now *I* finally understand!

>

> You do?

>

> > The homomorphism column vector resulting from your calculations

> > is also off for this scale; here, it should be:

> >

> > [24]

> > [39]

> > [56]

> >

> > where the middle value is [39] instead of your [38].

> > What does this mean? I can see that it's because in this

> > scale the 38th degree is much farther away from 3/2 than

> > is the 39th degree, whereas in 24-EDO it's the other

> > way around. Is there a way to adjust your formula so

> > that it works for this JI case?

>

> I think you're *not* understanding why this is "not a well-behaved

> periodicity-block". Sorry!

OK, I thought I was beginning to get it. Can you please explain

in a little more detail why what I wrote here is wrong? Gene

gave an example which resulted in "quartertones", and this

periodicity-block doesn't do that!

Lost in the fog...

-monz

_________________________________________________________

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--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> OK, I thought I was beginning to get it. Can you please explain

> in a little more detail why what I wrote here is wrong? Gene

> gave an example which resulted in "quartertones", and this

> periodicity-block doesn't do that!

First of all, Gene was talking about the case where the two unison

vectors are _tempered out_. He initially thought that the result of

doing that, for this periodicity block (the SAME ONE as your

srutiblock), would be 24-tET. But you _aren't_ tempering out the

unison vectors, so you wouldn't see quartertones anyway, even if Gene

were right.

Secondly, you need to continue where you left off in the archives. I

ended up convincing Gene that this does not in fact result in

quartertones. He then realized that his error was because of

something called "torsion". If you temper out the unison vectors, you

actually get 12-tone equal temperament, not 24-tone equal temperament.

Another reason that this is not a well-behaved periodicity block is

that the 3:2, for example, is not subtended by the same number of PB

steps everywhere it occurs -- even if you only look at 3:2s within

the Indian Diatonic scale. Thus, I feel that your PB interpretation

of the Indian sruti system is quite poor, because the sruti system

prides itself on being able to represent all the 3:2s (at least the

commonly used ones) by the same number of srutis.

>

> Lost in the fog...

>

>

>

> -monz

>

>

>

>

>

> _________________________________________________________

> Do You Yahoo!?

> Get your free @yahoo.com address at http://mail.yahoo.com

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

> First of all, Gene was talking about the case where the two unison

> vectors are _tempered out_. He initially thought that the result of

> doing that, for this periodicity block (the SAME ONE as your

> srutiblock), would be 24-tET.

And I was wrong--I turned the crank of an alorithm, and the 5-limit 24-et popped out, but it wasn't correct. You can't define it in terms of commas in the same way as the 12-et, which leads to the question (which we got into again lately) if we even want to bother with it. This example is where the whole business of "torsion" came from--if you try to define something in terms of commas, you don't get the

24-et, you get a goofball system with two 12-ets separated by an interval whose square is a unison--in other words, it isn't a comma but the square of it is.

> From: paulerlich <paul@stretch-music.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Wednesday, December 26, 2001 1:14 PM

> Subject: [tuning-math] Re: Microtemperament and scale structure

>

>

> First of all, Gene was talking about the case where the two unison

> vectors are _tempered out_. He initially thought that the result of

> doing that, for this periodicity block (the SAME ONE as your

> srutiblock), would be 24-tET. But you _aren't_ tempering out the

> unison vectors, so you wouldn't see quartertones anyway, even if Gene

> were right.

Well... I understood that. But I think the reason I got confused

is because...

> Secondly, you need to continue where you left off in the archives. I

> ended up convincing Gene that this does not in fact result in

> quartertones. He then realized that his error was because of

> something called "torsion". If you temper out the unison vectors, you

> actually get 12-tone equal temperament, not 24-tone equal temperament.

OK, I'm only up to late August, so there's obviously some more

important stuff coming up.

> Another reason that this is not a well-behaved periodicity block is

> that the 3:2, for example, is not subtended by the same number of PB

> steps everywhere it occurs -- even if you only look at 3:2s within

> the Indian Diatonic scale. Thus, I feel that your PB interpretation

> of the Indian sruti system is quite poor, because the sruti system

> prides itself on being able to represent all the 3:2s (at least the

> commonly used ones) by the same number of srutis.

Thanks, Paul, this helps a lot.

-monz

Do You Yahoo!?

Get your free @yahoo.com address at http://mail.yahoo.com

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

> OK, I'm only up to late August, so there's obviously some more

> important stuff coming up.

Monz, it means a lot to me that you are taking the time and trouble

to go through this stuff, and hopefully incorporate as much of it as

possible into your webpages. It all feels a little more "meaningful",

somehow . . .

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:

>

> > From: paulerlich <paul@s...>

> > To: <tuning-math@y...>

> > Sent: Wednesday, December 26, 2001 1:14 PM

> > Subject: [tuning-math] Re: Microtemperament and scale structure

> >

> >

> > First of all, Gene was talking about the case where the two

unison

> > vectors are _tempered out_. He initially thought that the result

of

> > doing that, for this periodicity block (the SAME ONE as your

> > srutiblock), would be 24-tET. But you _aren't_ tempering out the

> > unison vectors, so you wouldn't see quartertones anyway, even if

Gene

> > were right.

>

>

> Well... I understood that. But I think the reason I got confused

> is because...

>

>

> > Secondly, you need to continue where you left off in the

archives. I

> > ended up convincing Gene that this does not in fact result in

> > quartertones. He then realized that his error was because of

> > something called "torsion". If you temper out the unison vectors,

you

> > actually get 12-tone equal temperament, not 24-tone equal

temperament.

>

>

> OK, I'm only up to late August, so there's obviously some more

> important stuff coming up.

>

>

> > Another reason that this is not a well-behaved periodicity block

is

> > that the 3:2, for example, is not subtended by the same number of

PB

> > steps everywhere it occurs -- even if you only look at 3:2s

within

> > the Indian Diatonic scale. Thus, I feel that your PB

interpretation

> > of the Indian sruti system is quite poor, because the sruti

system

> > prides itself on being able to represent all the 3:2s (at least

the

> > commonly used ones) by the same number of srutis.

>

>

> Thanks, Paul, this helps a lot.

Would you mind mentioning the torsion bit, and the point in the last

paragraph above, in your srutiblock page? It would seem appropriate.