Let's consider the situation where 81/80 is the commatic unison

vector, 25/24 is the chromatic unison vector, and 2 is what you might

call the repetition vector, defining the unit which is repeated

(since of course that doesn't actually need to be octaves.) Then if

each of these defines the row of a matrix, we get Graham's matrix

[ 1 0 0]

M = [-3 -1 2]

[-4 4 -1]

The inverse of this is

[ 7 0 0]

M^(-1) = 1/7 [11 1 2]

[16 4 1].

The question is how to interpret M^(-1). One use for it can be found

when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1)

and select all notes [a b c] such that

0 <= i < 1

-1/2 <= j < 1/2

-1/2 <= k < 1/2

we obtain the seven-note scale

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

with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone

system which equates 9/8 and 10/9 (which means we are using the fact

that 81/80 is a commatic vector) we get a PB scale with just two

sizes of intervals, LsLLLsL, ready to delight the fans of modal

melodies. Does this have anything to do with Graham's comments on

this matrix?

In-Reply-To: <9lv41h+ag6a@eGroups.com>

Gene wrote:

> The inverse of this is

>

> [ 7 0 0]

> M^(-1) = 1/7 [11 1 2]

> [16 4 1].

>

> The question is how to interpret M^(-1). One use for it can be found

> when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1)

> and select all notes [a b c] such that

>

> 0 <= i < 1

> -1/2 <= j < 1/2

> -1/2 <= k < 1/2

>

> we obtain the seven-note scale

>

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

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

[0 0 0]

[1 -2 1]

[1 1 -1]

[2 -1 0]

[-1 1 0]

[0 -1 1]

[0 2 -1]

[1 0 0]

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

[0 0 0]

[0.142857143 0.285714286 -0.428571429]

[0.285714286 -0.428571429 0.142857143]

[0.428571429 -0.142857143 -0.285714286]

[0.571428571 0.142857143 0.285714286]

[0.714285714 0.428571429 -0.142857143]

[0.857142857 -0.285714286 0.428571429]

[1 0 0]

Where, indeed, the first column lies between 0 and 1, and the others

between -1/2 and 1/2. Furthermore, the list is already ordered by pitch,

before we even worked out any pitches! That's because the left hand

column, being "the thing that's left over when all the unison vectors are

taken out", defines approximate pitch order.

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

> with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone

> system which equates 9/8 and 10/9 (which means we are using the fact

> that 81/80 is a commatic vector) we get a PB scale with just two

> sizes of intervals, LsLLLsL, ready to delight the fans of modal

> melodies. Does this have anything to do with Graham's comments on

> this matrix?

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

adjoint.

[ 7 0 0]

[11 1 2]

[16 4 1]

The first row defines the octave (which also happens to be the interval of

equivalence) the second defines the 3:1 approximation and the third the

5:4 approximation.

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

[1 2 4]

The first column tells you the number of steps in the scale LsLLLsL the

interval spans. 9:8 is L so spans 1 step. The second column tells you

the number of generators give the interval in the system where 81:80 is

tempered out (ie meantone). The generator is the fifth (or twelfth, etc),

so 9:8 is two fifths (octave reduced). The last column defines the

interval in the system where 25:24 is tempered out. I'll guess that's

correct.

10:9 is [1 -2 1]. Multiply by Adj(M) and we get [1 2 3]. The first two

columns are the same as for 9:8 which is no surprise because 9:8 and 10:9

are identical in meantone. That the third column differs by 1 could well

tell you that 9:8 and 10:9 differ by a comma.

16:15 is [4 -1 -1]. This multiplies out to [1 -5 -3]. So it's also one

scale step (s this time). And it's generated by -5 fifths, or 5 fourths.

That is C-F-Bb-Eb-Ab-Db.

The same process should work for all intervals.

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

[-1 2]

[ 4 -1]

the adjoint is

[-1 -2]

[-4 -1]

sign isn't important, so make that

[1 2]

[4 1]

This is identical to a subset of adj(M). This will always be the case for

correct unison vectors, but I don't have the criteria for those unison

vectors being "correct". It tells us the octave-equivalent mapping in

terms of fifths for all 5-limit intervals. The determinant tells us the

number of notes in the periodicity block, but nothing I can see tells us

what order they should be in. So that's all we lose by being octave

equivalent.

