Re: [tuning] Re: Proposal: a high-order septimal schisma

Gene,

I understand pieces of your explanation and am working to piece everything
together. I have inserted questions in the appropriate places. Many thanks.
Paul

genewardsmith
<genewardsmith@ju To: tuning@yahoogroups.com
no.com> cc:
Subject: [tuning] Re: Proposal: a high-order septimal schisma
09/04/2002 01:27
PM
Please respond to
tuning

--- In tuning@y..., paul.hjelmstad@u... wrote:
>
> Thanks. Hope this doesn't sound stupid, but could you tell me the
> significance of each number in the "wedge invariant"? (Being really
literal
> please) Are they the powers of 2,3,5,7 or something?

It's hardly stupid, and in fact it's complicated enough I suggest further
discussion should take place on tuning-math, and not here. There *are*
commas of various kinds hidden in it, for which the numbers are exponents,
and as we just saw, if the first number is "1" then 5 and 7 can be
expressed in terms of 2 and 3; however it really comes from multilinear
algebra, and is not such a good thing to discuss here. There's quite a lot
in the tuning-math archives about it.

Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then
>Why are there two "vectors?" What is p and what is q??

p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
u1*v3-v1*u3,u1*v4-v1*u4]
>Got it. It's like some kind of dot product, with every combination of
pairs of p and q?

Let r be the mapping to primes of an equal temperament given
by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
means r has u1 notes to the octave, u2 notes in the approximation of 3, and
so forth; hence [12, 19, 28, 24] would be the usual 12-equal, and [31, 49,
72, 87] the usual 31-et.
>Got it

The wedge now is

r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
u4*v2-u2*v4,u2*v3-v2*u3]
> OK, what is r and s (again?)
Whether we've computed in terms of commas or ets, the wedge product of the
linear temperament is exactly the same, up to sign.
>So signs can be reversed?

If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
> Is this the same wedgie as above? (Based on r ^ s for example)?

2^u6 3^(-u2) 5^u1
2^u5 3^u3 7^(-u1)
2^u4 5^(-u3) 7^u2
3^u4 5^u5 7^u6
> I can see that it is all the possible triples of 2,3,5,7. Also I see that
this uses both signs of u1, u2, and u3 and uses u4, u5 and u6 twice each. I
think if you can answer the above questions I should be able to work this
final part out myself. Thanks:)

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--- In tuning-math@y..., paul.hjelmstad@u... wrote:

> Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then
> >Why are there two "vectors?" What is p and what is q??

We need two of something to define a linear 7-limit temperament--two generators, two equal temperaments, two commas. In general two commas define something of codimension two, but in four dimensions this is the same. The above would be the two commas; so, for instance, we could define meantone using 81/80 and 126/125, and miracle using 225/224 and 1029/1024.

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

> >Got it. It's like some kind of dot product, with every combination of
> pairs of p and q?

So say it's the four-dimensional analog of the cross-product would be more correct. There's a web site which some people found useful for getting the gist of it:

http://www.ses.swin.edu.au/homes/browne/grassmannalgebra/book/

> Let r be the mapping to primes of an equal temperament given
> by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
> means r has u1 notes to the octave, u2 notes in the approximation of 3, and
> so forth; hence [12, 19, 28, 24] would be the usual 12-equal, and [31, 49,
> 72, 87] the usual 31-et.
> >Got it
>
> The wedge now is
>
> r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> u4*v2-u2*v4,u2*v3-v2*u3]
> > OK, what is r and s (again?)

Two "vals", or duals to intervals; in the above example, the 12 and 31 equal temperaments.

> Whether we've computed in terms of commas or ets, the wedge product of the
> linear temperament is exactly the same, up to sign.
> >So signs can be reversed?

Standardizing so the first non-zero entry is positive is what Graham and I have agreed on as a convention for the "wedgie", or wedge invariant of a temperament.

> If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
> > Is this the same wedgie as above? (Based on r ^ s for example)?

It is the same numerically, and so for our purposes identical; in theory you get into Poincare duality, or you might do things as in the Grassman book I gave a url for.

Did you find what you wanted re the Riemann Zeta function? Can I ask what your math background is?

