Tuning-Math messages 775 - 799

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Message: 775

Date: Tue, 21 Aug 2001 21:59:06

Subject: Re: The hypothesis

From: genewardsmith@j...

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

> So I ask you again -- why not leave the question of how to tune the 
> octave as an outside question, and deal with scales as if they 
exist 
> in a cyclic continuum, modulo the octave, in the majority of our 
> manipulations?

That's certainly the way to deal with scales--I was still considering 
tuning and temperament.


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Message: 776

Date: Tue, 21 Aug 2001 22:25:51

Subject: Re: The hypothesis

From: genewardsmith@j...

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

> Yes and no. I suppose here we've hit the limits of Gene's 
formalism. 
> The chromatic unison vector is not a true equivalence, but is 
> considered an interval too small (or too whatever) to keep in the 
> resulting scale. It's an abrupt boundary in the lattice.

The dawn breaks! In other words, the difference between chromatic and 
commatic unison vectors has nothing to do with the kernel, it's 
purely a matter of tuning!


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Message: 777

Date: Tue, 21 Aug 2001 23:45:01

Subject: Scales

From: genewardsmith@j...

Let's see if I understand the situation. In a p-limit system of 
harmony G, containing n primes, we select n-1 unison vectors and the 
prime 2, which is a sort of special unison vector. This defines a 
kernel K and so congruence classes on G, and a homomorphic image H = 
G/K which under the best circumstances is cyclic but which in any 
case is a finite abelian group of order h. We now select a set of 
coset representatives in G for H by taking elements in one period of 
the period lattice defined by the n generators of K.

Now we select out a certain number m+1 of the unison vectors, 
including 2, and call them (aside from the 2) chromatic. The 
remaining unison vectors now define a temperament whose codimension 
is m+1, and hence is an m-dimensional temperament. We now select a 
tuning for that temperament, and we have a scale with h tones in an 
octave. If m = 0 we have a just scale; if m = 1 we have the scales 
with one chromatic vector which people have been talking about.

I'm not sure if this helps anyone but me, but I think I am getting it.


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Message: 778

Date: Wed, 22 Aug 2001 00:02:20

Subject: Re: The hypothesis

From: Dave Keenan

--- In tuning-math@y..., graham@m... wrote:
> In this case, it happens that the 
> simplest tuning will always have one interval in the set under 
> consideration just.  As a mathematician, perhaps you can prove this.

I'm pretty sure I once found a counterexample (but the margin was too 
small to contain it :-) where minimising the max absolute error did 
not produce any rational n-limit interval because the two equal 
maximum errors were in intervals that had no common factor e.g. 2:3 
and 5:7.

-- Dave Keenan


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Message: 779

Date: Wed, 22 Aug 2001 00:23:55

Subject: Re: Chromatic = commatic?

From: Dave Keenan

--- In tuning-math@y..., graham@m... wrote:
> In-Reply-To: <9lt3q7+gtig@e...>
> In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:
> > Why do you say 
> > linear temperament, which we've just determined means rank 2?
> 
> Because the rest of the world says "linear temperament" and has done 
so 
> for longer than we've been alive.

Sorry if I'm responsible for confusing you, Gene.

Meantone remains a 5-limit linear temperament whether you are 
including 2s or ignoring them. When including 2s it appears 2 
dimensional, when ignoring 2s it appears 1 dimensional.

I really like your approach of treating 2s just like the other primes. 
We then need to switch our thinking to integer-limit rather than 
odd-limit when deciding which errors we care about. Or how about an 
a*b complexity limit for ratios a:b?

-- Dave Keenan


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Message: 780

Date: Wed, 22 Aug 2001 02:05:37

Subject: Interpreting Graham's matrix

From: genewardsmith@j...

Let's consider the situation where 81/80 is the commatic unison 
vector, 25/24 is the chromatic unison vector, and 2 is what you might 
call the repetition vector, defining the unit which is repeated 
(since of course that doesn't actually need to be octaves.) Then if 
each of these defines the row of a matrix, we get Graham's matrix

    [ 1  0  0]
M = [-3 -1  2]
    [-4  4 -1]

The inverse of this is

             [ 7 0 0]
M^(-1) = 1/7 [11 1 2]
             [16 4 1].

The question is how to interpret M^(-1). One use for it can be found 
when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1)
and select all notes [a b c] such that

   0 <= i < 1
-1/2 <= j < 1/2
-1/2 <= k < 1/2

we obtain the seven-note scale 

1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)

with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone 
system which equates 9/8 and 10/9 (which means we are using the fact 
that 81/80 is a commatic vector) we get a PB scale with just two 
sizes of intervals, LsLLLsL, ready to delight the fans of modal 
melodies. Does this have anything to do with Graham's comments on 
this matrix?


