hey gene

What's this:

># h2 scale blocks
>
>cm1 := [9/7, 6/5, 8/7];
>c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]];
>s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];

And did anything ever become of this:

http://www.freelists.org/archives/tuning-math/07-2002/msg00054.html

Oh, and I don't see anything explaining TM reduction on
your website, or anywhere else for that matter.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Oh, and I don't see anything explaining TM reduction on
> your website, or anywhere else for that matter.
>
> -Carl

it's been explained repeatedly here. t stands for tenney, m for
minkowski.

>> Oh, and I don't see anything explaining TM reduction on
>> your website, or anywhere else for that matter.
>>
>> -Carl
>
>it's been explained repeatedly here. t stands for tenney, m for
>minkowski.

Yes, Paul, I know that. And I've just re-read the threads
all through the LLL -> TM stuff, but I still haven't found any
of the "repeated" explanations.

-Carl

hey Carl (and Gene, even moreso),

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Oh, and I don't see anything explaining TM reduction on
> your website, or anywhere else for that matter.
>
> -Carl

i asked Gene for a good definition for TM reduction a
long time ago ... and Gene, if you gave it to me and i
lost it in the shuffle, i apologize. can you send it again?

-monz

>i asked Gene for a good definition for TM reduction a
>long time ago ... and Gene, if you gave it to me and i
>lost it in the shuffle, i apologize. can you send it again?

Here's a definion of Minkowski reduction:

http://www.farcaster.com/papers/sm-thesis/node6.html

Manuel

>>i asked Gene for a good definition for TM reduction a
>>long time ago ... and Gene, if you gave it to me and i
>>lost it in the shuffle, i apologize. can you send it again?
>
>Here's a definion of Minkowski reduction:
>
>http://www.farcaster.com/papers/sm-thesis/node6.html
>
>Manuel

I understand roughly what TM reduction is, and I even found this:

>>>In either case, could you please explain the mathematical
>>>criterion that defines "Minkowski reduced", as you did for LLL?
>>
>>Let p/q be reduced to lowest terms; then T(p/q) = pq. A pair of
>>intervals {p/q, r/s} with p/q>1, r/s>1, T(p/q) < T(r/s) and p/q
>>and r/s independent is Minkowski reduced iff the only numbers in
>>the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s) are powers
>>of p/q.
>
>That's astoundingly simple! Wouldn't it be quite reasonable
>to further require that the only ratio t/u in the set
>{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
>idea would be that otherwise, the two unison vectors are
>"mismatched".

Aside from the fact that I don't know what elements are in a
set notated like {(p/q)^i (r/s)^i}, and I can't fathom the
function of t/u in this definition, my question was more along
the lines of...

What goes in? Ratios? Vals?

What comes out? A list of commas that define a PB?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >>i asked Gene for a good definition for TM reduction a
> >>long time ago ... and Gene, if you gave it to me and i
> >>lost it in the shuffle, i apologize. can you send it again?
> >
> >Here's a definion of Minkowski reduction:
> >
> >http://www.farcaster.com/papers/sm-thesis/node6.html
> >
> >Manuel
>
> I understand roughly what TM reduction is, and I even found this:
>
> >>>In either case, could you please explain the mathematical
> >>>criterion that defines "Minkowski reduced", as you did for LLL?
> >>
> >>Let p/q be reduced to lowest terms; then T(p/q) = pq. A pair of
> >>intervals {p/q, r/s} with p/q>1, r/s>1, T(p/q) < T(r/s) and p/q
> >>and r/s independent is Minkowski reduced iff the only numbers in
> >>the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s) are powers
> >>of p/q.
> >
> >That's astoundingly simple! Wouldn't it be quite reasonable
> >to further require that the only ratio t/u in the set
> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
> >idea would be that otherwise, the two unison vectors are
> >"mismatched".
>
> Aside from the fact that I don't know what elements are in a
> set notated like {(p/q)^i (r/s)^i},

the last i should be a j. that's the set of ratios that can be
expressed as (p/q)^i *times* (r/s)^j.

