Yahoo Tuning Groups Ultimate Backup tuning-math Rank 3 scale modes

First off, much thanks to Keenan Pepper for the remarkably productive
discussion about all this. We had a great conversation about rank 3
scales on Facebook the other day and figured out a lot of stuff. These
are some general ideas that resulted from that discussion. It should
be noted that if any of these ideas are novel he deserves as much
credit as I do.

Firstly, it's often been stated that rank-3 periodicity blocks don't
extend "in a unique way" to scales. That is, whereas different
transpositions of the unison vectors over the lattice for a rank-2
scale will always lead to some modal permutation of the same scale,
this isn't true for rank-3. For example, if you move the chromatic
unison vectors {81/80, 25/24} around on the 5-limit lattice, you're
going to end up at different things that appear to be modes of the JI
major scale, but aren't, because one of the notes will be
comma-shifted in some way or another by 81/80. So it's often said that
these scales are "not modes" of one another.

What's actually going on are that these are still modes of one
another, but that rank-3 scales have a 2-dimensional system of modes,
so that cyclic permutations of one scale don't adequately describe all
of the modes available. Rank-2 modes are often thought of as being
cyclic permutations of one another, but more fundamentally what they
are is a generator chain differing in "up" vs "down" generators. In a
rank-3 scale, there are two separate generator chains, and hence there
are two separate sets of generators that can go up vs down.

The tricky thing is, however, that the interval between the generators
is also a generator, and in fact you can define your basis however you
want. So what I'm not clear on is whether or not the modal system
changes depending on how you define your basis. As the convexity of
the periodicity block is basis-invariant, I think it's likely that it
doesn't, but I don't know for sure.

Another question is - for the JI major scale, given {81/80, 25/24} as
a basis, there are two chains of fifths a 5/4 apart - one of them has
a length of 4, and the other has a length of 3. What does that mean?
We obviously can't alter each chain of 3/2's independently, or else
we're slicing the block in half machete-style. And is the 5/4 a
generator itself? At any rate, the fact that the generators can't just
move independently of one means that more's going on here, but once
it's sorted out this will be a very powerful tool in figuring out
rank-3 modal harmony. I have a hunch, maybe misguided, that all of
these different modes are going to be something similar to
"Z-relations" of one another, in a generalized sense, so the
permutations of every Z-relation end up describing all of the modes of
a rank-3 system... maybe. It would involve generalizing the concept of
a Z-relation to an infinite system.

Has anyone worked on this before?

-Mike

🔗Carl Lumma <carl@lumma.org>

Mike wrote:

>Firstly, it's often been stated that rank-3 periodicity blocks don't
>extend "in a unique way" to scales. That is, whereas different
>transpositions of the unison vectors over the lattice for a rank-2
>scale will always lead to some modal permutation of the same scale,
>this isn't true for rank-3. For example, if you move the chromatic
>unison vectors {81/80, 25/24} around on the 5-limit lattice, you're
>going to end up at different things that appear to be modes of the JI
>major scale, but aren't, because one of the notes will be
>comma-shifted in some way or another by 81/80. So it's often said that
>these scales are "not modes" of one another.

You lost me here. What do you mean by moving unison vectors
around?

>What's actually going on are that these are still modes of one
>another, but that rank-3 scales have a 2-dimensional system of modes,
>so that cyclic permutations of one scale don't adequately describe all
>of the modes available. Rank-2 modes are often thought of as being
>cyclic permutations of one another, but more fundamentally what they
>are is a generator chain differing in "up" vs "down" generators. In a
>rank-3 scale, there are two separate generator chains, and hence there
>are two separate sets of generators that can go up vs down.

The definition of a mode is a cyclic permutation. You need
a new term.

>The tricky thing is, however, that the interval between the generators
>is also a generator, and in fact you can define your basis however you
>want. So what I'm not clear on is whether or not the modal system
>changes depending on how you define your basis. As the convexity of
>the periodicity block is basis-invariant, I think it's likely that it
>doesn't, but I don't know for sure.

What's a "modal system" and when does one change?

>Another question is - for the JI major scale, given {81/80, 25/24} as
>a basis,

A basis for the kernel maybe; not for the scale.

-Carl

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Mike wrote:
>
> >Firstly, it's often been stated that rank-3 periodicity blocks don't
> >extend "in a unique way" to scales. That is, whereas different
> >transpositions of the unison vectors over the lattice for a rank-2
> >scale will always lead to some modal permutation of the same scale,
> >this isn't true for rank-3. For example, if you move the chromatic
> >unison vectors {81/80, 25/24} around on the 5-limit lattice, you're
> >going to end up at different things that appear to be modes of the JI
> >major scale, but aren't, because one of the notes will be
> >comma-shifted in some way or another by 81/80. So it's often said that
> >these scales are "not modes" of one another.
>
> You lost me here. What do you mean by moving unison vectors
> around?

A Fokker block consists of all the lattice points inside a parallelogram defined by some chromata. But depending on the specific position of the parallelogram with respect to the lattice, the arrangement of the lattice points inside can be different. That is, the chromata don't define one unique Fokker block, but several Fokker blocks of slightly different shapes.

For example, take {81/80, 25/24} as your chromata. This defines a parallelogram that can be moved around the 5-limit lattice. Assuming no points are exactly on the boundary, the parallelogram always contains 7 points. (It corresponds to a specific val, in this case <7 11 16|.) But the points don't always form the same scale.

One such Fokker block is {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}. If you translate the parallelogram by the 4/3 lattice vector, you get {1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 16/9, 2/1}, which is simply a different mode of the same scale. But if you translate the parallelogram by some *fraction* of the 4/3 lattice vector, you can get {1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}, which is *not* simply a different mode. (In this case it happens to be an inversion of the original scale, but there are other examples where it's not, for example {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 9/5, 2/1}.)

> The definition of a mode is a cyclic permutation. You need
> a new term.

See, Mike, I told you that people would object to this. The word "mode" is already used for non-MOS scales and simply means cyclic permutation. If you want to use the word "mode" at least say meta-mode or quasi-mode or something. You can argue that the correct generalization of modes to rank >2 is your thing, but don't just call your thing "modes" because that's too confusing.

> >Another question is - for the JI major scale, given {81/80, 25/24} as
> >a basis,
>
> A basis for the kernel maybe; not for the scale.

Yeah, this is a basis for the kernel of the val, or equivalently a basis for the chroma lattice.

Keenan

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>A Fokker block consists of all the lattice points inside a
>parallelogram defined by some chromata.

Some commas, yes.

>But depending on the specific
>position of the parallelogram with respect to the lattice, the
>arrangement of the lattice points inside can be different. That is,
>the chromata don't define one unique Fokker block, but several Fokker
>blocks of slightly different shapes.
>
>For example, take {81/80, 25/24} as your chromata. This defines a
>parallelogram that can be moved around the 5-limit lattice. Assuming
>no points are exactly on the boundary,

except the unison

>the parallelogram always
>contains 7 points. (It corresponds to a specific val, in this case <7
>11 16|.) But the points don't always form the same scale.
>
>One such Fokker block is {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}. If
>you translate the parallelogram by the 4/3 lattice vector, you get
>{1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 16/9, 2/1}, which is simply a
>different mode of the same scale.

It's the same scale. You transposed it and applied octave
equivalence, so it's not surprising you got a different mode.

>But if you translate the
>parallelogram by some *fraction* of the 4/3 lattice vector,

I don't think I'd call it Fokker block, but whatever it is,
it's interesting...

-Carl

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
> A Fokker block consists of all the lattice points inside a parallelogram defined by some chromata. But depending on the specific position of the parallelogram with respect to the lattice, the arrangement of the lattice points inside can be different. That is, the chromata don't define one unique Fokker block, but several Fokker blocks of slightly different shapes.
>
> For example, take {81/80, 25/24} as your chromata. This defines a parallelogram that can be moved around the 5-limit lattice. Assuming no points are exactly on the boundary, the parallelogram always contains 7 points. (It corresponds to a specific val, in this case <7 11 16|.) But the points don't always form the same scale.
>
> One such Fokker block is {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}. If you translate the parallelogram by the 4/3 lattice vector, you get {1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 16/9, 2/1}, which is simply a different mode of the same scale. But if you translate the parallelogram by some *fraction* of the 4/3 lattice vector, you can get {1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}, which is *not* simply a different mode. (In this case it happens to be an inversion of the original scale, but there are other examples where it's not, for example {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 9/5, 2/1}.)