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Message: 781

Date: Tue, 21 Aug 2001 20:26:30

Subject: Re: Tetrachordal alterations

From: David C Keenan

I've ignore tetrachordality for too long, it's time I figured out how to
make it happen in any linear temperament. Paul, I had hoped you had more of
a handle on it than I did. It seems not.

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:
> > Are "tetrachordal alterations" only possible when the interval of 
> > repetition is some whole-number fraction of octave?
> 
> In an MOS, the interval of repetition is _always_ some whole-number 
> fraction of the interval of equivalence.

I meant "fraction" in the popular sense, as not including the whole. But
no, I now believe tetrachordal (not necc. omnitetrachordal) alterations are
always possible, it's just a question of finding the minimal alteration
that makes it tetrachordal and just how big that alteration turns out to be
in each case.

It's not clear to me why a tetrachordal scale in some linear temperament,
must have the same number of notes as a MOS in that temperament, but I'll
assume it for now. Know any good reason(s)?

I'd measure the magnitude of a particular alteration (the alteredness?) by
how many notes would have to be added to the original MOS, by consecutive
generators, keeping all chains the same, before you can contain the
tetrachordal scale. e.g. Your tetrachordal decatonic has an alteredness of
2, the minimum possible with a half-octave period.

> > How do you do them, in general?
> 
> I don't know if there's a general way, but you understand what 
> omnitetrachorality is, right?

Yes. But I wasn't considering _omni_tetrachordality. That looks too hard
for now. Do any popular or historical scales have it?

> "Alteration" simply means re-shuffling 
> the step sizes in an MOS or hyper-MOS.
> > 
> > What would be a "tetrachordal alteration" of Blackjack?
> 
> Don't know if there is one! Can you make a blackjack-like scale 
> omnitetrachordal?

God knows!

But I had previously failed to appreciate that Blackjack (a 21 note chain
of Miracle generators = secors = 116.7c) is tetrachordal. duh! You get
disjunct 9-step tetrachords when you start from the point of symmetry.

Ok. Lets look at the 10 note Miracle MOS. What's the minimum alteration to
make it tetrachordal. This just amounts to asking: What is the 10 note
tetrachordal Miracle scale that spans the fewest secors.

I'll notate my tetrachords as the conventional C...F and G...C. Since the
tetrachords must share exactly one note (C) and must be melodically
identical, a 10-note tetrachordal scale must have 5 notes in each
tetrachord and 1 note outside the tetrachords. (The possibility of 3 notes
outside the tetrachord can be ignored because even Blackjack has only 2 and
we expect to get out of it with fewer generators than Blackjack).

Here's what we start with

             5                   1
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                                 5                   1

That's a long enough chain of secors (Blackjack) where we will number the
notes of each tetrachord of our 10-note scale in pitch order. One
tetrachord above and one below. We'll then mark the note that's outside the
tetrachords with an "X". We'll try to stay as close to the center of the
chain as possible.

Note that X can go on either Gb^ or F#v. So here are eight minimal
possibilities. They all span 16 notes in the chain.

             5                   1  2   3  4
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                X                5               X   1  2   3  4

          4  5                   1  2   3
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                X             4  5               X   1  2   3

      3   4  5                   1  2
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                X         3   4  5               X   1  2

   2  3   4  5                   1
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                X      2  3   4  5               X   1

Actually, not all eight are distinct. Some are simply transposed. There are
really only 5. 3 of which are disjunct-tetrachordal in 2 positions.

Here's what they look like melodically (in steps of 72-tET). Vertical bar
"|" is used to show tetrachords.

7 7 7 9|5 7|7 7 7 9
7 7 9 7|5 7|7 7 9 7     7 7 7 9|7 5|7 7 7 9
7 9 7 7|5 7|7 9 7 7     7 7 9 7|7 5|7 7 9 7
9 7 7 7|5 7|9 7 7 7     7 9 7 7|7 5|7 9 7 7
                        9 7 7 7|7 5|9 7 7 7

The middle one (the most even) looks like a very interesting detempering of
your tetrachordal (pentachordal) decatonic. 
It has 
2 of 4:5:6:7
2 of 1/(7:6:5:4)
2 of 4:5:6
2 of 1/(6:5:4)
2 of 3:7:9:21   ASS
1 of 3:9:11:33  ASS
and probably some necessarily-tempered ASSes.

