Otonal, utonal, and ztonal chords

Assume that we're working within the 11-limit lattice, and that we're
looking to find chords that are homometric to 4:5:6:7:9:11. The most
trivial case of a chord that would be homometric is 1/(4:5:6:7:9:11),
that is, the 11-limit utonality. Since this chord is homometric to the
first, it will have only 11-odd-limit dyads. In general, any
translation or reflection of a set is "trivially homometric" to the
original set.

Warren's work opens up the possibility that we can find some
"nontrivially homometric" cases - homometric sets that can't be
produced by some combination of translation or reflection on the
original set. Such a chord would contain all of the same 11-limit
dyads in 4:5:6:7:9:11, but would be neither otonal nor utonal.

Whether or not a chord is "otonal" or "utonal" is usually determined
up to octave-equivalence: 1:3:5:7:9:11 and 4:5:6:7:9:11 are both
usually considered to be "otonal" chords, and likewise the different
octave transpositions of 1/(4:5:6:7:9:11) are all usually considered
to be "utonal." All of these chords have dyads that stick strictly to
the 11-"odd"-limit, a measure that itself implies octave equivalence.
Thus it can be said that these chords are "trivially homometric up to
octave equivalence" with respect to one another.

The most interesting musical possibilities to emerge from this would
be to find hexads that are "nontrivially homometric up to octave
equivalence" to the 11-limit otonality, and hence also to the 11-limit
utonality. Such hexads would stick strictly to the 11-odd-limit, but
would be neither otonal nor utonal. Music set theory has long had a
name for sets that are nontrivially homometric up to octave
equivalence, and that name is "z-related," so we might call these
11-limit ztonal chords.

An interesting test would be to find 11-limit ztonal hexads, and see
if they sound xenharmonically "minor."

More to come...

-Mike

On Tue, Oct 18, 2011 at 7:40 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> More to come...

It can be shown from Warren's work that no non-trivial homometric sets
exist that have six incommensurate step sizes. In fact, they can only
exist if they have two incommensurate step sizes (at a maximum). By
Warren's theorem, 6-element homometric sets can only come in the two
following forms, for a, b in R:

(The reflections of these set are trivially also homometric to one another)
Form 1:
Original set: {0, 2a-b, b, a+b, 5a-2b, 4a-b}
Homometric set: {0, 3a-2b, a, 5a-3b, 5a-2b, 4a-b}

Form 2:
Original set: {0, a+b, 3a+5b, 4a+5b, 5a+7b, 5a+8b}
Homometric set: {0, a+2b, 3a+4b, 4a+7b, 5a+7b, 5a+8b}

Since each entry in this set is some linear combination of two
variables a and b, it can also be thought of as an interval that's a
combination of two types of generators, or as a point on a tempered 2d
lattice.

If we're concerned about ztonal chords, meaning if we're doing
everything up to octave equivalence, then we can allow any of the
elements in this set to vary by an octave. So we can think of this as
the addition of another axis representing 2/1 to our tempered space.
We'll then simply just discard the information on that axis when we
test to see if two sets are z-related to one another.

So we're dealing with, at a maximum, a rank-3 space, with the
condition that one of the generators has to be an octave. The question
of how to deal with tempered spaces in which the octave is subdivided
into two or more parts (such as if 50/49 vanishes) is more
complicated, and not dealt with here. Assuming that one of the
generators is 2/1, it's also trivial to see that any further tempering
of this space that maintains the set {1/1, 2/1} as being convex can
also support 6-note homometric sets. We'll work under these restraints
for now, so we'll assume that we're in a rank-3 tempered space which
defines a 2d tempered lattice.

If we take form 1, and lay it out as a bunch of ket vectors, we arrive
at a simple theorem, which is that the following sets of monzos will
be homometric to one another. Assume these monzos lay out coordinates
in the tempered space (tonzos?), that the first coordinate represents
the "a" generator from Warren's sets, and that the second coordinate
represents "b." (Assume that normally, the first coordinate would be
2, but we're ignoring it for the sake of octave equivalence).

Form 1:
{[0 0>, [2 -1>, [0 1>, [1 1>, [5 -2>, [4 -1>} is homometric to
{[0 0>, [3 -2>, [1 0>, [5 -3>, [5 -2>, [4 -1>}

Form 2:
{[0 0>, [1 1>, [3 5>, [4 5>, [5 7>, [5 8>} is homometric to
{[0 0>, [1 2>, [3 4>, [4 7>, [5 7>, [5 8>}

One obvious problem is for us to have both 3/2 and 9/4 (up to octave
equivalence) in this set, we'd need one of the monzos to be double the
other. Epic fail on that, so I guess that means that there's no rank-3
11-limit ztonal hexads (this might change if you temper down, but I'm
not going to deal with that yet).