I can get 7 different Fokker blocks for {81/80, 25/24}, and 5 different ones for {81/80, 16/15}.

I conjecture that for an N-note scale (and fixed chromata) there are always N different Fokker blocks, but I have no proof yet.

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> >For example, take {81/80, 25/24} as your chromata. This defines a
> >parallelogram that can be moved around the 5-limit lattice. Assuming
> >no points are exactly on the boundary,
>
> except the unison

Huh? I said exactly what I meant to say. If no points whatsoever are on the boundary, then there are 7 points inside.

If there are any points on the boundary, then you have to say which ones of those you're counting and everything gets more complicated?

> >the parallelogram always
> >contains 7 points. (It corresponds to a specific val, in this case <7
> >11 16|.) But the points don't always form the same scale.
> >
> >One such Fokker block is {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1}. If
> >you translate the parallelogram by the 4/3 lattice vector, you get
> >{1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 16/9, 2/1}, which is simply a
> >different mode of the same scale.
>
> It's the same scale. You transposed it and applied octave
> equivalence, so it's not surprising you got a different mode.
>
> >But if you translate the
> >parallelogram by some *fraction* of the 4/3 lattice vector,
>
> I don't think I'd call it Fokker block, but whatever it is,
> it's interesting...

What's your definition of a Fokker block?

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Keenan wrote:
>
> >A Fokker block consists of all the lattice points inside a
> >parallelogram defined by some chromata.
>
> Some commas, yes.

I just realized my use of the word "chroma" is horribly at odds with http://xenharmonic.wikispaces.com/Fokker+blocks , because I was using it to mean an interval c for which v(c) = 0 (where v is the val w/r/t which the scale is epimorphic), but that article uses it to mean an interval for which v(c) = 1, which I would have called a "step".

However, my usage was consistent with http://xenharmonic.wikispaces.com/MODMOS+Scales , and I believe with common sense, because a "chroma" sounds like something you'd use to make "chromatic" alterations to a scale, not something that's an integral part of your scale like a step.

If people really want to use "chroma" for an interval such that v(c) = 1, then I'm willing to change my usage to avoid confusion.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 8:14 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> > The definition of a mode is a cyclic permutation. You need
> > a new term.
>
> See, Mike, I told you that people would object to this. The word "mode" is already used for non-MOS scales and simply means cyclic permutation. If you want to use the word "mode" at least say meta-mode or quasi-mode or something. You can argue that the correct generalization of modes to rank >2 is your thing, but don't just call your thing "modes" because that's too confusing.

I don't remember you objecting, but either way the least important
part of this discussion is what we call it. If we want to call the
things that are permutations of one another "modes," then we need to
call the things that are comma-shifted versions but still Fokker-block
related versions something else. I suggest "domes," because it's the
word "mode" with the letters moved around in a way that's not a
permutation, kind of like the things that we're talking about. So then
all of these Fokker blocks are modes and domes of one another.

> > >Another question is - for the JI major scale, given {81/80, 25/24} as
> > >a basis,
> >
> > A basis for the kernel maybe; not for the scale.
>
> Yeah, this is a basis for the kernel of the val, or equivalently a basis for the chroma lattice.

I meant a comma basis for the periodicity block. But what would a
basis for the scale be? The basis {5/4, 3/2} doesn't uniquely specify
the JI major scale.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 8:22 PM, Carl Lumma <carl@lumma.org> wrote:
>
> >But depending on the specific
> >position of the parallelogram with respect to the lattice, the
> >arrangement of the lattice points inside can be different. That is,
> >the chromata don't define one unique Fokker block, but several Fokker
> >blocks of slightly different shapes.
> >
> >For example, take {81/80, 25/24} as your chromata. This defines a
> >parallelogram that can be moved around the 5-limit lattice. Assuming
> >no points are exactly on the boundary,
>
> except the unison

No, not even the unison. See Paul's "A Gentle Guide To Periodicity
Blocks" for a good visual diagram:

http://tonalsoft.com/enc/f/fokker-gentle-2.aspx

He performs the exact manipulation that I'm talking about on the
5-limit JI major scale and shows how the resulting pitch set isn't a
permutation of the original set.

-Mike

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>If there are any points on the boundary, then you have to say which
>ones of those you're counting and everything gets more complicated?

If you have more than one point on a boundary, I believe
you have torsion.

>> I don't think I'd call it Fokker block, but whatever it is,
>> it's interesting...
>
>What's your definition of a Fokker block?

Howabout this:

Given a discrete subgroup (lattice) of a real vector space, a
Fokker block is a finite subset of n lattice points enclosed
by a parallelepiped whose vertices are in the lattice group,
for which there exists a p-limit val mapping its elements to
the integers mod n.

I think the interesting things you get are MODMOS, where a
MODMOS is defined as a scale that can be produced from a
Fokker block by optionally replacing each of its elements with
the corresponding element in an *adjacent* block. Rothenberg
mean variety is non-decreasing with respect to this operation.

That fits with your conjecture of there being n "blocks".

-Carl

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>If people really want to use "chroma" for an interval such that
>v(c) = 1, then I'm willing to change my usage to avoid confusion.

Unison vectors (aka commas) are intervals for which v(c) = 0.

Commatic unison vectors (aka commas) are intervals for which
v(c) = 0 and V(c) = 0, where V() is a fractional val mapping
just intonation to the reals.

Chromatic unison vectors (aka chroma or chromata) are intervals
for which v(c) = 0 and V(c) != 0.

That's one way to put it anyway. Better would be to mention
that the usual way to get V(c) != 0 is by using an optimized
rank-r tuning, r > 1, whose kernel intersects that of v().

Clear as mud!

-Carl

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Keenan wrote:
>
> >If there are any points on the boundary, then you have to say which
> >ones of those you're counting and everything gets more complicated?
>
> If you have more than one point on a boundary, I believe
> you have torsion.

Ah. Took me a while to figure out what you were saying here.

I was speaking of "the boundary" of the parallelogram, which contains all four edges and all four vertices. This is the definition of "boundary" in topology. (You can work out the obvious generalization to higher dimensions.)

What you mean is that if there's more than one point on a single *edge* of the parallelogram (other than just having one lattice point at each vertex of that edge) then there's torsion. I agree with this.

> >> I don't think I'd call it Fokker block, but whatever it is,
> >> it's interesting...
> >
> >What's your definition of a Fokker block?
>
> Howabout this:
>
> Given a discrete subgroup (lattice) of a real vector space, a
> Fokker block is a finite subset of n lattice points enclosed
> by a parallelepiped whose vertices are in the lattice group,
> for which there exists a p-limit val mapping its elements to
> the integers mod n.

This definition is ambiguous, because you have lattice points at all four corners of the parallelogram but you're not saying which of those you're including. For the scale to be epimorphic you need to include exactly one, but any one of the four will work.

For example, if your commas are {81/80, 25/24}, then the four vertices are 1/1, 81/80, 25/24, and 135/128. Your definition doesn't say which of these to include. I guess you were assuming 1/1 would be included and not the others, but why should that be assumed?

> I think the interesting things you get are MODMOS, where a
> MODMOS is defined as a scale that can be produced from a
> Fokker block by optionally replacing each of its elements with
> the corresponding element in an *adjacent* block. Rothenberg
> mean variety is non-decreasing with respect to this operation.
>
> That fits with your conjecture of there being n "blocks".

No, it doesn't, because there are 4 adjacent blocks (or maybe 8 depending on the definition of "adjacent"). Also, this procedure gives you things that are not convex, but the construction I was originally talking about always gives you convex sets (because they're the interiors of parallelograms).

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Keenan wrote:
>
> >If people really want to use "chroma" for an interval such that
> >v(c) = 1, then I'm willing to change my usage to avoid confusion.
>
> Unison vectors (aka commas) are intervals for which v(c) = 0.
>
> Commatic unison vectors (aka commas) are intervals for which
> v(c) = 0 and V(c) = 0, where V() is a fractional val mapping
> just intonation to the reals.
>
> Chromatic unison vectors (aka chroma or chromata) are intervals
> for which v(c) = 0 and V(c) != 0.

Yeah, this is how I was using "chroma": an interval that maps to 0 scale steps under the val but is not tempered out completely. So if this is the terminology scheme we agree on, the only thing to change would be http://xenharmonic.wikispaces.com/Fokker+blocks .

