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Canonical generators for 7-limit planar temperaments

🔗Gene Ward Smith <gwsmith@svpal.org>

2/17/2004 6:14:55 PM

My ISP conked out, so I installed Juno's free connection, which I
recommend for temporary uses like this. It has an incredibly
obnoxious ad which can be placed over the top of Yahoo's obnoxious ad.

If we have a comma which does not (as 50/49, for instance, will) lead
to a part-octave period, we can find generators which are analogous
to the period-and-generator generator of a linear temperament, by
projecting orthogonally onto a plane in a way which makes the comma
dissapper (people may recall my lattice diagrams of this.) We can now
do a Minkowski reduction--pick the class representative in the range
1 < q < sqrt(2) which is closest to the origin, as projected, and
which has the smallest Tenney norm. Now do the same for second
closest. We end up with two rational numbers 1 < q, r < sqrt(2) which
serve as generators for the temperament. If we put these together
with 2 and the comma, we get a unimodular matrix; inverting this
gives a matrix whose columns are the four vals I list. This shows how
to map a 7-limit note to octave, the two generators, and the comma;
we get the temperament by dropping the comma and retuning.

Below you find the comma, then the three generators, followed by the
square of the distance from the origin of the two non-octave
generators when projected, followed by the ratio of second-nearest to
nearest square size. This can be all the way from the same (225/224
and 2401/2400) to quite large (schisma.) On the line below I give the
map.

64/63 {2, 3/2, 5/4} [1/7, 5/7] 5
[<1 1 2 4|, <0 -1 0 2|, <0 0 1 0|, <0 0 0 -1|]

81/80 {2, 4/3, 9/7} [1/13, 9/13] 9
[<1 1 0 2|, <0 -1 -4 -2|, <0 0 0 -1|, <0 0 -1 0|]

245/243 {2, 9/7, 7/6} [3/17, 5/17] 5/3
[<1 1 1 2|, <0 1 3 1|, <0 1 1 2|, <0 0 1 0|]

126/125 {2, 6/5, 5/4} [1/7, 5/7] 5
[<1 2 2 1|, <0 1 0 -2|, <0 1 1 1|, <0 0 0 1|]

225/224 {2, 4/3, 15/14} [1/3, 1/3] 1
[<1 1 3 3|, <0 -1 1 0|, <0 0 -1 -2|, <0 0 1 1|]

1728/1715 {2, 7/6, 36/35} [1/5, 1/2] 5/2
[<1 2 3 3|, <0 -2 -3 -1|, <0 1 0 1|, <0 -1 -1 -1|]

1029/1024 {2, 8/7, 35/32} [7/10, 37/10] 37/7
[<1 4 3 2|, <0 3 1 -1|, <0 0 1 0|, <0 1 0 0|]

3136/3125 {2, 28/25, 168/125} [1/19, 18/19] 18
[<1 1 2 2|, <0 1 2 5|, <0 1 0 0|, <0 -1 -1 -2|]

5120/5103 {2, 4/3, 27/20} [1/37, 25/37] 25
[<1 2 4 2|, <0 -1 -3 3|, <0 0 -1 -1|, <0 0 0 -1|]

6144/6125 {2, 35/32, 5/4} [1/5, 1/3] 5/3
[<1 1 2 3| <0 2 0 1|, <0 1 1 -1|, <0 1 0 0|]

32805/32768 {2, 4/3, 448/405} [1/73, 57/73] 57
[<1 2 -1 1|, <0 -1 8 4|, <0 0 0 1|, <0 0 1 1|]

2401/2400 {2, 7/5, 49/40} [3/11, 3/11] 1
[<1 1 3 3|, <0 0 -2 -1|, <0 2 1 1|, <0 -1 0 0|]

4375/4374 {2, 27/25, 10/9} [1/7, 6/35] 6/5
[<1 2 3 3|, <0 -1 -2 1|, <0 -2 -3 -2|, <0 0 0 1|]

🔗Gene Ward Smith <gwsmith@svpal.org>

2/17/2004 10:39:23 PM

Here is the same list, sorted by Graham-style complexity. By this I
mean we take the Graham complexities individually of the two
generators, and multiply them, giving the area of the minimum
rectangle which contains a tetrad, if the generators are arranged in
a square grid. It is striking that 81/80 and 2401/2400 have the same
complexity!

Complexity 3

64/63 {2, 3/2, 5/4} [1/7, 5/7] 5
[<1 1 2 4|, <0 -1 0 2|, <0 0 1 0|, <0 0 0 -1|]

126/125 {2, 6/5, 5/4} [1/7, 5/7] 5
[<1 2 2 1|, <0 1 0 -2|, <0 1 1 1|, <0 0 0 1|]

1728/1715 {2, 7/6, 36/35} [1/5, 1/2] 5/2
[<1 2 3 3|, <0 -2 -3 -1|, <0 1 0 1|, <0 -1 -1 -1|]

Complexity 4

81/80 {2, 4/3, 9/7} [1/13, 9/13] 9
[<1 1 0 2|, <0 -1 -4 -2|, <0 0 0 -1|, <0 0 -1 0|]

225/224 {2, 4/3, 15/14} [1/3, 1/3] 1
[<1 1 3 3|, <0 -1 1 0|, <0 0 -1 -2|, <0 0 1 1|]

1029/1024 {2, 8/7, 35/32} [7/10, 37/10] 37/7
[<1 4 3 2|, <0 3 1 -1|, <0 0 1 0|, <0 1 0 0|]

6144/6125 {2, 35/32, 5/4} [1/5, 1/3] 5/3
[<1 1 2 3| <0 2 0 1|, <0 1 1 -1|, <0 1 0 0|]

2401/2400 {2, 7/5, 49/40} [3/11, 3/11] 1
[<1 1 3 3|, <0 0 -2 -1|, <0 2 1 1|, <0 -1 0 0|]

Complexity 5

3136/3125 {2, 28/25, 168/125} [1/19, 18/19] 18
[<1 1 2 2|, <0 1 2 5|, <0 1 0 0|, <0 -1 -1 -2|]

Complexity 6

245/243 {2, 9/7, 7/6} [3/17, 5/17] 5/3
[<1 1 1 2|, <0 1 3 1|, <0 1 1 2|, <0 0 1 0|]

5120/5103 {2, 4/3, 27/20} [1/37, 25/37] 25
[<1 2 4 2|, <0 -1 -3 3|, <0 0 -1 -1|, <0 0 0 -1|]

Complexity 9

4375/4374 {2, 27/25, 10/9} [1/7, 6/35] 6/5
[<1 2 3 3|, <0 -1 -2 1|, <0 -2 -3 -2|, <0 0 0 1|]

32805/32768 {2, 4/3, 448/405} [1/73, 57/73] 57
[<1 2 -1 1|, <0 -1 8 4|, <0 0 0 1|, <0 0 1 1|]

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/18/2004 12:51:14 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> My ISP conked out, so I installed Juno's free connection, which I
> recommend for temporary uses like this. It has an incredibly
> obnoxious ad which can be placed over the top of Yahoo's obnoxious
ad.
>
> If we have a comma which does not (as 50/49, for instance, will)
lead
> to a part-octave period, we can find generators which are analogous
> to the period-and-generator generator of a linear temperament, by
> projecting orthogonally onto a plane in a way which makes the comma
> dissapper (people may recall my lattice diagrams of this.) We can
now
> do a Minkowski reduction--pick the class representative in the range
> 1 < q < sqrt(2) which is closest to the origin, as projected, and
> which has the smallest Tenney norm. Now do the same for second
> closest. We end up with two rational numbers 1 < q, r < sqrt(2)
which
> serve as generators for the temperament. If we put these together
> with 2 and the comma, we get a unimodular matrix; inverting this
> gives a matrix whose columns are the four vals I list. This shows
how
> to map a 7-limit note to octave, the two generators, and the comma;
> we get the temperament by dropping the comma and retuning.
>
(Please disregard my other post. Did not see this message)
Just one question: What is "a 7-limit note" that is mapped by the
four vals listed? Everything else makes sense. Also, how do you
get the Temperament - by dropping the comma and retuning. Say for
81/80

> Below you find the comma, then the three generators, followed by
the
> square of the distance from the origin of the two non-octave
> generators when projected, followed by the ratio of second-nearest
to
> nearest square size. This can be all the way from the same (225/224
> and 2401/2400) to quite large (schisma.) On the line below I give
the
> map.
>
> 64/63 {2, 3/2, 5/4} [1/7, 5/7] 5
> [<1 1 2 4|, <0 -1 0 2|, <0 0 1 0|, <0 0 0 -1|]
>
> 81/80 {2, 4/3, 9/7} [1/13, 9/13] 9
> [<1 1 0 2|, <0 -1 -4 -2|, <0 0 0 -1|, <0 0 -1 0|]
>
> 245/243 {2, 9/7, 7/6} [3/17, 5/17] 5/3
> [<1 1 1 2|, <0 1 3 1|, <0 1 1 2|, <0 0 1 0|]
>
> 126/125 {2, 6/5, 5/4} [1/7, 5/7] 5
> [<1 2 2 1|, <0 1 0 -2|, <0 1 1 1|, <0 0 0 1|]
>
> 225/224 {2, 4/3, 15/14} [1/3, 1/3] 1
> [<1 1 3 3|, <0 -1 1 0|, <0 0 -1 -2|, <0 0 1 1|]
>
> 1728/1715 {2, 7/6, 36/35} [1/5, 1/2] 5/2
> [<1 2 3 3|, <0 -2 -3 -1|, <0 1 0 1|, <0 -1 -1 -1|]
>
> 1029/1024 {2, 8/7, 35/32} [7/10, 37/10] 37/7
> [<1 4 3 2|, <0 3 1 -1|, <0 0 1 0|, <0 1 0 0|]
>
> 3136/3125 {2, 28/25, 168/125} [1/19, 18/19] 18
> [<1 1 2 2|, <0 1 2 5|, <0 1 0 0|, <0 -1 -1 -2|]
>
> 5120/5103 {2, 4/3, 27/20} [1/37, 25/37] 25
> [<1 2 4 2|, <0 -1 -3 3|, <0 0 -1 -1|, <0 0 0 -1|]
>
> 6144/6125 {2, 35/32, 5/4} [1/5, 1/3] 5/3
> [<1 1 2 3| <0 2 0 1|, <0 1 1 -1|, <0 1 0 0|]
>
> 32805/32768 {2, 4/3, 448/405} [1/73, 57/73] 57
> [<1 2 -1 1|, <0 -1 8 4|, <0 0 0 1|, <0 0 1 1|]
>
> 2401/2400 {2, 7/5, 49/40} [3/11, 3/11] 1
> [<1 1 3 3|, <0 0 -2 -1|, <0 2 1 1|, <0 -1 0 0|]
>
> 4375/4374 {2, 27/25, 10/9} [1/7, 6/35] 6/5
> [<1 2 3 3|, <0 -1 -2 1|, <0 -2 -3 -2|, <0 0 0 1|]

