Group Direct Product

Complete change of pace here. Been playing around with "GAP" and
have verified something I thought may be true. Not a big deal, but
it kind of brings group theory together with the necklace work I was
doing (okay, that's group theory anyway!) What I verified is that you
can kind of "see" the transpositional structure of these group direct
products...

For example, starting easy, take C4 X C3.

C4: x_1^4 + x_2^2 + 2x_4
C3: x_1^3 + 2x_3

I won't expand this out, just remember x_2^4 would be (x^2 + 1)^4 etc.

Now, put them together to obtain

x_1^12 + x_2^6 + 2x_4^3 + 2x_3^4 + 2x_6^2 + 4x_12

These correspond to T0, T6, T3/T9, T4/T8, T2/T10, T1/T5/T7/T11 !
One can construct a table for each cardinality, and each tranpose.

Kind of neat. Using a similar argument:

D4: x_1^4 + 2x_2x_1^2 + 3x_2^2 + 2x_4
C3: x_1^3 + 2x_3

Obtain:

x_1^12 + 2x_2^3x_1^6 + 3x_2^6 + 2x_4^3 + 2x_3^4 + 4x_6x_3^2 + 6x_6^2
+ 4x_12

Let's focus on the second term. It is the product of x_1^3 which
fixes (1)(2)(3) in C3 and 2x_2x_1^2 which does (13)(2)(4) or (24)(1)
(3) in D4. Suffice it to say this represents T_6I and T_0I in Z12
when applied to D4 X C3. A table can also be constructed. Every term
can be broken into T0-T11 and T0I-T11I, 24 pieces. (with Tx occuring
in various places of course)

D4 X S3 gives us 48 parts. You can construct a table with 12
transposes as columns, and 4 rows corresponding to (identity),
Inverse, D4 itself, and S3 itself. Also interesting that D4 itself
is just the inverse of S3 in Z12.

I know I don't have this completely worked out, and that I am not
even mentioning generators, like I should. I should add that
in the two-row table for C3 X D4 you have (identity) and D4 as rows,
but you can consider D4 as (Inverse) anyway in this context.

For fun I might make tables for hexachords in C3 X C4, C3 X D4,
S3 X C4 and S3 X D4 and upload them to my files section.

Not a big deal, but I thought it was kind of fun that one can
see the transpositional structure in these group direct products

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Complete change of pace here. Been playing around with "GAP" and
> have verified something I thought may be true. Not a big deal, but
> it kind of brings group theory together with the necklace work I
was
> doing (okay, that's group theory anyway!) What I verified is that
you
> can kind of "see" the transpositional structure of these group
direct
> products...
>
> For example, starting easy, take C4 X C3.
>
> C4: x_1^4 + x_2^2 + 2x_4
> C3: x_1^3 + 2x_3
>
> I won't expand this out, just remember x_2^4 would be (x^2 + 1)^4
etc.
>
> Now, put them together to obtain
>
> x_1^12 + x_2^6 + 2x_4^3 + 2x_3^4 + 2x_6^2 + 4x_12
>
> These correspond to T0, T6, T3/T9, T4/T8, T2/T10, T1/T5/T7/T11 !
> One can construct a table for each cardinality, and each tranpose.
>
> Kind of neat. Using a similar argument:
>
> D4: x_1^4 + 2x_2x_1^2 + 3x_2^2 + 2x_4
> C3: x_1^3 + 2x_3
>
> Obtain:
>
> x_1^12 + 2x_2^3x_1^6 + 3x_2^6 + 2x_4^3 + 2x_3^4 + 4x_6x_3^2 +
6x_6^2
> + 4x_12
>
> Let's focus on the second term. It is the product of x_1^3 which
> fixes (1)(2)(3) in C3 and 2x_2x_1^2 which does (13)(2)(4) or (24)(1)
> (3) in D4. Suffice it to say this represents T_6I and T_0I in Z12
> when applied to D4 X C3. A table can also be constructed. Every term
> can be broken into T0-T11 and T0I-T11I, 24 pieces. (with Tx
occuring
> in various places of course)
>
> D4 X S3 gives us 48 parts. You can construct a table with 12
> transposes as columns, and 4 rows corresponding to (identity),
> Inverse, D4 itself, and S3 itself. Also interesting that D4 itself
> is just the inverse of S3 in Z12.
>
> I know I don't have this completely worked out, and that I am not
> even mentioning generators, like I should. I should add that
> in the two-row table for C3 X D4 you have (identity) and D4 as
rows,
> but you can consider D4 as (Inverse) anyway in this context.