What do you think of these terms, Gene?

> That's one way to put it anyway. Better would be to mention
> that the usual way to get V(c) != 0 is by using an optimized
> rank-r tuning, r > 1, whose kernel intersects that of v().

This should say "whose kernel is a subset of that of v()". Any two kernels always intersect, because they both contain 1/1.

> Clear as mud!

Yup.

Keenan

🔗Carl Lumma <carl@lumma.org>

Sorry for the copy my son sent, if you got it.

Keenan wrote:

>What you mean is that if there's more than one point on a single
>*edge* of the parallelogram (other than just having one lattice point
>at each vertex of that edge) then there's torsion. I agree with this.

Yes, edges, sorry.

>> Given a discrete subgroup (lattice) of a real vector space, a
>> Fokker block is a finite subset of n lattice points enclosed
>> by a parallelepiped whose vertices are in the lattice group,
>> for which there exists a p-limit val mapping its elements to
>> the integers mod n.
>
>This definition is ambiguous, because you have lattice points at all
>four corners of the parallelogram but you're not saying which of those
>you're including. For the scale to be epimorphic you need to include
>exactly one, but any one of the four will work.
>For example, if your commas are {81/80, 25/24}, then the four vertices
>are 1/1, 81/80, 25/24, and 135/128. Your definition doesn't say which
>of these to include. I guess you were assuming 1/1 would be included
>and not the others, but why should that be assumed?

There can be an arbitrary number of vertices, but only one is
included in the set, yes. *Which* one is irrelevant, as the
whole point of a lattice is that there is translational symmetry
within the lattice group. As a matter of standardization, make
each vertex 1/1 in turn and choose the one which gives the
pitches of the block the lowest average Tenney height.

>> I think the interesting things you get are MODMOS, where a
>> MODMOS is defined as a scale that can be produced from a
>> Fokker block by optionally replacing each of its elements with
>> the corresponding element in an *adjacent* block. Rothenberg
>> mean variety is non-decreasing with respect to this operation.
>>
>> That fits with your conjecture of there being n "blocks".
>
>No, it doesn't, because there are 4 adjacent blocks (or maybe 8
>depending on the definition of "adjacent"). Also, this procedure
>gives you things that are not convex, but the construction I was
>originally talking about always gives you convex sets (because
>they're the interiors of parallelograms).

You're right (there are 8 adjacent blocks in R^2). But the
procedure clearly generates convex MODMOS.

-Carl

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>> Unison vectors (aka commas) are intervals for which v(c) = 0.
>> Commatic unison vectors (aka commas) are intervals for which
>> v(c) = 0 and V(c) = 0, where V() is a fractional val mapping
>> just intonation to the reals.
>> Chromatic unison vectors (aka chroma or chromata) are intervals
>> for which v(c) = 0 and V(c) != 0.
>
>Yeah, this is how I was using "chroma": an interval that maps to 0
>scale steps under the val but is not tempered out completely.

I think the Fokker block construction is useful regardless of
the rank of the temperament one finally uses. One can always
call the unison vectors commas to be safe. Or uvs. They're
only chromata if they are M(c) = 1 where M is the reduced
mapping of the temperament guiding the tuning.

>> That's one way to put it anyway. Better would be to mention
>> that the usual way to get V(c) != 0 is by using an optimized
>> rank-r tuning, r > 1, whose kernel intersects that of v().
>
>This should say "whose kernel is a subset of that of v()". Any two
>kernels always intersect, because they both contain 1/1.

Very well. Does it have to be a subset though? Like, can't
I tune a block conforming to <19 30 44 53| with dominant
temperament? Perhaps there's some contradiction there...

-Carl

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> There can be an arbitrary number of vertices, but only one is
> included in the set, yes. *Which* one is irrelevant, as the
> whole point of a lattice is that there is translational symmetry
> within the lattice group. As a matter of standardization, make
> each vertex 1/1 in turn and choose the one which gives the
> pitches of the block the lowest average Tenney height.

But my whole point is that it's not irrelevant. You get four different scales this way, and in fact they're four out of the seven scales I'm talking about (the ones I consider seven different Fokker blocks).

The four scales are

1/1, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
(or transposed, 1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8)

81/80, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
(or transposed, 1/1, 9/8, 5/4, 27/20, 3/2, 5/3, 15/8)

25/24, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
(or transposed, 1/1, 9/8, 5/4, 25/18, 3/2, 5/3, 15/8)

135/128, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
(or transposed, 1/1, 9/8, 5/4, 45/32, 3/2, 5/3, 15/8)

You're trying to make there be one unique Fokker block (which of course you can do by defining it so that's the case), but *naturally* there are many of them.

> >> I think the interesting things you get are MODMOS, where a
> >> MODMOS is defined as a scale that can be produced from a
> >> Fokker block by optionally replacing each of its elements with
> >> the corresponding element in an *adjacent* block. Rothenberg
> >> mean variety is non-decreasing with respect to this operation.
> >>
> >> That fits with your conjecture of there being n "blocks".
> >
> >No, it doesn't, because there are 4 adjacent blocks (or maybe 8
> >depending on the definition of "adjacent"). Also, this procedure
> >gives you things that are not convex, but the construction I was
> >originally talking about always gives you convex sets (because
> >they're the interiors of parallelograms).
>
> You're right (there are 8 adjacent blocks in R^2). But the
> procedure clearly generates convex MODMOS.

I must be missing something. Take the JI major scale {1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8}. Replace 3/2 by 40/27, which is the corresponding pitch in an adjacent block. Now you have {1/1, 9/8, 5/4, 4/3, 40/27, 5/3, 15/8}, which is not convex (10/9 is in the convex hull but the scale doesn't contain 10/9).

What part of your procedure am I misunderstanding?

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 10:07 PM, Carl Lumma <carl@lumma.org> wrote:
>
> >> I think the interesting things you get are MODMOS, where a
> >> MODMOS is defined as a scale that can be produced from a
> >> Fokker block by optionally replacing each of its elements with
> >> the corresponding element in an *adjacent* block. Rothenberg
> >> mean variety is non-decreasing with respect to this operation.
> >>
> >> That fits with your conjecture of there being n "blocks".
> >
> >No, it doesn't, because there are 4 adjacent blocks (or maybe 8
> >depending on the definition of "adjacent"). Also, this procedure
> >gives you things that are not convex, but the construction I was
> >originally talking about always gives you convex sets (because
> >they're the interiors of parallelograms).
>
> You're right (there are 8 adjacent blocks in R^2). But the
> procedure clearly generates convex MODMOS.

No, this is wrong. A counterexample would be to look at a
rank-2 periodicity block, e.g. one operating on something like the
Pythagorean chain of fifths. The only way to create a convex set is
for all of the fifths to be adjacent; a "hole" clearly presents itself
as soon as you change anything.

For some diagrams:
/tuning-math/message/19017?var=0&l=1

Here are some 2d diagrams:
/tuning-math/message/19029?var=0&l=1

One of the example MODMOS's presented here is convex, and the other is
not convex.

It is possible to construct MODMOS's that are convex, but I'm not
entirely sure why they'd be more important than those that aren't. But
at any rate, if you ever temper down to rank-2, MODMOS blocks being
non-convex are more the exception than the rule.

It's also possible for MODMOS's that are convex on the rank-3 lattice
to lead to equivalent non-convex rank-2 blocks when one of the commas
is tempered out. For example, consider 1/1 9/8 5/4 4/3 3/2 27/16 15/8
2/1, which forms a hexagon on the lattice and hence is convex. If you
temper out 25/24, you end up with dicot[7] with one of the notes
altered, and if you lay this out as a chain of tempered 5/4's, you can
see that the resulting block takes 81/80 as a chromatic unison vector
and is not convex.

It's similarly possible to have non-convex MODMOS's that become convex
when tempered, a trivial example of which would occur if you shift one
of the notes in the rank-3 case by one of the chromatic vectors in
such a way that the resulting set is not convex, and then you
transform the same chromatic vector into a commatic vector. The
resulting alteration will nullify itself and you'll end up with a
convex rank-2 block.

I'm now going by the affine definition of convexity, if there's any confusion.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 10:22 PM, Keenan Pepper <keenanpepper@gmail.com>
wrote:
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...>
wrote:
> > --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@>
wrote:
> >
> > > I conjecture that for an N-note scale (and fixed chromata) there are
always N different Fokker blocks, but I have no proof yet.
> >
> > Not true, I'm afraid.
>
> Counterexample?