> and I can't fathom the
> function of t/u in this definition,

the definition should have read, "the only numbers t and u in the
set . . ." instead of "the only numbers in the set . . ."

> What goes in? Ratios? Vals?

ratios.

> What comes out? A list of commas that define a PB?

more generally, a set of ratios which are a basis for the same
lattice as the ratios you put in are a basis for, but which are, if
possible, simpler (in a certain sense) than the ones you put in.

apparently minkowski is not the only reasonable reduction definition,
so we could also have TKZ reduction, at least.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:

> the definition should have read, "the only numbers t and u in the
> set . . ." instead of "the only numbers in the set . . ."

let me try again -- it should have read "the only ratios t/u in the
set . . ."

there.

>> >>Let p/q be reduced to lowest terms; then T(p/q) = pq. A pair of
>> >>intervals {p/q, r/s} with p/q>1, r/s>1, T(p/q) < T(r/s) and p/q
>> >>and r/s independent is Minkowski reduced iff the only numbers in
>> >>the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s) are powers
>> >>of p/q.
>> >
>> >That's astoundingly simple! Wouldn't it be quite reasonable
>> >to further require that the only ratio t/u in the set
>> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
>> >idea would be that otherwise, the two unison vectors are
>> >"mismatched".
>>
>> Aside from the fact that I don't know what elements are in a
>> set notated like {(p/q)^i (r/s)^i},
>
>the last i should be a j.

Whoops, finger failure.

>that's the set of ratios that can be expressed
>as (p/q)^i *times* (r/s)^j.

I figured it was a multiply, but didn't realize this was what
it meant.

>> and I can't fathom the
>> function of t/u in this definition,
>
>the definition should have read, "the only numbers t and u in the
>set . . ." instead of "the only numbers in the set . . ."

Uh...

>> What goes in? Ratios? Vals?
>
>ratios.
>
>> What comes out? A list of commas that define a PB?
>
>more generally, a set of ratios which are a basis for the same
>lattice as the ratios you put in are a basis for, but which are, if
>possible, simpler (in a certain sense) than the ones you put in.

Ok.

So how can we get this into monz' dictionary?

>apparently minkowski is not the only reasonable reduction definition,
>so we could also have TKZ reduction, at least.

http://www.farcaster.com/papers/sm-thesis/node6.html

What's bit here...

>In the definition of Minkowski reduction, successive basis vectors
>b_i are added to the lattice basis only if b_i is the shortest vector
>in the lattice which will allow the basis to be extended. In
>Korkin-Zolotarev reduction, though, successive basis vectors b_i are
>chosen based on their length in the orthogonal complement of the space
>spanned by the previous basis vectors b_1, ..., b_(i-1).

I wonder if this has anything to do with straightness.

Oh, and I don't follow this...

>> >That's astoundingly simple! Wouldn't it be quite reasonable
>> >to further require that the only ratio t/u in the set
>> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
>> >idea would be that otherwise, the two unison vectors are
>> >"mismatched".

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> I wonder if this has anything to do with straightness.

yes, a reduced basis will have good straightness, because the set of
basis vectors is, in some sense, as short as possible. and, as we
discussed before, shortness implies straightness. the "block" always
has the same "area", so if the vectors are close to parallel, they'll
have to be long to compensate. remember that whole confusing
discussion?

> Oh, and I don't follow this...
>
> >> >That's astoundingly simple! Wouldn't it be quite reasonable
> >> >to further require that the only ratio t/u in the set
> >> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
> >> >idea would be that otherwise, the two unison vectors are
> >> >"mismatched".
>
> -Carl

that would mean that otherwise, r/s is much longer than p/q, and the
basis itself is an odd one to choose because it necessarily involves
two unison vectors of such different proportions.

>> the definition should have read, "the only numbers t and u in the
>> set . . ." instead of "the only numbers in the set . . ."
>
>let me try again -- it should have read "the only ratios t/u in the
>set . . ."
>
>there.

Now it makes sense...