Wrong! Brain cell out of place - Sorry. It's just D4.
>
> For fun I might make tables for hexachords in C3 X C4, C3 X D4,
> S3 X C4 and S3 X D4 and upload them to my files section.
>
> Not a big deal, but I thought it was kind of fun that one can
> see the transpositional structure in these group direct products
>
> Paul Hj
>

First off, here is what we know at the start;

An unknown set of frequencies exists . Each frequency
in the set is the product of two separate values
(tones if viewed as music) and each frequency
is presented as a decimal ratio (interval) of the two
values.

If the frequency set were viewed as being musical then
it is unknown which frequency is the unison interval.

Two frequencies from the set are known, for example;

1.833333333
4.333333333

If we were to view the frequency set as a scale, it is
unknown at the beginning what the musical interval
either of these are within the scale.

Now here is the frequency set to which the above two
actually belong;

0.076923077
0.142857143
0.2
0.25
0.294117647
0.333333333
0.368421053
0.4
0.428571429
0.454545455
1.833333333
1.909090909
2
2.111111111
2.25
2.428571429
2.666666667
3
3.5
4.333333333
6
11

We see that the lowest value is ; 0.076923077
the highest value is 11

Now I realize the nature of this question still
remains a bit "odd" and that there is likely more than
one answer that could address
it and perhaps offer a variety of other values other
than the ones above. But just for the sake of
brainstorming, how might we generate
the above values (using logarithms or other
approaches) to aquire the above frequency set from the
given intial two frequencies;

1.833333333
4.333333333

Could this also be done if we only knew the first
frequency of 1.833333333 ?

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--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Complete change of pace here. Been playing around with "GAP" and
> have verified something I thought may be true. Not a big deal, but
> it kind of brings group theory together with the necklace work I
was
> doing (okay, that's group theory anyway!) What I verified is that
you
> can kind of "see" the transpositional structure of these group
direct
> products...
>
> For example, starting easy, take C4 X C3.
>
> C4: x_1^4 + x_2^2 + 2x_4
> C3: x_1^3 + 2x_3
>
> I won't expand this out, just remember x_2^4 would be (x^2 + 1)^4
etc.
>
> Now, put them together to obtain
>
> x_1^12 + x_2^6 + 2x_4^3 + 2x_3^4 + 2x_6^2 + 4x_12
>
> These correspond to T0, T6, T3/T9, T4/T8, T2/T10, T1/T5/T7/T11 !
> One can construct a table for each cardinality, and each tranpose.
>
> Kind of neat. Using a similar argument:
>
> D4: x_1^4 + 2x_2x_1^2 + 3x_2^2 + 2x_4
> C3: x_1^3 + 2x_3
>
> Obtain:
>
> x_1^12 + 2x_2^3x_1^6 + 3x_2^6 + 2x_4^3 + 2x_3^4 + 4x_6x_3^2 +
6x_6^2
> + 4x_12
>
> Let's focus on the second term. It is the product of x_1^3 which
> fixes (1)(2)(3) in C3 and 2x_2x_1^2 which does (13)(2)(4) or (24)(1)
> (3) in D4. Suffice it to say this represents T_6I and T_0I in Z12
> when applied to D4 X C3. A table can also be constructed. Every term
> can be broken into T0-T11 and T0I-T11I, 24 pieces. (with Tx
occuring in various places of course)

* Actually, in this case the 24 pieces are just 12 transposes of C3
and 12 transposes of D4, (just regular ones, not TnI)

> D4 X S3 gives us 48 parts. You can construct a table with 12
> transposes as columns, and 4 rows corresponding to (identity),
> Inverse, D4 itself, and S3 itself. Also interesting that D4 itself
> is just the inverse of S3 in Z12.

* This is a little trivial, but for anyone who's interested here's
how it works:

S3 is (123), (231), (312), (132), (321) and (213)
D4 is (1234), (2341), (3412), (4123), (1432), (4321), (3214) and
(2143)

S3 has three "forward" permutations and three "backwards" ones
D4 has four of each.