One counterexample can be found in the block delimited by 128/125 and 81/80,
as seen on this page:

http://tonalsoft.com/enc/f/fokker-gentle-2.aspx

By my count, there are only three possible Fokker blocks, each corresponding
to this shape (view fixed width):

x.x.x.x
x.x.x.x
x.x.x.x
x.x.x.x

x.x.x
x.x.x.x
x.x.x.x
x

x.x
x.x.x.x
x.x.x.x
x.x
x
x.x.x.x
x.x.x.x
x.x.x

If we're changing the definition of a Fokker block such that all the
vertices of the block have to be elements in the block/the adjacent blocks,
then I think that only the top one is actually a Fokker block. The rest are
what I was calling "domes" of the original block.

-Mike

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>> You're right (there are 8 adjacent blocks in R^2). But the
>> procedure clearly generates convex MODMOS.
>
>I must be missing something. Take the JI major scale {1/1, 9/8, 5/4,
>4/3, 3/2, 5/3, 15/8}. Replace 3/2 by 40/27, which is the corresponding
>pitch in an adjacent block. Now you have {1/1, 9/8, 5/4, 4/3, 40/27,
>5/3, 15/8}, which is not convex (10/9 is in the convex hull but the
>scale doesn't contain 10/9).
>
>What part of your procedure am I misunderstanding?

I meant *your* (translation) procedure.

>> There can be an arbitrary number of vertices, but only one is
>> included in the set, yes. *Which* one is irrelevant, as the
>> whole point of a lattice is that there is translational symmetry
>> within the lattice group. As a matter of standardization, make
>> each vertex 1/1 in turn and choose the one which gives the
>> pitches of the block the lowest average Tenney height.
>
>But my whole point is that it's not irrelevant. You get four different
>scales this way, and in fact they're four out of the seven scales I'm
>talking about (the ones I consider seven different Fokker blocks).
>
>The four scales are
>1/1, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
>(or transposed, 1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8)

The simplest mode is 1/1 9/8 5/4 4/3 3/2 5/3 15/8,
avg tenney height 35.

>81/80, 9/8, 5/4, 45/32, 3/2, 27/16, 15/8
>(or transposed, 1/1, 9/8, 5/4, 27/20, 3/2, 5/3, 15/8)

The simplest mode is 1/1 10/9 5/4 4/3 3/2 5/3 9/5,
avg tenney height 27.

OK, howabout this instead:
1. all uvs point away from 1/1
2. all uvs must be positive
3. the other (three) scales are MODMOS

-Carl

🔗Mike Battaglia <battaglia01@gmail.com>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, Carl Lumma <carl@> wrote:
> > If you have more than one point on a boundary, I believe
> > you have torsion.
>
> What you mean is that if there's more than one point on a single *edge* of the parallelogram (other than just having one lattice point at each vertex of that edge) then there's torsion. I agree with this.

Actually, on second thought, this is false. Counterexample: {81/80, 128/125}. Each edge that goes along the 5/4 axis is split into thirds, so it has 4 lattice points on it. But there is no torsion.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 11:57 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
> >
> > --- In tuning-math@yahoogroups.com, Carl Lumma <carl@> wrote:
> > > If you have more than one point on a boundary, I believe
> > > you have torsion.
> >
> > What you mean is that if there's more than one point on a single *edge* of the parallelogram (other than just having one lattice point at each vertex of that edge) then there's torsion. I agree with this.
>
> Actually, on second thought, this is false. Counterexample: {81/80, 128/125}. Each edge that goes along the 5/4 axis is split into thirds, so it has 4 lattice points on it. But there is no torsion.

Mmmm, tricky! There's no torsion if you temper out 128/125, but there
is if you temper out 8/125, the latter being (2/5)^3. They share the
same representation in an octave-equivalent space, however. Maybe we
should start treating 2 as its own axis.

-Mike

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> >--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@> wrote:
>
> >Actually, on second thought, this is false. Counterexample: {81/80,
> >128/125}. Each edge that goes along the 5/4 axis is split into thirds,
> >so it has 4 lattice points on it. But there is no torsion.
>
> Yeah, those are consecutive points. I think I was thinking of
> non-consecutive points. -Carl

What do you mean by "consecutive"?

If 2048/2025 is a UV, are 1/1, 45/32, and 2045/1024 consecutive or not?

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

Okay, I need something to call these things that I had previously been calling simply "Fokker blocks", now that I know there's disagreement about the definition of that term. Since Paul's "gentle introduction" is the earliest thing I've seen that defines them exactly the way I want to (and explicitly describes how they can have different shapes even for the same UVs), I suggest "Fokker-Erlich periodicity block", or "F-E block" for short.

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> Mmmm, tricky! There's no torsion if you temper out 128/125, but there
> is if you temper out 8/125, the latter being (2/5)^3. They share the
> same representation in an octave-equivalent space, however. Maybe we
> should start treating 2 as its own axis.

Aha! This leads me to a revised version of my conjecture:

For an interval c, let f(c) be the greatest integer p such that (c')^p = c, up to octave equivalence, for some other JI interval c'. For example, f(128/125) = 3 because (8/5)^3 = 128/125 up to octave equivalence. f(81/80) = 1 because it's not expressible as a nontrivial power.

An equivalent definition is that f(c) is the number of periods per octave for the codimension-1 temperament tempering out c.

Conjecture: The number of distinct Fokker-Erlich blocks for a given UV set is equal to N / prod(f(c_i)), where N is the number of notes in the block, and {c_i} are the UVs.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Sun, Sep 11, 2011 at 11:40 PM, Mike Battaglia <battaglia01@gmail.com>
wrote:
>
> If we're changing the definition of a Fokker block such that all the
> vertices of the block have to be elements in the block/the adjacent
blocks,
> then I think that only the top one is actually a Fokker block. The rest
are
> what I was calling "domes" of the original block.

No, this is wrong. Three of the four satisfy this definition, and it's
these:

x.x.x.x
x.x.x.x
x.x.x.x

x
x.x.x.x
x.x.x.x
x.x.x

x.x.x
x.x.x.x
x.x.x.x
x

Think about it this way: if you state that a Fokker block has to have
vertices that are in the set, then that means each vertex, and every point
on an edge, is ambiguous as to which block it's a part of. The different
shapes result from which of the neighboring blocks you assign different
edges and vertices to.

This one doesn't look to be constructible that way, unless I'm missing
something:

x.x
x.x.x.x
x.x.x.x
x.x

If this version has radically different properties than the other ones, i.e.
if it behaves more like a MODMOS, then that would be a very notable find.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 12:32 AM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> Okay, I need something to call these things that I had previously been calling simply "Fokker blocks", now that I know there's disagreement about the definition of that term. Since Paul's "gentle introduction" is the earliest thing I've seen that defines them exactly the way I want to (and explicitly describes how they can have different shapes even for the same UVs), I suggest "Fokker-Erlich periodicity block", or "F-E block" for short.

As a nice gesture to Paul, I wouldn't mind just calling them Erlich
blocks for short.

> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > Mmmm, tricky! There's no torsion if you temper out 128/125, but there
> > is if you temper out 8/125, the latter being (2/5)^3. They share the
> > same representation in an octave-equivalent space, however. Maybe we
> > should start treating 2 as its own axis.
>
> Aha! This leads me to a revised version of my conjecture:
>
> For an interval c, let f(c) be the greatest integer p such that (c')^p = c, up to octave equivalence, for some other JI interval c'. For example, f(128/125) = 3 because (8/5)^3 = 128/125 up to octave equivalence. f(81/80) = 1 because it's not expressible as a nontrivial power.
>
> An equivalent definition is that f(c) is the number of periods per octave for the codimension-1 temperament tempering out c.

I think you mean the octave-reduction of c in your equivalent definition above?

I think another equivalent definition is that f(c) is the GCD of the
all the non-2 coefficients in the ket/monzo representation of c.

> Conjecture: The number of distinct Fokker-Erlich blocks for a given UV set is equal to N / prod(f(c_i)), where N is the number of notes in the block, and {c_i} are the UVs.

So for {128/125, 81/80}, that's 12 / (3*1), which is 4. Looks like it
holds for that case, but it also looks like the number of distinct,
let's say, Fokker-Lumma blocks as defined by Carl is only 3. The one
that doesn't fit is the one I described in my last post.