But I don't see any sense in which any of these tetrachordal scales are an
"alteration" of the 10 note MOS. It seems to me that tetrachordal scale
creation (in a given temperament) is not related to MOS scale creation (in
the same temperament) in any way.

Here's the most compact (on the chain) inversionally-symmetric 7-note
tetrachordal scale in Miracle.

          3  4                   1  2
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                              3  4                   1  2
Steps 7 16 7|12|7 16 7

Here's the most even one that fits in Blackjack.

   2         4                   1         3
C> D[ Eb< Ev F  Gb^ G> A[ Bb< Bv C  Db^ D> E[ F< F#v G  Ab^ A> B[ C<
                       2         4                   1         3
Steps 9 12 9|12|9 12 9

Which is of course the neutral thirds MOS, or Mohajira.

These are even less of an alteration of a MOS.

-- Dave Keenan
Brisbane, Australia
Dave Keenan's Home Page *


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Message: 782

Date: Wed, 22 Aug 2001 07:16:40

Subject: A counterexample to the hypothesis?

From: genewardsmith@j...

Let's see if I understand the hypothesis well enough to have found a 
counterexample.

Suppose I decide to construct a scale with 15 notes to the octave in 
7-limit harmony. I therefore turn my attention to the homomorphic map 
[15, 24, 35, 42] which in spite of the way I have written it I may 
consider to be a column vector. The kernel of this homomorphism is 
generated by 28/27, 64/63 and 126/125, of which the latter two seem 
to be good candidates for commatic unisons and the first for our 
chromatic unison. Setting up a Graham-type matrix and inverting it, I 
find 15 notes within an octave forming a scale:

1 - 21/20 - 10/9 - 9/8 - 6/5 - 5/4 - 4/3 - 7/5 -
10/7 - 3/2 - 8/5 - 5/3 - 16/9 - 9/5 - 40/21 - (2)

This scale has intervals of sizes 81/80, 50/49, 25/24, 21/20,
16/15 and 200/189. Since six intervals of widely differing sizes will 
surely drive us nuts, we decide to temper out the two unison vectors, 
namely 64/63 and 126/125. One simple way to do this is to use the 27-
et, which has both of them in its kernel (like the 15-et.) When we do 
this, we find that 81/80 and 50/49 go to intervals of one scale step, 
while the rest go to intervals of two scale steps. We now have the 
pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a 
third of an octave. Isn't this a counterexample?


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Message: 783

Date: Wed, 22 Aug 2001 08:14:07

Subject: Re: About CS

From: genewardsmith@j...

--- In tuning-math@y..., Pierre Lamothe <plamothe@a...> wrote:

> The CS property corresponds to the main axiom in the gammoid theory 
(the
> _congruity_ condition) and subtends the _periodicity_ concept in Z-
module.
> If you flush CS you loss the mathematical sense of _interval class_ 
(or do
> you have a _fuzzy class_ definition to replace it?) 
> 
> I can't see how you could define mathematically such terms as step, 
degree,
> class, mode . . . in a system S without the existence of a 
primordial
> epimorphism S --> Z which gives its consistence.

Hmmm. A Z-module is the same as an abelian group, which I've been 
raving about, and an epimorphism to Z sounds like an ET to me. Other 
than that I don't know what this is about, but I wonder if Pierre 
shows up here from time to time?


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Message: 784

Date: Wed, 22 Aug 2001 12:56 +0

Subject: Re: A counterexample to the hypothesis?

From: graham@m...

In-Reply-To: <9lvm8o+9v8r@e...>
Gene wrote:

> This scale has intervals of sizes 81/80, 50/49, 25/24, 21/20,
> 16/15 and 200/189. Since six intervals of widely differing sizes will 
> surely drive us nuts, we decide to temper out the two unison vectors, 
> namely 64/63 and 126/125. One simple way to do this is to use the 27-
> et, which has both of them in its kernel (like the 15-et.) When we do 
> this, we find that 81/80 and 50/49 go to intervals of one scale step, 
> while the rest go to intervals of two scale steps. We now have the 
> pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a 
> third of an octave. Isn't this a counterexample?

No, it's an MOS with a period of a third of an octave for the purposes of 
the hypothesis.  The generator is the interval L.


            Graham


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Message: 785

Date: Wed, 22 Aug 2001 12:56 +0

Subject: Re: Interpreting Graham's matrix

From: graham@m...

