Yahoo Tuning Groups Ultimate Backup tuning-math The hypothesis

I found a posting by Paul over on the tuning group, and it seems I
may be closing on a statement of the Paul Hypothesis.

"In fact, a few months ago I posted my Hypothesis, which states that
if you temper out all but one of the unison vectors of a Fokker
periodicity block, you end up with an MOS scale. We're discussing
this Hypothesis on tuning-math@y..."

Sounds like we may be getting there, but there seems to be some
confusion as to whether 2 counts as a prime, and so whether for
instance the 5-limit is 2D or 3D. Most of the time it makes sense to
treat 2 like any other prime.

"A temperament can be
constructed by tempering out anywhere from 1 to n unison vectors. If
you temper out n (and do it uniformly), you have an ET. If you temper
out n-1, you have a linear temperament. If you temper out n-2, you
have a planar temperament (Dave Keenan has created some examples of
those)."

From my point of view, the 5-limit is rank (dimension) 3, and the 7-
limit 4, and so forth. If you temper out n-1 unison vectors which
generate a well-behaved kernel, then you map onto a rank-1 group, and
get an equal temperment. So "codimension" 1 (one less than the full
number of dimensions) leads to a rank-1 group. In the same way,
codimension 2 for the kernel leads to a rank 2 group, etc. If for
instance you temper out 81/80 in the 5 limit, the kernel has
dimension 1 and codimension 2, and leads to a rank 2 image group.

We can tune the rank 1 group any way we like so long as the steps are
of the same size, which means that our ET can have stretched or
squashed octaves if we so choose. In the same way, we can tune the
rank 2 group any way we like, except that we need to retain
incommensurability of two generators (or at least to ignore the fact
if they are not.) If we make the octaves pure in our example where
the kernel is generated by a comma, we could for instance make the
fifths pure also, leading to Pythagorean tuning. Alternatively, we
could make the major thirds pure, leading to 1/4 comma mean tone
temperment. (Pythagorean tuning is not considered a temperment, since
the fifth isn't tempered, but it is the same sort of thing
mathematically as 1/4 comma mean-tone temperment.) Other choices lead
to other results, and all we need to do is to ensure the circle of
fifths does not close--or at least to pretend otherwise it if it does.

A rank 3 image group, coming from a kernel of codimension 3, is what
people have been calling a 2D temperment. I hope that clarifies
things (as it does for me) rather than further confuses them!

🔗graham@microtonal.co.uk

In-Reply-To: <9lqhhc+s6ru@eGroups.com>
In article <9lqhhc+s6ru@eGroups.com>, genewardsmith@juno.com () wrote:

> Sounds like we may be getting there, but there seems to be some
> confusion as to whether 2 counts as a prime, and so whether for
> instance the 5-limit is 2D or 3D. Most of the time it makes sense to
> treat 2 like any other prime.

2 is certainly prime, but most of the time we consider octave-invariant
scales.

> "A temperament can be
> constructed by tempering out anywhere from 1 to n unison vectors. If
> you temper out n (and do it uniformly), you have an ET. If you temper
> out n-1, you have a linear temperament. If you temper out n-2, you
> have a planar temperament (Dave Keenan has created some examples of
> those)."

So this is all octave invariant.

> From my point of view, the 5-limit is rank (dimension) 3, and the 7-
> limit 4, and so forth. If you temper out n-1 unison vectors which
> generate a well-behaved kernel, then you map onto a rank-1 group, and
> get an equal temperment. So "codimension" 1 (one less than the full
> number of dimensions) leads to a rank-1 group. In the same way,
> codimension 2 for the kernel leads to a rank 2 group, etc. If for
> instance you temper out 81/80 in the 5 limit, the kernel has
> dimension 1 and codimension 2, and leads to a rank 2 image group.

So the kernel has dimension 1 because it contains 1 unison vector? I
think I see the codimension.

So in octave-invariant terms, 5-limit is rank 2, but cyclic about the
octave. An ET would be rank 0 I suppose, but you've already given the
real name for that case. Tempering out 81/80 would than be rank-1, hence
"linear" temperament. I think that terminology goes back to either Ellis
or Bosanquet.

> We can tune the rank 1 group any way we like so long as the steps are
> of the same size, which means that our ET can have stretched or
> squashed octaves if we so choose.

In the octave-invariant case the octave lies outside the system, so you
can't say anything about it.