Some stats on those:

Mean variety 3.09091:
x.x.x.x
x.x.x.x
x.x.x.x

Mean variety 3.45455:
x
x.x.x.x
x.x.x.x
x.x.x

Mean variety 3.27273:
x.x
x.x.x.x
x.x.x.x
x.x

Mean variety 3.45455:
x.x.x
x.x.x.x
x.x.x.x
x

Hm...!

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 12:13 AM, Carl Lumma <carl@lumma.org> wrote:
>
> I wrote:
> >OK, howabout this instead:
> >1. all uvs point away from 1/1
> >2. all uvs must be positive
> >3. the other (three) scales are MODMOS
>
> Or if you prefer, 2. all uvs must be between 0 and 600 cents. -C.

If your goal is to arrive at only one block, this still runs into
problems because of the four Fokker-Erlich blocks for {128/125,
81/80}, three of them end up fitting this definition. It becomes
ambiguous as to which block every vertex is in, and depending on which
blocks you assign the vertices to, the shape of the block changes.

-Mike

🔗Carl Lumma <carl@lumma.org>

Mike wrote:

>> You're right (there are 8 adjacent blocks in R^2). But the
>> procedure clearly generates convex MODMOS.
>
>No, this is wrong. A counterexample would be to look at a
>rank-2 periodicity block, e.g. one operating on something like the
>Pythagorean chain of fifths. The only way to create a convex set is
>for all of the fifths to be adjacent; a "hole" clearly presents itself
>as soon as you change anything.

I didn't say all MODMOS are convex, I said that translating a
Fokker block off the lattice subgroup produces convex MODMOS.
As you note, in the 1D case there are no convex MODMOS and
hence these translations do not produce different scales.

-Carl

🔗Carl Lumma <carl@lumma.org>

Mike wrote:

>> >OK, howabout this instead:
>> >1. all uvs point away from 1/1
>> >2. all uvs must be positive
>> >3. the other (three) scales are MODMOS
>>
>> Or if you prefer, 2. all uvs must be between 0 and 600 cents. -C.
>
>If your goal is to arrive at only one block, this still runs into
>problems because of the four Fokker-Erlich blocks for {128/125,
>81/80}, three of them end up fitting this definition.

Look again. The vertices are 1/1, 128/125, 81/80, and 648/625.
If 128/125 were in the block instead of 1/1, it would become 1/1
(rule #1) and the uv would go to 125/128 ("negative") or 125/64
(> 600 cents). The same is true for the other two commas.

-Carl

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@> wrote:
> > --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@> wrote:
> >
> > > I conjecture that for an N-note scale (and fixed chromata) there are always N different Fokker blocks, but I have no proof yet.
> >
> > Not true, I'm afraid.
>
> Counterexample?

Try any composite number. For example, {16/15, 648/625} with 8 notes. Kind of nice, one starling tempers to the star scale, the other to star2. The two semimag scales. LOTS more examples.

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:
>
> Okay, I need something to call these things that I had previously been calling simply "Fokker blocks", now that I know there's disagreement about the definition of that term. Since Paul's "gentle introduction" is the earliest thing I've seen that defines them exactly the way I want to (and explicitly describes how they can have different shapes even for the same UVs), I suggest "Fokker-Erlich periodicity block", or "F-E block" for short.

I suggest we not pollute the world with bad definitions of things. I think mathematicians do this stuff right--define intelligently, and redefine if it's not the best definition you can contrive.\

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
> --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@> wrote:
> >
> > Okay, I need something to call these things that I had previously been calling simply "Fokker blocks", now that I know there's disagreement about the definition of that term. Since Paul's "gentle introduction" is the earliest thing I've seen that defines them exactly the way I want to (and explicitly describes how they can have different shapes even for the same UVs), I suggest "Fokker-Erlich periodicity block", or "F-E block" for short.
>
> I suggest we not pollute the world with bad definitions of things. I think mathematicians do this stuff right--define intelligently, and redefine if it's not the best definition you can contrive.\

So what are you suggesting then? That the things Mike and I are talking about should be called "Fokker blocks"?

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Keenan wrote:
>
> >What do you mean by "consecutive"?
> >
> >If 2048/2025 is a UV, are 1/1, 45/32, and 2045/1024 consecutive or not?
>
> Non-consecutive; when a block edge intersects a lattice point
> whose taxicab distance to either vertex is > 1. In the
> 5-limit, the shortest such vectors have taxicab length 4 as
> shown in the attached (let's see if attachments work these days).

I don't understand the usefulness of this definition. Clearly even non-consecutive (as you've defined it) lattice points on an edge of the parallelogram do not imply torsion. Take a {81/80, 2048/2025} block for example.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 2:30 AM, Carl Lumma <carl@lumma.org> wrote:
>
> Mike wrote:
>
> >> >OK, howabout this instead:
> >> >1. all uvs point away from 1/1
> >> >2. all uvs must be positive
> >> >3. the other (three) scales are MODMOS
> >>
> >> Or if you prefer, 2. all uvs must be between 0 and 600 cents. -C.
> >
> >If your goal is to arrive at only one block, this still runs into
> >problems because of the four Fokker-Erlich blocks for {128/125,
> >81/80}, three of them end up fitting this definition.
>
> Look again. The vertices are 1/1, 128/125, 81/80, and 648/625.
> If 128/125 were in the block instead of 1/1, it would become 1/1
> (rule #1) and the uv would go to 125/128 ("negative") or 125/64
> (> 600 cents). The same is true for the other two commas.

OK, that makes sense. I didn't get what you meant by #1.

Pondering this more, I think you're onto something. I think that every
type of block except the ones that fit your criteria can also be
described as a hexagonal periodicity block, but I'm not sure how
significant that is. I also note that the ones that fit your criteria
for {81/80, 128/125} do seem to have the lowest mean variety as well,
although that's the only case I've examined so far.

So there are three categories of blocks then:
1) blocks that fit your stricter definition of a Fokker block
2) blocks that fit both Paul's definition a Fokker block, and also fit
your stricter definition of MODMOS blocks (hexagonal in the cases I've
seen so far)
3) blocks that can only be viewed as convex MODMOS blocks

So the million dollar question is, do #2 blocks result in any scalar
properties that #3 blocks don't?

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 3:08 AM, Carl Lumma <carl@lumma.org> wrote:
>
> [Attachment(s) from Carl Lumma included below]
>
> Keenan wrote:
>
> >What do you mean by "consecutive"?
> >
> >If 2048/2025 is a UV, are 1/1, 45/32, and 2045/1024 consecutive or not?
>
> Non-consecutive; when a block edge intersects a lattice point
> whose taxicab distance to either vertex is > 1. In the
> 5-limit, the shortest such vectors have taxicab length 4 as
> shown in the attached (let's see if attachments work these days).

I don't think this definition of "consecutive" is basis-invariant, as
the taxicab distance is going to vary depending on which two vectors
you decide are supposed to be orthonormal. Also, consider that the
periodicity block tempering out 81/80 and 1/125 is going to exhibit
torsion, as 1/125 is (1/5)^3, whereas the periodicity block tempering
out 81/80 and 128/125 won't. The two will appear identical on an
octave-equivalent lattice.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 4:59 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> I don't think this definition of "consecutive" is basis-invariant, as
> the taxicab distance is going to vary depending on which two vectors
> you decide are supposed to be orthonormal. Also, consider that the
> periodicity block tempering out 81/80 and 1/125 is going to exhibit
> torsion, as 1/125 is (1/5)^3, whereas the periodicity block tempering
> out 81/80 and 128/125 won't. The two will appear identical on an
> octave-equivalent lattice.

Actually, I think our problem is that ALL of these things are torsion
blocks, at least in this quirky octave-equivalent space we keep
working in.

If we're tempering out two vectors we call "128/125" and "81/80," then
the vector we're assigning the name "5/4" to is a torsion element in
this space. That vector has an order of 3, and our block is a torsion
block, no different than if we were tempering out (81/80)^2 or
something. Similarly with 81/80 and 2048/2025, the thing we're calling
"2048/2025" is reachable by two of the things we're calling "45/32",
so it also defines a torsion block.

I'm sure that there's a way to formally define the space so that these
aren't actually torsion elements, but that information isn't reflected
in the geometric patterns that we're trying to find in the lattice. So
maybe it just makes more sense to think of these as really being
torsion blocks that consist of smaller Fokker blocks, until we're
ready to graduate out of this strange space and start looking at the
full 2.3.5 lattice.