In-Reply-To: <9lv41h+ag6a@e...>
Gene wrote:

> The inverse of this is
> 
>              [ 7 0 0]
> M^(-1) = 1/7 [11 1 2]
>              [16 4 1].
> 
> The question is how to interpret M^(-1). One use for it can be found 
> when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1)
> and select all notes [a b c] such that
> 
>    0 <= i < 1
> -1/2 <= j < 1/2
> -1/2 <= k < 1/2
> 
> we obtain the seven-note scale 
> 
> 1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)

That is clever!  The list of [a b c] is

[0 0 0]
[1 -2 1]
[1 1 -1]
[2 -1 0]
[-1 1 0]
[0 -1 1]
[0 2 -1]
[1 0 0]

So you multiply that by the inverse (*not* the adjoint) and get

[0 0 0]
[0.142857143 0.285714286 -0.428571429]
[0.285714286 -0.428571429 0.142857143]
[0.428571429 -0.142857143 -0.285714286]
[0.571428571 0.142857143 0.285714286]
[0.714285714 0.428571429 -0.142857143]
[0.857142857 -0.285714286 0.428571429]
[1 0 0]

Where, indeed, the first column lies between 0 and 1, and the others 
between -1/2 and 1/2.  Furthermore, the list is already ordered by pitch, 
before we even worked out any pitches!  That's because the left hand 
column, being "the thing that's left over when all the unison vectors are 
taken out", defines approximate pitch order.

Over to Paul: is this a general method for finding a periodicity block?

> with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone 
> system which equates 9/8 and 10/9 (which means we are using the fact 
> that 81/80 is a commatic vector) we get a PB scale with just two 
> sizes of intervals, LsLLLsL, ready to delight the fans of modal 
> melodies. Does this have anything to do with Graham's comments on 
> this matrix?

I'll bring the inverse down again, and lose the 1/7 to make it the 
adjoint.

[ 7 0 0]
[11 1 2]
[16 4 1]

The first row defines the octave (which also happens to be the interval of 
equivalence) the second defines the 3:1 approximation and the third the 
5:4 approximation.

The interval 9:8 is [-3 2 0].  So multiply [-3 2 0] by adj(M) to get

[1 2 4]

The first column tells you the number of steps in the scale LsLLLsL the 
interval spans.  9:8 is L so spans 1 step.  The second column tells you 
the number of generators give the interval in the system where 81:80 is 
tempered out (ie meantone).  The generator is the fifth (or twelfth, etc), 
so 9:8 is two fifths (octave reduced).  The last column defines the 
interval in the system where 25:24 is tempered out.  I'll guess that's 
correct.

10:9 is [1 -2 1].  Multiply by Adj(M) and we get [1 2 3].  The first two 
columns are the same as for 9:8 which is no surprise because 9:8 and 10:9 
are identical in meantone.  That the third column differs by 1 could well 
tell you that 9:8 and 10:9 differ by a comma.

16:15 is [4 -1 -1].  This multiplies out to [1 -5 -3].  So it's also one 
scale step (s this time).  And it's generated by -5 fifths, or 5 fourths.  
That is C-F-Bb-Eb-Ab-Db.

The same process should work for all intervals.

Now, the clever part is that if we take m, the octave equivalent of M,

[-1  2]
[ 4 -1]

 the adjoint is

[-1 -2]
[-4 -1]

sign isn't important, so make that

[1 2]
[4 1]

This is identical to a subset of adj(M).  This will always be the case for 
correct unison vectors, but I don't have the criteria for those unison 
vectors being "correct".  It tells us the octave-equivalent mapping in 
terms of fifths for all 5-limit intervals.  The determinant tells us the 
number of notes in the periodicity block, but nothing I can see tells us 
what order they should be in.  So that's all we lose by being octave 
equivalent.

                     Graham


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Message: 787

Date: Wed, 22 Aug 2001 15:05 +0

Subject: Re: Generators and unison vectors

From: graham@m...

In-Reply-To: <9lug7f+8e65@e...>
Paul wrote:

> > C   D   E   F   G   A   B   C
> >  t+p  t   s  t+p  t  p+t  s  
> 
> How did you get this, without appealing to a parallelogram or some 
> such construction?

It's an arbitrary scale, not generated from anything more fundamental.  

> > Making (4 -1 -1)H or 16:15 another commatic unison vector gives us 
> > 5-equal:
> > 
> >      ( 5)(t)
> > H' = ( 8)
> >      (12)
> > 
> > Gene's already explained this in algebraic terms, but I think the 
> example 
> > makes it clearer.  It means I can now interpret
> > 
> > ( 1 -2  1) 
> > ( 4 -1 -1)
> > (-4  4 -1) 
> > 
> > as the octave-specific equivalent of a periodicity block.  
> Unfortunately 
> > it only has 1 note,
> 
> How so?