> In the same way, we can tune the
> rank 2 group any way we like, except that we need to retain
> incommensurability of two generators (or at least to ignore the fact
> if they are not.) If we make the octaves pure in our example where
> the kernel is generated by a comma, we could for instance make the
> fifths pure also, leading to Pythagorean tuning. Alternatively, we
> could make the major thirds pure, leading to 1/4 comma mean tone
> temperment. (Pythagorean tuning is not considered a temperment, since
> the fifth isn't tempered, but it is the same sort of thing
> mathematically as 1/4 comma mean-tone temperment.) Other choices lead
> to other results, and all we need to do is to ensure the circle of
> fifths does not close--or at least to pretend otherwise it if it does.

For the octave invariant case, the fourth or fifth is the generator, which
I think agrees with both meanings of "generator".

> A rank 3 image group, coming from a kernel of codimension 3, is what
> people have been calling a 2D temperment. I hope that clarifies
> things (as it does for me) rather than further confuses them!

It's also thought of as repeating every octave. You'll have to come up
with the group theoretic description of that. Although the same tuning
can be described the way you did, octave repetition is another constraint
or simplification.

Graham

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

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

> 2 is certainly prime, but most of the time we consider octave-
invariant
> scales.

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.

Of course, unless you have an infinite number of notes in your scale,
which you may have conceptually but not in practice, you don't have
octave invariance anyway.

> So the kernel has dimension 1 because it contains 1 unison vector?

Because it is generated by one unison vector. I'm not clear yet if a
unison vector is supposed to be an element of the kernel or a
generator of the kernel, as I mentioned.

> So in octave-invariant terms, 5-limit is rank 2, but cyclic about
the
> octave.

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".

>An ET would be rank 0 I suppose, but you've already given the
> real name for that case.

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.

> In the octave-invariant case the octave lies outside the system, so
you
> can't say anything about it.

Whether to tune the octave exactly or not is a question which lies at
a more specific, less abstract level than that created by defining
certain things to be unison vectors. As a general rule, you only
confuse things by insisting on concrete particulars when they are not
required. About all one can say for certain is that you can't toss 2
out of a discussion involving unison vectors, because without 2 we
can't tell what is a small interval and what is not.

> For the octave invariant case, the fourth or fifth is the
generator, which
> I think agrees with both meanings of "generator".

You can generate by fifths and octaves if you want, but you don't
need to.

You have octaves on the brain, which is the usual situation in music
theory; however when discussing tuning and temperment it really is
just another interval.

Suppose I decide to have a mean-tone system, so that 81/80 is a
unison vector. I could tune things so that octaves were pure 2's, but
I don't have to. Suppose instead I decide that I want the major sixth
to be exact. Now I can look at the circle of octaves, and notice that
it approximately returns after 14 octaves--14 octaves is almost the
same as 19 major sixths; 2^14 = (5/3)^18.9968... Suppose I decide to
tune octaves so that I represent 2 by (5/3)^(19/14); this is equal to
2.000232... and is sharp by about 1/5 of a cent. Since I have fixed
two values and I am making 81/80 a unison vector, major thirds are
now determined also. Since 2 and 5/3 are not now incommensurable, I
actually have a rank 1 group. It is the 14 equal division of the
major sixth, with a very slightly sharp octave; it is in practice
more or less indistinguishable from the 19 equal division of the
octave, with very slightly flat major sixths.

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.

🔗Paul Erlich <paul@stretch-music.com>

🔗Dave Keenan <D.KEENAN@UQ.NET.AU>

🔗genewardsmith@juno.com

🔗graham@microtonal.co.uk

In-Reply-To: <9ls5st+gcan@eGroups.com>
In article <9ls5st+gcan@eGroups.com>, genewardsmith@juno.com () wrote:

> > 2 is certainly prime, but most of the time we consider octave-
> invariant
> > scales.
>
> 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."

> Of course, unless you have an infinite number of notes in your scale,
> which you may have conceptually but not in practice, you don't have
> octave invariance anyway.

Then we'll add a new category:

1) just intonation

2) approximations

3) tuning

4) scale

5) scale in practice

In general, just intonation will have an infinite number of notes in
multiple dimensions. Approximations will have an infinite number of notes
in fewer dimensions. A tuning is a special case of either a just
intonation of approximation. A scale is a subset of a tuning that has a
finite number of notes, either overall or to the interval of repetition.
A scale in practice will have a finite total number of notes.

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.

> > So the kernel has dimension 1 because it contains 1 unison vector?
>
> Because it is generated by one unison vector. I'm not clear yet if a
> unison vector is supposed to be an element of the kernel or a
> generator of the kernel, as I mentioned.

It looks like a commatic UV would be in the kernel, and a chromatic UV
would not.

> > So in octave-invariant terms, 5-limit is rank 2, but cyclic about
> the
> > octave.
>
> 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.

> >An ET would be rank 0 I suppose, but you've already given the
> > real name for that case.
>
> 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?