> >>Let p/q be reduced to lowest terms; then T(p/q) = pq. A pair of
> >>intervals {p/q, r/s} with p/q>1, r/s>1, T(p/q) < T(r/s) and p/q
> >>and r/s independent is Minkowski reduced iff the only ratios t/u
> >>in the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s) are powers
> >>of p/q.

So IOW, if you have a pair of unison vectors for a PB, you shouldn't
be able to stack them both in some way to get an interval that's
simpler than the more complex of the pair is by itself.

That is simple.

How does this work when there are more than 2 vectors in the basis?

> >That's astoundingly simple! Wouldn't it be quite reasonable
> >to further require that the only ratio t/u in the set
> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
> >idea would be that otherwise, the two unison vectors are
> >"mismatched".

So you want to rule out cases where one uv is way longer than
the other?

-Carl

>> I wonder if this has anything to do with straightness.
>
>yes, a reduced basis will have good straightness, because the set of
>basis vectors is, in some sense, as short as possible. and, as we
>discussed before, shortness implies straightness. the "block" always
>has the same "area", so if the vectors are close to parallel, they'll
>have to be long to compensate. remember that whole confusing
>discussion?

Yes, but I meant does the difference between KZ and M have to do
with straightness?

-Carl

>> >That's astoundingly simple! Wouldn't it be quite reasonable
>> >to further require that the only ratio t/u in the set
>> >{(p/q)^i (r/s)^j} such that T(t/u) < T(r/s), is p/q itself? The
>> >idea would be that otherwise, the two unison vectors are
>> >"mismatched".
>
>So you want to rule out cases where one uv is way longer than
>the other?

I see you've just answered affirmatively.

-C.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> I wonder if this has anything to do with straightness.
> >
> >yes, a reduced basis will have good straightness, because the set
of
> >basis vectors is, in some sense, as short as possible. and, as we
> >discussed before, shortness implies straightness. the "block"
always
> >has the same "area", so if the vectors are close to parallel,
they'll
> >have to be long to compensate. remember that whole confusing
> >discussion?
>
> Yes, but I meant does the difference between KZ and M have to do
> with straightness?
>
> -Carl

dunno, haven't thought about it . . .

>> Yes, but I meant does the difference between KZ and M have to do
>> with straightness?
>>
>> -Carl
>
>dunno, haven't thought about it . . .

In case you missed it, it's this bit here, which I don't quite
follow...

>In the definition of Minkowski reduction, successive basis vectors
>b_i are added to the lattice basis only if b_i is the shortest vector
>in the lattice which will allow the basis to be extended. In
>Korkin-Zolotarev reduction, though, successive basis vectors b_i are
>chosen based on their length in the orthogonal complement of the space
>spanned by the previous basis vectors b_1, ..., b_(i-1).

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> Yes, but I meant does the difference between KZ and M have to do
> >> with straightness?
> >>
> >> -Carl
> >
> >dunno, haven't thought about it . . .
>
> In case you missed it, it's this bit here, which I don't quite
> follow...
>
> >In the definition of Minkowski reduction, successive basis vectors
> >b_i are added to the lattice basis only if b_i is the shortest
vector
> >in the lattice which will allow the basis to be extended. In
> >Korkin-Zolotarev reduction, though, successive basis vectors b_i
are
> >chosen based on their length in the orthogonal complement of the
space
> >spanned by the previous basis vectors b_1, ..., b_(i-1).
>
> -Carl

i'm confused about that, because wouldn't b_2, b_2 + b_1, b_2 +
2*b_1, b_2 - b+1, etc., all have the same length in the orthogonal
complement of the space spanned by b_1?

>i'm confused about that, because wouldn't b_2, b_2 + b_1, b_2 +
>2*b_1, b_2 - b+1, etc., all have the same length in the orthogonal
>complement of the space spanned by b_1?

I've heard of orthogonal and complement, but never
"orthogonal complement".