Everything's a unison!

> > so the analogy breaks down.  But whatever, I'll call 
> > all the intervals unison vectors anyway.
> 
> _All_ the intervals?

Yep!

> > Now, we can also define meantone as
> > 
> > ( 1  0  0)        (1 0)
> > ( 0  1  0)C   =   (0 1)
> > (-4  4 -1)        (0 0)
> > 
> > Here, the octave and twelfth are taking the place of the chromatic 
> unison 
> > vectors.  So, can we call them unison vectors?
> 
> A unison vector has to come out to a musical _unison_! An augmented 
> unison maybe, but still a unison.

Fokker says, in "Unison Vectors and Periodicity Blocks" 
<A.D. Fokker: Unison Vectors and Periodicity Blocks *> "All pairs of notes 
differing by octaves only are considered unisons."  That's not far from 
"an octave is a unison vector".

> > I think not, because C 
> > evaluates to
> > 
> > ( 1 0)
> > ( 0 1)
> > (-4 4)
> > 
> > That has negative numbers in it, which means the generators (for so 
> they 
> > are) don't add up to give all the primary consonances.
> 
> Don't follow.

You can't define the 5:4 approximation by adding octaves and twelfths.  
This gives us a way of defining "small" intervals without needing interval 
sizes.  We need to construct a scale somehow or other, and say that scale 
has to be constructed by adding together a set of linearly independent 
small intervals.  The only problem is that they can be smaller than a 
unison.  So adding a major and minor tone can be replaced by a major third 
and a descending minor tone.  So the next constraint is that the scale has 
to have monotonically increasing pitch, which you need the metric for.

> > So now we have a 
> > definition for what interval qualifies as a "unison vector".  
> > Interestingly, a fifth would work as a generator here, but not for 
> a 
> > diatonic scale.
> 
> You've completely lost me.

Four fifths add up to a 5:1 approximation.  But you can't define a 
diatonic scale by adding fifths and octaves.

> > An alternative to "unison vector" would be "melodic generator".
> 
> Ouch! Are you sure? The steps in the scale can't be unison vectors, 
> nor can the generator of the scale be a unison vector.

I don't know what to call them.

> > Usually we call the octave the period and a fifth (which is 
> therefore 
> > equivalent to the twelfth) the generator.  It isn't much of a 
> stretch to 
> > think of two generators.
> 
> Gene has referred to that concept, I believe.

Yes.

> > In octave-equivalent terms, a unison is:
> > 
> > v v v v v v v v v v v w
> > 
> > where v is s fifth and w is a wolf.  The same algebra applies, but 
> it's 
> > harder to visualise.  I think the fifth here does count as a unison 
> > vector.
> 
> Please, please, please don't refer to a fifth as a unison vector. 
> Musical intervals go: unison, second, third, fourth, fifth. There's a 
> long way between a unison and a fifth.

Probably it'd be better written as 

v v v v v v v v v v v v p

where v is a fifth and p is a Pythagorean comma.  That means p is the 
(chromatic) unison vector.  That leaves v as "the thing you have left when 
you take out all the unison vectors".  That's what we need a name for.

> > Defining an MOS in octave-equivalent terms is easy, once you 
> realize that 
> > the generators that pop out needn't be unison vectors in octave-
> specific 
> > space.
> 
> They can't be unison vectors in any space! If the generator is a 
> unison vector, you never get beyond the first note of the scale and 
> its chromatic alterations.

But the generator does have a close relationship to the chromatic unison 
vector, in an octave equivalent system.  You're taking out everything 
except the chromatic unison vector, and describing what you have left in 
terms of the generator.


                  Graham


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Message: 788

Date: Wed, 22 Aug 2001 15:05 +0

Subject: Re: The hypothesis

From: graham@m...

In-Reply-To: <9luhvn+n2lb@e...>
Gene wrote:

> > Oh, right, but whatever they are, they're octave invariant.  
> CLAMPITT.PDF 
> > is a relevant place for definitions, as it's already been 
> referenced: "By 
> > /scale/ we refer to a set of pitches ordered according to ascending 
> > frequencies (pitch height) bounded by an interval of periodicity."
> 
> This definition of a scale does not assume it repeats octaves, so it 
> does not assume octave invariance.

It assumes it repeats, so it may as well repeat at the octave.

> OK, we can consider a conceptual scale to be something defined on the 
> level of the homomorphic image (including the identity map.) Then C 
> major (or mean tone diatonic) is a conceptual scale in the mean-tone 
> linear temperament, which becomes a tuned scale if you pick a tuning. 
> How's that?