🔗Gene Ward Smith <gwsmith@svpal.org>

2/18/2004 2:19:26 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> Just one question: What is "a 7-limit note" that is mapped by the
> four vals listed? Everything else makes sense. Also, how do you
> get the Temperament - by dropping the comma and retuning. Say for
> 81/80

The "7-limit note" is just a positive rational number of the 7-limit.
It can be written as a product of prime numbers, but it can also be
written in terms of the octave, generator or generators, and comma or
commas. So, for instance, for 81/80-planar, we could use any product
2^a (4/3)^b (9/7)^c (81/80)^d to represent a 7-limit note. We temper
by dropping the 81/80 out of the product, and retuning 2, 4/3 and
9/7. The same comment applies to linear temperaments, as for instance
2^a (4/3)^b (81/80)^c (126/125)^d, where we drop both commas.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/18/2004 2:38:14 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > Just one question: What is "a 7-limit note" that is mapped by the
> > four vals listed? Everything else makes sense. Also, how do you
> > get the Temperament - by dropping the comma and retuning. Say for
> > 81/80
>
> The "7-limit note" is just a positive rational number of the 7-
limit.
> It can be written as a product of prime numbers, but it can also be
> written in terms of the octave, generator or generators, and comma
or
> commas. So, for instance, for 81/80-planar, we could use any
product
> 2^a (4/3)^b (9/7)^c (81/80)^d to represent a 7-limit note. We
temper
> by dropping the 81/80 out of the product, and retuning 2, 4/3 and
> 9/7. The same comment applies to linear temperaments, as for
instance
> 2^a (4/3)^b (81/80)^c (126/125)^d, where we drop both commas.

Thanks! I'm slowly getting this material (I hate the word "stuff"
Paul Hj

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/19/2004 3:30:42 AM

So, is a,b,c,d a column or a row? (1,1,2,4) or (1,0,0,0) to use the
first one...

🔗Gene Ward Smith <gwsmith@svpal.org>

2/19/2004 11:39:56 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> So, is a,b,c,d a column or a row? (1,1,2,4) or (1,0,0,0) to use the
> first one...

I'm not sure which mapping you are looking at--64/63 planar? For
81/80 planar, the val <1 1 0 2| gives you the exponent "a" in the
product 2^a (3/2)^b (9/7)^c (81/80)^d, <0 -1 -4 -2| gives "b",
<0 0 0 -1| gives you "c", and <0 0 -1 0| gives "d". These vals can be
found by inverting the matrix whose rows are the four monzos for the
generators.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/19/2004 1:30:47 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > So, is a,b,c,d a column or a row? (1,1,2,4) or (1,0,0,0) to use
the
> > first one...
>
> I'm not sure which mapping you are looking at--64/63 planar?

Yes

> For 81/80 planar, the val <1 1 0 2| gives you the exponent "a" in
> the product 2^a (3/2)^b (9/7)^c (81/80)^d, <0 -1 -4 -2| gives "b",
> <0 0 0 -1| gives you "c", and <0 0 -1 0| gives "d".

How does a whole val like <1 1 0 2| become an exponent. (Sorry I
really thought I understood this)

> These vals can be found by inverting the matrix whose rows are the
> four monzos for the generators.

That I get...I crosschecked a bunch in Excel

🔗Gene Ward Smith <gwsmith@svpal.org>

2/19/2004 2:30:24 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> How does a whole val like <1 1 0 2| become an exponent. (Sorry I
> really thought I understood this)

That should have been <1 2 4 4|. Suppose we want the product for 5/4;
as a monzo, it is |-2 0 1 0>, and we find the exponent for 2 by

<1 2 4 4|-2 0 1 0> = 2

Similarly, we get

<0 -1 -4 -2|-2 0 1 0> = -4
<0 0 0 -1|-2 0 1 0> = 0
<0 0 -1 0|-2 0 1 0> = -1

and the product is 2^2 (4/3)^(-4) (81/80)^(-1) = 5/4

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

2/19/2004 2:35:29 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > How does a whole val like <1 1 0 2| become an exponent. (Sorry I
> > really thought I understood this)
>
> That should have been <1 2 4 4|. Suppose we want the product for
5/4;
> as a monzo, it is |-2 0 1 0>, and we find the exponent for 2 by
>
> <1 2 4 4|-2 0 1 0> = 2
>
> Similarly, we get
>
> <0 -1 -4 -2|-2 0 1 0> = -4
> <0 0 0 -1|-2 0 1 0> = 0
> <0 0 -1 0|-2 0 1 0> = -1
>
> and the product is 2^2 (4/3)^(-4) (81/80)^(-1) = 5/4

Thanks!!!

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/1/2004 10:30:47 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

Gene, could you point me to this lattice diagram/message?

now
> do a Minkowski reduction--pick the class representative in the range
> 1 < q < sqrt(2) which is closest to the origin, as projected, and
> which has the smallest Tenney norm. Now do the same for second
> closest. We end up with two rational numbers 1 < q, r < sqrt(2)
which
> serve as generators for the temperament. If we put these together
> with 2 and the comma, we get a unimodular matrix; inverting this
> gives a matrix whose columns are the four vals I list. This shows
how
> to map a 7-limit note to octave, the two generators, and the comma;
> we get the temperament by dropping the comma and retuning.
>
So projecting onto the plane, in such a way as to make the comma
disappear, gives us the generators, based on the projected values,
which are minimized, as close to the origin as possible? Does the
lattice diagram mentioned above explain all?

Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

3/1/2004 11:17:53 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
>
> > If we have a comma which does not (as 50/49, for instance, will)
> lead
> > to a part-octave period, we can find generators which are
analogous
> > to the period-and-generator generator of a linear temperament, by
> > projecting orthogonally onto a plane in a way which makes the
comma
> > dissapper (people may recall my lattice diagrams of this.) We can
>
>
> Gene, could you point me to this lattice diagram/message?

You can find the plots in the files section of this newsgroup, in the
planarplots directory.

> So projecting onto the plane, in such a way as to make the comma
> disappear, gives us the generators, based on the projected values,
> which are minimized, as close to the origin as possible? Does the
> lattice diagram mentioned above explain all?

The diagrams are just jpg files, but you can see the result of
projecting, and what lies near the origin.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/1/2004 11:45:39 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > wrote:
> >
> > > If we have a comma which does not (as 50/49, for instance,
will)
> > lead
> > > to a part-octave period, we can find generators which are
> analogous
> > > to the period-and-generator generator of a linear temperament,
by
> > > projecting orthogonally onto a plane in a way which makes the
> comma
> > > dissapper (people may recall my lattice diagrams of this.) We
can
> >
> >
> > Gene, could you point me to this lattice diagram/message?
>
> You can find the plots in the files section of this newsgroup, in
the
> planarplots directory.
>
> > So projecting onto the plane, in such a way as to make the comma
> > disappear, gives us the generators, based on the projected values,
> > which are minimized, as close to the origin as possible? Does the
> > lattice diagram mentioned above explain all?
>
> The diagrams are just jpg files, but you can see the result of
> projecting, and what lies near the origin.

Thanks. So for example - 81/80, has generators 4/3 and 9/7. How does
one project onto the plane to obtain 1/7 and 5/7 respectively?

Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

3/1/2004 12:21:06 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> Thanks. So for example - 81/80, has generators 4/3 and 9/7. How does
> one project onto the plane to obtain 1/7 and 5/7 respectively?

You lost me. Where did 1/7 and 5/7 come from? Do you mean project to
exact coordinates on the 3/2-9/7 plane, or simply give them in terms
of 3/2 and 9/7?

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/1/2004 12:31:07 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > Thanks. So for example - 81/80, has generators 4/3 and 9/7. How
does
> > one project onto the plane to obtain 1/7 and 5/7 respectively?
>
> You lost me. Where did 1/7 and 5/7 come from? Do you mean project
to
> exact coordinates on the 3/2-9/7 plane, or simply give them in
terms
> of 3/2 and 9/7?

I was referring back to your (excellent) message 9767. 1/7 and 5/7
are "the square of the distance from the origin of the two non-octave
generators when projected" Just how does that work?

🔗Gene Ward Smith <gwsmith@svpal.org>

3/1/2004 4:00:22 PM

I transformed coordinates first to face centered cubic when I did this
before, but you don't have to.

Define

Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc

Then this is a positive definite quadratic form on note classes

http://164.8.13.169/Enciklopedija/math/math/p/p520.htm

Taking the square root defines the symmetric lattice distance to the
origin. We get a bilinear form, and hence a dot product, from B(a,b)=
(Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in terms
of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)

I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by
dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
product of this with [a,b,c] and substracting, I get the orthogonal
projection of [a,b,c]:

[-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]

I now get a positive semidefinite quadratic form by substituting this
in Q; if I normalize by clearing denominators, I get

Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc

Positive semidefinite:
http://164.8.13.169/Enciklopedija/math/math/p/p524.htm

Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
definite form, and hence a lattice, by going to two dimensions with
basis 3/2 and 9/7:

Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2

If we take the TM reduction of intervals in the octave we can sort
them by size according to the above distance measure. Below I give
the intervals in order of size, the scales you get by combining them
(which should be tempered by "didymus", or 81/80 planar) and the
meantone reduction of the scales, where 7 is mapped to ten generators
of a fifth.

