The triple wedge product of 7-limit intervals is basically a val. We

can also take wedge products of vals, and the triple product can be

identified with an interval. Taking a wedge product of two intervals

or two vals gives us something in the Twilight Zone halfway between

an interval and a val, and we can compare them directly.

The {64/63, 49/48} system has 5, 10, and 15 for vals--in other words,

these are common kernel elements. If we write them as vectors, we get

h5: [ 5] h10: [10] h15: [15]

[ 8] [16] [35]

[12] [23] [35]

[14] [28] [42]

Taking wedge products gives us:

h5^h10 = [0, -5, 0, -8, 0, 14]

h15^h10 = [0, -5, 0, -8, 0, 14]

h5^h15 = [0, -5, 0, -8, 0, 14]

On the other hand, we also have

64/63^49/48 = [-14, 0, 8, 0, -5, 0]

This is the the same as the previous wedge products read backwards,

up to a change of sign, and this is no accident. The basis element

v2^v3 for vals corresponds to e5^e7 for intervals, v2^v7 corresponds

to e3^e5, and so forth. These wedge products tell us that neither the

vals nor the intervals are giving us torsion, and that they are all

giving us the same linear temperament. The wedge product, in fact,

could be used to identify linear temperaments, rather than Minkowski

reduced pairs of intervals.

In-Reply-To: <9tvaik+ehul@eGroups.com>

In article <9tvaik+ehul@eGroups.com>, genewardsmith@juno.com () wrote:

> The triple wedge product of 7-limit intervals is basically a val. We

> can also take wedge products of vals, and the triple product can be

> identified with an interval. Taking a wedge product of two intervals

> or two vals gives us something in the Twilight Zone halfway between

> an interval and a val, and we can compare them directly.

What's a val?

These wedge products seem to behave like the vector products we use in

physics. Are they similar? The same thing?

> The {64/63, 49/48} system has 5, 10, and 15 for vals--in other words,

> these are common kernel elements. If we write them as vectors, we get

>

> h5: [ 5] h10: [10] h15: [15]

> [ 8] [16] [35]

> [12] [23] [35]

> [14] [28] [42]

So these are the normal ET mappings. How did you get them? I notice that

h15 isn't the same as h5+h10. Is that right?

> Taking wedge products gives us:

>

> h5^h10 = [0, -5, 0, -8, 0, 14]

> h15^h10 = [0, -5, 0, -8, 0, 14]

> h5^h15 = [0, -5, 0, -8, 0, 14]

>

> On the other hand, we also have

>

> 64/63^49/48 = [-14, 0, 8, 0, -5, 0]

>

> This is the the same as the previous wedge products read backwards,

> up to a change of sign, and this is no accident. The basis element

> v2^v3 for vals corresponds to e5^e7 for intervals, v2^v7 corresponds

> to e3^e5, and so forth. These wedge products tell us that neither the

> vals nor the intervals are giving us torsion, and that they are all

> giving us the same linear temperament. The wedge product, in fact,

> could be used to identify linear temperaments, rather than Minkowski

> reduced pairs of intervals.

Could the wedge product be used to remove the torsion, or only recognize

it? I'd still like a routing for working backwards from the mapping to

the unison vectors.

What algorithm are you using to find the temperaments? I don't remember

seeing it. It might be useful to see your wedge product routine as well,

if they turn out to be important. My unison vectors to temperament

routine has been able to handle all the torsion I've thrown at it so far,

so it doesn't really matter if I don't spot it in time.

Graham

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

> What's a val?

It's something which takes an interval and sends it to an integer.

Examples would be:

(1) An et; the val h12 for instance sends 2/1 to 12, 3/1 to 19,

5/4 to 4, etc.

(2) A mapping of generators to primes, where the val tells you how

many generator steps you need for a particular prime. From this you

know how many generator steps to get to a p-limit interval, and that

defines a val.

(3) A p-adic valuation is a val. For instance, the valuation "v3"

counts how many powers of 3, positive or negative, we find in a

particular rational number, so that v3(9/8)=2, v3(5/4)=0, v3(5/3)=-1

and so forth.

Vals are dual to intervals, since an interval takes a val and sends

it to an integer; therefore, if intervals are represented by row

vectors of integers, a val is a column vector of integers.

> These wedge products seem to behave like the vector products we use

in

> physics. Are they similar? The same thing?

The three dimensional wedge product sends a vector with basis e1, e2,

e3 to one with basis e1^e2, e2^e3, e3^e1. Using the inner product, we

can identify the two, which results in the vector product. Hence the

vector product is just the 3-dimensional version of the wedge product.

> > The {64/63, 49/48} system has 5, 10, and 15 for vals--in other

words,

> > these are common kernel elements. If we write them as vectors, we

get

> >

> > h5: [ 5] h10: [10] h15: [15]

> > [ 8] [16] [35]

> > [12] [23] [35]

> > [14] [28] [42]

>

> So these are the normal ET mappings. How did you get them?

I checked to see what was dual to {64/63,49/48}, or in other words

for what values of n I got hn(64/63) = hn(49/48) = 0.

I notice that

> h15 isn't the same as h5+h10. Is that right?

Certainly.

> Could the wedge product be used to remove the torsion, or only

recognize

> it? I'd still like a routing for working backwards from the

mapping to

> the unison vectors.

You can recognize torsion easily with the wedge product. Removing

torsion using it would also be possible, since you could look for

vectors which when wedged with the wedge product give 0.

> What algorithm are you using to find the temperaments? I don't

remember

> seeing it.

I find two linearly independent vals, and LLL reduce them. I then

convert this (which I call "map") to a version where one of the

generators is an nth part of an octave, by column operations, and

call this "reduced map". I usually put the second generator within

the first, and always make them inside an octave.

It might be useful to see your wedge product routine as well,

> if they turn out to be important.

I gave the definitions, that's about all the routine you need. But

need to fix that again, I see!