Sounds okay.

> > It looks like a commatic UV would be in the kernel, and a chromatic 
> UV 
> > would not.
> 
> So far as I can see, they are both in the kernel and refer to the 
> same thing.

They aren't the same thing.  I'm not sure if they'd be in the kernel.  I'd 
need to understand what "kernel" means a little better.

> > You certainly can tell a small interval if 2 is taken out.  Of 
> course, you 
> > need a way of calculating interval size.  As I see it, that lies 
> outside 
> > of the group theory we've been discussing so far.  There are two 
> ways of 
> > doing it, corresponding to the two interpretations above.
> 
> I think your two ways in effect put the 2 back in by calculating what 
> it must be.

Oh.  Well, we can take factors of 2 out of the algebra.

> >If 
> > it doesn't work out that way perhaps you, as the mathematician, can 
> tell 
> > us the size constraints on the unison vectors.
> 
> So far as I can see, there aren't any. A unison vector is simply 
> anything you've decided to regard as a unison, and so lies in (or 
> generates?) the kernel.

When working on my MOS finding program, I found some unison vectors didn't 
work.  For example, take the meantone matrix

[ 1  0  0]
[-3 -1  2]
[-4  4 -1]

The (negative) adjoint is

[ 7 0 0]
[11 1 2]
[16 4 1]

The left hand column defines 7-equal.  Now, if you replace 81:80 by 
160:81,

[ 1  0  0]
[-3 -1  2]
[ 5 -4  1]

you get

[ 7 0  0]
[13 1 -2]
[17 4 -1]

so it doesn't work.  Which means the size of the unison vectors must be 
important.  However, for octave-equivalent matrices it doesn't seem to 
matter at all.  Which is good, because "interval size" is harder to define 
anyway.  The adjoint of

[-1  2]
[ 4 -1]

is

[-1 -2]
[-4 -1]

And the adjoint of 

[-1 2]
[-4 1]

is

[1 -2]
[4 -1]

both define the same linear temperaments.

> > > You can generate by fifths and octaves if you want, but you don't 
> > > need to.
> 
> > You do need to if you want to enforce octave equivalence. 
> 
> I don't see why. Why not by octaves and major thirds instead?

It won't work for meantone temperament.  The temperament it does work for 
is excellent.  I have an example in it at 
<http://www.microtonal.co.uk/magicpump.mp3 - Ok *>.

>  > So the nature of this problem is such that I do need to make two 
> intervals 
> > exact.  There are plenty of other problems that have different 
> natures, 
> > but I'm puzzled as to why you insist on bringing them up when we're 
> > discussing octave-invariant systems.
> 
> I was discussing, among other things, tuning and temperament, where 
> the question of octave tuning is relevant. As I pointed out, tuning 
> an octave as a 2 is a choice of tuning.

The subject line says "The hypothesis" and Paul's hypothesis most 
definitely concerns an octave-equivalent system.


                          Graham


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Message: 789

Date: Wed, 22 Aug 2001 15:05 +0

Subject: Re: Chromatic = commatic?

From: graham@m...

In-Reply-To: <9lufbl+8f7m@e...>
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


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Message: 790

Date: Wed, 22 Aug 2001 15:05 +0

Subject: Re: Chromatic = commatic?

From: graham@m...

In-Reply-To: <9lul29+57gl@e...>
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


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Message: 791

Date: Wed, 22 Aug 2001 19:19:22

Subject: Re: Chromatic = commatic?

From: Paul Erlich

--- In tuning-math@y..., genewardsmith@j... wrote:
> 
> If you are looking *only* at diatonic music and not allowing any 
> accidentals,

We _are_ allowing accidentals, and they play a very specific role in 
diatonic music (as opposed to, say, atonal serial music).


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Message: 792

Date: Wed, 22 Aug 2001 19:24:53

Subject: Re: Scales

From: Paul Erlich

--- In tuning-math@y..., genewardsmith@j... wrote:
> Let's see if I understand the situation. In a p-limit system of 
> harmony G, containing n primes, we select n-1 unison vectors and 
the 
> prime 2, which is a sort of special unison vector. This defines a 
> kernel K and so congruence classes on G, and a homomorphic image H 
= 
> G/K which under the best circumstances is cyclic but which in any 
> case is a finite abelian group of order h.

What differentiates the "best circumstances" and just "any case" here?