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >i'm confused about that, because wouldn't b_2, b_2 + b_1, b_2 +
> >2*b_1, b_2 - b+1, etc., all have the same length in the orthogonal
> >complement of the space spanned by b_1?
>
> I've heard of orthogonal and complement, but never
> "orthogonal complement".
>
> -Carl

you can always look it up!

http://mathworld.wolfram.com/OrthogonalComplement.html

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> What's this:
>
> ># h2 scale blocks
> >
> >cm1 := [9/7, 6/5, 8/7];
> >c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]];
> >s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];

These are what I called "chord blocks", which are 7-limit scales
analogous to Fokker blocks. This works because 7-limit tetrads,
uniquely among prime limits, form a lattice. The resulting scales
have the nice property of having a lot of chords to work with.

> And did anything ever become of this:
>
> http://www.freelists.org/archives/tuning-math/07-2002/msg00054.html

This is it, I think.

>
> Oh, and I don't see anything explaining TM reduction on
> your website, or anywhere else for that matter.

Maybe it's time to add more stuff there.

--- In tuning-math@yahoogroups.com, "monz" <monz@a...> wrote:

> i asked Gene for a good definition for TM reduction a
> long time ago ... and Gene, if you gave it to me and i
> lost it in the shuffle, i apologize. can you send it again?

First we need to define Tenney height: if p/q is a positive rational
number in reduced form, then the Tenney height is TH(p/q) = p q.

Now suppose {q1, ..., qn} are n multiplicatively linearly independent
positive rational numbers. Linear independence can be equated, for
instance, with the condition that rank of the matrix whose rows are
the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice
L, consisting of every positive rational number of the form q1^e1 ...
qn^en where the ei are integers and where the log of the Tenney
height defines a norm. Let t1>1 be the shortest (in terms of Tenney
height) rational number in L greater than 1. Define ti>1 inductively
as the shortest number in L independent of {t1, ... t_{i-1}} and such
that {t1, ..., ti} can be extended to be a basis for L. In this way
we obtain {t1, ..., tn}, the TM reduced basis of L.

what's the point of defining tenney height as p*q if you're only
going to use the log anyway, and tenney harmonic distance is already
log(p*q)?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "monz" <monz@a...> wrote:
>
> > i asked Gene for a good definition for TM reduction a
> > long time ago ... and Gene, if you gave it to me and i
> > lost it in the shuffle, i apologize. can you send it again?
>
> First we need to define Tenney height: if p/q is a positive
rational
> number in reduced form, then the Tenney height is TH(p/q) = p q.
>
> Now suppose {q1, ..., qn} are n multiplicatively linearly
independent
> positive rational numbers. Linear independence can be equated, for
> instance, with the condition that rank of the matrix whose rows are
> the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice
> L, consisting of every positive rational number of the form
q1^e1 ...
> qn^en where the ei are integers and where the log of the Tenney
> height defines a norm. Let t1>1 be the shortest (in terms of Tenney
> height) rational number in L greater than 1. Define ti>1
inductively
> as the shortest number in L independent of {t1, ... t_{i-1}} and
such
> that {t1, ..., ti} can be extended to be a basis for L. In this way
> we obtain {t1, ..., tn}, the TM reduced basis of L.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Yes, but I meant does the difference between KZ and M have to do
> with straightness?

One difference is that KZ requires a Euclidean norm, which Tenney
doesn't give us.

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:

> what's the point of defining tenney height as p*q if you're only
> going to use the log anyway, and tenney harmonic distance is
already
> log(p*q)?

It's simpler--it doesn't involve any transcendental functions. Aside
from that logs do have advantages, because you get a norm.

>> What's this:
>>
>> ># h2 scale blocks
>> >
>> >cm1 := [9/7, 6/5, 8/7];
>> >c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]];
>> >s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];

What are cm1, c1, and s1?

>These are what I called "chord blocks", which are 7-limit scales
>analogous to Fokker blocks. This works because 7-limit tetrads,
>uniquely among prime limits, form a lattice. The resulting scales
>have the nice property of having a lot of chords to work with.

I remember this stuff. But I don't remember the bit about the
7-limit being unique in this. What is it that makes, say, the
5-limit triads not form a "lattice"?