In fact, there are two different ways of treating octave equivalence:

1) Consider the scale repeating about the octave.

2) Consider the scale as only existing within the octave.

And (2) is actually closer to the way octave-invariant matrices work.

> > In the octave-invariant case the octave lies outside the system, so
> you
> > can't say anything about it.
>
> Whether to tune the octave exactly or not is a question which lies at
> a more specific, less abstract level than that created by defining
> certain things to be unison vectors. As a general rule, you only
> confuse things by insisting on concrete particulars when they are not
> required. About all one can say for certain is that you can't toss 2
> out of a discussion involving unison vectors, because without 2 we
> can't tell what is a small interval and what is not.

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.

For (1), you calculate the pitch, and allow an arbitrary number of octaves
to be added or subtracted.

For (2), you calculate the pitch modulo the octave.

If you want a small interval, you take the smallest option for (1). That
is actually more liberal than (2). [4 -1] will give 81:80 either way.
But [-4 1] could be 80:81 for (1) but has to be 160:81 for (2). My
program works with (2).

You seem to be saying that specifying the tuning of the octave is
confusing when *you* are the one who wants to specify it! By taking it
out of the system, I don't care either way. The size of the octave
becomes a property of the metric, not the matrices. I do consider the
metric to be less abstract than the algebra. It happens that in the
octave-specific case, the metric is itself a column matrix, so that is a
simplification. It means you can define the tuning within the system.
With octave invariant matrices, you can only comment on intervals within
the octave.

So now we come to consider unison vectors. It sort of looks like unison
vectors have to be small intervals. But I haven't seen a definition of
*how* small. I'm hoping that it doesn't matter at all for the octave
invariant case, so that the tuning of that octave becomes irrelevant. If
it doesn't work out that way perhaps you, as the mathematician, can tell
us the size constraints on the unison vectors.

> > For the octave invariant case, the fourth or fifth is the
> generator, which
> > I think agrees with both meanings of "generator".
>
> 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. For other
approximations, other generators will be needed. I've even done
calculations with a different interval of equivalence. See

<http://x31eq.com/tritave.py>
<http://x31eq.com/tritave.txt>
<http://x31eq.com/tritave.nonoct.py>
<http://x31eq.com/tritave.nonoct.txt>

I haven't considered any systems without an interval of equivalence. Your
idea of using a maximal number of ETs may be a way forward.

> You have octaves on the brain, which is the usual situation in music
> theory; however when discussing tuning and temperment it really is
> just another interval.

Yes, it is the usual situation in music theory.

> Suppose I decide to have a mean-tone system, so that 81/80 is a
> unison vector. I could tune things so that octaves were pure 2's, but
> I don't have to. Suppose instead I decide that I want the major sixth
> to be exact. Now I can look at the circle of octaves, and notice that
> it approximately returns after 14 octaves--14 octaves is almost the
> same as 19 major sixths; 2^14 = (5/3)^18.9968... Suppose I decide to
> tune octaves so that I represent 2 by (5/3)^(19/14); this is equal to
> 2.000232... and is sharp by about 1/5 of a cent. Since I have fixed
> two values and I am making 81/80 a unison vector, major thirds are
> now determined also. Since 2 and 5/3 are not now incommensurable, I
> actually have a rank 1 group. It is the 14 equal division of the
> major sixth, with a very slightly sharp octave; it is in practice
> more or less indistinguishable from the 19 equal division of the
> octave, with very slightly flat major sixths.

Yes, I think that would work.

> 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. You need a harmonic metric, so that you
can decide how well approximated a given interval is. That needs to
include the rule for adding together a number of approximations. And you
need to decide the ideal tuning for each interval. Here are the choices I
made:

The set of intervals is an odd limit. That means, all ratios within in
octave that don't contain and odd number larger than the chosen limit.
All are weighted equally.

The closeness of approximation is measured in cents relative to just
intonation. This assumes octaves are already just.

The rule for adding together approximations is to take the poorest
alternative.

The "method of least squares" is totally irrelevant here. Indeed, it
seems to be part of the harmonic metric, and not a method at all. You
still need a minimisation algorithm. 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. But
it means all I have to do is consider each interval in term, which is a
very simple method.

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.

Graham

🔗Paul Erlich <paul@stretch-music.com>

🔗genewardsmith@juno.com

--- 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.

🔗genewardsmith@juno.com

🔗Dave Keenan <D.KEENAN@UQ.NET.AU>

🔗graham@microtonal.co.uk

In-Reply-To: <9luhvn+n2lb@eGroups.com>
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]

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

And the adjoint of

[-1 2]
[-4 1]

[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://x31eq.com/magicpump.mp3>.

> > 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