Thanks. I am going to discuss your Riemann Zeta Function/ tuning system
postings with a PhD math friend. I also have been corresponding with
Matthew Watkins, of Exeter, England. My math background is a BA Math major
from St.Olaf College, Northfield MN, with some study of my own since
college. I am especially interested in the math of tuning systems
(obviously). I also received a BM in Piano from St. Olaf. I am also
interested in music theory, especially combinatorics (Allen Forte, John
Rahn, etc.)

genewardsmith
<genewardsmith@ju To: tuning-math@yahoogroups.com
no.com> cc:
Subject: [tuning-math] [tuning] Re: Proposal: a high-order septimal schisma
09/05/2002 09:58
PM
Please respond to
tuning-math

--- In tuning-math@y..., paul.hjelmstad@u... wrote:

> Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then
> >Why are there two "vectors?" What is p and what is q??

We need two of something to define a linear 7-limit temperament--two
generators, two equal temperaments, two commas. In general two commas
define something of codimension two, but in four dimensions this is the
same. The above would be the two commas; so, for instance, we could define
meantone using 81/80 and 126/125, and miracle using 225/224 and 1029/1024.

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

> >Got it. It's like some kind of dot product, with every combination of
> pairs of p and q?

So say it's the four-dimensional analog of the cross-product would be more
correct. There's a web site which some people found useful for getting the
gist of it:

http://www.ses.swin.edu.au/homes/browne/grassmannalgebra/book/

> Let r be the mapping to primes of an equal temperament given
> by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
> means r has u1 notes to the octave, u2 notes in the approximation of 3,
and
> so forth; hence [12, 19, 28, 24] would be the usual 12-equal, and [31,
49,
> 72, 87] the usual 31-et.
> >Got it
>
> The wedge now is
>
> r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> u4*v2-u2*v4,u2*v3-v2*u3]
> > OK, what is r and s (again?)

Two "vals", or duals to intervals; in the above example, the 12 and 31
equal temperaments.

> Whether we've computed in terms of commas or ets, the wedge product of
the
> linear temperament is exactly the same, up to sign.
> >So signs can be reversed?

Standardizing so the first non-zero entry is positive is what Graham and I
have agreed on as a convention for the "wedgie", or wedge invariant of a
temperament.

> If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
> > Is this the same wedgie as above? (Based on r ^ s for example)?

It is the same numerically, and so for our purposes identical; in theory
you get into Poincare duality, or you might do things as in the Grassman
book I gave a url for.

Did you find what you wanted re the Riemann Zeta function? Can I ask what
your math background is?

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I've been studying the Grassman Algebra book-in-progress. So far, I
understand the calculation for r ^ s, and how it creates the wedgie at the
bottom, and the commas that are given at the very bottom make sense, but
alas, I am still not clear on p ^ q (wedge of commas?) Could you be so kind
as to plug in values for u1 to u4 and v1 to v4 so I could seem how the
resultant wedgie for p ^ q resembles the wedgie for r ^ s? I obtain (-1.
-4, -10, -12, -13, 4) for it.

genewardsmith
<genewardsmith@ju To: tuning-math@yahoogroups.com
no.com> cc:
Subject: [tuning-math] [tuning] Re: Proposal: a high-order septimal schisma
09/05/2002 09:58
PM
Please respond to
tuning-math

--- In tuning-math@y..., paul.hjelmstad@u... wrote:

> Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then
> >Why are there two "vectors?" What is p and what is q??

We need two of something to define a linear 7-limit temperament--two
generators, two equal temperaments, two commas. In general two commas
define something of codimension two, but in four dimensions this is the
same. The above would be the two commas; so, for instance, we could define
meantone using 81/80 and 126/125, and miracle using 225/224 and 1029/1024.

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

> >Got it. It's like some kind of dot product, with every combination of
> pairs of p and q?

So say it's the four-dimensional analog of the cross-product would be more
correct. There's a web site which some people found useful for getting the
gist of it:

http://www.ses.swin.edu.au/homes/browne/grassmannalgebra/book/

> Let r be the mapping to primes of an equal temperament given
> by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
> means r has u1 notes to the octave, u2 notes in the approximation of 3,
and
> so forth; hence [12, 19, 28, 24] would be the usual 12-equal, and [31,
49,
> 72, 87] the usual 31-et.
> >Got it
>
> The wedge now is
>
> r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> u4*v2-u2*v4,u2*v3-v2*u3]
> > OK, what is r and s (again?)

Two "vals", or duals to intervals; in the above example, the 12 and 31
equal temperaments.

> Whether we've computed in terms of commas or ets, the wedge product of
the
> linear temperament is exactly the same, up to sign.
> >So signs can be reversed?