Graham

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

> Let's consider the situation where 81/80 is the commatic unison

> vector, 25/24 is the chromatic unison vector, and 2 is what you

might

> call the repetition vector, defining the unit which is repeated

> (since of course that doesn't actually need to be octaves.) Then if

> each of these defines the row of a matrix, we get Graham's matrix

>

> [ 1 0 0]

> M = [-3 -1 2]

> [-4 4 -1]

>

> The inverse of this is

>

> [ 7 0 0]

> M^(-1) = 1/7 [11 1 2]

> [16 4 1].

>

> The question is how to interpret M^(-1). One use for it can be

found

> when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-

1)

> and select all notes [a b c] such that

>

> 0 <= i < 1

> -1/2 <= j < 1/2

> -1/2 <= k < 1/2

>

> we obtain the seven-note scale

>

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

This is exactly how my PB program works, as explained in part 3 of

the _Gentle Introduction_.

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

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

block?

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

Introduction_.

In-Reply-To: <9m121j+2t5n@eGroups.com>

Paul wrote:

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

> Introduction_.

Oh yes. That's

<http://www.ixpres.com/interval/td/erlich/intropblock3.htm>. The

differences are that you take all elements between 0 and 1 instead of some

between -0.5 and 0.5, sometimes, and you don't order them by the

generator. As the PB repeats about the half octave, that means each

generator occurs twice.

Actually, no, it's more serious than that. You (Paul) use

octave-equivalent vectors, and so the generator isn't there, or rather

it's a different generator (MOS rather than ET). Would it work in general

to take the generator to increase in steps until you hit the period, and

find the only values for the unison vectors that work for each generator?

And would this make all the scale steps members of a unitary matrix?

I note that the left hand column of Paul's would cover "spin" as well.

Well, this leads to a new interpretation of the matrix. Let's take the

canonical example

[ 4 -1 -1]-1 [ 7 5 3]

[-3 -1 -2] = [11 8 5]

[-4 4 -1] [16 12 7]

The vectors are written with the most important at the top. That means

the left hand column is the first approximation to the scale -- an equal

temperament. The first two columns define the second approximation, a

scale with two steps, or a linear temperament. And all three columns

define the scale exactly -- or define the prime vectors in terms of these

three intervals.

Now, replace the melodic generator with the octave.

[ 1 0 0]-1 1[ 7 0 0]

[-3 -1 2] = -[11 1 2]

[-4 4 -1] 7[16 4 1]

The octave is still a generator here. The algebra doesn't know anything

about "interval of equivalence". The left hand column is still the first

approximation, but it works a bit better if we set the generator (the

octave) to be exact. The other columns are correction factors which still

work if the octave is exact. Hence this system can work in

octave-equivalent space.

The things that make this system different to the one before is that it

isn't unitary, and only one column of the inverse depends on the first

generator. It's the second criterion that allows us to draw the

non-arbitrary distinction between "interval of equivalence" and "unison

vector", and so throw away the former.

The non-unitariness is because the original 16:15 generator can only be

derived from the vectors we have now as

[4 -1 -1] = ([1 0 0] - 5*[-3 -1 2] - 3*[-4 4 -1])/7

(I'd like a magic way of deducing this vector from the other three.)

Each interval can be defined in terms of a fraction of the octave, a

fraction of a chromatic semitone, and a fraction of a syntonic comma. As

the first is clearly the largest, the others can be thought of as small

corrections. Unlike the original system, each new approximation makes a

small alteration to the previous one, instead of replacing it.

The periodicity block works like that. You start out by taking all the

notes of the first approximation. Then you add a correction between +/-

0.5 (or 0 and 1 which amounts to the same thing) for each other dimension

to get a just interval. All but the largest correction can be taken out

to leave a linear temperament.

The connection between that first correction and the MOS generator can be

seen when you think of practical tuning. Say you tune a keyboard to

meantone with no deviation for D. The mistuning of A will be equal and

opposite to that of G, relative to 12-equal. Those for E and C will be

twice as much, and so on. The correction factors for the second

approximation follow the generator.

Graham