0 {1}
[1]
[0]

3 {3/2, 4/3}
[1, 4/3, 3/2]
[-1, 0, 1]

12 {9/5, 9/8}
[1, 9/8, 4/3, 3/2, 9/5]
[-2, -1, 0, 1, 2]

27 {5/3, 6/5}
[1, 9/8, 6/5, 4/3, 3/2, 5/3, 9/5]
[-3, -2, -1, 0, 1, 2, 3]

35 {14/9, 9/7}
[1, 9/8, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 9/5]
[-8, -3, -2, -1, 0, 1, 2, 3, 8]

36 {7/6, 12/7}
[1, 9/8, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 9/5]
[-9, -8, -3, -2, -1, 0, 1, 2, 3, 8, 9]

40 {21/20, 27/14}
[1, 21/20, 9/8, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 9/5, 27/14]
[-9, -8, -7, -3, -2, -1, 0, 1, 2, 3, 7, 8, 9]

43 {8/7, 7/4}
[1, 21/20, 9/8, 8/7, 7/6, 6/5, 9/7, 4/3, 3/2, 14/9, 5/3, 12/7, 7/4,
9/5, 27/14]
[-10, -9, -8, -7, -3, -2, -1, 0, 1, 2, 3, 7, 8, 9, 10]

48 {5/4, 8/5}
[1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 4/3, 3/2, 14/9, 8/5, 5/3,
12/7, 7/4, 9/5, 27/14]
[-10, -9, -8, -7, -4, -3, -2, -1, 0, 1, 2, 3, 4, 7, 8, 9, 10]

51 {7/5, 10/7}
[1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 4/3, 7/5, 10/7, 3/2, 14/9,
8/5, 5/3, 12/7, 7/4, 9/5, 27/14]
[-10, -9, -8, -7, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 7, 8, 9, 10]

56 {21/16, 32/21}
[1, 21/20, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7, 3/2,
32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 27/14]
[-11, -10, -9, -8, -7, -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6, 7, 8, 9,
10, 11]

68 {15/14, 28/15}
[1, 21/20, 15/14, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16, 4/3, 7/5, 10/7,
3/2, 32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 28/15, 27/14]
[-11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11]

75 {15/8, 35/18, 16/15, 36/35}
[1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 5/4, 9/7, 21/16,
4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 5/3, 12/7, 7/4, 9/5, 28/15,
15/8, 27/14, 35/18]
[-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12]

91 {56/45, 45/28}
[1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7,
21/16, 4/3, 7/5, 10/7, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 12/7, 7/4,
9/5, 28/15, 15/8, 27/14, 35/18]
[-12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5,
6, 7, 8, 9, 10, 11, 12]

100 {48/35, 35/24}
[1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7,
21/16, 4/3, 48/35, 7/5, 10/7, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28,
5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18]
[-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

108 {36/25, 25/18}
[1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 6/5, 56/45, 5/4, 9/7,
21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21, 14/9,
8/5, 45/28, 5/3, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18]
[-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

120 {25/21, 42/25}
[1, 36/35, 21/20, 16/15, 15/14, 9/8, 8/7, 7/6, 25/21, 6/5, 56/45, 5/4,
9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24, 3/2, 32/21,
14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5, 28/15, 15/8, 27/14, 35/18]
[-13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, 10, 11, 12, 13]

131 {35/32, 64/35}
[1, 36/35, 21/20, 16/15, 15/14, 35/32, 9/8, 8/7, 7/6, 25/21, 6/5,
56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25, 35/24,
3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5, 64/35,
28/15, 15/8, 27/14, 35/18]
[-14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2,
3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]

139 {49/27, 54/49}
[1, 36/35, 21/20, 16/15, 15/14, 35/32, 54/49, 9/8, 8/7, 7/6, 25/21,
6/5, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7, 36/25,
35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 5/3, 42/25, 12/7, 7/4, 9/5,
49/27, 64/35, 28/15, 15/8, 27/14, 35/18]
[-17, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 17]

140 {49/40, 80/49}
[1, 36/35, 21/20, 16/15, 15/14, 35/32, 54/49, 9/8, 8/7, 7/6, 25/21,
6/5, 49/40, 56/45, 5/4, 9/7, 21/16, 4/3, 48/35, 25/18, 7/5, 10/7,
36/25, 35/24, 3/2, 32/21, 14/9, 8/5, 45/28, 80/49, 5/3, 42/25, 12/7,
7/4, 9/5, 49/27, 64/35, 28/15, 15/8, 27/14, 35/18]
[-17, -16, -14, -13, -12, -11, -10, -9, -8, -7, -6, -5, -4, -3, -2,
-1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 16, 17]

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/2/2004 10:21:22 AM

Gene, just a few quick questions. Thanks.

> > > > Thanks. So for example - 81/80, has generators 4/3 and 9/7.
How
> > does
> > > > one project onto the plane to obtain 1/7 and 5/7 respectively?
>

Oops, I meant 1/13 and 9/13 of course!

So, which one are you using?

> I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by
> dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> product of this with [a,b,c] and substracting, I get the orthogonal
> projection of [a,b,c]:
>

Subtracting what?

> [-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]
>
> I now get a positive semidefinite quadratic form by substituting
this
> in Q; if I normalize by clearing denominators, I get
>
> Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
>
> Positive semidefinite:
> http://164.8.13.169/Enciklopedija/math/math/p/p524.htm
>
> Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> definite form, and hence a lattice, by going to two dimensions with
> basis 3/2 and 9/7:
>
> Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2
>
> If we take the TM reduction of intervals in the octave we can sort
> them by size according to the above distance measure. Below I give
> the intervals in order of size, the scales you get by combining them
> (which should be tempered by "didymus", or 81/80 planar) and the
> meantone reduction of the scales, where 7 is mapped to ten
generators
> of a fifth.

This I understand...
>
> 0 {1}
> [1]
> [0]
>
> 3 {3/2, 4/3}
> [1, 4/3, 3/2]
> [-1, 0, 1]
>
So, would this one be Q81((3/2)^a (4/3)^b)? Sorry if this is an
obvious question...

🔗Gene Ward Smith <gwsmith@svpal.org>

3/2/2004 11:03:58 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > Taking the square root defines the symmetric lattice distance to
the
> > origin. We get a bilinear form, and hence a dot product, from B
(a,b)
> =
> > (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in
> terms
> > of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
> >
>
> So, which one are you using?

It depends on whether you feed the program a monzo or a ratio. You
should get the same result using either <-4 4 -1| or 81/80, since
they refer to the same thing.

> > I have Q(81/80)=13, so I get a unit vector in the 81/80 direction
by
> > dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> > product of this with [a,b,c] and substracting, I get the
orthogonal
> > projection of [a,b,c]:
> >
>
> Subtracting what?

The unit vector times its dot product with the other vector. In other
words, if u is a unit vector and v is any vector, we can take
v - (u.v)u and then the dot product of this with u will be zero, since
u.(v - (u.v)u) = u.v - (u.v)(u.u) = 0 since u.u = 1. Hence, it will
be in the subspace orthogonal to u, and be the orthogonal projection
to that subspace.

> > 0 {1}
> > [1]
> > [0]
> >
> > 3 {3/2, 4/3}
> > [1, 4/3, 3/2]
> > [-1, 0, 1]
> >
> So, would this one be Q81((3/2)^a (4/3)^b)? Sorry if this is an
> obvious question...

These are just scales of 81/80 planar, along with their reduction to
septimal meantone. They are whatever lies inside of balls of various
sizes around the unison, according to the Q81 metric.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/2/2004 12:12:26 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@u...> wrote:
> > > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > > wrote:
> > > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > > <paul.hjelmstad@u...> wrote:
> > > >
>
> Gene, just a few quick questions. Thanks.
>
> > > > > Thanks. So for example - 81/80, has generators 4/3 and 9/7.
> How
> > > does
> > > > > one project onto the plane to obtain 1/7 and 5/7
respectively?
> >
>
> Oops, I meant 1/13 and 9/13 of course!
>
> I transformed coordinates first to face centered cubic when I did
this
> > before, but you don't have to.
> >
> > Define
> >
> > Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc
> >
> > Then this is a positive definite quadratic form on note classes
> >
> > http://164.8.13.169/Enciklopedija/math/math/p/p520.htm
> >
> > Taking the square root defines the symmetric lattice distance to
the
> > origin. We get a bilinear form, and hence a dot product, from B
(a,b)
> =
> > (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in
> terms
> > of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
> >
>
> So, which one are you using?
>
> > I have Q(81/80)=13, so I get a unit vector in the 81/80 direction
by
> > dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> > product of this with [a,b,c] and substracting, I get the
orthogonal
> > projection of [a,b,c]:
> >
>
> Subtracting what?
>
> > [-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]
> >
> > I now get a positive semidefinite quadratic form by substituting
> this
> > in Q; if I normalize by clearing denominators, I get
> >
> > Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
> >
> > Positive semidefinite:
> > http://164.8.13.169/Enciklopedija/math/math/p/p524.htm
> >
> > Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> > definite form, and hence a lattice, by going to two dimensions
with
> > basis 3/2 and 9/7:
> >
> > Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2
> >
> > If we take the TM reduction of intervals in the octave we can sort
> > them by size according to the above distance measure. Below I
give
> > the intervals in order of size, the scales you get by combining
them
> > (which should be tempered by "didymus", or 81/80 planar) and the
> > meantone reduction of the scales, where 7 is mapped to ten
> generators
> > of a fifth.
>
> This I understand...
> >
> > 0 {1}
> > [1]
> > [0]
> >
> > 3 {3/2, 4/3}
> > [1, 4/3, 3/2]
> > [-1, 0, 1]
> >
> So, would this one be Q81((3/2)^a (4/3)^b)? Sorry if this is an
> obvious question...
Scratch this last question!

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/2/2004 12:22:04 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > > Taking the square root defines the symmetric lattice distance
to
> the
> > > origin. We get a bilinear form, and hence a dot product, from B
> (a,b)
> > =
> > > (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively,
in
> > terms
> > > of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
> > >
> >
> > So, which one are you using?
>
> It depends on whether you feed the program a monzo or a ratio. You
> should get the same result using either <-4 4 -1| or 81/80, since
> they refer to the same thing.
>
> > > I have Q(81/80)=13, so I get a unit vector in the 81/80
direction
> by
> > > dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the
dot
> > > product of this with [a,b,c] and substracting, I get the
> orthogonal
> > > projection of [a,b,c]:
> > >
> >
> > Subtracting what?
>
> The unit vector times its dot product with the other vector. In
other
> words, if u is a unit vector and v is any vector, we can take
> v - (u.v)u and then the dot product of this with u will be zero,
since
> u.(v - (u.v)u) = u.v - (u.v)(u.u) = 0 since u.u = 1. Hence, it will
> be in the subspace orthogonal to u, and be the orthogonal
projection
> to that subspace.
>
> > > 0 {1}
> > > [1]
> > > [0]
> > >
> > > 3 {3/2, 4/3}
> > > [1, 4/3, 3/2]
> > > [-1, 0, 1]
> > >
> > So, would this one be Q81((3/2)^a (4/3)^b)? Sorry if this is an
> > obvious question...
>
> These are just scales of 81/80 planar, along with their reduction
to
> septimal meantone. They are whatever lies inside of balls of
various
> sizes around the unison, according to the Q81 metric.

Thanks, I'm asymptotically approaching 100% understanding...

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/2/2004 3:26:28 PM

**Afraid I am stuck again. How is this fed back into Q? Sorry to be
dense, I've stared at this too long...
>
> Positive semidefinite:
> http://164.8.13.169/Enciklopedija/math/math/p/p524.htm
>
> Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> definite form, and hence a lattice, by going to two dimensions with
> basis 3/2 and 9/7:
>
> Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2

This would seem to follow if I could grasp **
>

🔗Gene Ward Smith <gwsmith@svpal.org>

3/2/2004 4:00:15 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > ***[-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]
> >
> > I now get a positive semidefinite quadratic form by substituting
> this
> > in Q; if I normalize by clearing denominators, I get
> >
> > Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
>
> **Afraid I am stuck again. How is this fed back into Q? Sorry to be
> dense, I've stared at this too long...

It's the result of feeding *** into Q.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/3/2004 6:11:00 AM

With a b c ?

> > > I now get a positive semidefinite quadratic form by
substituting
> > this
> > > in Q; if I normalize by clearing denominators, I get
> > >
> > > Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
> >
> > **Afraid I am stuck again. How is this fed back into Q? Sorry to
be
> > dense, I've stared at this too long...
>
> It's the result of feeding *** into Q.