Another way to put this is that while working within the simplified
octave-equivalent space, we should focus more on temperaments that
have a full-octave period, and save the semi-octave periods for when
we start looking at the full 2.3.5 lattice. That's how I did it when
figuring out MODMOS's and it made everything a lot easier.

-Mike

🔗Graham Breed <gbreed@gmail.com>

Mike Battaglia <battaglia01@gmail.com> wrote:

> Another way to put this is that while working within the
> simplified octave-equivalent space, we should focus more
> on temperaments that have a full-octave period, and save
> the semi-octave periods for when we start looking at the
> full 2.3.5 lattice. That's how I did it when figuring out
> MODMOS's and it made everything a lot easier.

You can think about period-equivalent space. With a
non-octave period, the Fokker block should be periodic
about the period. So you can re-scale the period to look
like the octave, and treat it as an octave-equivalent
torsion-free block.

In group theory terms, yes, blocks where the period divides
the octave have torsion in octave-equivalent space. That
makes them mixed groups.

Graham

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 6:08 AM, Graham Breed <gbreed@gmail.com> wrote:
>
> Mike Battaglia <battaglia01@gmail.com> wrote:
>
> > Another way to put this is that while working within the
> > simplified octave-equivalent space, we should focus more
> > on temperaments that have a full-octave period, and save
> > the semi-octave periods for when we start looking at the
> > full 2.3.5 lattice. That's how I did it when figuring out
> > MODMOS's and it made everything a lot easier.
>
> You can think about period-equivalent space. With a
> non-octave period, the Fokker block should be periodic
> about the period. So you can re-scale the period to look
> like the octave, and treat it as an octave-equivalent
> torsion-free block.
>
> In group theory terms, yes, blocks where the period divides
> the octave have torsion in octave-equivalent space. That
> makes them mixed groups.
>
> Graham

Yes, this is right. This is how we have to do things. It's all much clearer now.

From this perspective, Carl's approach is clearly right, at least
where torsional blocks are concerned. The periodicity block formed by
{81/80, 128/125} is actually the same as the block formed by {81/80,
5/4}, or even more simply, {81/5, 5/1}. So our 3x4 block is now a 1x4
block, and the only block that's not obviously a MODMOS is the one
that corresponds to the x.x.x.x shape. There are four possible
MODMOS's for this block, and hence Keenan's conjecture holds. This
also takes care of Gene's counterexample tempering 648/625, which
would similarly be a torsion block in octave-equivalent space (I'm not
sure what the semimag scales are, and they're not on the wiki, so I
don't know how to test them).

So is in fact true that Paul's definition is too permissive where
torsion blocks are concerned. What remains to be seen, however, is if
Paul's definition will fail for larger non-torsion blocks, and if so,
if Carl's will fail for larger blocks.

Given all this, Keenan's conjecture doesn't need to be reformulated
with the torsion constraint, so it's a lot simpler. However, I came up
with a trivial way to disprove it anyway - given a rank-2 Fokker block
with any amount of notes you'd like, there's only one block that you
get if you move the vectors around fractionally.

I think that the number of blocks that you can obtain using fractional
chroma lattice shifts will be equal to the total number of convex
MODMOS blocks that you can produce from a particular periodicity
block, plus one for the original block itself. So in this case, the
rank-2 case yields 1, because you can't produce any convex MODMOS's
from it, and you can only get the block.

This should correlate in some intuitive way that I'm not yet sure how
to meaningfully define to the number of notes on the "outside" of the
block, e.g. those closest to the uvs. I think that the number of
fractionally shifted blocks = number of notes will fail for larger
blocks that have lots of "inside" elements in the block. This is just
a shameless intuitive statement and I'm not sure how exactly it'll
look.

-Mike

🔗Carl Lumma <carl@lumma.org>

Mike wrote:

>Pondering this more, I think you're onto something. I think that every
>type of block except the ones that fit your criteria can also be
>described as a hexagonal periodicity block, but I'm not sure how
>significant that is. I also note that the ones that fit your criteria
>for {81/80, 128/125} do seem to have the lowest mean variety as well,
>although that's the only case I've examined so far.

Hexagonal "blocks" are also MODMOS (you'll have to tell me
if the definition of MODMOS I provided isn't right), and *always*
have equal or greater mean variety than the corresponding
Fokker block.

>So there are three categories of blocks then:
>1) blocks that fit your stricter definition of a Fokker block
>2) blocks that fit both Paul's definition a Fokker block, and also fit
>your stricter definition of MODMOS blocks (hexagonal in the cases I've
>seen so far)
>3) blocks that can only be viewed as convex MODMOS blocks
>
>So the million dollar question is, do #2 blocks result in any scalar
>properties that #3 blocks don't?

Of all lattice structures, Fokker blocks have the tightest
connection to regular temperament. I would therefore investigate
these other structures as MODMOS (or "uv-transformed" if you
prefer) variants. Transformations which increase the 3-component
of a block, for instance, are useful because 3s are very strong
concordances (even melodically). But these transformations never
decrease mean variety or increase the chord count (e.g.
pentachordal decatonic scale vs. symmetric decatonic).

-Carl

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> So is in fact true that Paul's definition is too permissive where
> torsion blocks are concerned. What remains to be seen, however, is if
> Paul's definition will fail for larger non-torsion blocks, and if so,
> if Carl's will fail for larger blocks.

I don't understand what you mean by this. I thought the only difference between Paul's definition and Carl's was that in Paul's you could move the parallelogram around and get different blocks, whereas in Carl's definition for "the" Fokker block the parallelogram is fixed to the lattice.

What does it mean for a definition to "fail"?

> Given all this, Keenan's conjecture doesn't need to be reformulated
> with the torsion constraint, so it's a lot simpler. However, I came up
> with a trivial way to disprove it anyway - given a rank-2 Fokker block
> with any amount of notes you'd like, there's only one block that you
> get if you move the vectors around fractionally.

I should have said this explicitly, but my conjecture was supposed to be for the rank-3 case only. It's pretty clear to me that in rank 4, for example, the number of F-E blocks will be a lot greater than the number of notes.

> I think that the number of blocks that you can obtain using fractional
> chroma lattice shifts will be equal to the total number of convex
> MODMOS blocks that you can produce from a particular periodicity
> block, plus one for the original block itself. So in this case, the
> rank-2 case yields 1, because you can't produce any convex MODMOS's
> from it, and you can only get the block.

Counterexample (use fixed width font):

. . . . . .
. x x x x x .
. . x x . .
. . . . . . .

10/9 5/3 5/4 15/8 45/32
. 1/1 3/2 .

This is a convex MODMOS for the JI major scale (replace 9/8 with 10/9 and 4/3 with 45/32), but it is not one of the seven Fokker-Erlich blocks.

Keenan

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>> Non-consecutive; when a block edge intersects a lattice point
>> whose taxicab distance to either vertex is > 1. In the
>> 5-limit, the shortest such vectors have taxicab length 4 as
>> shown in the attached (let's see if attachments work these days).
>
>I don't understand the usefulness of this definition. Clearly even
>non-consecutive (as you've defined it) lattice points on an edge of
>the parallelogram do not imply torsion. Take a {81/80, 2048/2025}
>block for example.

You're right

http://bit.ly/ps7Ofv

forget I mentioned it. -C.

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> At 11:28 AM 9/12/2011, you wrote:
>
> >Why does the thing you defined (as "Fokker block") deserve a name?
> >
> >Keenan
>
> Fokker defined them, not me. I stated why here:
> /tuning-math/message/19543

But I'm still uncertain what the actual definition is.

I believe the most recent definition you gave was

* All UVs are between 0 and 600 cents.
* 1/1 is included, but none of the other vertices of the parallelepiped.

Is this correct? Is this equivalent to a precise definition that Fokker gave, or is it your interpretation?

It's important because I want to find some counterexamples to statements you've made, but I want to make sure I have the right definition first.

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> I think the interesting things you get are MODMOS, where a
> MODMOS is defined as a scale that can be produced from a
> Fokker block by optionally replacing each of its elements with
> the corresponding element in an *adjacent* block. Rothenberg
> mean variety is non-decreasing with respect to this operation.

I believe I have a concrete counterexample to this last statement, which I'll give as soon as you verify the precise definition of "Fokker block" you're using here.

