Yahoo Tuning Groups Ultimate Backup tuning-math Polya Polynomials

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> For Gene, or anyone else who knows how!
>
> I would like to start generating my own Polya Polynomials
> for temperaments other than 12-tET, for various group
> symmetries. I know how to do the Cyclic Group and the
> Dihedral Group. If you could show me how you did, for
> example, D4 X S3, or just give me a clue, I am sure I could
> extend this to other compositions and temperaments. I take it that
> is has to do with the permutations in, for example, D4 X S3, but
> I need to know how to set this up using Polya's method.
>

Well, lemme dig out my lectures...

A general method to get the Polya polynomial for an arbitraty
permutation group is to go through all of its elements and determine
its cycle structure - i.e. the orbits repeated actions of the
permutation on its basic set create. From the cycle structure of
each permutation, a term for the polynomial ist formed in the
following way.

As a simple example, S3 consists of the unit element (with 3 fix
points, i.e. cycles of size 1), 3 transpositions (with one fix
element and one 2-cycle) and two permutations with one 3-cycle. From
the unit element you get the term x1^3, from each transposition the
term x1 * x2 and from each 3-cycle transposition the term x3.
(The xn are just variables - there is one for every cycle size that
appears, and the number of cycles is the exponent.)

All the terms are added, and the whole thing is divided through the
number of elements of the group. For S3, the resulting polynomial is

1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )

Enumerations of many sorts can then be calculated by setting in
various values for the variables xn. E.g., to get the number of
subset patterns with a given number of elements, you replace every
xn by (1+x^n), calculate it out, and the wanted number for patterns
with n elements will be the coefficient of x^n.

All not basically difficult, but for big groups tedious - if done by
hand. But computer programs should do it quite easily - I have been
having in mind writing such a program for quite some time, but have
not found the time yet...
--
Hans Straub

🔗Gene Ward Smith <gwsmith@svpal.org>

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...> wrote:

> A general method to get the Polya polynomial for an arbitraty
> permutation group is to go through all of its elements and determine
> its cycle structure - i.e. the orbits repeated actions of the
> permutation on its basic set create. From the cycle structure of
> each permutation, a term for the polynomial ist formed in the
> following way.

All true, but I mentioned McKay because that had something you really
need also--namely, the names of the groups in question.

> As a simple example, S3 consists of the unit element (with 3 fix
> points, i.e. cycles of size 1), 3 transpositions (with one fix
> element and one 2-cycle) and two permutations with one 3-cycle.

That's what it is as a permutation on three elements, but we want
groups represented as permutation groups on 12 elements. D3 x S3 isn't
just an abstract group, but a particular representation of that group
as a unique up to conjugacy permutation group.

> All not basically difficult, but for big groups tedious - if done by
> hand. But computer programs should do it quite easily - I have been
> having in mind writing such a program for quite some time, but have
> not found the time yet...

There are already two programs--Cayley and GAP--which do that, and
both have huge libraries of group theoretic functions. GAP is
freeware, and can be found here:

http://www-gap.mcs.st-and.ac.uk/Download/index.html

🔗hstraub64 <hstraub64@telesonique.net>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> > As a simple example, S3 consists of the unit element (with 3 fix
> > points, i.e. cycles of size 1), 3 transpositions (with one fix
> > element and one 2-cycle) and two permutations with one 3-cycle.
>
> That's what it is as a permutation on three elements, but we want
> groups represented as permutation groups on 12 elements. D3 x S3
> isn't just an abstract group, but a particular representation of
> that group as a unique up to conjugacy permutation group.
>

That is indeed a key point - it is not (the isomorphy class of) the
group itself that matters, but the effects on the set it actsupon as
a permutation group. (I was using S3 just as an example hwo to get
to the formula.)

>
> There are already two programs--Cayley and GAP--which do that, and
> both have huge libraries of group theoretic functions. GAP is
> freeware, and can be found here:
>
> http://www-gap.mcs.st-and.ac.uk/Download/index.html
>

Thanks for the link! I have been looking for group theroy freeware
software for quite some time.

The question is always how quickly you get familiar with a software -
sometimes doing it by hand may be faster. I once tried to get into
Cayley and left it again, fot this reason.