🔗Gene Ward Smith <gwsmith@svpal.org>

3/3/2004 10:01:11 AM

3^a 5^b 7^c defining the note class.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/3/2004 11:34:58 AM

I actually wrote the whole thing out long hand and checked
it with a calculator. (Too much time on my hands) Surprised to find a
common denominator of 52.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/3/2004 3:33:13 PM

Gene, just a couple loose ends, whenever you get a chance Thanks!

I transformed coordinates first to face centered cubic when I did
this
> before, but you don't have to.
>
> Define
>
> Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc
>
> Then this is a positive definite quadratic form on note classes
>
> http://164.8.13.169/Enciklopedija/math/math/p/p520.htm
>
> Taking the square root defines the symmetric lattice distance to the
> origin. We get a bilinear form, and hence a dot product, from B(a,b)
=
> (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in
terms
> of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
>
> I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by
> dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> product of this with [a,b,c] and substracting, I get the orthogonal
> projection of [a,b,c]:

Message 9767, you have 1/13, 9/13 as the squares of the projected
generators. I can see where the denominator comes from, but where
do 1 and 9 come from?

Once again, I don't see how (3/2) and (9/7) replace 3 and 5

I should say I really enjoy learning these techniques.

🔗Gene Ward Smith <gwsmith@svpal.org>

3/3/2004 9:47:12 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> Message 9767, you have 1/13, 9/13 as the squares of the projected
> generators. I can see where the denominator comes from, but where
> do 1 and 9 come from?

From a mistake somewhere.

> > Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> > definite form, and hence a lattice, by going to two dimensions with
> > basis 3/2 and 9/7:
> >
> > Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2
>
> Once again, I don't see how (3/2) and (9/7) replace 3 and 5

Do you see that you can get every 7-limit interval of the 81/80-planar
system from 2, 3/2 and 9/7? We have 5~(3/2)^4, and 7 ~ 4 (3/2)^2 (7/9)

> I should say I really enjoy learning these techniques.

Great!

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/4/2004 5:02:03 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > Message 9767, you have 1/13, 9/13 as the squares of the projected
> > generators. I can see where the denominator comes from, but where
> > do 1 and 9 come from?
>
> From a mistake somewhere.
>
> > > Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> > > definite form, and hence a lattice, by going to two dimensions
with
> > > basis 3/2 and 9/7:
> > >
> > > Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2
> >
> > Once again, I don't see how (3/2) and (9/7) replace 3 and 5
>
> Do you see that you can get every 7-limit interval of the 81/80-
planar
> system from 2, 3/2 and 9/7? We have 5~(3/2)^4, and 7 ~ 4 (3/2)^2
(7/9)

Yes. I still need to figure out how you get 3a^2 + 2ab +35b^2. Give
me a couple days...
>
> > I should say I really enjoy learning these techniques.
>
> Great!

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/5/2004 4:15:04 PM

Thanx, I know it hasn't been a couple days, but its Friday...

🔗Gene Ward Smith <gwsmith@svpal.org>

3/5/2004 5:32:02 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > Yes. I still need to figure out how you get 3a^2 + 2ab +35b^2. Give
> > me a couple days...
> > >
> Hmm. Do you feed a unit vector in the 81/80 direction (based on 3/2
> and 9/7) into an orthogonal projection, and then feed that back
> into Q? (Also based on 3/2 and 9/7) or I am going back too far?
>
> Thanx, I know it hasn't been a couple days, but its Friday...

I gave a semidefinite form in terms of 3, 5 and 9; you feed 3/2 and
9/7 into that, and get a definite form in terms of those two.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/5/2004 5:46:01 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > > Yes. I still need to figure out how you get 3a^2 + 2ab +35b^2.
Give
> > > me a couple days...
> > > >
> > Hmm. Do you feed a unit vector in the 81/80 direction (based on
3/2
> > and 9/7) into an orthogonal projection, and then feed that back
> > into Q? (Also based on 3/2 and 9/7) or I am going back too far?
> >
> > Thanx, I know it hasn't been a couple days, but its Friday...
>
> I gave a semidefinite form in terms of 3, 5 and 9; you feed 3/2 and
> 9/7 into that, and get a definite form in terms of those two.

Okay. I have to figure out how you killed off "c", and how "b" is
based on 9/7...
on 9/7, and so forth

🔗Gene Ward Smith <gwsmith@svpal.org>

3/5/2004 5:57:56 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > I gave a semidefinite form in terms of 3, 5 and 9; you feed 3/2
and
> > 9/7 into that, and get a definite form in terms of those two.
>
> Okay. I have to figure out how you killed off "c", and how "b" is
> based on 9/7...
> on 9/7, and so forth

The new a and b are different from the old a and b; I killed off c
because I never fed a c in in the first place.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/5/2004 6:10:58 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > > I gave a semidefinite form in terms of 3, 5 and 9; you feed 3/2
> and
> > > 9/7 into that, and get a definite form in terms of those two.
> >
> > Okay. I have to figure out how you killed off "c", and how "b"
is
> > based on 9/7...
> > on 9/7, and so forth
>
> The new a and b are different from the old a and b; I killed off c
> because I never fed a c in in the first place.

True, I should know better. I just need to figure out how it is
*calculated*. I certainly don't expect you to crunch numbers for me...
that I can do myself (hopefully) once I understand things
theoretically. Here's a more theoretical question: Why does doing
an orthogonal projection make the comma vanish?

-Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

3/5/2004 6:51:36 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> True, I should know better. I just need to figure out how it is
> *calculated*. I certainly don't expect you to crunch numbers for me...
> that I can do myself (hopefully) once I understand things
> theoretically. Here's a more theoretical question: Why does doing
> an orthogonal projection make the comma vanish?

Because you are projecting orthogonally from the comma.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/5/2004 6:59:20 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
>
> > True, I should know better. I just need to figure out how it is
> > *calculated*. I certainly don't expect you to crunch numbers for
me...
> > that I can do myself (hopefully) once I understand things
> > theoretically. Here's a more theoretical question: Why does doing
> > an orthogonal projection make the comma vanish?
>
> Because you are projecting orthogonally from the comma.

Wow, it's really that simple. So plugging 81/80 into (v-(u.v)u) makes
zero...

🔗Paul Erlich <perlich@aya.yale.edu>

3/6/2004 3:44:11 PM

I think I can jump in with the answer here -- because it makes all
comma-separated pairs of notes overlap exactly, thus representing the
comma by a zero length in the lattice.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/6/2004 4:46:30 PM

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@u...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > <paul.hjelmstad@u...> wrote:
> > >
> > > > > I gave a semidefinite form in terms of 3, 5 and 9; you feed
3/2
> > > and
> > > > > 9/7 into that, and get a definite form in terms of those
two.
> > > >
> > > > Okay. I have to figure out how you killed off "c", and
how "b"
> > is
> > > > based on 9/7...
> > > > on 9/7, and so forth
> > >
> > > The new a and b are different from the old a and b; I killed
off c
> > > because I never fed a c in in the first place.
> >
> > True, I should know better. I just need to figure out how it is
> > *calculated*. I certainly don't expect you to crunch numbers for
me...
> > that I can do myself (hopefully) once I understand things
> > theoretically. Here's a more theoretical question: Why does doing
> > an orthogonal projection make the comma vanish?
>
> I think I can jump in with the answer here -- because it makes all
> comma-separated pairs of notes overlap exactly, thus representing
the
> comma by a zero length in the lattice.

Now I just need to understand how the lattice is formed from (in this
case) 3a^2+2ab+35b^2...

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/8/2004 8:55:45 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@u...> wrote:
> > > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > > wrote:
> > > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > > <paul.hjelmstad@u...> wrote:
> > > >
>
> Now I just need to understand how the lattice is formed from (in
this
> case) 3a^2+2ab+35b^2...

Can someone tell me how this is used to form a lattice based on 3/2
and 9/7? Thanks!

Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

3/8/2004 12:36:20 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > Now I just need to understand how the lattice is formed from (in
> this
> > case) 3a^2+2ab+35b^2...
>
> Can someone tell me how this is used to form a lattice based on 3/2
> and 9/7? Thanks!

If we define a norm by

||(3/2)^a (9/7)^b|| = sqrt(3a^2+2ab+35b^2)

then we get a lattice, since 3a^2+2ab+35b^2 is positive-definite--the
only way to get a zero distance from the unison is to be the unison.

🔗Gene Ward Smith <gwsmith@svpal.org>

3/8/2004 12:42:57 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> If we define a norm by
>
> ||(3/2)^a (9/7)^b|| = sqrt(3a^2+2ab+35b^2)
>
> then we get a lattice, since 3a^2+2ab+35b^2 is positive-definite--the
> only way to get a zero distance from the unison is to be the unison.

To see 3a^2+2ab+35b^2 is positive-definite, we can take the
corresponding symmetric matrix [[3 1], [1 35]]. The characteristic
polynomial for this is x^2 - 38x + 104, which has two positive real
roots, 19+sqrt(257) and 19-sqrt(257), so it is positive-definite.

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/8/2004 2:13:55 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
>
> > If we define a norm by
> >
> > ||(3/2)^a (9/7)^b|| = sqrt(3a^2+2ab+35b^2)
> >
> > then we get a lattice, since 3a^2+2ab+35b^2 is positive-definite--
the
> > only way to get a zero distance from the unison is to be the
unison.
>
> To see 3a^2+2ab+35b^2 is positive-definite, we can take the
> corresponding symmetric matrix [[3 1], [1 35]]. The characteristic
> polynomial for this is x^2 - 38x + 104, which has two positive real
> roots, 19+sqrt(257) and 19-sqrt(257), so it is positive-definite.

Thanks. Is the example lattice for this in Planarplots? I'd let to
look at it.

-Paul Hj

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

3/8/2004 2:32:45 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> >
> > > If we define a norm by
> > >
> > > ||(3/2)^a (9/7)^b|| = sqrt(3a^2+2ab+35b^2)
> > >
> > > then we get a lattice, since 3a^2+2ab+35b^2 is positive-
definite--
> the
> > > only way to get a zero distance from the unison is to be the
> unison.
> >
> > To see 3a^2+2ab+35b^2 is positive-definite, we can take the
> > corresponding symmetric matrix [[3 1], [1 35]]. The characteristic
> > polynomial for this is x^2 - 38x + 104, which has two positive
real
> > roots, 19+sqrt(257) and 19-sqrt(257), so it is positive-definite.
>
> Thanks. Is the example lattice for this in Planarplots? I'd let to
> look at it.
>
Oops, Found it. Didymus of course! Since I am posting a new message,
would it be too much trouble if you could enlighten me on how Q81
((3/2)^a,(9/7)^b) equals 3a^2 + 2ab +35b^2? You change basis with
such ease, but it still isn't that easy for me to follow:(
> -Paul Hj

🔗Gene Ward Smith <gwsmith@svpal.org>

3/8/2004 2:38:19 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@u...> wrote:

> > To see 3a^2+2ab+35b^2 is positive-definite, we can take the
> > corresponding symmetric matrix [[3 1], [1 35]]. The characteristic
> > polynomial for this is x^2 - 38x + 104, which has two positive real
> > roots, 19+sqrt(257) and 19-sqrt(257), so it is positive-definite.
>
> Thanks. Is the example lattice for this in Planarplots? I'd let to
> look at it.