Combining 3 and 4 forward permutations just give 12 plain
transpositions

Combining 3 forward and 4 backwards gives 12 "D4" transpositions
Combining 3 backwards and 4 forwards gives 12 "S3" transpositions
Combining 3 backwards and 4 backwards gives 12 Inverse transpositions

The values of each that map into themselves under each of these is
just their values in the polynomial. You can of course break
everything out by cardinality too. So a table can be constructed
for hexachords, for example, with values in these 48 cells. (Some
being zero, however, for all cardinalities together, I believe
the cells are always completed filled up)

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > Complete change of pace here. Been playing around with "GAP" and
> > have verified something I thought may be true. Not a big deal, but
> > it kind of brings group theory together with the necklace work I
> was
> > doing (okay, that's group theory anyway!) What I verified is that
> you
> > can kind of "see" the transpositional structure of these group
> direct
> > products...
> >
> > For example, starting easy, take C4 X C3.
> >
> > C4: x_1^4 + x_2^2 + 2x_4
> > C3: x_1^3 + 2x_3
> >
> > I won't expand this out, just remember x_2^4 would be (x^2 + 1)^4
> etc.
> >
> > Now, put them together to obtain
> >
> > x_1^12 + x_2^6 + 2x_4^3 + 2x_3^4 + 2x_6^2 + 4x_12
> >
> > These correspond to T0, T6, T3/T9, T4/T8, T2/T10, T1/T5/T7/T11 !
> > One can construct a table for each cardinality, and each tranpose.
> >
> > Kind of neat. Using a similar argument:
> >
> > D4: x_1^4 + 2x_2x_1^2 + 3x_2^2 + 2x_4
> > C3: x_1^3 + 2x_3
> >
> > Obtain:
> >
> > x_1^12 + 2x_2^3x_1^6 + 3x_2^6 + 2x_4^3 + 2x_3^4 + 4x_6x_3^2 +
> 6x_6^2
> > + 4x_12
> >
> > Let's focus on the second term. It is the product of x_1^3 which
> > fixes (1)(2)(3) in C3 and 2x_2x_1^2 which does (13)(2)(4) or (24)
(1)
> > (3) in D4. Suffice it to say this represents T_6I and T_0I in Z12
> > when applied to D4 X C3. A table can also be constructed. Every
term
> > can be broken into T0-T11 and T0I-T11I, 24 pieces. (with Tx
> occuring in various places of course)
>
> * Actually, in this case the 24 pieces are just 12 transposes of C3
> and 12 transposes of D4, (just regular ones, not TnI)
>
> > D4 X S3 gives us 48 parts. You can construct a table with 12
> > transposes as columns, and 4 rows corresponding to (identity),
> > Inverse, D4 itself, and S3 itself. Also interesting that D4
itself
> > is just the inverse of S3 in Z12.
>
> * This is a little trivial, but for anyone who's interested here's
> how it works:
>
> S3 is (123), (231), (312), (132), (321) and (213)
> D4 is (1234), (2341), (3412), (4123), (1432), (4321), (3214) and
> (2143)
>
> S3 has three "forward" permutations and three "backwards" ones
> D4 has four of each.
>
> Combining 3 and 4 forward permutations just give 12 plain
> transpositions
>
> Combining 3 forward and 4 backwards gives 12 "D4" transpositions
> Combining 3 backwards and 4 forwards gives 12 "S3" transpositions
> Combining 3 backwards and 4 backwards gives 12 Inverse
transpositions
>
> The values of each that map into themselves under each of these is
> just their values in the polynomial. You can of course break
> everything out by cardinality too. So a table can be constructed
> for hexachords, for example, with values in these 48 cells. (Some
> being zero, however, for all cardinalities together, I believe
> the cells are always completed filled up)
>
> Paul Hj

This is mostly for my own reference, to keep track of my work.
One can go to 96 cells, and include S2-complementability, D4 X S2(C),
S3 X S2(C), (Inverse) X S2(C). Only hexachords map this way
(complementability), even though complementability will collapse
the whole structure (septachords/pentachords for example). There is a
complicated way to find S2(C) hexachords (Gilbert and Riordan, 1961)
but combining S3 X D4 X S2(C) I have only been able to do by
inspection. (No formula yet). The point is, the grids are now
complete, either a master grid for all cardinalities or a special
one just for hexachords - which produces the 26 A - Z hexachords
of my system...

In summary: 12 transposes of each of the following:

C4 X C3 (Formula)
D4 (Formula)
S3 (Formula)
D12 (Inverse) (Formula)
S2-comp (Formula)
D4 X S2-comp (No formula yet)
S3 X S2-comp (No formula yet)
Reverse comp (D12 X S2-comp) (Actually have a formula for this one)

This makes 96 cells....