-Carl

hi paul and Carl,

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:

> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> > I wonder if this has anything to do with straightness.
>
> yes, a reduced basis will have good straightness, because
> the set of basis vectors is, in some sense, as short as
> possible. and, as we discussed before, shortness implies
> straightness. the "block" always has the same "area", so
> if the vectors are close to parallel, they'll have to be
> long to compensate. remember that whole confusing
> discussion?
>
> > Oh, and I don't follow this...
> >
> > > That's astoundingly simple! Wouldn't it be quite
> > > reasonable to further require that the only ratio t/u
> > > in the set {(p/q)^i (r/s)^j} such that T(t/u) < T(r/s),
> > > is p/q itself? The idea would be that otherwise, the
> > > two unison vectors are "mismatched".
> >
> > -Carl
>
> that would mean that otherwise, r/s is much longer than p/q,
> and the basis itself is an odd one to choose because it
> necessarily involves two unison vectors of such different
> proportions.

i understand this, in a nutshell, to mean that the reduction
process places the bounding vectors of the periodicity-block
as close as possible to the center/origin, and as short as
possible, thus ensuring that the entire block is compacted
as much as possible towards the center/origin.

yes?

-monz

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >> What's this:
> >>
> >> ># h2 scale blocks
> >> >
> >> >cm1 := [9/7, 6/5, 8/7];
> >> >c1 := [[-1, 0, 0], [-1, 0, 1], [-1, 1, 1], [0, 0, 0]];
> >> >s1 := [1, 15/14, 6/5, 5/4, 9/7, 3/2, 12/7, 7/4];
>
> What are cm1, c1, and s1?

s1 is the scale. If I recall correctly, cm1 is the comma basis, and
c1 is something obtained from cm1 and used to calculate s1.

> I remember this stuff. But I don't remember the bit about the
> 7-limit being unique in this. What is it that makes, say, the
> 5-limit triads not form a "lattice"?

5-limit triads can be represented as the vertices of a hexagonal
tiling; this isn't a lattice in the sense of the word I use since it
isn't a group. If you extend to the group it generates, you get a
triangular lattice by adding the centers of the hexagons. These
centers don't represent major or minor triads, but the unique note in
common to all of the chords of a hexagon. We can decree that they
represent something--for instance the augmented triad q--5/4 q--8/5 q
where "q" is the central note. However, this is not symmetrical and
rather artificial.

>5-limit triads can be represented as the vertices of a hexagonal
>tiling; this isn't a lattice in the sense of the word I use since it
>isn't a group.

Why isn't it a group?

-Carl

thanks, Gene!!!!

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "monz" <monz@a...> wrote:
>
> > i asked Gene for a good definition for TM reduction a
> > long time ago ... and Gene, if you gave it to me and i
> > lost it in the shuffle, i apologize. can you send it again?
>
> First we need to define Tenney height: if p/q is a positive
rational
> number in reduced form, then the Tenney height is TH(p/q) = p q.
>
> Now suppose {q1, ..., qn} are n multiplicatively linearly
independent
> positive rational numbers. Linear independence can be equated, for
> instance, with the condition that rank of the matrix whose rows are
> the monzos for qi is n. Then {q1, ..., qn} is a basis for a lattice
> L, consisting of every positive rational number of the form
q1^e1 ...
> qn^en where the ei are integers and where the log of the Tenney
> height defines a norm. Let t1>1 be the shortest (in terms of Tenney
> height) rational number in L greater than 1. Define ti>1
inductively
> as the shortest number in L independent of {t1, ... t_{i-1}} and
such
> that {t1, ..., ti} can be extended to be a basis for L. In this way
> we obtain {t1, ..., tn}, the TM reduced basis of L.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> >5-limit triads can be represented as the vertices of a hexagonal
> >tiling; this isn't a lattice in the sense of the word I use since
it
> >isn't a group.
>
> Why isn't it a group?

One step takes a C minor chord to a C major chord. Where does the
next step go? It doesn't go to a chord at all--we don't have a group,
since we don't have closure under addition.

