Yahoo Tuning Groups Ultimate Backup tuning-math Re: Microtemperament and scale structure

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.

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗genewardsmith@juno.com

--- 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.

🔗Paul Erlich <paul@stretch-music.com>

--- 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).

🔗genewardsmith@juno.com

--- 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.

🔗genewardsmith@juno.com

🔗graham@microtonal.co.uk

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

🔗genewardsmith@juno.com

--- 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.

🔗graham@microtonal.co.uk

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

🔗monz <joemonz@yahoo.com>

🔗genewardsmith@juno.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.

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- 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?)

🔗genewardsmith@juno.com

🔗carl@lumma.org

🔗genewardsmith@juno.com

🔗carl@lumma.org

🔗graham@microtonal.co.uk

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

🔗graham@microtonal.co.uk

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗Paul Erlich <paul@stretch-music.com>

🔗carl@lumma.org

🔗Paul Erlich <paul@stretch-music.com>

🔗carl@lumma.org

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

🔗Paul Erlich <paul@stretch-music.com>

🔗Dave Keenan <D.KEENAN@UQ.NET.AU>

🔗genewardsmith@juno.com

--- 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?

🔗genewardsmith@juno.com

--- 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?

🔗graham@microtonal.co.uk

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

🔗graham@microtonal.co.uk

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 )

(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

🔗graham@microtonal.co.uk

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

🔗Paul Erlich <paul@stretch-music.com>

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

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.

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- 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!

🔗genewardsmith@juno.com

--- 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?

🔗Dave Keenan <D.KEENAN@UQ.NET.AU>

🔗graham@microtonal.co.uk

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

🔗graham@microtonal.co.uk

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

🔗Paul Erlich <paul@stretch-music.com>

🔗graham@microtonal.co.uk

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

🔗Paul Erlich <paul@stretch-music.com>

--- 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?

🔗Graham Breed <graham@microtonal.co.uk>

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

🔗Paul Erlich <paul@stretch-music.com>

--- 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.

🔗monz <joemonz@yahoo.com>

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

_________________________________________________________
Do You Yahoo!?
Get your free @yahoo.com address at http://mail.yahoo.com

🔗paulerlich <paul@stretch-music.com>

--- 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!

🔗monz <joemonz@yahoo.com>

> 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

_________________________________________________________
Do You Yahoo!?
Get your free @yahoo.com address at http://mail.yahoo.com

🔗paulerlich <paul@stretch-music.com>

--- 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

🔗genewardsmith <genewardsmith@juno.com>

🔗monz <joemonz@yahoo.com>

> 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

🔗paulerlich <paul@stretch-music.com>

--- 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.