> We now select a set of 
> coset representatives in G for H by taking elements in one period 
of 
> the period lattice defined by the n generators of K.
> 
> Now we select out a certain number m+1 of the unison vectors, 
> including 2, and call them (aside from the 2) chromatic. The 
> remaining unison vectors now define a temperament whose codimension 
> is m+1, and hence is an m-dimensional temperament. We now select a 
> tuning for that temperament, and we have a scale with h tones in an 
> octave. If m = 0 we have a just scale;

You mean an equal temperament?

> if m = 1 we have the scales 
> with one chromatic vector which people have been talking about.

Sounds right.


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Message: 793

Date: Wed, 22 Aug 2001 19:27:00

Subject: Re: Chromatic = commatic?

From: Paul Erlich

--- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:
> --- In tuning-math@y..., graham@m... wrote:
> > In-Reply-To: <9lt3q7+gtig@e...>
> > In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:
> > > Why do you say 
> > > linear temperament, which we've just determined means rank 2?
> > 
> > Because the rest of the world says "linear temperament" and has 
done 
> so 
> > for longer than we've been alive.
> 
> Sorry if I'm responsible for confusing you, Gene.
> 
> Meantone remains a 5-limit linear temperament whether you are 
> including 2s or ignoring them. When including 2s it appears 2 
> dimensional, when ignoring 2s it appears 1 dimensional.
> 
> I really like your approach of treating 2s just like the other 
primes. 
> We then need to switch our thinking to integer-limit rather than 
> odd-limit when deciding which errors we care about. Or how about an 
> a*b complexity limit for ratios a:b?

When discussing _tuning_, then these can all be useful 
considerations. But when discussing _scales which use octave-
equivalence_, then a*b turns into odd-limit, as I've demonstrated on 
harmonic_entropy.


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Message: 794

Date: Wed, 22 Aug 2001 19:28:51

Subject: Re: Interpreting Graham's matrix

From: Paul Erlich

--- In tuning-math@y..., genewardsmith@j... wrote:
> Let's consider the situation where 81/80 is the commatic unison 
> vector, 25/24 is the chromatic unison vector, and 2 is what you 
might 
> call the repetition vector, defining the unit which is repeated 
> (since of course that doesn't actually need to be octaves.) Then if 
> each of these defines the row of a matrix, we get Graham's matrix
> 
>     [ 1  0  0]
> M = [-3 -1  2]
>     [-4  4 -1]
> 
> The inverse of this is
> 
>              [ 7 0 0]
> M^(-1) = 1/7 [11 1 2]
>              [16 4 1].
> 
> The question is how to interpret M^(-1). One use for it can be 
found 
> when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-
1)
> and select all notes [a b c] such that
> 
>    0 <= i < 1
> -1/2 <= j < 1/2
> -1/2 <= k < 1/2
> 
> we obtain the seven-note scale 
> 
> 1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)

This is exactly how my PB program works, as explained in part 3 of 
the _Gentle Introduction_.


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Message: 795

Date: Wed, 22 Aug 2001 19:39:35

Subject: Re: Tetrachordal alterations

From: Paul Erlich

--- In tuning-math@y..., David C Keenan <D.KEENAN@U...> wrote:
> I've ignore tetrachordality for too long, it's time I figured out 
how to
> make it happen in any linear temperament. Paul, I had hoped you had 
more of
> a handle on it than I did. It seems not.
> 
> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> > --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:
> > > Are "tetrachordal alterations" only possible when the interval 
of 
> > > repetition is some whole-number fraction of octave?
> > 
> > In an MOS, the interval of repetition is _always_ some whole-
number 
> > fraction of the interval of equivalence.
> 
> I meant "fraction" in the popular sense, as not including the 
whole. But
> no, I now believe tetrachordal (not necc. omnitetrachordal) 
alterations are
> always possible, it's just a question of finding the minimal 
alteration
> that makes it tetrachordal and just how big that alteration turns 
out to be
> in each case.
> 
> It's not clear to me why a tetrachordal scale in some linear 
temperament,
> must have the same number of notes as a MOS in that temperament, 
but I'll
> assume it for now. Know any good reason(s)?

Because the unison vectors define a set of N equivalence classes, 
which we typically map to letters of the alphabet, etc. Melodies, and 
even harmonies, will be understood in terms of patterns in this set 
of N. If you add or eliminate notes, you'll be ruining these patterns.

In other words, the tetrachordal scale should be a periodicity block 
(not necessarily of the Fokker parallelogram type).
> 
> Yes. But I wasn't considering _omni_tetrachordality. That looks too 
hard
> for now. Do any popular or historical scales have it?