So we'll next look at 4:5:6:7:11:13 instead, which is
octave-equivalent to 1:3:5:7:11:13.

-Mike

On Tue, Oct 18, 2011 at 9:16 PM, Mike Battaglia <battaglia01@gmail.com>
wrote:
> On Tue, Oct 18, 2011 at 7:40 PM, Mike Battaglia <battaglia01@gmail.com>
wrote:
>
> Form 1:
> {[0 0>, [2 -1>, [0 1>, [1 1>, [5 -2>, [4 -1>} is homometric to
> {[0 0>, [3 -2>, [1 0>, [5 -3>, [5 -2>, [4 -1>}

So we'll take a look at Form 1 first. If we take the top set of monzos,
interpret them as column vectors and put them into a matrix, we get (view
fixed width)

? ? ? ? ?
|< 2 0 1 5 4|,
<-1 1 1 -2 -1|>

If we throw 2 back in there, and we allow the * glyph to specify that we
don't care what the mapping is, we get

Form 1a:
2 ? ? ? ? ?
|< 1 * * * * *|,
< 0 2 0 1 5 4|,
< 0 -1 1 1 -2 -1|>

Doing the same with the other entry in Form 1 gives us:

Form 1b:
2 ? ? ? ? ?
|< 1 * * * * *|,
< 0 3 1 5 5 4|,
< 0 -2 0 -3 -2 -1|>

If we repeat with Form 2, we get these two entries:

Form 2a:
2 ? ? ? ? ?
|< 1 * * * * *|,
< 0 1 3 4 5 5|,
< 0 1 5 5 7 8|>

Form 2b:
2 ? ? ? ? ?
|< 1 * * * * *|,
< 0 1 3 4 5 5|,
< 0 2 4 7 7 8|>

So the goal is to find vals that match one of the following, where * could
be anything, and the basis is some permutation of {2, 3, 5, 7, 11, 13}.

This seems like it'd be incredibly hard to brute force, so I'll leave it
here for now. If anyone has any clever insights into how to simplify this
problem, let's hear it!

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> One obvious problem is for us to have both 3/2 and 9/4 (up to octave
> equivalence) in this set, we'd need one of the monzos to be double the
> other. Epic fail on that, so I guess that means that there's no rank-3
> 11-limit ztonal hexads (this might change if you temper down, but I'm
> not going to deal with that yet).

I looked at this case, and I believe that it's only possible if the difference set of the hexad is one of the following (in terms of generators):

(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11)
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13)
(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14)

In particular, this means that the simplest 11 intervals must *all* be 11-limit consonances for it to work. I think no matter what the generator is, one of these intervals is always going to be at most 1\11, i.e. 109 cents. But this has to represent something like 12/11 or 11/10. Eugh.

So another epic fail on that front.

Next I'm going to look at non-octave periods. For you math people, this means homometric sets that don't live in R, or Z^2 or anything, but they live in Z x G where G is a finite cyclic group.

For example, for half-octave period temperaments like pajara, you'd look for homometric sets in Z x (Z/2Z).

Keenan

On Wed, Oct 19, 2011 at 2:46 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> Next I'm going to look at non-octave periods. For you math people, this means homometric sets that don't live in R, or Z^2 or anything, but they live in Z x G where G is a finite cyclic group.
>
> For example, for half-octave period temperaments like pajara, you'd look for homometric sets in Z x (Z/2Z).
>
> Keenan

So if anyone's confused by that, you're going to have to take one of
Warren's generators and mod it out by 2. So you'd get

Original homometric sets, where * is anything:
{[* 0 0>, [* 2 -1>, [* 0 1>, [* 1 1>, [* 5 -2>, [* 4 -1>} is
homometric to {[* 0 0>, [* 3 -2>, [* 1 0>, [* 5 -3>, [* 5 -2>, [* 4
-1>}

Let's say we temper out 2/a^2. We'll end up with

{[0 0>, [0 -1>, [0 1>, [1 1>, [1 -2>, [0 -1>} is homometric to {[0 0>,
[1 -2>, [1 0>, [1 -3>, [1 -2>, [0 -1>}

I'm not enjoying too much the fact that [0 -1> appears twice in that
first set. Anyway, more to come later, in a rush for now.

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Assume that we're working within the 11-limit lattice, and that we're
> looking to find chords that are homometric to 4:5:6:7:9:11. The most
> trivial case of a chord that would be homometric is 1/(4:5:6:7:9:11),
> that is, the 11-limit utonality. Since this chord is homometric to the
> first, it will have only 11-odd-limit dyads. In general, any
> translation or reflection of a set is "trivially homometric" to the
> original set.

