--- In tuning@y..., "Paul H. Erlich" <PERLICH@A...> wrote:

On October 29, 1999, Paul Hahn wrote,

>If you look at the way the truncated

>octa fills space, you'll find that its symmetry group is that of a

>body-centered cubic lattice. The rhombic dodec, OTOH, fills space

such

>that is _its_ symmetry group is that of a _face_-centered cubic

>lattice.

>Why is this important? Well, let's think about why we use FCC (= oc-

tet

>or triangulated lattice) for pitch diagrams in the first place. If

you

>ignore the edges and just look at the lattice points, it's

equivalent to

>the cubic lattice--it's just been subjected to a couple of affine

>(shear) transformations. (Translation: we squished it a bit so that

it

>slants.) This makes sense because the lattice is actually a space

whose

>basis is the three vectors representing the 3/2, the 5/4, and the

7/4.

>Right?

>Now look at the way the shapes you want to use to tesselate space

with.

>They _also_ are related to each other by three basic vectors, it's

just

>that this time, it's the three unison vectors. Other than that, the

>relationship is the same. But there's no way you can map the BCC to

a

>straightforward cubic lattice--you either have to leave some points

out

>of the cubic lattice, or interlock two cubic lattices together.

Conjecture: the bizarre, double-vision periodicity block I found

could even

happen in 3 dimensions, if there are 4 unison vectors defining

truncated-octahedron equivalence regions, but due to the

parallelopiped

basis of the periodicity block construction from three unison

vectors, these

truncated-octahedron regions could only come up two at a time.

What I found may be some sort of higher-dimensional analogue with 5

dimensions and 7(?) operative unison vectors.

Paul H., does this make any sense?

--- End forwarded message ---

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Paul H., does this make any sense?

It certainly makes sense to look at the lattice-pair one gets by

taking the lattice of utonal tetrads together with the lattice of

otonal tetrads. If we look at triples [a,b,c] with a quadradic form

Q(a,b,c) = a^2+b^2+c62+ab+ac+bc, we have the symmetric lattice of 7-

limit note-classes.

The tetrads are defined as four lattice points, each of which is at

unit distance from the other three. The tetrad centroids are simply

the means of the lattice points, if we omit the division by 4, we get

for the 1-3-5-7 otonal tetrad

[0 0 0]+[1 0 0]+[0 1 0]+[0 0 1] = [1 1 1], where 1+1+1=(-1) (mod 4).

For the utonal tetrad which is its inversion, we get

[0 0 0]+[-1 0 0]+[0 -1 0]+[0 0 -1] = [-1 -1 -1], and -1-1-1=1 (mod 4).

If we translate either of these tetrads by an arbitary [a b c] we end

up with the same result mod 4 (since we add four each of a, b, and

c.) Hence the otonal tetrads can be considered as a lattice with base

point [1 1 1], defined as [4a+1, 4b+1, 4c+1], or equivalently as

those triples [u v w] such that u+v+w=-1 (mod 4); the same goes for

the utonal lattice, with base point [-1 -1 -1] and u+v+w=1 (mod 4).

This generalizes easily to any p-limit, with the caveat that treating

all odd primes the same makes progressively less sense. I've been

intending to bring this up, partly because it allows us to do strange

things of the kind Robert seems fond of doing.

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

> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

>

> > Paul H., does this make any sense?

Gene, I have no idea what your response has to do with whether this:

"Conjecture: the bizarre, double-vision periodicity block I found

could even

happen in 3 dimensions, if there are 4 unison vectors defining

truncated-octahedron equivalence regions, but due to the

parallelopiped

basis of the periodicity block construction from three unison

vectors, these

truncated-octahedron regions could only come up two at a time."

makes any sense or not, but I'll try to follow what you wrote anyway.

>

> It certainly makes sense to look at the lattice-pair one gets by

> taking the lattice of utonal tetrads together with the lattice of

> otonal tetrads.

What does that mean?

> If we look at triples [a,b,c] with a quadradic form

> Q(a,b,c) = a^2+b^2+c62+ab+ac+bc, we have the symmetric lattice of 7-

> limit note-classes.