The most popular ones do -- the diatonic and pentatonic.
> 
> Here's what they look like melodically (in steps of 72-tET). 
Vertical bar
> "|" is used to show tetrachords.
> 
> 7 7 7 9|5 7|7 7 7 9
> 7 7 9 7|5 7|7 7 9 7     7 7 7 9|7 5|7 7 7 9
> 7 9 7 7|5 7|7 9 7 7     7 7 9 7|7 5|7 7 9 7
> 9 7 7 7|5 7|9 7 7 7     7 9 7 7|7 5|7 9 7 7
>                         9 7 7 7|7 5|9 7 7 7

I might not call these tetrachordal. According to my paper, the 
disjunction is supposed to contain some pattern of scale steps found 
with the tetrachords themselves. But in this case, the 5/72 interval 
isn't found in the tetrachords.

Perhaps we can call these "weakly tetrachordal".

> But I don't see any sense in which any of these tetrachordal scales 
are an
> "alteration" of the 10 note MOS.

Each note should either agree with the corresponding note in the MOS 
or be a chromatic unison vector different.

> It seems to me that tetrachordal scale
> creation (in a given temperament) is not related to MOS scale 
creation (in
> the same temperament) in any way.

I see the periodicity block concept as underlying both.


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Message: 796

Date: Wed, 22 Aug 2001 19:41:01

Subject: Re: A counterexample to the hypothesis?

From: Paul Erlich

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

> We now have the 
> pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a 
> third of an octave. Isn't this a counterexample?

Why would this be a counterexample if LssssLssss wasn't?


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Message: 797

Date: Wed, 22 Aug 2001 19:41:59

Subject: Re: About CS

From: Paul Erlich

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

> but I wonder if Pierre 
> shows up here from time to time?

Not often enough, sadly. How's your French?


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Message: 798

Date: Wed, 22 Aug 2001 19:43:47

Subject: Re: Interpreting Graham's matrix

From: Paul Erlich

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

> Over to Paul: is this a general method for finding a periodicity 
block?

This is how I've been doing it all along -- see part 3 of the _Gentle 
Introduction_.


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Message: 799

Date: Wed, 22 Aug 2001 19:51:02

Subject: Re: Generators and unison vectors

From: Paul Erlich

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

> > > Here, the octave and twelfth are taking the place of the 
chromatic 
> > unison 
> > > vectors.  So, can we call them unison vectors?
> > 
> > A unison vector has to come out to a musical _unison_! An 
augmented 
> > unison maybe, but still a unison.
> 
> Fokker says, in "Unison Vectors and Periodicity Blocks" 
> <A.D. Fokker: Unison Vectors and Periodicity Blocks *> "All pairs of 
notes 
> differing by octaves only are considered unisons."  That's not far 
from 
> "an octave is a unison vector".

The octave, OK, but the twelfth?
> 
> You can't define the 5:4 approximation by adding octaves and 
twelfths.

Why not? We've been doing so all along. The 5:4 approximation is four 
twelfths minus six octaves.
> 
> Four fifths add up to a 5:1 approximation.  But you can't define a 
> diatonic scale by adding fifths and octaves.

Why not? We've been doing so all along, right?

> > > An alternative to "unison vector" would be "melodic generator".
> > 
> > Ouch! Are you sure? The steps in the scale can't be unison 
vectors, 
> > nor can the generator of the scale be a unison vector.
> 
> I don't know what to call them.

Something else, I beseech you!
> 
> Probably it'd be better written as 
> 
> v v v v v v v v v v v v p
> 
> where v is a fifth and p is a Pythagorean comma.  That means p is 
the 
> (chromatic) unison vector.  That leaves v as "the thing you have 
left when 
> you take out all the unison vectors".  That's what we need a name 
for.

The generator?
> 
> > > Defining an MOS in octave-equivalent terms is easy, once you 
> > realize that 
> > > the generators that pop out needn't be unison vectors in octave-
> > specific 
> > > space.
> > 
> > They can't be unison vectors in any space! If the generator is a 
> > unison vector, you never get beyond the first note of the scale 
and 
> > its chromatic alterations.
> 
> But the generator does have a close relationship to the chromatic 
unison 
> vector, in an octave equivalent system.  You're taking out 
everything 
> except the chromatic unison vector, and describing what you have 
left in 
> terms of the generator.

Great! That's a valuable insight! But that doesn't make the generator 
a unison!

Look, the unison vectors together define N equivalence classes in the 
infinite lattice. All that unison vectors can do is to move you 
around within a single equivalence class. You'll never get to the 
other N-1 equivalence classes using unison vectors.


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