It's p81.jpg

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

5/13/2004 2:19:14 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

Gene, please see question below. Thanks.

> If we have a comma which does not (as 50/49, for instance, will)
lead
> to a part-octave period, we can find generators which are analogous
> to the period-and-generator generator of a linear temperament, by
> projecting orthogonally onto a plane in a way which makes the comma
> dissapper (people may recall my lattice diagrams of this.) We can
now
> do a Minkowski reduction--pick the class representative in the range
> 1 < q < sqrt(2) which is closest to the origin, as projected, and
> which has the smallest Tenney norm. Now do the same for second
> closest. We end up with two rational numbers 1 < q, r < sqrt(2)
which
> serve as generators for the temperament. If we put these together
> with 2 and the comma, we get a unimodular matrix; inverting this
> gives a matrix whose columns are the four vals I list. This shows
how
> to map a 7-limit note to octave, the two generators, and the comma;
> we get the temperament by dropping the comma and retuning.
>
> Below you find the comma, then the three generators, followed by
the
> square of the distance from the origin of the two non-octave
> generators when projected,

*Gene, a while back you said there were errors in the [1/7, 5/7] items
(the squares of the commas when projected). Is there any way you could
rerun these calculations with the correct values?

- Paul Hj

followed by the ratio of second-nearest to
> nearest square size. This can be all the way from the same (225/224
> and 2401/2400) to quite large (schisma.) On the line below I give
the
> map.
>
> 64/63 {2, 3/2, 5/4} [1/7, 5/7] 5
> [<1 1 2 4|, <0 -1 0 2|, <0 0 1 0|, <0 0 0 -1|]
>
> 81/80 {2, 4/3, 9/7} [1/13, 9/13] 9
> [<1 1 0 2|, <0 -1 -4 -2|, <0 0 0 -1|, <0 0 -1 0|]
>
> 245/243 {2, 9/7, 7/6} [3/17, 5/17] 5/3
> [<1 1 1 2|, <0 1 3 1|, <0 1 1 2|, <0 0 1 0|]
>
> 126/125 {2, 6/5, 5/4} [1/7, 5/7] 5
> [<1 2 2 1|, <0 1 0 -2|, <0 1 1 1|, <0 0 0 1|]
>
> 225/224 {2, 4/3, 15/14} [1/3, 1/3] 1
> [<1 1 3 3|, <0 -1 1 0|, <0 0 -1 -2|, <0 0 1 1|]
>
> 1728/1715 {2, 7/6, 36/35} [1/5, 1/2] 5/2
> [<1 2 3 3|, <0 -2 -3 -1|, <0 1 0 1|, <0 -1 -1 -1|]
>
> 1029/1024 {2, 8/7, 35/32} [7/10, 37/10] 37/7
> [<1 4 3 2|, <0 3 1 -1|, <0 0 1 0|, <0 1 0 0|]
>
> 3136/3125 {2, 28/25, 168/125} [1/19, 18/19] 18
> [<1 1 2 2|, <0 1 2 5|, <0 1 0 0|, <0 -1 -1 -2|]
>
> 5120/5103 {2, 4/3, 27/20} [1/37, 25/37] 25
> [<1 2 4 2|, <0 -1 -3 3|, <0 0 -1 -1|, <0 0 0 -1|]
>
> 6144/6125 {2, 35/32, 5/4} [1/5, 1/3] 5/3
> [<1 1 2 3| <0 2 0 1|, <0 1 1 -1|, <0 1 0 0|]
>
> 32805/32768 {2, 4/3, 448/405} [1/73, 57/73] 57
> [<1 2 -1 1|, <0 -1 8 4|, <0 0 0 1|, <0 0 1 1|]
>
> 2401/2400 {2, 7/5, 49/40} [3/11, 3/11] 1
> [<1 1 3 3|, <0 0 -2 -1|, <0 2 1 1|, <0 -1 0 0|]
>
> 4375/4374 {2, 27/25, 10/9} [1/7, 6/35] 6/5
> [<1 2 3 3|, <0 -1 -2 1|, <0 -2 -3 -2|, <0 0 0 1|]

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

5/17/2004 4:00:27 PM

Hello. Looks like I am stuck on something else, much more rudimentary.
Please see question below. Thanks.

> Define
>
> Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc
>
> Then this is a positive definite quadratic form on note classes
>
> http://164.8.13.169/Enciklopedija/math/math/p/p520.htm
>
> Taking the square root defines the symmetric lattice distance to the
> origin. We get a bilinear form, and hence a dot product, from B(a,b)
=
> (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in
terms
> of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
>
> I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by
> dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> product of this with [a,b,c] and substracting, I get the orthogonal
> projection of [a,b,c]:
>
> [-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]

Would anyone be so kind as to show how this is (v-(u.v)u), showing
all the steps? Thanks!!!!!

🔗Paul G Hjelmstad <paul.hjelmstad@us.ing.com>

6/1/2004 5:07:49 PM

Hate to be a pain, but I obtain:

[(a+4b)/17, (4a+16b)/17, c] Am I missing something?

This is with w=v-(v.u/u.u)u

Paul Hj

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/15/2005 10:04:17 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

Please see question below...

Could you please show me how you project the generators so the comma
disappears? (For example, 4/3 and 9/7 projected so 81/80 disappears)

to
> nearest square size. This can be all the way from the same (225/224
> and 2401/2400) to quite large (schisma.) On the line below I give
the
> map.
>
> 64/63 {2, 3/2, 5/4} [1/7, 5/7] 5
> [<1 1 2 4|, <0 -1 0 2|, <0 0 1 0|, <0 0 0 -1|]
>
> 81/80 {2, 4/3, 9/7} [1/13, 9/13] 9
> [<1 1 0 2|, <0 -1 -4 -2|, <0 0 0 -1|, <0 0 -1 0|]
>
> 245/243 {2, 9/7, 7/6} [3/17, 5/17] 5/3
> [<1 1 1 2|, <0 1 3 1|, <0 1 1 2|, <0 0 1 0|]
>
> 126/125 {2, 6/5, 5/4} [1/7, 5/7] 5
> [<1 2 2 1|, <0 1 0 -2|, <0 1 1 1|, <0 0 0 1|]
>
> 225/224 {2, 4/3, 15/14} [1/3, 1/3] 1
> [<1 1 3 3|, <0 -1 1 0|, <0 0 -1 -2|, <0 0 1 1|]
>
> 1728/1715 {2, 7/6, 36/35} [1/5, 1/2] 5/2
> [<1 2 3 3|, <0 -2 -3 -1|, <0 1 0 1|, <0 -1 -1 -1|]
>
> 1029/1024 {2, 8/7, 35/32} [7/10, 37/10] 37/7
> [<1 4 3 2|, <0 3 1 -1|, <0 0 1 0|, <0 1 0 0|]
>
> 3136/3125 {2, 28/25, 168/125} [1/19, 18/19] 18
> [<1 1 2 2|, <0 1 2 5|, <0 1 0 0|, <0 -1 -1 -2|]
>
> 5120/5103 {2, 4/3, 27/20} [1/37, 25/37] 25
> [<1 2 4 2|, <0 -1 -3 3|, <0 0 -1 -1|, <0 0 0 -1|]
>
> 6144/6125 {2, 35/32, 5/4} [1/5, 1/3] 5/3
> [<1 1 2 3| <0 2 0 1|, <0 1 1 -1|, <0 1 0 0|]
>
> 32805/32768 {2, 4/3, 448/405} [1/73, 57/73] 57
> [<1 2 -1 1|, <0 -1 8 4|, <0 0 0 1|, <0 0 1 1|]
>
> 2401/2400 {2, 7/5, 49/40} [3/11, 3/11] 1
> [<1 1 3 3|, <0 0 -2 -1|, <0 2 1 1|, <0 -1 0 0|]
>
> 4375/4374 {2, 27/25, 10/9} [1/7, 6/35] 6/5
> [<1 2 3 3|, <0 -1 -2 1|, <0 -2 -3 -2|, <0 0 0 1|]

🔗Gene Ward Smith <gwsmith@svpal.org>

8/17/2005 4:35:53 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Could you please show me how you project the generators so the comma
> disappears? (For example, 4/3 and 9/7 projected so 81/80 disappears)

We first need a Euclidean metric on note classes; this means in effect
that we want a positive definate quadratic form on note classes
u=|* a b c>. Here I was projecting with respect to the symmetrical
lattice of note classes, with form Q(u)=a^2+b^2+c^2+ab+bc+ca.

If u is a unit vector, so that u.u=1, then I can project the vector v
onto the subspace of vectors orthogonal to u by finding

w = v - (v.u)u

Then w.u=v.u - (v.u)(u.u) = 0. The dot product with respect to a given
quadratic form Q is the bilinear form defined by

(u.v) = (Q(u+v)-Q(u)-Q(v))/2

Applying that to the case |* 4 -1 0> we find the corresponding unit
vector is |* 4/sqrt(13) -1/sqrt(13) 0>, and computing w as defined
above for 9/7, corresponding to v=|* 2 0 -1> gives us v.u equal to
11sqrt(13)/26, and hence w=|* 4/13 11/26 -1>. Now w.u=0, so it is
orthogonal to u.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/18/2005 6:58:10 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Could you please show me how you project the generators so the
comma
> > disappears? (For example, 4/3 and 9/7 projected so 81/80
disappears)
>
> We first need a Euclidean metric on note classes; this means in
effect
> that we want a positive definate quadratic form on note classes
> u=|* a b c>. Here I was projecting with respect to the symmetrical
> lattice of note classes, with form Q(u)=a^2+b^2+c^2+ab+bc+ca.
>
> If u is a unit vector, so that u.u=1, then I can project the vector
v
> onto the subspace of vectors orthogonal to u by finding
>
> w = v - (v.u)u
>
> Then w.u=v.u - (v.u)(u.u) = 0. The dot product with respect to a
given
> quadratic form Q is the bilinear form defined by
>
> (u.v) = (Q(u+v)-Q(u)-Q(v))/2
>
> Applying that to the case |* 4 -1 0> we find the corresponding unit
> vector is |* 4/sqrt(13) -1/sqrt(13) 0>, and computing w as defined
> above for 9/7, corresponding to v=|* 2 0 -1> gives us v.u equal to
> 11sqrt(13)/26, and hence w=|* 4/13 11/26 -1>. Now w.u=0, so it is
> orthogonal to u.

