Tuning-Math Digests messages 1350 - 1374

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Message: 1350

Date: Tue, 21 Aug 2001 20:27:27

Subject: Re: Generators and unison vectors

From: Paul Erlich

--- In tuning-math@y..., graham@m... wrote:
> I was thinking about what n "unison vectors" in n-dimensional space 
could 
> mean.  In fact, it's what I've already called a "basis".  See 
> <404 Not Found * Search for http://www.microtonal.co.uk/matrix.htm in Wayback Machine>.  I assume that's 
analogous to 
> Gene's "kernel".  Let's revisit the example there:
> 
> (t)   ( 1 -2  1) 
> (s) = ( 4 -1 -1)H
> (p)   (-4  4 -1) 
> 
> Which can be used to define this scale
> 
> 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?
> 
> The inverse of the matrix is
> 
> ( 5 2 3)
> ( 8 3 5)
> (12 4 7)
> 
> It defines H in terms of t, s and p.  That I already knew.  Now, 
the 
> interesting thing is, taking p out of the scale
> 
> C   D   E   F   G   A   B   C
>   t   t   s   t   t   t  s  
> 
> is the same as making p a commatic unison vector.

By definition.

> And we can define the 
> approximate H in terms of t and s
> 
>      ( 5 2)(t)
> H' = ( 8 3)(s)
>      (12 4)
> 
> I already knew this, but I didn't think of it as "designating a 
commatic 
> unison vector" or whatever.  Before we've been thinking about 
starting 
> with commatic unison vectors and making them chromatic.  It 
actually makes 
> more sense the other way around.

If you mean, start with the unison vectors as non-zero intervals and 
then selectively bring some of them to zero, yes, that's how I think 
of it in the lattice/PB regime.

> 
> 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?

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

_All_ the intervals?

> 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.

> 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.

> 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.
> 
> 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.
> 
> 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.
> 
> 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.

> 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.
> 
> That's the way I see it now.  I've also realized that I've been 
saying 
> "octave invariant" when "octave equivalent" would probably be more 
to the 
> point.

I suppose -- but you got me doing it too :)


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Message: 1351

Date: Tue, 21 Aug 2001 20:57:27

Subject: Re: The hypothesis

From: genewardsmith@xxxx.xxx

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

> > Considering scales is another level of generality altogether--
first 
> > we have approximations (kernels, unison vectors, and so forth) 
then 
> > we have tuning, and finally we select a subset and have scales.

> 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.

> I don't agree with this anyway.  You can have a scale without a 
tuning.  

> For example, a C major scale can be tuned to either 12- or 19-equal.

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?

> 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.

> > If we consider equivalence classes modulo octaves, the 5-limit is 
> > free of rank two, but I don't know what you mean by "cyclic 
around 
> > the octave".

> You describe an ET as cyclic below.

I said an ET modulo ocatves was cyclic. In the case above, where we 
are considering equivalence classes of notes with representative 
elements 3^a * 5^b, we have't modded out and we have something which 
is free, not cyclic. By setting h(3^a*5^b) = 7*a+2*b (mod 12) we do 
get a circle of note equivalence classes. 

> > An ET would be free of rank 1, or "cyclic of infinite order". If 
we 
> > mod out by octaves, it would no longer be free but would (still) 
be 
> > cyclic, which implies one generator.

> But if we "mod out" by one scale step, it would still have rank 1?

It would have one generator, but the rank is the rank of the group 
modulo torsion, so in this case the rank is 0. There is a structure 
theorem for finitely generated abelian groups (which is what we have 
been considering) which gives a complete set of invariants specifying 
the group; the rank being one of them and perhaps the most important.
In the case of C(12), it is in terms of groups of prime power order
C(3) x C(4), which specifies it.

> 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.

>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.

> > 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?

> > However, there is nothing in the nature of the problem to suggest 
I 
> > need to make any interval exact. One obvious way to decide would 
be 
> > to pick a set of intervals {t1, ... , tn} which I want to be well 
> > approximated, and a corresponding set of weights {w1, ... , wn} 
> > defining how important I think it is to have that interval 
> > approximate nicely. Perhaps I could do this using harmonic 
entropy? 
> > In any case, having done this I now have an optimization problem 
> > which I can decide using the method of least squares. If I have 
two 
> > generators, which I have in the case of the 5-limit with 81/80 a 
> > unison vector, then solving this will give me tunings for the 
> > generators and hence tunings for the entire system. There is no 
> > special treatment given to the octave in this method, but I see 
no 
> > reason in terms of psychological acoustics why there needs to be.

> Hold on, you need more than that. 

Some one else said people around here have done such computations 
many times, so there seems to be some confusion on that score. I 
could certainly do one using the method above, at any rate.

 > 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.


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Message: 1353

Date: Tue, 21 Aug 2001 21:50:01

Subject: Re: Chromatic = commatic?

From: genewardsmith@xxxx.xxx

--- 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?


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Message: 1354

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

Subject: Re: A counterexample to the hypothesis?

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9lvm8o+9v8r@xxxxxxx.xxx>
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: 1355

Date: Wed, 22 Aug 2001 23:50:00

Subject: Re: About CS

From: genewardsmith@xxxx.xxx

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

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

Even worse than his English. :(


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Message: 1356

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

Subject: Re: Interpreting Graham's matrix

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9lv41h+ag6a@xxxxxxx.xxx>
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: 1357

Date: Wed, 22 Aug 2001 23:59:26

Subject: Re: A counterexample to the hypothesis?

From: genewardsmith@xxxx.xxx

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

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

I had somehow gotten the idea such a thing would be a counterexample. 
I didn't look at 10 or 12 because they would have already been picked 
over, so 15 looked like a good composite to try. I think by the way 
that 15 notes out of a 27 scale would be a fine system to try for 
anyone who likes sharp systems and who wants to wash the 12-et out of 
their head for a while.

I have the idea that what we are talking about is some sort of 
equidistribution property, but I can't seem to pin down the property. 
Suppose I define a two-value scale structure to be a periodic 
sequence of L and s values, containing both values. I could try the 
following:

(1) Every contiguous set of n values has the same number of L's and 
s's as every other such set.

The trouble with this is that nothing passes this condition.

(2) Contiguous sets of n values have on average the same number of 
L's and s's.

The trouble with this is that everything passes this condition.


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Message: 1359

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

Subject: Re: Generators and unison vectors

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9lug7f+8e65@xxxxxxx.xxx>
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: 1360

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: 1361

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

Subject: Re: The hypothesis

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9luhvn+n2lb@xxxxxxx.xxx>
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: 1362

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: 1363

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

Subject: Re: Chromatic = commatic?

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9lufbl+8f7m@xxxxxxx.xxx>
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: 1364

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

Subject: Interpreting Graham's matrix

From: genewardsmith@xxxx.xxx

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: 1365

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

Subject: Re: Chromatic = commatic?

From: graham@xxxxxxxxxx.xx.xx

In-Reply-To: <9lul29+57gl@xxxxxxx.xxx>
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: 1366

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: 1367

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

Subject: A counterexample to the hypothesis?

From: genewardsmith@xxxx.xxx

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: 1368

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: 1369

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

Subject: Re: About CS

From: genewardsmith@xxxx.xxx

--- 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: 1370

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: 1371

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: 1372

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: 1373

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: 1374

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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