Standardizing so the first non-zero entry is positive is what Graham and I
have agreed on as a convention for the "wedgie", or wedge invariant of a
temperament.

> If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
> > Is this the same wedgie as above? (Based on r ^ s for example)?

It is the same numerically, and so for our purposes identical; in theory
you get into Poincare duality, or you might do things as in the Grassman
book I gave a url for.

Did you find what you wanted re the Riemann Zeta function? Can I ask what
your math background is?

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--- In tuning-math@y..., paul.hjelmstad@u... wrote:
>
> I've been studying the Grassman Algebra book-in-progress. So far, I
> understand the calculation for r ^ s, and how it creates the wedgie at the
> bottom, and the commas that are given at the very bottom make sense, but
> alas, I am still not clear on p ^ q (wedge of commas?) Could you be so kind
> as to plug in values for u1 to u4 and v1 to v4 so I could seem how the
> resultant wedgie for p ^ q resembles the wedgie for r ^ s? I obtain (-1.
> -4, -10, -12, -13, 4) for it.

Here are the two formulas for wedge product:

Wedge of two intervals:
> > p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
> > u1*v3-v1*u3,u1*v4-v1*u4]

For example p = 126/125 and q=81/80, then p = 2^1 3^2 5^(-3) 7^1,
so in vector form it is [1,2,-3,1]. Similarly,
q=2^(-4) 3^4 5^(-1) 7^0, which in vector form is [-4,4,-1,0].
Wedging the two gives the wedgie for meantone, but 126/125 ^ 225/224, for example, will work also.

Wedge of two vals (e.g., equal temperament mappings)
> > r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> > u4*v2-u2*v4,u2*v3-v2*u3]
> > > OK, what is r and s (again?)

For instance h19=[19,30,44,53] and h12=[12,19,28,34] will work, the wedge product h19 ^ h12 being the same as the above, and the same as, for example, h50 ^ h31. We can also wedge together the mappings to primes of two generators of the temperament and get the wedgie (up to sign again.) For instance, the matrix

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

represents the mapping to primes of meantone in with generators of octave (first column) and fifth (second column.) Then

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

which is again the meantone wedgie.

hi Gene,

From: "genewardsmith" <genewardsmith@juno.com>
To: <tuning@yahoogroups.com>
Sent: Wednesday, September 04, 2002 11:27 AM
Subject: [tuning] Re: Proposal: a high-order septimal schisma

> --- In tuning@y..., paul.hjelmstad@u... wrote:
> >
> > Thanks. Hope this doesn't sound stupid, but could you tell me the
> > significance of each number in the "wedge invariant"? (Being really
literal
> > please) Are they the powers of 2,3,5,7 or something?
>
> It's hardly stupid, and in fact it's complicated enough I suggest further
discussion should take place on tuning-math, and not here. There *are*
commas of various kinds hidden in it, for which the numbers are exponents,
and as we just saw, if the first number is "1" then 5 and 7 can be expressed
in terms of 2 and 3; however it really comes from multilinear algebra, and
is not such a good thing to discuss here. There's quite a lot in the
tuning-math archives about it.
>
> Let p = 2^u1 3^u2 5^u3 7^u4 and q = 2^v1 3^v2 5^v3 7^v4, then
>
> p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
> u1*v3-v1*u3,u1*v4-v1*u4]
>
> Let r be the mapping to primes of an equal temperament given
> by r = [u1, u2, u3, u4], and s be given by [v1, v2, v3, v4]. This
> means r has u1 notes to the octave, u2 notes in the approximation of 3,
and so forth; hence [12, 19, 28, 24] would be the usual 12-equal, and [31,
49, 72, 87] the usual 31-et. The wedge now is
>
> r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> u4*v2-u2*v4,u2*v3-v2*u3]
>
> Whether we've computed in terms of commas or ets, the wedge product of the
linear temperament is exactly the same, up to sign.
>
> If the wedgie is [u1,u2,u3,u4,u5,u6] then we have commas given by
>
> 2^u6 3^(-u2) 5^u1
> 2^u5 3^u3 7^(-u1)
> 2^u4 5^(-u3) 7^u2
> 3^u4 5^u5 7^u6

at last, i finally understand how you're calculating wedgies!

but that last bit has me a little confused.

from the example meantone wedgie [1,4,10,12,-13,4] :