Thanks! This is interesting. I am still not sure where [1/13, 9/13]
comes from in Message 9767 (For 81/80 and (2, 4/3, 9/7))

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/18/2005 9:29:40 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul.hjelmstad@m...> wrote:
> >
> > > Could you please show me how you project the generators so the
> comma
> > > disappears? (For example, 4/3 and 9/7 projected so 81/80
> disappears)
> >
> > We first need a Euclidean metric on note classes; this means in
> effect
> > that we want a positive definate quadratic form on note classes
> > u=|* a b c>. Here I was projecting with respect to the symmetrical
> > lattice of note classes, with form Q(u)=a^2+b^2+c^2+ab+bc+ca.
> >
> > If u is a unit vector, so that u.u=1, then I can project the
vector
> v
> > onto the subspace of vectors orthogonal to u by finding
> >
> > w = v - (v.u)u
> >
> > Then w.u=v.u - (v.u)(u.u) = 0. The dot product with respect to a
> given
> > quadratic form Q is the bilinear form defined by
> >
> > (u.v) = (Q(u+v)-Q(u)-Q(v))/2
> >
> > Applying that to the case |* 4 -1 0> we find the corresponding
unit
> > vector is |* 4/sqrt(13) -1/sqrt(13) 0>, and computing w as defined
> > above for 9/7, corresponding to v=|* 2 0 -1> gives us v.u equal to
> > 11sqrt(13)/26, and hence w=|* 4/13 11/26 -1>. Now w.u=0, so it is
> > orthogonal to u.
>
> Thanks! This is interesting. I am still not sure where [1/13, 9/13]
> comes from in Message 9767 (For 81/80 and (2, 4/3, 9/7))

Never mind, you've already answered that. However, could you pretty
please explain how you get the result for u.v using ((Q(u+v)-Q(u)-Q
(v))/2. Also, I found a small error: w should be |* 44/13 11/26 -1>

pAUL

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/18/2005 10:10:20 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > <paul.hjelmstad@m...> wrote:
> > >
> > > > Could you please show me how you project the generators so
the
> > comma
> > > > disappears? (For example, 4/3 and 9/7 projected so 81/80
> > disappears)
> > >
> > > We first need a Euclidean metric on note classes; this means in
> > effect
> > > that we want a positive definate quadratic form on note classes
> > > u=|* a b c>. Here I was projecting with respect to the
symmetrical
> > > lattice of note classes, with form Q(u)=a^2+b^2+c^2+ab+bc+ca.
> > >
> > > If u is a unit vector, so that u.u=1, then I can project the
> vector
> > v
> > > onto the subspace of vectors orthogonal to u by finding
> > >
> > > w = v - (v.u)u
> > >
> > > Then w.u=v.u - (v.u)(u.u) = 0. The dot product with respect to
a
> > given
> > > quadratic form Q is the bilinear form defined by
> > >
> > > (u.v) = (Q(u+v)-Q(u)-Q(v))/2
> > >
> > > Applying that to the case |* 4 -1 0> we find the corresponding
> unit
> > > vector is |* 4/sqrt(13) -1/sqrt(13) 0>, and computing w as
defined
> > > above for 9/7, corresponding to v=|* 2 0 -1> gives us v.u equal
to
> > > 11sqrt(13)/26, and hence w=|* 4/13 11/26 -1>. Now w.u=0, so it
is
> > > orthogonal to u.
> >
> > Thanks! This is interesting. I am still not sure where [1/13,
9/13]
> > comes from in Message 9767 (For 81/80 and (2, 4/3, 9/7))
>
> Never mind, you've already answered that. However, could you pretty
> please explain how you get the result for u.v using ((Q(u+v)-Q(u)-Q
> (v))/2. Also, I found a small error: w should be |* 44/13 11/26 -1>
>
> pAUL

Sorry. It's right (4/13.)The reason I am asking about u.v is in part
to understand some of your earlier posts. Thanks.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/18/2005 2:10:20 PM

Gene wrote:

> > > > We first need a Euclidean metric on note classes; this means
in
> > > effect
> > > > that we want a positive definate quadratic form on note
classes
> > > > u=|* a b c>. Here I was projecting with respect to the
> symmetrical
> > > > lattice of note classes, with form Q(u)=a^2+b^2+c^2+ab+bc+ca.
> > > >
> > > > If u is a unit vector, so that u.u=1, then I can project the
> > vector
> > > v
> > > > onto the subspace of vectors orthogonal to u by finding
> > > >
> > > > w = v - (v.u)u
> > > >
> > > > Then w.u=v.u - (v.u)(u.u) = 0. The dot product with respect
to
> a
> > > given
> > > > quadratic form Q is the bilinear form defined by
> > > >
> > > > (u.v) = (Q(u+v)-Q(u)-Q(v))/2
> > > >
> > > > Applying that to the case |* 4 -1 0> we find the
corresponding
> > unit
> > > > vector is |* 4/sqrt(13) -1/sqrt(13) 0>, and computing w as
> defined
> > > > above for 9/7, corresponding to v=|* 2 0 -1> gives us v.u
equal
> to
> > > > 11sqrt(13)/26, and hence w=|* 4/13 11/26 -1>. Now w.u=0, so
it
> is
> > > > orthogonal to u.

I actually got the numbers to work out, even though I don't really
understand the use of positive definite quadratic form and bilinear
form. I take it that sqrt(13) in the denominator of the unit vector
is a 'unitizing' factor. Could you instead of answering my trivial
number-crunching question expand a little on the meaning of these
forms? I apologize for all the posts, I'll try to consolidate my
thoughts going forward. Also, doesn't u.u=17/13 (or does it equal 1
because of these forms?)

Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

8/18/2005 6:29:38 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I actually got the numbers to work out, even though I don't really
> understand the use of positive definite quadratic form and bilinear
> form.

The following articles may be helpful:

http://en.wikipedia.org/wiki/Quadratic_form

http://en.wikipedia.org/wiki/Bilinear_form

http://en.wikipedia.org/wiki/Inner_product_space

Tell me if you have questions after looking at these; the first
article, I hope, would be especially useful.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/19/2005 6:25:13 AM

Gene wrote:
> Define
>
> Q(3^a 5^b 7^c) = a^2+b^2+c^2+ab+ac+bc
>
> Then this is a positive definite quadratic form on note classes
>
> http://164.8.13.169/Enciklopedija/math/math/p/p520.htm
>
> Taking the square root defines the symmetric lattice distance to the
> origin. We get a bilinear form, and hence a dot product, from B(a,b)
=
> (Q(ab)-Q(a)-Q(b))/2 (writing it multiplicitively, additively, in
terms
> of row vectors, (Q(a+b)-Q(a)-Q(b))/2.)
>
> I have Q(81/80)=13, so I get a unit vector in the 81/80 direction by
> dividing by sqrt(13): [4/sqrt(13), -1/sqrt(13), 0]. Taking the dot
> product of this with [a,b,c] and substracting, I get the orthogonal
> projection of [a,b,c]:
>
> [-1/13*a-4/13*b-6/13*c, 14/13*b+3/26*c+7/26*a, c]
>
> I now get a positive semidefinite quadratic form by substituting
this
> in Q; if I normalize by clearing denominators, I get
>
> Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
>
> Positive semidefinite:
> http://164.8.13.169/Enciklopedija/math/math/p/p524.htm
>
> Q81 is semidefinite, since Q81(81/80)=0. We can get a positive
> definite form, and hence a lattice, by going to two dimensions with
> basis 3/2 and 9/7:
>
> Q81((3/2)^a (9/7)^b) = 3a^2 + 2ab + 35b^2
>
> If we take the TM reduction of intervals in the octave we can sort
> them by size according to the above distance measure. Below I give
> the intervals in order of size, the scales you get by combining them
> (which should be tempered by "didymus", or 81/80 planar) and the
> meantone reduction of the scales, where 7 is mapped to ten
generators
> of a fifth.

I'm very close. The only step I don't understand is going to two
dimensions with Q81 above. Could you please show how you obtain
3a^2 + 2ab + 35b^2?

Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

8/19/2005 3:04:48 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> > Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc

> I'm very close. The only step I don't understand is going to two
> dimensions with Q81 above. Could you please show how you obtain
> 3a^2 + 2ab + 35b^2?

We have that 3/2 corresponds to |* 1 0 0> and 9/7 to |* 2 0 -1>,
so (3/2)^u (9/7)^v corresponds to |* u+2v 0 -v>. Subsituting
a=u+2v, b=0, c=-v in Q81 above gives 3u^2 + 2uv + 35v^2.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/22/2005 6:19:13 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > > Q81(3^a 5^b 7^c) = 3a^2+48b^2+43c^2+24ab+10ac+40bc
>
> > I'm very close. The only step I don't understand is going to two
> > dimensions with Q81 above. Could you please show how you obtain
> > 3a^2 + 2ab + 35b^2?
>
> We have that 3/2 corresponds to |* 1 0 0> and 9/7 to |* 2 0 -1>,
> so (3/2)^u (9/7)^v corresponds to |* u+2v 0 -v>. Subsituting
> a=u+2v, b=0, c=-v in Q81 above gives 3u^2 + 2uv + 35v^2.

Thanks - I should have known that...

🔗Gene Ward Smith <gwsmith@svpal.org>

8/22/2005 12:04:44 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> > We have that 3/2 corresponds to |* 1 0 0> and 9/7 to |* 2 0 -1>,
> > so (3/2)^u (9/7)^v corresponds to |* u+2v 0 -v>. Subsituting
> > a=u+2v, b=0, c=-v in Q81 above gives 3u^2 + 2uv + 35v^2.
>
> Thanks - I should have known that...

There's an equivalency lattice basis <==> quadratic form <==> bilinear
form, so once you get the quadratic form corresponding to (for
instance) 81/80 planar, you also get a lattice on pitch classes, which
can be depicted visually. Back in 2002 I did some plots of projected
lattices, projecting a different lattice; these are in the files
section here under planar plots. Even though the lattice I started
from was not the symmetric lattice, the plot "p81" for 81/80 looks
about right--nearly orthogonal 3/2 and 9/7 generators.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/22/2005 2:26:22 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > > We have that 3/2 corresponds to |* 1 0 0> and 9/7 to |* 2 0 -1>,
> > > so (3/2)^u (9/7)^v corresponds to |* u+2v 0 -v>. Subsituting
> > > a=u+2v, b=0, c=-v in Q81 above gives 3u^2 + 2uv + 35v^2.
> >
> > Thanks - I should have known that...
>
> There's an equivalency lattice basis <==> quadratic form <==> bilinear
> form, so once you get the quadratic form corresponding to (for
> instance) 81/80 planar, you also get a lattice on pitch classes, which
> can be depicted visually. Back in 2002 I did some plots of projected
> lattices, projecting a different lattice; these are in the files
> section here under planar plots. Even though the lattice I started
> from was not the symmetric lattice, the plot "p81" for 81/80 looks
> about right--nearly orthogonal 3/2 and 9/7 generators.