Keenan

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>> Fokker defined them, not me. I stated why here:
>> /tuning-math/message/19543
>
>But I'm still uncertain what the actual definition is.
>I believe the most recent definition you gave was
>* All UVs are between 0 and 600 cents.
>* 1/1 is included, but none of the other vertices of the parallelepiped.
>Is this correct?

Yes, as a modification to the original definition I posted.

>Is this equivalent to a precise definition that
>Fokker gave, or is it your interpretation?

I tried to give a precise definition for what Fokker
described and dealt with in his papers. It's been about
ten years since I've read them I think, and I'm not a
mathematician, so YMMV.

>It's important because I want to find some counterexamples to
>statements you've made, but I want to make sure I have the right
>definition first.

Fire away!

-Carl

🔗Carl Lumma <carl@lumma.org>

Keenan wrote:

>> I think the interesting things you get are MODMOS, where a
>> MODMOS is defined as a scale that can be produced from a
>> Fokker block by optionally replacing each of its elements with
>> the corresponding element in an *adjacent* block. Rothenberg
>> mean variety is non-decreasing with respect to this operation.
>
>I believe I have a concrete counterexample to this last statement,
>which I'll give as soon as you verify the precise definition of
>"Fokker block" you're using here.

Note that by "adjacent" I mean transpositions by linear
combinations of the commas with coefficients of at most 1.

I'll be very surprised if there is a counterexample to
this statement (which I've made maybe a dozen times on this
list in the past couple months). But I am willing to
be surprised!

-Carl

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, "genewardsmith" <genewardsmith@...> wrote:
> --- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@> wrote:
>
> > So what are you suggesting then? That the things Mike and I are talking about should be called "Fokker blocks"?
>
> Maybe I don't understand what you are doing, but it sounds as if you want to arbitrarily call one of the Fokker blocks "the" Fokker block and the rest Erlich blocks. If so, bad idea.

No, I don't want to do that. I want to treat them all on an equal footing, same as you. But Carl maintains that he has the "real" definition of "Fokker block", as in the one Fokker meant, and that there's only one of them. I'm just trying to cooperate and avoid confusion.

I'm happy to call them all "Fokker-Erlich blocks"; I'm happy to call them all "Fokker blocks" (as long as nobody's using that term to mean something different); I'm happy to call them all "wakalixes".

Keenan

P.S. I changed http://xenharmonic.wikispaces.com/Fokker+blocks so that it no longer uses the word "chroma"; I hope you don't mind. The rest of us seem to agree that a "chroma" should be an interval that maps to 0 scale steps, but does not necessarily vanish.

For example, in the meantone diatonic scale, 81/80 is a comma (because it vanishes), 25/24 is a chroma (because it maps to 0 scale steps but does not vanish), and 16/15 is a scale step (because it maps to 1 scale step).

🔗Carl Lumma <carl@lumma.org>

>Maybe I don't understand what you are doing, but it sounds as if you
>want to arbitrarily call one of the Fokker blocks "the" Fokker block
>and the rest Erlich blocks. If so, bad idea.
>
>No, I don't want to do that. I want to treat them all on an equal
>footing, same as you.

He does?

>But Carl maintains that he has the "real"
>definition of "Fokker block", as in the one Fokker meant, and that
>there's only one of them. I'm just trying to cooperate and avoid confusion.

I didn't say anything like that. To the best of my knowledge,
Fokker never considered the case where the vectors don't have
endpoints in the lattice. Nor have I ever seen a such a beast
in all the investigations done into Fokker blocks (are there
any so labeled in the Scala archive?). And it does seem to make
sense to view them as pitchwise transformations of a usual
block -- what property, other than parallelogram-convexity, do
they have over other MODMOS?

>I'm happy to call them all "Fokker-Erlich blocks";

Now that would be a bad idea.

>I'm happy to call them all "Fokker blocks"

A slightly better idea.

>I'm happy to call them all "wakalixes".

Now this I like. If you don't use it for something, I want to.

-Carl

🔗Mike Battaglia <battaglia01@gmail.com>

Aggregated reply to Carl

On Mon, Sep 12, 2011 at 2:00 PM, Carl Lumma <carl@lumma.org> wrote:
>
> >But this thing we're talking about, Fokker-Erlich blocks, is not at
> >all the same thing as MODMOS.
> >
> >Even in rank 2, they're different concepts. In higher ranks, still different.
>
> They are MODMOS according to the definition I gave.
> Not the same thing, no, but why do they deserve a name?
//
> Hexagonal "blocks" are also MODMOS (you'll have to tell me
> if the definition of MODMOS I provided isn't right), and *always*
> have equal or greater mean variety than the corresponding
> Fokker block.

They're the same thing. A hexagon is a parallelogram with some parts
cut off and reassembled. So it's a MODMOS.

> >So there are three categories of blocks then:
> >1) blocks that fit your stricter definition of a Fokker block
> >2) blocks that fit both Paul's definition a Fokker block, and also fit
> >your stricter definition of MODMOS blocks (hexagonal in the cases I've
> >seen so far)
> >3) blocks that can only be viewed as convex MODMOS blocks
> >
> >So the million dollar question is, do #2 blocks result in any scalar
> >properties that #3 blocks don't?
>
> Of all lattice structures, Fokker blocks have the tightest
> connection to regular temperament. I would therefore investigate
> these other structures as MODMOS (or "uv-transformed" if you
> prefer) variants. Transformations which increase the 3-component
> of a block, for instance, are useful because 3s are very strong
> concordances (even melodically). But these transformations never
> decrease mean variety or increase the chord count (e.g.
> pentachordal decatonic scale vs. symmetric decatonic).

What do you mean by increase the chord count?

-Mike

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> >No, I don't want to do that. I want to treat them all on an equal
> >footing, same as you.
>
> He does?

Yes. The construction given in http://xenharmonic.wikispaces.com/Fokker+blocks in terms of the floor function and offsets is equivalent to arbitrarily translating a parallelepiped.

> >But Carl maintains that he has the "real"
> >definition of "Fokker block", as in the one Fokker meant, and that
> >there's only one of them. I'm just trying to cooperate and avoid confusion.
>
> I didn't say anything like that. To the best of my knowledge,
> Fokker never considered the case where the vectors don't have
> endpoints in the lattice. Nor have I ever seen a such a beast
> in all the investigations done into Fokker blocks (are there
> any so labeled in the Scala archive?). And it does seem to make
> sense to view them as pitchwise transformations of a usual
> block -- what property, other than parallelogram-convexity, do
> they have over other MODMOS?

What do you mean you've never seen it? It's right there in Paul's "gentle introduction": http://sonic-arts.org/td/erlich/intropblock2.htm

> >I'm happy to call them all "Fokker-Erlich blocks";
>
> Now that would be a bad idea.
>
> >I'm happy to call them all "Fokker blocks"
>
> A slightly better idea.
>
> >I'm happy to call them all "wakalixes".
>
> Now this I like. If you don't use it for something, I want to.

"Wakalixes" is actually a nonsense word Feynman used.

What about some adjective modifying "Fokker block"? "Free Fokker block"? "Floating Fokker block"?

I'll keep you in suspense about my counterexamples until tomorrow.

Keenan

🔗Mike Battaglia <battaglia01@gmail.com>

Actually, I'm not sure this'll work, because what's the period if
128/125 and 81/80 are unison vectors? If you treat the 128/125 as
commatic, then the period is 5/4, but if you don't, then the period is
still 2/1. Are we to look at different spaces depending on whether the
vectors of our block are commatic or chromatic?

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Mon, Sep 12, 2011 at 2:09 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> > So is in fact true that Paul's definition is too permissive where
> > torsion blocks are concerned. What remains to be seen, however, is if
> > Paul's definition will fail for larger non-torsion blocks, and if so,
> > if Carl's will fail for larger blocks.
>
> I don't understand what you mean by this. I thought the only difference between Paul's definition and Carl's was that in Paul's you could move the parallelogram around and get different blocks, whereas in Carl's definition for "the" Fokker block the parallelogram is fixed to the lattice.
>
> What does it mean for a definition to "fail"?

It means that the resulting blocks don't yield all of the properties
that one would intuitively expect out of a periodicity block. Until
those intuitions are shown to be inconsistent they should be
respected. However, when I wrote this, I thought that the other ones
were rank-3 MODMOS's, but now I realize that we can't definitively say
that unless we know what a rank-3 MOS is.