Paul Hj

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> > >
> > > Complete change of pace here. Been playing around with "GAP" and
> > > have verified something I thought may be true. Not a big deal,
but
> > > it kind of brings group theory together with the necklace work
I
> > was
> > > doing (okay, that's group theory anyway!) What I verified is
that
> > you
> > > can kind of "see" the transpositional structure of these group
> > direct
> > > products...
> > >
> > > For example, starting easy, take C4 X C3.
> > >
> > > C4: x_1^4 + x_2^2 + 2x_4
> > > C3: x_1^3 + 2x_3
> > >
> > > I won't expand this out, just remember x_2^4 would be (x^2 + 1)
^4
> > etc.
> > >
> > > Now, put them together to obtain
> > >
> > > x_1^12 + x_2^6 + 2x_4^3 + 2x_3^4 + 2x_6^2 + 4x_12
> > >
> > > These correspond to T0, T6, T3/T9, T4/T8, T2/T10, T1/T5/T7/T11 !
> > > One can construct a table for each cardinality, and each
tranpose.
> > >
> > > Kind of neat. Using a similar argument:
> > >
> > > D4: x_1^4 + 2x_2x_1^2 + 3x_2^2 + 2x_4
> > > C3: x_1^3 + 2x_3
> > >
> > > Obtain:
> > >
> > > x_1^12 + 2x_2^3x_1^6 + 3x_2^6 + 2x_4^3 + 2x_3^4 + 4x_6x_3^2 +
> > 6x_6^2
> > > + 4x_12
> > >
> > > Let's focus on the second term. It is the product of x_1^3 which
> > > fixes (1)(2)(3) in C3 and 2x_2x_1^2 which does (13)(2)(4) or
(24)
> (1)
> > > (3) in D4. Suffice it to say this represents T_6I and T_0I in
Z12
> > > when applied to D4 X C3. A table can also be constructed. Every
> term
> > > can be broken into T0-T11 and T0I-T11I, 24 pieces. (with Tx
> > occuring in various places of course)
> >
> > * Actually, in this case the 24 pieces are just 12 transposes of
C3
> > and 12 transposes of D4, (just regular ones, not TnI)
> >
> > > D4 X S3 gives us 48 parts. You can construct a table with 12
> > > transposes as columns, and 4 rows corresponding to (identity),
> > > Inverse, D4 itself, and S3 itself. Also interesting that D4
> itself
> > > is just the inverse of S3 in Z12.
> >
> > * This is a little trivial, but for anyone who's interested
here's
> > how it works:
> >
> > S3 is (123), (231), (312), (132), (321) and (213)
> > D4 is (1234), (2341), (3412), (4123), (1432), (4321), (3214) and
> > (2143)
> >
> > S3 has three "forward" permutations and three "backwards" ones
> > D4 has four of each.
> >
> > Combining 3 and 4 forward permutations just give 12 plain
> > transpositions
> >
> > Combining 3 forward and 4 backwards gives 12 "D4" transpositions
> > Combining 3 backwards and 4 forwards gives 12 "S3" transpositions
> > Combining 3 backwards and 4 backwards gives 12 Inverse
> transpositions
> >
> > The values of each that map into themselves under each of these
is
> > just their values in the polynomial. You can of course break
> > everything out by cardinality too. So a table can be constructed
> > for hexachords, for example, with values in these 48 cells. (Some
> > being zero, however, for all cardinalities together, I believe
> > the cells are always completed filled up)
> >
> > Paul Hj
>
> This is mostly for my own reference, to keep track of my work.
> One can go to 96 cells, and include S2-complementability, D4 X S2
(C),
> S3 X S2(C), (Inverse) X S2(C). Only hexachords map this way
> (complementability), even though complementability will collapse
> the whole structure (septachords/pentachords for example). There is
a
> complicated way to find S2(C) hexachords (Gilbert and Riordan,
1961)
> but combining S3 X D4 X S2(C) I have only been able to do by
> inspection. (No formula yet). The point is, the grids are now
> complete, either a master grid for all cardinalities or a special
> one just for hexachords - which produces the 26 A - Z hexachords
> of my system...
>
> In summary: 12 transposes of each of the following:
>
> C4 X C3 (Formula)
> D4 (Formula)
> S3 (Formula)
> D12 (Inverse) (Formula)
> S2-comp (Formula)
> D4 X S2-comp (No formula yet)
> S3 X S2-comp (No formula yet)
> Reverse comp (D12 X S2-comp) (Actually have a formula for this one)
>
> This makes 96 cells....
>
> Paul Hj

I promise this is my last post on this! Here we go, at least for
totals:

Full Grid: (352 + 100 + 84 + 96 + 8 + 12 + 20 + 32)/8 = 88
Hexachords only: (80 + 20 + 16 + 20 + 8 + 12 + 20 + 32)/8 = 26

I should really say C3 X D4 for D4 and S3 X C4 for S3 of course, in
the above list.