I'll look at it. I'm finally getting better with technique, but theory
is more difficult - I'm taking it on faith and hoping it will sink in.
I looked at the Wikipedia entries, very interesting. I'm still
struggling to see how/why you apply the theory in this case, I should
have some questions tomorrow.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/23/2005 11:14:57 AM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > I actually got the numbers to work out, even though I don't really
> > understand the use of positive definite quadratic form and bilinear
> > form.
>
> The following articles may be helpful:
>
> http://en.wikipedia.org/wiki/Quadratic_form
>
> http://en.wikipedia.org/wiki/Bilinear_form
>
> http://en.wikipedia.org/wiki/Inner_product_space
>
> Tell me if you have questions after looking at these; the first
> article, I hope, would be especially useful.

I'm going to ask something really simple: Why is quadratic form for two
dimensions ax^2+by^2+cxy? I looked at the articles on distance and
Euclidean distance and get different formulas. Where does cxy come from?
I also am stuck on the fact that Euclidean distance is PX, PY. (How can
they use the same P for X and Y)...Am I barking up the wrong tree?

Paul

🔗Gene Ward Smith <gwsmith@svpal.org>

8/24/2005 12:33:49 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I'm going to ask something really simple: Why is quadratic form for two
> dimensions ax^2+by^2+cxy? I looked at the articles on distance and
> Euclidean distance and get different formulas.

A quadratic form is expressed via a homogenous polynomial of degree
two, and that is the general form of a homogenous polynomial in two
variables of degree two. What sort of formulas are you getting?

> Where does cxy come from?

It's nonzero when the basis vectors are not perpendicular.

> I also am stuck on the fact that Euclidean distance is PX, PY. (How can
> they use the same P for X and Y)...Am I barking up the wrong tree?

I dunno--what's P?

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/24/2005 1:30:48 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > I'm going to ask something really simple: Why is quadratic form
for two
> > dimensions ax^2+by^2+cxy? I looked at the articles on distance
and
> > Euclidean distance and get different formulas.
>
> A quadratic form is expressed via a homogenous polynomial of degree
> two, and that is the general form of a homogenous polynomial in two
> variables of degree two. What sort of formulas are you getting?

I ended up with ax^2-c(x+y)=by^2
>
> > Where does cxy come from?
>
> It's nonzero when the basis vectors are not perpendicular.

That makes sense
>
> > I also am stuck on the fact that Euclidean distance is PX, PY.
(How can
> > they use the same P for X and Y)...Am I barking up the wrong tree?
>
> I dunno--what's P?

http://en.wikipedia.org/wiki/Euclidean_distance

It must be different values for X, Y ?

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/29/2005 7:56:07 AM

I guess I just don't understand your application of these forms.
For example, why do you apply the three-dimensional quadratic form and
assume a=b=c=d=e=f=1?

🔗Gene Ward Smith <gwsmith@svpal.org>

8/29/2005 6:07:26 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> I guess I just don't understand your application of these forms.
> For example, why do you apply the three-dimensional quadratic form and
> assume a=b=c=d=e=f=1?

Because that gives the symmetrical lattice of 7-limit pitch
classes--that is, the lattice where 3, 5, and 7 are treated as the
same, so that tetrads are regular tetrahedrons and hexanys regular
octahedrons.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/29/2005 6:44:45 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > I guess I just don't understand your application of these forms.
> > For example, why do you apply the three-dimensional quadratic form
and
> > assume a=b=c=d=e=f=1?
>
> Because that gives the symmetrical lattice of 7-limit pitch
> classes--that is, the lattice where 3, 5, and 7 are treated as the
> same, so that tetrads are regular tetrahedrons and hexanys regular
> octahedrons.

I remember tetrads as being dominant-seventh and half-diminished
sevenths. Are hexanies 3/5, 3/7, 5/7, 5/3, 7/3, 7/5?

🔗Paul Erlich <perlich@aya.yale.edu>

8/29/2005 6:57:04 PM

No, a hexany is much more harmonically compact than that, and can be
expressed as 1*3, 1*5, 1*7, 3*5, 3*7, 5*7, but usually is transposed
so that one of the notes becomes 1.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/30/2005 1:55:49 PM

Right. However, in Erv Wilson's 1-3-5-7 Diamond (which is a
cubeoctohedron) you have tetrahedrons for 1-3-5-7 and 1-1/3-1/5-1/7
and 7/5, 3/5, 3/7, 5/7, 5/3 and 7/3 for the remaining 6 points. What
would you call this? (If not a hexany?)

p.10 of http://www.anaphoria.com/dal.PDF

🔗Gene Ward Smith <gwsmith@svpal.org>

8/30/2005 3:29:34 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> Right. However, in Erv Wilson's 1-3-5-7 Diamond (which is a
> cubeoctohedron) you have tetrahedrons for 1-3-5-7 and 1-1/3-1/5-1/7
> and 7/5, 3/5, 3/7, 5/7, 5/3 and 7/3 for the remaining 6 points. What
> would you call this? (If not a hexany?)

That's the 7-limit tonality diamond.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/30/2005 6:56:57 PM

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Right. However, in Erv Wilson's 1-3-5-7 Diamond (which is a
> > cubeoctohedron) you have tetrahedrons for 1-3-5-7 and 1-1/3-1/5-1/7
> > and 7/5, 3/5, 3/7, 5/7, 5/3 and 7/3 for the remaining 6 points.
What
> > would you call this? (If not a hexany?)
>
> That's the 7-limit tonality diamond.

It also uses a cubeoctohedron?

Okay, Now that I know what hexanies really are, could you point me to
a diagram which shows how the "deep holes" are hexanies and
the "shallow holes" are tetrads, in the cubeoctohedron...or perhaps
just tell me how many different hexany-sets there are (I take it there
are 2 tetrads)in this geometric figure. Or more simply, a diagram
which labels all 12 vertices. Thanx

🔗Carl Lumma <ekin@lumma.org>

8/30/2005 7:29:08 PM

>It also uses a cubeoctohedron?
>
>Okay, Now that I know what hexanies really are, could you point me to
>a diagram which shows how the "deep holes" are hexanies and
>the "shallow holes" are tetrads, in the cubeoctohedron...

There are no complete 2|4 [1 3 5 7] hexanies in the cuboctahedron.
But if you build a large enough section of the face-centered cubic
lattice, you'll see only 2 kinds of holes -- octahera and
tetrahedra.

>perhaps just tell me how many different hexany-sets there are (I
>take it there are 2 tetrads)in this geometric figure. Or more
>simply, a diagram which labels all 12 vertices. Thanx

Hexanies are loosely defined as any 6-tone combination product set,
but the term has come to be used to mean the most famous (aside from
the 11-limit hexad, perhaps) of these: the 2|4 [1 3 5 7].

See:
http://anaphoria.com/dal.PDF

-Cral!

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/30/2005 8:40:03 PM

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >It also uses a cubeoctohedron?
> >
> >Okay, Now that I know what hexanies really are, could you point
me to
> >a diagram which shows how the "deep holes" are hexanies and
> >the "shallow holes" are tetrads, in the cubeoctohedron...
>
> There are no complete 2|4 [1 3 5 7] hexanies in the cuboctahedron.
(Could you show me an octahedron that corresponds to a partial-
hexany?)
> But if you build a large enough section of the face-centered cubic
> lattice, you'll see only 2 kinds of holes -- octahera and
> tetrahedra.

I'm having trouble visualizing octahedra as holes. Is there anything
in the Erv Wilson article that shows this? I may have missed it-
even though I went through the whole thing earlier today. How do
you "build a large enough section of the FC cubic lattice?"

Thanx
Pual

🔗Carl Lumma <ekin@lumma.org>

8/30/2005 9:00:57 PM

>> There are no complete 2|4 [1 3 5 7] hexanies in the cuboctahedron.
>
>(Could you show me an octahedron that corresponds to a partial-
>hexany?)

Octahedra have six vertices, which correspond to the pitch classes
of a hexany in the visualization. (The faces are triangles whose 3
points correspond to the pitch classes of the hexany's triads.) An
incomplete octahedron is an incomplete hexany.

>> But if you build a large enough section of the face-centered cubic
>> lattice, you'll see only 2 kinds of holes -- octahera and
>> tetrahedra.
>
>I'm having trouble visualizing octahedra as holes. Is there anything
>in the Erv Wilson article that shows this? I may have missed it-
>even though I went through the whole thing earlier today. How do
>you "build a large enough section of the FC cubic lattice?"

See figures 6c & 6d.

And...

http://www.rwgrayprojects.com/synergetics/s04/figs/f7002a.html

http://www.tabletoptelephone.com/~hopspage/Fuller.html

And you might like...

http://lumma.org/tuning/erlich/erlich-tFoT.pdf

-Carl

🔗Carl Lumma <ekin@lumma.org>

8/30/2005 9:02:37 PM

>And...
>
>http://www.rwgrayprojects.com/synergetics/s04/figs/f7002a.html
>
>http://www.tabletoptelephone.com/~hopspage/Fuller.html

Also...

http://www.channel1.com/users/bobwb/synergetics/bucky/fig17.html

-Carl

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/31/2005 6:32:48 AM

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> There are no complete 2|4 [1 3 5 7] hexanies in the
cuboctahedron.
> >
> >(Could you show me an octahedron that corresponds to a partial-
> >hexany?)
>
> Octahedra have six vertices, which correspond to the pitch classes
> of a hexany in the visualization. (The faces are triangles whose 3
> points correspond to the pitch classes of the hexany's triads.) An
> incomplete octahedron is an incomplete hexany.
>
> >> But if you build a large enough section of the face-centered
cubic
> >> lattice, you'll see only 2 kinds of holes -- octahera and
> >> tetrahedra.
> >
> >I'm having trouble visualizing octahedra as holes. Is there
anything
> >in the Erv Wilson article that shows this? I may have missed it-
> >even though I went through the whole thing earlier today. How do
> >you "build a large enough section of the FC cubic lattice?"
>
> See figures 6c & 6d.
>
I actually printed it out yesterday. The key is that the hexanies are
incomplete: The deep holes are:

7/1 5/1 7/3 5/3 1/1 missing: 35/3
7/5 3/5 3/1 7/1 1/1 missing: 21/5
3/1 3/7 5/7 5/1 1/1 missing: 15/7

Thanks for the Bucky links. I also have read Paul's paper - very
thorough!