Can you give the "for musician dummies" version of this statement?

>

> The tetrads are defined as four lattice points, each of which is at

> unit distance from the other three. The tetrad centroids are simply

> the means of the lattice points, if we omit the division by 4, we

get

> for the 1-3-5-7 otonal tetrad

> [0 0 0]+[1 0 0]+[0 1 0]+[0 0 1] = [1 1 1], where 1+1+1=(-1) (mod

4).

> For the utonal tetrad which is its inversion, we get

> [0 0 0]+[-1 0 0]+[0 -1 0]+[0 0 -1] = [-1 -1 -1], and -1-1-1=1 (mod

4).

>

> If we translate either of these tetrads by an arbitary [a b c] we

end

> up with the same result mod 4 (since we add four each of a, b, and

> c.) Hence the otonal tetrads can be considered as a lattice with

base

> point [1 1 1], defined as [4a+1, 4b+1, 4c+1], or equivalently as

> those triples [u v w] such that u+v+w=-1 (mod 4); the same goes for

> the utonal lattice, with base point [-1 -1 -1] and u+v+w=1 (mod 4).

OK, I think I understand this, but what do tetrads have to do with

what I was talking about?

> This generalizes easily to any p-limit, with the caveat that

treating

> all odd primes the same makes progressively less sense.

Well you know I agree with you on that!

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Gene, I have no idea what your response has to do with whether this:

You left out the most relevant portion, which talked about rhombic

dodecahedra, the angles between 3/2, 5/4, and 7/4, and interlocking

lattices. That was all by Paul Hahn.

> > It certainly makes sense to look at the lattice-pair one gets by

> > taking the lattice of utonal tetrads together with the lattice of

> > otonal tetrads.

> What does that mean?

The tetrads form the dual honeycomb (or tesselation) to the lattice

of 7-limit note classes, but it consists of two interlocking lattices

of utonal and otonal tetrads. In the 5-limit, the dual tesselation to

the triangular one of note classes is the hexagonal tiling, and also

consists of two interlocking triangular lattices, one of major and

one of minor triads. This situation generalizes to the p-limit.

> > If we look at triples [a,b,c] with a quadradic form

> > Q(a,b,c) = a^2+b^2+c^2+ab+ac+bc, we have the symmetric lattice of

7-

> > limit note-classes.

> Can you give the "for musician dummies" version of this statement?

We can define a Euclidean metric on a real vector space in three

equivalent ways:

(1) A Euclidean metric; in three dimensions this might be

d(u, v) = sqrt((u1-v1)^2 + (u2-v2)^2 + (u3-v3)^2)

(2) A bilnear form, or dot product, for instance

B(u, v) = u.v = u1 v1 + u2 v2 + u3 v3

(3) A quadratic form, such as Q(u) = u1^2 + u2^2 + u3^2.

Each of the other two can be defined in terms of one of these, and in

particular if Q is a positive definite quadratic form (meaning it is

homogenous of the second order in the variables u1, u2, u3 in three

dimensions, etc., and Q(u)>=0, with Q(u)=0 iff u=0) then we can

define a corresponding bilinear form, or dot product, by

B(u,v) = (Q(u+v)-Q(u)-Q(v))/2

We can also define the metric, by

d(u,v) = sqrt(Q(u-v))

If you are going to draw the lattice diagrams you do, it could be

helpful to realize that the Euclidean geometric structure is defined

by the quadratic form. Hence for instance the distance between 4/3 and

15/8 is

d([-1 0], [1 1]) = sqrt(Q([-2 -1]) = sqrt(7),

and the cosine of the angle between the vector to 3/2 and the vector

to 7/5 can be determined using the appropriate dot product, which

would be

u.v = ((u1+v1)^2 + (u2+v2)^2 + (u3+v3)^2 + (u1+v1)(u2+v2)+(u1+v1)

(u3+v3) + (u2+v2)(u3+v3) - u1^2-u2^2-u3^2-u1u2-u1u3-u2u3-v1^2-v2^2-

v3^2-v1v2-v1v3-v2v3)/2 =

u1v1+u2v2+u3v3+(u1v2+u1v3+u2v1+u2v3+u3v1+u3v2)/2

We then have

cos t = u.v/(||u||*||v||) = 0/(1*1) = 0,

so that the classes defined by 3/2 and 7/5 are unit vectors at right

angles, just as you've been drawing them.