BTW, what has become of Cayley nowadays? Haven't heard of it for 15
years...
--
Hans Straub

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
> >
> > For Gene, or anyone else who knows how!
> >
> > I would like to start generating my own Polya Polynomials
> > for temperaments other than 12-tET, for various group
> > symmetries. I know how to do the Cyclic Group and the
> > Dihedral Group. If you could show me how you did, for
> > example, D4 X S3, or just give me a clue, I am sure I could
> > extend this to other compositions and temperaments. I take it
that
> > is has to do with the permutations in, for example, D4 X S3, but
> > I need to know how to set this up using Polya's method.
> >
>
> Well, lemme dig out my lectures...
>
> A general method to get the Polya polynomial for an arbitraty
> permutation group is to go through all of its elements and
determine
> its cycle structure - i.e. the orbits repeated actions of the
> permutation on its basic set create. From the cycle structure of
> each permutation, a term for the polynomial ist formed in the
> following way.
>
> As a simple example, S3 consists of the unit element (with 3 fix
> points, i.e. cycles of size 1), 3 transpositions (with one fix
> element and one 2-cycle) and two permutations with one 3-cycle.
From
> the unit element you get the term x1^3, from each transposition the
> term x1 * x2 and from each 3-cycle transposition the term x3.
> (The xn are just variables - there is one for every cycle size that
> appears, and the number of cycles is the exponent.)
>
> All the terms are added, and the whole thing is divided through the
> number of elements of the group. For S3, the resulting polynomial is
>
> 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )
>
> Enumerations of many sorts can then be calculated by setting in
> various values for the variables xn. E.g., to get the number of
> subset patterns with a given number of elements, you replace every
> xn by (1+x^n), calculate it out, and the wanted number for patterns
> with n elements will be the coefficient of x^n.

So, you can convert the above formula to a formula in one variable?
(x). I don't see why xn becomes (1+x^n). Thanks! Does this formula
then "mix in" with the polynomial for transpostions (Cyclic group)
and inverses (dihedral group)?

Oh, BTW, I got "From Polychords to Polya" this weekend. Quite good,
some new formulas, including a general method for necklaces with
3 colors or more that I have been working with.

🔗hstraub64 <hstraub64@telesonique.net>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
> wrote:
> >

> > For S3, the resulting polynomial is
> >
> > 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )
> >
> > Enumerations of many sorts can then be calculated by setting in
> > various values for the variables xn. E.g., to get the number of
> > subset patterns with a given number of elements, you replace every
> > xn by (1+x^n), calculate it out, and the wanted number for
> > patterns with n elements will be the coefficient of x^n.
>
> So, you can convert the above formula to a formula in one variable?
> (x). I don't see why xn becomes (1+x^n).

(1+x^n) is just one special example of what to fill in.

Taking the example of the necklaces , the value (1+x^n) is for
measuring the number of necklacese with 2 colors (black and white,
e.g.), giving not only the total number but also the numbers of
necklaces with exactly n black beads.

For other measurements, you will use other values to replace. If you
want just the number, you put in the number of coloers into each
variable - but you can measure a lot of stuff, like the number with so
and so many elements of a certain color, the number of those with
exactly k colors and so on. It's a really amazing formula - just a
litlle difficult to understand. To derive what to fill in is a theory
of its own that I do not fully understand at the moment.

> Thanks! Does this formula
> then "mix in" with the polynomial for transpostions (Cyclic group)
> and inverses (dihedral group)?
>

I assume it does mix in some way - but AFAIK there is no
straightforward connection or formula - unfortunately! Every case must
be treated separately.
--
Hans Straub

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
> > wrote:
> > >
>
> > > For S3, the resulting polynomial is
> > >
> > > 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )

> > >
> > > Enumerations of many sorts can then be calculated by setting in
> > > various values for the variables xn. E.g., to get the number of
> > > subset patterns with a given number of elements, you replace
every
> > > xn by (1+x^n), calculate it out, and the wanted number for
> > > patterns with n elements will be the coefficient of x^n.
>

So for this one I get 1/6[(1+x^1)^3 + 3(1+x^1)(1+x^2)+2(1+x^3)] ?

And that's the group S3?

My Polya book has a number of neat formulas, I will have to bring it
in and post some of them at least.

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

It works out to 1 1 1 1 which is the correct expansion. Now if someone
could be super generous and give me the the first polynomial for D4 X
S3, I can convert it to the second polynomial with the help of Octave.

I would look at that other software, but don't really have time. Luckily
Octave is adequate for doing conv(z,z) expansions. Thanks all.

Oh BTW, I am calling the polynomial above the "first" polynomial. The
one expanding with (x^n + 1) I am calling the "second" polynomial.

