C4 X C3 decomposition

Gene,

If you have a second, can you explain the connection between
C4 X C3 and the comma basis (128/125, 648/625) using infinite
groups?

Also do you think there is any connection at all between this and
(3,4,5,6), (4,5,6,7,8) -> (12,15,16,18,20,21,24) and if so could
you expound upon such a connection.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Gene,
>
> If you have a second, can you explain the connection between
> C4 X C3 and the comma basis (128/125, 648/625) using infinite
> groups?

I'm not sure what you are asking. There are various things I could
say, from different points of view, related to this. Most obvious is
that with generators of [2, 5/4, 6/5] we get the whole 5-limit, and in
these terms 128/125 = 2 (5/4)^(-3) and 648/625 = 2^(-1) (6/5)^4, so
that in terms of octave classes (5/4) is of order three if 128/125 is
tempered out, and 6/5 is of order four if 648/625 is tempered out.
Tempering both out gives 12-et, where both order statements are true,
and hence C12 as C4 X C3.

> Also do you think there is any connection at all between this and
> (3,4,5,6), (4,5,6,7,8) -> (12,15,16,18,20,21,24) and if so could
> you expound upon such a connection.

I don't know what this is. Are these chords?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > Gene,
> >
> > If you have a second, can you explain the connection between
> > C4 X C3 and the comma basis (128/125, 648/625) using infinite
> > groups?
>
> I'm not sure what you are asking. There are various things I could
> say, from different points of view, related to this. Most obvious is
> that with generators of [2, 5/4, 6/5] we get the whole 5-limit, and
in
> these terms 128/125 = 2 (5/4)^(-3) and 648/625 = 2^(-1) (6/5)^4, so
> that in terms of octave classes (5/4) is of order three if 128/125
is
> tempered out, and 6/5 is of order four if 648/625 is tempered out.
> Tempering both out gives 12-et, where both order statements are
true,
> and hence C12 as C4 X C3.

Thanks. I take it is infinite because you can create an infinite
lattice...

> > Also do you think there is any connection at all between this and
> > (3,4,5,6), (4,5,6,7,8) -> (12,15,16,18,20,21,24) and if so could
> > you expound upon such a connection.
>
> I don't know what this is. Are these chords?
>

Yes. I'm trying to prove that C12 is inherently 7-limit, because
of the connection between arithmetic, geometric and harmonic means.
(Both in three parts and in four parts). Just like you can
generate all 12 tones with 3^(0) to 3^(3) and 5^(0) to 5^(2),
which transforms to 6/5 and 5/4, I propose that you can go to
three dimensions and use 3, 5 and 7 transformed to 6/5, 5/4 and 10/7
where (6/5)^2 -> 1/7*25 -> 10/7 by taking the 128/125 comma.

The structure would be C2 X C2 X C3, is that valid? C2 X C2 would
look like

1 6/5
10/7 12/7

Just like 648/625 is 128/125 * 81/80, You obtain (6/5)^2/(256/175)
=63/64, and also 128/125 * 36/35 for all of C2 X C2, and you
pick up 126/125 by taking (36/25)/(10/7) (Third term after tempering
out 128/125) so this is 63/64 * 128/125 which has a nice
correspondence with 648/625.

I won't pretend this has anything to do with 35
pentachords/hexachords/septachords...

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <genewardsmith@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> > >
> > > Gene,
> > >
> > > If you have a second, can you explain the connection between
> > > C4 X C3 and the comma basis (128/125, 648/625) using infinite
> > > groups?
> >
> > I'm not sure what you are asking. There are various things I could
> > say, from different points of view, related to this. Most obvious
is
> > that with generators of [2, 5/4, 6/5] we get the whole 5-limit,
and
> in
> > these terms 128/125 = 2 (5/4)^(-3) and 648/625 = 2^(-1) (6/5)^4,
so
> > that in terms of octave classes (5/4) is of order three if
128/125
> is
> > tempered out, and 6/5 is of order four if 648/625 is tempered out.
> > Tempering both out gives 12-et, where both order statements are
> true,
> > and hence C12 as C4 X C3.
>
> Thanks. I take it is infinite because you can create an infinite
> lattice...
>
> > > Also do you think there is any connection at all between this
and
> > > (3,4,5,6), (4,5,6,7,8) -> (12,15,16,18,20,21,24) and if so could
> > > you expound upon such a connection.
> >
> > I don't know what this is. Are these chords?
> >
>
> Yes. I'm trying to prove that C12 is inherently 7-limit, because
> of the connection between arithmetic, geometric and harmonic means.
> (Both in three parts and in four parts). Just like you can
> generate all 12 tones with 3^(0) to 3^(3) and 5^(0) to 5^(2),
> which transforms to 6/5 and 5/4, I propose that you can go to
> three dimensions and use 3, 5 and 7 transformed to 6/5, 5/4 and 10/7
> where (6/5)^2 -> 1/7*25 -> 10/7 by taking the 128/125 comma.
>
> The structure would be C2 X C2 X C3, is that valid? C2 X C2 would
> look like

* This is not C2 X C2. I'm not sure what to call it. Does it
correspond with anything legitimate in mathematics? It is however
based on {(0,0),(0,1),(1,0),(1,1)}

>
> 1 6/5
> 10/7 12/7
>
> Just like 648/625 is 128/125 * 81/80, You obtain (6/5)^2/(256/175)
> =63/64, and also 128/125 * 36/35 for all of C2 X C2, and you
> pick up 126/125 by taking (36/25)/(10/7) (Third term after tempering
> out 128/125) so this is 63/64 * 128/125 which has a nice
> correspondence with 648/625.
>
> I won't pretend this has anything to do with 35
> pentachords/hexachords/septachords...
>