For now, I'm going to call the original things "Fokker blocks," aka
that allow for fractional translations of the chroma lattice, and I'm
going to call the thing Carl's talking about the singular "Lumma
block" for some set of uvs. The Lumma block does seem to have special
properties in that it tends to have lower mean variety than the other
blocks when put in scale form. However, it is noteworthy that ALL the
Fokker blocks for an MOS seem to temper down to legitimate MOS when
either of the uvs is tempered out, which isn't true of MODMOS in
general - any decent JI rendition of harmonic minor, for instance,
only becomes MODMOS if 25/24 vanishes, and not if 81/80 vanishes. So
that's an interesting property. Keenan's counterexample, which isn't
one of the 7 Fokker blocks for the JI major scale, doesn't have this
property, as it's only MOS if 81/80 vanishes, not if 25/24 vanishes,
at which point you get something looking more like maqam rast mode.

Lastly, compare the scales 1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1 and
1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1. The former is the Lumma
block, whereas the latter is one of the other Fokker blocks. The
former can be generated by the following sequence of generators,
stacked on top of one another: 5/4-6/5-5/4-6/5-5/4-6/5-5/4, and the
latter is 6/5-5/4-6/5-5/4-6/5-5/4-6/5. Given this, I wonder if the
definition of the Lumma block, as it exists, is basis-invariant.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

On Tue, Sep 13, 2011 at 3:29 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
> On Mon, Sep 12, 2011 at 2:09 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>>
>> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>> > So is in fact true that Paul's definition is too permissive where
>> > torsion blocks are concerned. What remains to be seen, however, is if
>> > Paul's definition will fail for larger non-torsion blocks, and if so,
>> > if Carl's will fail for larger blocks.
>>
>> I don't understand what you mean by this. I thought the only difference between Paul's definition and Carl's was that in Paul's you could move the parallelogram around and get different blocks, whereas in Carl's definition for "the" Fokker block the parallelogram is fixed to the lattice.
>>
>> What does it mean for a definition to "fail"?
>
> It means that the resulting blocks don't yield all of the properties
> that one would intuitively expect out of a periodicity block. Until
> those intuitions are shown to be inconsistent they should be
> respected. However, when I wrote this, I thought that the other ones
> were rank-3 MODMOS's, but now I realize that we can't definitively say
> that unless we know what a rank-3 MOS is.

Eh, this got screwed up. I copied and pasted some paragraph around and
the last sentence here is now out of context. These mysterious "other
ones" were supposed to be the Fokker blocks that aren't the Lumma
block.

-Mike

🔗Graham Breed <gbreed@gmail.com>

Mike Battaglia <battaglia01@gmail.com> wrote:
> On Mon, Sep 12, 2011 at 6:08 AM, Graham Breed
> <gbreed@gmail.com> wrote:
> >
> > You can think about period-equivalent space. With a
> > non-octave period, the Fokker block should be periodic
> > about the period. So you can re-scale the period to look
> > like the octave, and treat it as an octave-equivalent
> > torsion-free block.
>
> Actually, I'm not sure this'll work, because what's the
> period if 128/125 and 81/80 are unison vectors? If you
> treat the 128/125 as commatic, then the period is 5/4,
> but if you don't, then the period is still 2/1. Are we to
> look at different spaces depending on whether the vectors
> of our block are commatic or chromatic?

If you temper out 128/125, the tuning will be periodic.
That doesn't mean the periodicity block itself is
periodic. You could express the tempered block in a
tempered space, though, and then it would have the correct
period.

Graham

🔗genewardsmith <genewardsmith@sbcglobal.net>

--- In tuning-math@yahoogroups.com, "Keenan Pepper" <keenanpepper@...> wrote:

> No, I don't want to do that. I want to treat them all on an equal footing, same as you. But Carl maintains that he has the "real" definition of "Fokker block", as in the one Fokker meant, and that there's only one of them. I'm just trying to cooperate and avoid confusion.

I feel it is a bad idea to introduce a useless definition.

> P.S. I changed http://xenharmonic.wikispaces.com/Fokker+blocks so that it no longer uses the word "chroma"; I hope you don't mind.

I saw that; looks OK.

🔗genewardsmith <genewardsmith@sbcglobal.net>

🔗Carl Lumma <carl@lumma.org>

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> Note that by "adjacent" I mean transpositions by linear
> combinations of the commas with coefficients of at most 1.
>
> I'll be very surprised if there is a counterexample to
> this statement (which I've made maybe a dozen times on this
> list in the past couple months). But I am willing to
> be surprised!

Okay, here goes the counterexample:

It's a 10-note Fokker block in the 2.7.13 subgroup. It's UVs are 343/338 (|-1, 3, -2>) and 28672/28561 (|12, 1, -4>).

Here's "the Fokker block" as you've defined it:

! fokkerblock.scl
!
2.7.13 Fokker block (Carl Lumma's definition) with UVs 343/338, 28672/28561
10
!
14/13
196/169
2744/2197
224/169
3136/2197
43904/28561
3584/2197
50176/28561
702464/371293
2/1

Scala tells me "Average number of different intervals per interval class: 3.33333 = 10/3"

Now, here's a "free Fokker block" / "MODMOS of the block" / "wakalix":

! freefokkerblock.scl
!
2.7.13 Fokker block (free-floating parallelogram definition) with UVs 343/338, 28672/28561
10
!
14/13
1183/1024
637/512
169/128
91/64
49/32
13/8
7/4
49/26
2/1

This is a MODMOS of the original block where 8 of the notes have been altered by 28672/28561 to put them in the adjacent block. (You could also say just 2 have been altered, but then one of the altered ones is 1/1. Either way it's definitely a MODMOS.)

For this one, scala tells me "Average number of different intervals per interval class: 3.11111 = 28/9", which is less than fokkerblock.scl.

For these scala files and some diagrams, please see the 5 files I just uploaded to /tuning-math/files/Keenan%20Pepper/

I chose this counterexample rather than a simpler one because for this example, the four possible blocks you get from including each of the vertices in turn *all* have greater mean variety than the free Fokker block / MODMOS / wakalix I exhibit.

Keenan

🔗Keenan Pepper <keenanpepper@gmail.com>

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> For now, I'm going to call the original things "Fokker blocks," aka
> that allow for fractional translations of the chroma lattice, and I'm
> going to call the thing Carl's talking about the singular "Lumma
> block" for some set of uvs. The Lumma block does seem to have special
> properties in that it tends to have lower mean variety than the other
> blocks when put in scale form.

Check out my counterexample to this.

> However, it is noteworthy that ALL the
> Fokker blocks for an MOS seem to temper down to legitimate MOS when
> either of the uvs is tempered out, which isn't true of MODMOS in
> general

This is an excellent observation, and gives a characterization of free Fokker blocks that Carl was asking for.

If we want a block to become meantone[7] when we temper out 81/80, it has to lie in between two parallel lines both parallel to 81/80. For example if we want the block to temper down to F-B we can draw one line anywhere between Bb and F, and the other line anywhere between B and F#. The lines don't have to intersect any lattice points.

If we want the same block also to become dicot[7] when we temper out 25/24, it has to be between another pair of parallel lines parallel to 25/24. The locus of points between both pairs of parallel lines is obviously a parallelogram.

So a block tempers down to both MOSes if and only if it's in the interior of a parallelogram, but that parallelogram does not have to have vertices on lattice points. In other words it tempers down to both MOSes if and only if it's a free Fokker block.

> Lastly, compare the scales 1/1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1 and
> 1/1, 10/9, 5/4, 4/3, 3/2, 5/3, 15/8, 2/1. The former is the Lumma
> block, whereas the latter is one of the other Fokker blocks. The
> former can be generated by the following sequence of generators,
> stacked on top of one another: 5/4-6/5-5/4-6/5-5/4-6/5-5/4, and the
> latter is 6/5-5/4-6/5-5/4-6/5-5/4-6/5. Given this, I wonder if the
> definition of the Lumma block, as it exists, is basis-invariant.

Something's wrong here, because each chain should only have 6 generators, not 7. If 5/4-6/5-5/4-6/5-5/4-6/5-5/4 equals 4/1, then you're already in mavila, and if 6/5-5/4-6/5-5/4-6/5-5/4-6/5 equals 4/1, then you're already in meantone.

Keenan