>>>5-limit triads can be represented as the vertices of a hexagonal
>>>tiling; this isn't a lattice in the sense of the word I use since
>>>it isn't a group.
>>
>> Why isn't it a group?
>
>One step takes a C minor chord to a C major chord. Where does the
>next step go? It doesn't go to a chord at all--we don't have a group,
>since we don't have closure under addition.

Doesn't go to a chord? Aren't you connecting the centers of the
triangles? Then I get Cm->CM->C#m. If you connect the roots, or
any of the vertices, I get Cm->CM-Am. In fact, since there's so
much symmetry in this thing, I can't imagine ending up anywhere
other than on the corresponding part of some chord.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Doesn't go to a chord? Aren't you connecting the centers of the
> triangles? Then I get Cm->CM->C#m. If you connect the roots, or
> any of the vertices, I get Cm->CM-Am. In fact, since there's so
> much symmetry in this thing, I can't imagine ending up anywhere
> other than on the corresponding part of some chord.

I'm connecting the centers of the triangles with a line whenever
there is a common line between two of the triangles. This gives
hexagons, where you have a line from Cm to CM, and lines *in
different directions* from CM to Am and CM to Em. If you head in the
*same* direction, you end up in the center of a hexagon, which does
not correspond to either a major or a minor triad.

I could do this algebraically instead if it would help.

>> Doesn't go to a chord? Aren't you connecting the centers of the
>> triangles? Then I get Cm->CM->C#m. If you connect the roots, or
>> any of the vertices, I get Cm->CM-Am. In fact, since there's so
>> much symmetry in this thing, I can't imagine ending up anywhere
>> other than on the corresponding part of some chord.
>
>I'm connecting the centers of the triangles with a line whenever
>there is a common line between two of the triangles.

So CM and C#m aren't connected then? Why wouldn't you connect
them?

>This gives hexagons,

So does connecting the centers of all the triangles.

>where you have a line from Cm to CM, and lines *in
>different directions* from CM to Am and CM to Em. If you head in the
>*same* direction, you end up in the center of a hexagon, which does
>not correspond to either a major or a minor triad.

If you continue in the same direction and distance as from Cm -> CM,
you wind up at C#m. Is there some sort of reasoning behind not
including it because it only shares one pitch (instead of two) with
CM?

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> I remember this stuff. But I don't remember the bit about the
> 7-limit being unique in this. What is it that makes, say, the
> 5-limit triads not form a "lattice"?
>
> -Carl

the major triads do, and the minor triads do, but if you want 1 point
for each otonal *or* utonal chord, you only get a lattice in the 3-d
case. in 2-d (usually 5-limit), you would get an array corresponding
to the vertices of a hexagonal tiling -- not a lattice.

--- In tuning-math@yahoogroups.com, "monz" <monz@a...> wrote:

> i understand this, in a nutshell, to mean that the reduction
> process places the bounding vectors of the periodicity-block
> as close as possible to the center/origin, and as short as
> possible, thus ensuring that the entire block is compacted
> as much as possible towards the center/origin.
>
> yes?

if you choose to use the center/origin as one of the vertices of your
fokker periodicity block, then yes.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
> >>>5-limit triads can be represented as the vertices of a hexagonal
> >>>tiling; this isn't a lattice in the sense of the word I use
since
> >>>it isn't a group.
> >>
> >> Why isn't it a group?
> >
> >One step takes a C minor chord to a C major chord. Where does the
> >next step go? It doesn't go to a chord at all--we don't have a
group,
> >since we don't have closure under addition.
>
> Doesn't go to a chord? Aren't you connecting the centers of the
> triangles? Then I get Cm->CM->C#m.

no, you end up at the note E, because each step is supposed to be the
same distance.

> If you connect the roots, or
> any of the vertices, I get Cm->CM-Am.

that doesn't make sense. how do you see that?

>the major triads do, and the minor triads do, but if you want 1
>point for each otonal *or* utonal chord

http://lumma.org/tuning/doh.png

I finally realized the length on the right is too much. Sorry
Gene.