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

> > Just like you can
> > generate all 12 tones with 3^(0) to 3^(3) and 5^(0) to 5^(2),
> > which transforms to 6/5 and 5/4, I propose that you can go to
> > three dimensions and use 3, 5 and 7 transformed to 6/5, 5/4 and 10/7
> > where (6/5)^2 -> 1/7*25 -> 10/7 by taking the 128/125 comma.
> >
> > The structure would be C2 X C2 X C3, is that valid? C2 X C2 would
> > look like
>
> * This is not C2 X C2. I'm not sure what to call it. Does it
> correspond with anything legitimate in mathematics? It is however
> based on {(0,0),(0,1),(1,0),(1,1)}

What, exactly, is "this"? I'm having trouble finding your idea.
If you say (6/5)^2 ~ 7/5 then you are tempering out 36/35, not 128/125.
(6/5)^2 ~ 10/7 involves tempering out 126/125. All of these, of
course, are true in 12-et.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > > Just like you can
> > > generate all 12 tones with 3^(0) to 3^(3) and 5^(0) to 5^(2),
> > > which transforms to 6/5 and 5/4, I propose that you can go to
> > > three dimensions and use 3, 5 and 7 transformed to 6/5, 5/4 and
10/7
> > > where (6/5)^2 -> 1/7*25 -> 10/7 by taking the 128/125 comma.
> > >
> > > The structure would be C2 X C2 X C3, is that valid? C2 X C2
would
> > > look like
> >
> > * This is not C2 X C2. I'm not sure what to call it. Does it
> > correspond with anything legitimate in mathematics? It is however
> > based on {(0,0),(0,1),(1,0),(1,1)}
>
> What, exactly, is "this"? I'm having trouble finding your idea.
> If you say (6/5)^2 ~ 7/5 then you are tempering out 36/35, not
128/125.

> (6/5)^2 ~ 10/7 involves tempering out 126/125. All of these, of
> course, are true in 12-et.

The core of my idea is if you map 9/8 -> 8/7 (In the model based
on 3^n * 5^m) then you merely add 5 the denominator for the 3^n part
to go to (6/5)^n * (5/4)^m. 8/7 would therefore -> (256/7*25) or
256/175. Taking the diesis, we go to 10/7 (Not 7/5).

"This" is an attempt to split C4 into 2 pieces.

I thought "this" was C2 X C2 but I am wrong. My goal is to show
that the structure of C12 (Based on splitting up C4, whatever that is)
X C3 is 3-dimensional and corresponds to a 7-limit system.

I am posting to sci.math to see if I could find a mathematical
structure for how I am splitting up C4.

Apparently, I am using 2^{(2x+y)/4} which corresponds to the
set {0,1} X {0,1} and not a group-theoretical structure.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <genewardsmith@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> >
> > > > Just like you can
> > > > generate all 12 tones with 3^(0) to 3^(3) and 5^(0) to 5^(2),
> > > > which transforms to 6/5 and 5/4, I propose that you can go to
> > > > three dimensions and use 3, 5 and 7 transformed to 6/5, 5/4
and
> 10/7
> > > > where (6/5)^2 -> 1/7*25 -> 10/7 by taking the 128/125 comma.
> > > >
> > > > The structure would be C2 X C2 X C3, is that valid? C2 X C2
> would
> > > > look like
> > >
> > > * This is not C2 X C2. I'm not sure what to call it. Does it
> > > correspond with anything legitimate in mathematics? It is
however
> > > based on {(0,0),(0,1),(1,0),(1,1)}
> >
> > What, exactly, is "this"? I'm having trouble finding your idea.
> > If you say (6/5)^2 ~ 7/5 then you are tempering out 36/35, not
> 128/125.
>
> > (6/5)^2 ~ 10/7 involves tempering out 126/125. All of these, of
> > course, are true in 12-et.
>
> The core of my idea is if you map 9/8 -> 8/7 (In the model based
> on 3^n * 5^m) then you merely add 5 the denominator for the 3^n
part
> to go to (6/5)^n * (5/4)^m. 8/7 would therefore -> (256/7*25) or
> 256/175. Taking the diesis, we go to 10/7 (Not 7/5).
>
> "This" is an attempt to split C4 into 2 pieces.
>
> I thought "this" was C2 X C2 but I am wrong. My goal is to show
> that the structure of C12 (Based on splitting up C4, whatever that
is)
> X C3 is 3-dimensional and corresponds to a 7-limit system.
>
> I am posting to sci.math to see if I could find a mathematical
> structure for how I am splitting up C4.
>
> Apparently, I am using 2^{(2x+y)/4} which corresponds to the
> set {0,1} X {0,1} and not a group-theoretical structure.
>

Well, it might not correspond to group theory, but I am going to use
it. [(0,0),(0,1),(1,0),(1,1)] x [(1/2),(1/4)] generates [(0),(1/4),
(1/2),(3/4)]

Map 2^(1/2) to 10/7
Map 2^(1/4) to 6/5

Or you can map 2^(1/2) to 7/5, and temper out 36/35 on this step
(and 126/125 on the fourth step, instead of using 10/7, where you
temper out 126/125 on the second step and 36/35 on the fourth step
of C4)

Then you obtain a 2*2*3 shoebox