What about a slightly more lax definition?

"Homometric" requires the *multiset* of differences to be the same, that is, each difference must appear the same number of times. What if we only require the *set* of differences to be the same?

Example: [1, 2, 4, 5, 3] and [1, 5, 2, 4, 3]
(so the actual sets are {0, 1, 3, 7, 12, 15} and {0, 1, 6, 8, 12, 15})

These have the same set of differences {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 14, 15}, but each difference may appear a different nonzero number of times in each of the two sets.

If some chord had this relationship to 4:5:6:7:9:11 (a "weakly homometric" relationship, I guess...), then its dyads would still consist of all the 11-odd-limit consonances, and only 11-odd-limit consonances. Should such chords also be called ztonal?

A brief computer search tells me that you still need at least 6 elements for two sets to be nontrivially "weakly homometric".

Keenan

On Thu, Oct 20, 2011 at 5:39 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> >
> > Assume that we're working within the 11-limit lattice, and that we're
> > looking to find chords that are homometric to 4:5:6:7:9:11. The most
> > trivial case of a chord that would be homometric is 1/(4:5:6:7:9:11),
> > that is, the 11-limit utonality. Since this chord is homometric to the
> > first, it will have only 11-odd-limit dyads. In general, any
> > translation or reflection of a set is "trivially homometric" to the
> > original set.
>
> What about a slightly more lax definition?
>
> "Homometric" requires the *multiset* of differences to be the same, that is, each difference must appear the same number of times. What if we only require the *set* of differences to be the same?
>
> Example: [1, 2, 4, 5, 3] and [1, 5, 2, 4, 3]
> (so the actual sets are {0, 1, 3, 7, 12, 15} and {0, 1, 6, 8, 12, 15})
>
> These have the same set of differences {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 14, 15}, but each difference may appear a different nonzero number of times in each of the two sets.

Yeah... that would be rather interesting.

> If some chord had this relationship to 4:5:6:7:9:11 (a "weakly homometric" relationship, I guess...), then its dyads would still consist of all the 11-odd-limit consonances, and only 11-odd-limit consonances. Should such chords also be called ztonal?

I wouldn't call them ztonal, because they're not z-related -
z-relations are defined specifically with respect to the set's
"interval vector," which tells you specifically how many times the
interval, up to octave equivalence, appears. Two sets are z-related
iff they have the same interval vector, and are not just translations
or reflections of the original set.

Your metric states that two sets are related if the elements in the
interval vector have the same sign, basically. I think it's a neat
idea, it might do really well in situations where like, 6/5*6/5 ≈
16/11, and also 7/6*7/6≈11/8, which I guess means that 50/49 vanishes
too.

It looks like "weakly homometric" and "isometric" both already have
names in the literature:

http://www.sciencedirect.com/science/article/pii/S0195669808000322

I can't get my hands on this paper though, so maybe you can find it.

The terminology will get confusing if we don't sort it out. I was
going to use the "strongly" vs "weakly" prefix for something that has
to do with z-relations too. So far there's a heirarchy of
homometricism, and this just another layer on top of it:

1) At the top of the food chain, there's what you called "weakly
homometric" sets. That might be in conflict with some existing name,
but whatever you call it, that's the parent concept.
2) As a subset of that, we have homometric sets.
3) Homometric splits into "trivially" homometric and "nontrivially" homometric.
4) "Homometric" is a superset of "nontrivially homometric," but so is
"z-related." The latter defines a set that's nontrivially homometric
up to octave equivalence.
5) There are "homometric" sets that aren't z-related, because
homometric includes things like reflection and translation, but
"z-related" is defined specifically to exclude those things.
6) There are also sets that are z-related but not homometric, because
z-related allows for octave equivalence, but homometric doesn't.
7) x-limit ztonal chords are chords that are z-related to the z-limit
o/utonalities.

So then my property was

7) I think that there are z-related sets for which some octave
permutation of the notes yields a "nontrivially homometric"
relationship - one that doesn't depend on octave equivalence. But, I
also think there are z-related sets for which that can never be the
case; e.g. 5/4 in one set is 8/5 in the other, and transposing the 8/5
to be 5/4 will break some other relationship, so that you can never
work all the dyads out right.

I think that sums it up. I was going to call 7 "strongly z-related" vs
"weakly z-related," but it's going to make no sense if we have things
like "weakly strongly z-related" to describe strongly z-related chords
that make use of weak homometricism vs strong homometricism.

-Mike