More from 4/21/00: does this make sense?

--- 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.