> From: "Gene Ward Smith" <genewardsmith@juno.com>
> To: <tuning-math@yahoogroups.com>
> Sent: Monday, September 09, 2002 10:50 PM
> Subject: [tuning-math] [tuning] Re: Proposal: a high-order septimal
schisma
>
> <snip>
>
> Wedge of two intervals:
> > > p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
> > > u1*v3-v1*u3,u1*v4-v1*u4]
>
> For example p = 126/125 and q=81/80, then p = 2^1 3^2 5^(-3) 7^1,
> so in vector form it is [1,2,-3,1]. Similarly,
> q=2^(-4) 3^4 5^(-1) 7^0, which in vector form is [-4,4,-1,0].
> Wedging the two gives the wedgie for meantone, but 126/125 ^ 225/224, for
example, will work also.

i calculated these commas

[ 4 -4 1] = 80 / 81
[-13 10 -1] = 59049 / 57344
[ 12 -10 4] = 9834496 / 9765625
[ 12 -13 4] = 1275989841 / 1220703125

OK, so the syntonic comma (81/80) is there ...
but what happened to

This is really interesting. What is the magic that connects (126/125 and
81/80) with (12et and 19et)? And also with the wedging [1,1,0,-3] with
[0,1,4,10]? I guess its a loaded question - but - why does this work?

"Gene Ward Smith"
<genewardsmith@ju To: tuning-math@yahoogroups.com
no.com> cc:
Subject: [tuning-math] [tuning] Re: Proposal: a high-order septimal schisma
09/10/2002 12:50
AM
Please respond to
tuning-math

--- In tuning-math@y..., paul.hjelmstad@u... wrote:
>
> I've been studying the Grassman Algebra book-in-progress. So far, I
> understand the calculation for r ^ s, and how it creates the wedgie at
the
> bottom, and the commas that are given at the very bottom make sense, but
> alas, I am still not clear on p ^ q (wedge of commas?) Could you be so
kind
> as to plug in values for u1 to u4 and v1 to v4 so I could seem how the
> resultant wedgie for p ^ q resembles the wedgie for r ^ s? I obtain (-1.
> -4, -10, -12, -13, 4) for it.

Here are the two formulas for wedge product:

Wedge of two intervals:
> > p ^ q = [u3*v4-v3*u4,u4*v2-v4*u2,u2*v3-v2*u3,u1*v2-v1*u2,
> > u1*v3-v1*u3,u1*v4-v1*u4]

For example p = 126/125 and q=81/80, then p = 2^1 3^2 5^(-3) 7^1,
so in vector form it is [1,2,-3,1]. Similarly,
q=2^(-4) 3^4 5^(-1) 7^0, which in vector form is [-4,4,-1,0].
Wedging the two gives the wedgie for meantone, but 126/125 ^ 225/224, for
example, will work also.

Wedge of two vals (e.g., equal temperament mappings)
> > r ^ s = [u1*v2-v1*u2,u1*v3-v1*u3,u1*v4-v1*u4,u3*v4-v3*u4,
> > u4*v2-u2*v4,u2*v3-v2*u3]
> > > OK, what is r and s (again?)

For instance h19=[19,30,44,53] and h12=[12,19,28,34] will work, the wedge
product h19 ^ h12 being the same as the above, and the same as, for
example, h50 ^ h31. We can also wedge together the mappings to primes of
two generators of the temperament and get the wedgie (up to sign again.)
For instance, the matrix

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

represents the mapping to primes of meantone in with generators of octave
(first column) and fifth (second column.) Then

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

which is again the meantone wedgie.

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--- In tuning-math@y..., paul.hjelmstad@u... wrote:
>
> This is really interesting. What is the magic that connects
>(126/125 and
> 81/80) with (12et and 19et)?

i don't know about magic, but 12-equal and 19-equal are both tunings
where 126;125 and 81;80 vanish, i.e., are tempered out. you can think
of the septimal meantone temperament (which is defined by the wedgie
in question) as defined either by the family of temperaments where
both 126;125 and 81;80 are tempered out, or as the family of
temperaments lying along the straight line connecting 12-equal and 19-
equal in 7-limit space.

> And also with the wedging [1,1,0,-3] with
> [0,1,4,10]?

the second of these tells you that the generator of septimal meantone
temperament catches the third harmonic with 1 generator, the fifth
harmonic with 4 generators, and the seventh harmonic with 10
generators. the first tells you the number of octaves you need to
correct for register.