-Carl

>>the major triads do, and the minor triads do, but if you want 1
>>point for each otonal *or* utonal chord
>
>http://lumma.org/tuning/doh.png
>
>I finally realized the length on the right is too much. Sorry
>Gene.

Just one more question. Can you prove that the 7-limit is the
only one that works? What a strange thing.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> So CM and C#m aren't connected then? Why wouldn't you connect
> them?

They don't have a common interval. You can get to C#m by a common-
interval path, of course--CM-Am-C#m

> >This gives hexagons,
>
> So does connecting the centers of all the triangles.

The two are equivalent.

> >where you have a line from Cm to CM, and lines *in
> >different directions* from CM to Am and CM to Em. If you head in
the
> >*same* direction, you end up in the center of a hexagon, which
does
> >not correspond to either a major or a minor triad.
>
> If you continue in the same direction and distance as from Cm -> CM,
> you wind up at C#m.

No you don't. You end up at the *note* E; C#m is twice as far away.

Is there some sort of reasoning behind not
> including it because it only shares one pitch (instead of two) with
> CM?

See above. If you like, you can join both notes and chords into an
equilateral triangular lattice, but it seems a little weird to do so.
However, it might be a way to construct scales.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Just one more question. Can you prove that the 7-limit is the
> only one that works? What a strange thing.

Yes, but it hardly matters, since higher limit versions make
progressively less sense. This sort of strange thing happens often in
math, where two infinite classes of thing have an isomorphism between
two of the things in each class. So, for example, the group
PSL2(7) ~ PGL3(2). Here we have the lattices An (triangles,
tetrahedra, etc) and the lattices Dn (which can be described as the
cubic lattice Zn colored checkerboard fashion, and then taking only
the red lattice points.) The note lattices are An, but in the 7-limit
case it happens that A3 ~ D3, and so the 7-limit note lattice is the
face-centered cubic lattice, where the centers of the tetrads give us
a Z3 (integer-coordinate cubic lattice.) Actually showing the 7-limit
case is unique is best done algebraically.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:
>
> > Just one more question. Can you prove that the 7-limit is the
> > only one that works? What a strange thing.
>
> Yes, but it hardly matters, since higher limit versions make
> progressively less sense.

The algebraic proof is simple, so I'm going to give it despite the
fact that it is both algebra and a proof.

Ignoring the metric issue, we can represent octave classes of p-limit
intervals by reduced monzos--tuples of prime exponents with the 2
ignored. If n is the number of odd primes in the p-limit, then these
are n-1-tuples. We can represent (otonal or utonal) simplex chords by
taking the sum of the vertices, which gives as an n-tuple such that
the sum of the coefficients is +-1 mod n. If we add [-1, ..., -1] to
all these tuples we move [1, ..., 1] to the origin, and now the sum
of coefficients is 0 or 2 mod n. Only if n is 2 or 4 does this reduce
to a group (0 or 2 mod 4 is just the even numbers, and of course 0
mod 2 means even), and since summing the coordinates is a
homomorphism, only the 3-limit and the 7-limit can possibly have
chords which form a group, and it is easily checked they both do.

>> Just one more question. Can you prove that the 7-limit is the
>> only one that works? What a strange thing.
>
>Yes, but it hardly matters, since higher limit versions make
>progressively less sense. This sort of strange thing happens often in
>math, where two infinite classes of thing have an isomorphism between
>two of the things in each class. So, for example, the group
>PSL2(7) ~ PGL3(2). Here we have the lattices An (triangles,
>tetrahedra, etc) and the lattices Dn (which can be described as the
>cubic lattice Zn colored checkerboard fashion, and then taking only
>the red lattice points.) The note lattices are An, but in the 7-limit
>case it happens that A3 ~ D3, and so the 7-limit note lattice is the
>face-centered cubic lattice, where the centers of the tetrads give us
>a Z3 (integer-coordinate cubic lattice.) Actually showing the 7-limit
>case is unique is best done algebraically.

Thanks Gene!!

-Carl