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

> --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

>

> > Gene, I have no idea what your response has to do with whether

this:

>

> You left out the most relevant portion, which talked about rhombic

> dodecahedra, the angles between 3/2, 5/4, and 7/4, and interlocking

> lattices. That was all by Paul Hahn.

I think you're misunderstanding the context of that, as well as

misunderstanding that the original "does this make sense" was not

about any thing Paul Hahn wrote, but only about the other part.

>

> > > It certainly makes sense to look at the lattice-pair one gets

by

> > > taking the lattice of utonal tetrads together with the lattice

of

> > > otonal tetrads.

>

> > What does that mean?

>

> The tetrads form the dual honeycomb (or tesselation) to the lattice

> of 7-limit note classes, but it consists of two interlocking

lattices

> of utonal and otonal tetrads. In the 5-limit, the dual tesselation

to

> the triangular one of note classes is the hexagonal tiling, and

also

> consists of two interlocking triangular lattices, one of major and

> one of minor triads. This situation generalizes to the p-limit.

>

> > > If we look at triples [a,b,c] with a quadradic form

> > > Q(a,b,c) = a^2+b^2+c^2+ab+ac+bc, we have the symmetric lattice

of

> 7-

> > > limit note-classes.

>

> > Can you give the "for musician dummies" version of this statement?

>

> We can define a Euclidean metric on a real vector space in three

> equivalent ways:

>

> (1) A Euclidean metric; in three dimensions this might be

>

> d(u, v) = sqrt((u1-v1)^2 + (u2-v2)^2 + (u3-v3)^2)

>

> (2) A bilnear form, or dot product, for instance

>

> B(u, v) = u.v = u1 v1 + u2 v2 + u3 v3

>

> (3) A quadratic form, such as Q(u) = u1^2 + u2^2 + u3^2.

>

> Each of the other two can be defined in terms of one of these, and

in

> particular if Q is a positive definite quadratic form (meaning it

is

> homogenous of the second order in the variables u1, u2, u3 in three

> dimensions, etc., and Q(u)>=0, with Q(u)=0 iff u=0) then we can

> define a corresponding bilinear form, or dot product, by

>

> B(u,v) = (Q(u+v)-Q(u)-Q(v))/2

>

> We can also define the metric, by

>

> d(u,v) = sqrt(Q(u-v))

>

> If you are going to draw the lattice diagrams you do, it could be

> helpful to realize that the Euclidean geometric structure is

defined

> by the quadratic form. Hence for instance the distance between 4/3

and

> 15/8 is

>

> d([-1 0], [1 1]) = sqrt(Q([-2 -1]) = sqrt(7),

>

> and the cosine of the angle between the vector to 3/2 and the

vector

> to 7/5 can be determined using the appropriate dot product, which

> would be

>

> u.v = ((u1+v1)^2 + (u2+v2)^2 + (u3+v3)^2 + (u1+v1)(u2+v2)+(u1+v1)

> (u3+v3) + (u2+v2)(u3+v3) - u1^2-u2^2-u3^2-u1u2-u1u3-u2u3-v1^2-v2^2-

> v3^2-v1v2-v1v3-v2v3)/2 =

> u1v1+u2v2+u3v3+(u1v2+u1v3+u2v1+u2v3+u3v1+u3v2)/2

>

> We then have

>

> cos t = u.v/(||u||*||v||) = 0/(1*1) = 0,

>

> so that the classes defined by 3/2 and 7/5 are unit vectors at

right

> angles, just as you've been drawing them.

I may be coming back to this soon for other purposes, but I don't see

the connection any of it has to the truncated octahedron question.