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/31/2005 6:42:32 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> > >> There are no complete 2|4 [1 3 5 7] hexanies in the
> cuboctahedron.
> > >
> > >(Could you show me an octahedron that corresponds to a partial-
> > >hexany?)
> >
> > Octahedra have six vertices, which correspond to the pitch classes
> > of a hexany in the visualization. (The faces are triangles whose
3
> > points correspond to the pitch classes of the hexany's triads.)
An
> > incomplete octahedron is an incomplete hexany.
> >
> > >> But if you build a large enough section of the face-centered
> cubic
> > >> lattice, you'll see only 2 kinds of holes -- octahera and
> > >> tetrahedra.
> > >
> > >I'm having trouble visualizing octahedra as holes. Is there
> anything
> > >in the Erv Wilson article that shows this? I may have missed it-
> > >even though I went through the whole thing earlier today. How do
> > >you "build a large enough section of the FC cubic lattice?"
> >
> > See figures 6c & 6d.
> >
> I actually printed it out yesterday. The key is that the hexanies
are
> incomplete: The deep holes are:
>
> 7/1 5/1 7/3 5/3 1/1 missing: 35/3
> 7/5 3/5 3/1 7/1 1/1 missing: 21/5
> 3/1 3/7 5/7 5/1 1/1 missing: 15/7
>
> Thanks for the Bucky links. I also have read Paul's paper - very
> thorough!

For the sake of my own completeness, here are the remaining three
rectangular deep holes:

7/5 1/5 1/3 7/3 1/1 missing 7/15
3/5 1/5 3/7 1/7 1/1 missing 3/35
1/7 1/3 5/7 5/3 1/1 missing 5/21

Would these be antihexanies?

🔗Paul Erlich <perlich@aya.yale.edu>

8/31/2005 9:13:45 AM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > incomplete: The deep holes are:
> >
> > 7/1 5/1 7/3 5/3 1/1 missing: 35/3
> > 7/5 3/5 3/1 7/1 1/1 missing: 21/5
> > 3/1 3/7 5/7 5/1 1/1 missing: 15/7
> >
> > Thanks for the Bucky links. I also have read Paul's paper - very
> > thorough!
>
> For the sake of my own completeness, here are the remaining three
> rectangular

You mean octahedral?

> deep holes:
>
> 7/5 1/5 1/3 7/3 1/1 missing 7/15
> 3/5 1/5 3/7 1/7 1/1 missing 3/35
> 1/7 1/3 5/7 5/3 1/1 missing 5/21
>
> Would these be antihexanies?

As you can see in the lattice, all the octahedra are oriented the
same way, so there's no possible way of distinguishing 'hexanies'
from 'antihexanies'. Your supposed antihexanies can be transposed and
they'll coincide with your original hexanies. This is in
contradistinction from the situation with the tetrahedra, which do
come in two opposite forms (right-side up and upside-down, or major
and minor, or otonal and utonal), and you can't get from one to the
other by mere transposition.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

8/31/2005 9:57:18 AM

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > incomplete: The deep holes are:
> > >
> > > 7/1 5/1 7/3 5/3 1/1 missing: 35/3
> > > 7/5 3/5 3/1 7/1 1/1 missing: 21/5
> > > 3/1 3/7 5/7 5/1 1/1 missing: 15/7
> > >
> > For the sake of my own completeness, here are the remaining three
> > rectangular
>
> You mean octahedral?

Well they are octahedrons missing one point (on the outside)
Four corners are around the rectangles and 1/1 is in the center.
>
> > deep holes:
> >
> > 7/5 1/5 1/3 7/3 1/1 missing 7/15
> > 3/5 1/5 3/7 1/7 1/1 missing 3/35
> > 1/7 1/3 5/7 5/3 1/1 missing 5/21
> >
> > Would these be antihexanies?
>
> As you can see in the lattice, all the octahedra are oriented the
> same way, so there's no possible way of distinguishing 'hexanies'
> from 'antihexanies'. Your supposed antihexanies can be transposed
and
> they'll coincide with your original hexanies. This is in
> contradistinction from the situation with the tetrahedra, which do
> come in two opposite forms (right-side up and upside-down, or major
> and minor, or otonal and utonal), and you can't get from one to the
> other by mere transposition.

Hmm. They look like inverses to me. My source is Erv Wilson's
1-3-5-7 Diamond Figure 6c. Is this the wrong figure to use?
Actually, we may both be right. Transposing the last one by 21
give 3, 7, 15, 35, 21 and 5 missing. That's kind of neat.

🔗Carl Lumma <ekin@lumma.org>

8/31/2005 12:13:52 PM

>> >I'm having trouble visualizing octahedra as holes. Is there
>> >anything in the Erv Wilson article that shows this? I may
>> >have missed it- even though I went through the whole thing
>> >earlier today. How do you "build a large enough section of
>> >the FC cubic lattice?"
>>
>> See figures 6c & 6d.
>>
>I actually printed it out yesterday. The key is that the hexanies
>are incomplete: The deep holes are:
>
>7/1 5/1 7/3 5/3 1/1 missing: 35/3
>7/5 3/5 3/1 7/1 1/1 missing: 21/5
>3/1 3/7 5/7 5/1 1/1 missing: 15/7

It sounds like there's still a terminology misunderstanding here.
"Deep holes" just means the large species (out of the two total
species) of holes in the *entire* A1 lattice, *not* the cuboctahedron.

The cuboctahedron corresponds to the [1 3 5 7] tonality diamond.
There are 6 incomplete hexanies in it.

>Thanks for the Bucky links.

Just pictures of the lattice to look at.

-Carl

🔗Carl Lumma <ekin@lumma.org>

8/31/2005 12:14:14 PM

>For the sake of my own completeness, here are the remaining three
>rectangular deep holes:
>
>7/5 1/5 1/3 7/3 1/1 missing 7/15
>3/5 1/5 3/7 1/7 1/1 missing 3/35
>1/7 1/3 5/7 5/3 1/1 missing 5/21

Ah, you found them.

9/1/2005 6:27:38 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > > incomplete: The deep holes are:
> > > >
> > > > 7/1 5/1 7/3 5/3 1/1 missing: 35/3
> > > > 7/5 3/5 3/1 7/1 1/1 missing: 21/5
> > > > 3/1 3/7 5/7 5/1 1/1 missing: 15/7
> > > >
> > > For the sake of my own completeness, here are the remaining
three
> > > rectangular
> >
> > You mean octahedral?
>
> Well they are octahedrons missing one point (on the outside)
> Four corners are around the rectangles

Squares?

> and 1/1 is in the center.
> >
> > > deep holes:
> > >
> > > 7/5 1/5 1/3 7/3 1/1 missing 7/15
> > > 3/5 1/5 3/7 1/7 1/1 missing 3/35
> > > 1/7 1/3 5/7 5/3 1/1 missing 5/21
> > >
> > > Would these be antihexanies?
> >
> > As you can see in the lattice, all the octahedra are oriented the
> > same way, so there's no possible way of distinguishing 'hexanies'
> > from 'antihexanies'. Your supposed antihexanies can be transposed
> and
> > they'll coincide with your original hexanies. This is in
> > contradistinction from the situation with the tetrahedra, which
do
> > come in two opposite forms (right-side up and upside-down, or
major
> > and minor, or otonal and utonal), and you can't get from one to
the
> > other by mere transposition.
>
> Hmm. They look like inverses to me. My source is Erv Wilson's
> 1-3-5-7 Diamond Figure 6c. Is this the wrong figure to use?
> Actually, we may both be right. Transposing the last one by 21
> give 3, 7, 15, 35, 21 and 5 missing. That's kind of neat.

So we're agreed then?

🔗Paul Erlich <perlich@aya.yale.edu>

9/1/2005 6:30:26 PM

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> > >For the sake of my own completeness, here are the remaining
three
> > >rectangular deep holes:
> > >
> > >7/5 1/5 1/3 7/3 1/1 missing 7/15
> > >3/5 1/5 3/7 1/7 1/1 missing 3/35
> > >1/7 1/3 5/7 5/3 1/1 missing 5/21
> >
> > Ah, you found them.
> >
> > >Would these be antihexanies?
> >
> > No such thing.
>
> But opposite points of the cubeoctohedron are inverses of each other
> so that's why I dubbed them thus.

But the names chords and scales (such as 'hexany') are defined such
as to be invariant under transposition.

> Of course, you can find them by
> transposition also, but with different values missing.

Nothing is missing when you're talking about actual, complete.
hexanies.

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

9/2/2005 7:41:33 AM

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > > > incomplete: The deep holes are:
> > > > >
> > > > > 7/1 5/1 7/3 5/3 1/1 missing: 35/3
> > > > > 7/5 3/5 3/1 7/1 1/1 missing: 21/5
> > > > > 3/1 3/7 5/7 5/1 1/1 missing: 15/7
> > > > >
> > > > For the sake of my own completeness, here are the remaining
> three
> > > > rectangular
> > >
> > > You mean octahedral?
> >
> > Well they are octahedrons missing one point (on the outside)
> > Four corners are around the rectangles
>
> Squares?
>
> > and 1/1 is in the center.
> > >
> > > > deep holes:
> > > >
> > > > 7/5 1/5 1/3 7/3 1/1 missing 7/15
> > > > 3/5 1/5 3/7 1/7 1/1 missing 3/35
> > > > 1/7 1/3 5/7 5/3 1/1 missing 5/21
> > > >
> > > > Would these be antihexanies?
> > >
> > > As you can see in the lattice, all the octahedra are oriented
the
> > > same way, so there's no possible way of
distinguishing 'hexanies'
> > > from 'antihexanies'. Your supposed antihexanies can be
transposed
> > and
> > > they'll coincide with your original hexanies. This is in
> > > contradistinction from the situation with the tetrahedra, which
> do
> > > come in two opposite forms (right-side up and upside-down, or
> major
> > > and minor, or otonal and utonal), and you can't get from one to
> the
> > > other by mere transposition.
> >
> > Hmm. They look like inverses to me. My source is Erv Wilson's
> > 1-3-5-7 Diamond Figure 6c. Is this the wrong figure to use?
> > Actually, we may both be right. Transposing the last one by 21
> > give 3, 7, 15, 35, 21 and 5 missing. That's kind of neat.
>
> So we're agreed then?

Indeed. No need for antihexanies (It was only because in the
cubeoctohedron, the missing hexany items on opposite sides are
inverse of each other)

🔗Paul G Hjelmstad <paul.hjelmstad@medtronic.com>

9/2/2005 7:43:01 AM

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
> > --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> > > >For the sake of my own completeness, here are the remaining
> three
> > > >rectangular deep holes:
> > > >
> > > >7/5 1/5 1/3 7/3 1/1 missing 7/15
> > > >3/5 1/5 3/7 1/7 1/1 missing 3/35
> > > >1/7 1/3 5/7 5/3 1/1 missing 5/21
> > >
> > > Ah, you found them.
> > >
> > > >Would these be antihexanies?
> > >
> > > No such thing.
> >
> > But opposite points of the cubeoctohedron are inverses of each
other
> > so that's why I dubbed them thus.
>
> But the names chords and scales (such as 'hexany') are defined such
> as to be invariant under transposition.
>
> > Of course, you can find them by
> > transposition also, but with different values missing.
>
> Nothing is missing when you're talking about actual, complete.
> hexanies.

Oops missed this message. Yes, true, now that I understand the
tiling of the space with tetrahedrons and octohedrons :)