Standardizing our wedge product

To get into the standization fun over on the other list, I decided we
could standardize a basis for the 7-limit wedge product, getting the
product of vals and intervals to correspond. I propose:

Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals in the
prime-power basis. Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u1*v2-v1*u2, u1*v3-v1*u3, u1*v4-v1*u4,
u2*v3-v2*u3, u2*v4-v2*u4, u3*v4-v3*u4]

Let [u1,u2,u3,u4] and [v1,v2,v3,v4] be two vals in the prime-
valuation basis (that is, h12 would be [12,19,28,34] and so forth.)
Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u3*v4-v3*u4, v2*u4-u2*v4, u2*v3-v2*u3,
u1*v4-v1*u4, v1*u3-u1*v3, u1*v2-v1*u2]

This now means that a given linear temperament gives the same wedge
product, up to sign, whether defined in terms of vals or intervals.
For instance

2401/2400^2375/4374 = h270^h441 = [34, -22, -1, -18, 27, -18]

Since this is an invariant of the temperament, it would be a good
thing to use to refer to it, but for the fact that it is opaque and
does not immediately tell us how to define the temperament. However,
it does allow us to put vals and intervals to the test, and see right
away if they are giving us the same system, or any torsion.

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

Since no one is likely to complain about my proposed standard, I'd
better do it myself. I think it should be changed to make it more
useful from the point of view of music theory, where it makes sense
to have the first entry 3-limit, the next two 5-limit, the next three
7-limit, and so forth. This allows a more easy comparison from one
prime limit to another, which should help for extending temperaments
and classifying them. Hence I now will propose instead:

> Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals in the
> prime-power basis. Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u3*v4-v3*u4, v2*u4-u2*v4, u2*v3-v2*u3,
u1*v4-v1*u4, v1*u3-u1*v3, u1*v2-v1*u2]

> Let [u1,u2,u3,u4] and [v1,v2,v3,v4] be two vals in the prime-
> valuation basis (that is, h12 would be [12,19,28,34] and so forth.)
> Then

[u1,u2,u3,u4]^[v1,v2,v3,v4] = [u1*v2-v1*u2, u1*v3-v1*u3, u1*v4-v1*u4,
u2*v3-v2*u3, u2*v4-v2*u4, u3*v4-v3*u4]

Thus for example for paultone, we get

64/63^50/49 = [2,-4,-4,-11,-12,2]

h10^h12 = [-2,4,4,11,12,-2]

We can also use this on the maps of generators to primes, since they
are also vals; from

[ 0 2]
[-1 4]
[ 2 3]
[ 2 4]

we get [0,-1,2,2]^[2,4,3,4] = [2,-4,-4,-11,-12,2]. We see that three
seemingly very different defintions, in terms of two intervals, two
equal temperaments, or two maps of generators to primes, give us the
same wedge invariant. This strikes me as useful. It is also not too
difficult to extract the temperament from the wedge invariant, using
the fact that u^u=0 and testing for vals or intervals which give 0 in
a wedge product with the invariant, so it does work, though not
transparently, as a means of classifying temperaments.

We can see how the change in the definition works to our advantage
with the septimal version of meantone, which of course extends the 5-
limit version. We have

126/125^81/80 = h19^h31 = [1,4,10,4,13,12]

Here the [1,4,10] at the beginning is the 5-limit part.

I gave the following as "map" and "adjusted map"

[ 1 1] [ 0 1]
[ 1 2] [-1 2]
[ 0 4] [-4 4]
[-3 7] [-10 7]

If we take the wedge product here, we get

[1,1,0,-3]^[1,2,4,7] = [0,-1,-4,-10]^[1,2,4,7] = [1,4,10,4,13,12]

Graham is familiar with the cross-product, and chances are Paul and
others are as well, so I think we should incorporate them as much as
possible into a standard for the wedge product. Let's see if I can
finally agree with myself on this!

Definition of wedge product of 7-limit intervals

Let [u1, u2, u3, u4] and [v1, v2, v3, v4] be two intervals r and s in
the prime-power basis, so that r = 2^u1 3^u2 5^u3 7^u4 and
s = 2^v1 3^v1 5^v3 7^v4. Then the wedge product
r^s = [u1, u2, u3, u4]^[v1, v2, v3, v4] is defined as:

[[u2,u3,u4] X [v2,v3,v4], u1*[v2,v3,v4]-v1*[u2,u3,u4]],

where "X" is the cross-product and "*" is the scalar product. Written
out in full, this would be

r^s = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
u1*v3-v1*u3,u1*v4-v1*u4]

Definition of wedge product of 7-limit vals

Let h = [u1,u2,u3,u4] and g = [v1,v2,v3,v4] be two vals in the
prime-valuation basis, so that h(2) = u1, h(3)=u2, h(5)=u3, h(7)=u4
and g(2) = v1, g(3) = v2, g(5) = v3 and g(7) = v4. Then the wedge
product h^g = [u1,u2,u3,u4]^[v1,v2,v3,v4] is defined as:

[u1*[v2,v3,v4]-v1*[u2,u3,u4], [u2,u3,u4] X [v2,v3,v4]]

Written out in full, this is

h^g = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
u4*v2-v4*u2,u2*v3-v2*u3]

Note that the first three elements of the wedge invariant, defined
from any starting place, gives the mapping to primes of the non-
octave generator of the pair of generators. From the interval point
of view, it is the cross-product of Graham's beloved octave
equivalence classes. From the val point of view, it is the
combination of vals I have been using for various purposes.

--- In tuning-math@y..., genewardsmith@j... wrote:
> Graham is familiar with the cross-product,

Is that what you guys were referring to as the "vector product"? It
seemed to be.

> and chances are Paul

Yes . . .