In lighter news - not that it means anything in this context, but I
finally figured out how to do complementable necklaces
with three colors. I am only mentioning it because I have been stuck
for a couple days (using the formulas in the 1961 paper by Riordan
and Gilbert). I like to know why the formulas work, not just how to use
them. So although I am not doing proofs, I am at least validating them.

This is kind of funny - take necklaces with 4 beads and three colors.
Use red white and blue. There are 3 necklaces which are either all
red white or blue. Part of the expansion for complementability is
adding in (1^1+1^2+1^3+1^4) * 3, (complementable equivalences).
You need to divide the total by 4 transpositions and 6 color
combinations, you have to add 12 necklaces (regular) + 12 necklaces
(complementable equivalence) and divide by 24, so you get 1 necklace.
(The generic necklace, all one color). Yes, you obtain the equivalency
of an all-red necklace because you are swapping (the non-existent)
blue-white. I think this is right. It must break down to simple
two-way color swaps. Sound right? I can't believe I can get stuck
on something this easy. Feast or famine I guess.

Paul Hj

🔗hstraub64 <hstraub64@telesonique.net>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> > >
> > > > > For S3, the resulting polynomial is
> > > > >
> > > > > 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )
>
> It works out to 1 1 1 1 which is the correct expansion. Now if
> someone could be super generous and give me the the first
> polynomial for D4 X S3, I can convert it to the second polynomial
> with the help of Octave.
>
> I would look at that other software, but don't really have time.
> Luckily Octave is adequate for doing conv(z,z) expansions. Thanks
> all.
>
> Oh BTW, I am calling the polynomial above the "first" polynomial.

The official term is "cycle index polynomial" or "cycle index".

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

> The one expanding with (x^n + 1) I am calling the "second"
> polynomial.
>
> I like to know why the formulas work, not just how to use
> them.

Yes, that is my way, too. It is a quite fascinating theory, halfway
between group theory and combinatorics.

As for D4 X S3, I just tried that by hand (still affordable), and got

1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 + 4*x12 )

No warranty...
--
Hans Straub

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@a...> wrote:
> >
> > > >
> > > > > > For S3, the resulting polynomial is
> > > > > >
> > > > > > 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )
> >
> > It works out to 1 1 1 1 which is the correct expansion. Now if
> > someone could be super generous and give me the the first
> > polynomial for D4 X S3, I can convert it to the second polynomial
> > with the help of Octave.
> >
> > I would look at that other software, but don't really have time.
> > Luckily Octave is adequate for doing conv(z,z) expansions. Thanks
> > all.
> >
> > Oh BTW, I am calling the polynomial above the "first" polynomial.
>
> The official term is "cycle index polynomial" or "cycle index".
>
> http://mathworld.wolfram.com/CycleIndex.html
>
> > The one expanding with (x^n + 1) I am calling the "second"
> > polynomial.
> >
> > I like to know why the formulas work, not just how to use
> > them.
>
> Yes, that is my way, too. It is a quite fascinating theory, halfway
> between group theory and combinatorics.
>
> As for D4 X S3, I just tried that by hand (still affordable), and
got
>
> 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 + 4*x12 )
>
> No warranty...

Hey thanks. And I'll try to use the proper terminology in the future.
I know how to get D4, and you gave me S3. It will be fun to see what
one does with D4 X S3... once I take this apart. (I take it that you
gave me the cycle index, from which I need to create the Polya
Polynomial? Thanks - my job keeps me pretty busy so I can't spend
the time I would like on this.
> --
> Hans Straub
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@a...> wrote:
>
> --- In tuning-math@yahoogroups.com, "hstraub64" <hstraub64@t...>
> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@a...> wrote:
> > >
> > > > >
> > > > > > > For S3, the resulting polynomial is
> > > > > > >
> > > > > > > 1/6 * ( x1^3 + 3 * x1 * x2 + 2 * x3 )
> > >
> > > It works out to 1 1 1 1 which is the correct expansion. Now if
> > > someone could be super generous and give me the the first
> > > polynomial for D4 X S3, I can convert it to the second
polynomial
> > > with the help of Octave.
> > >
> > > I would look at that other software, but don't really have
time.
> > > Luckily Octave is adequate for doing conv(z,z) expansions.
Thanks
> > > all.
> > >
> > > Oh BTW, I am calling the polynomial above the "first"
polynomial.
> >
> > The official term is "cycle index polynomial" or "cycle index".
> >
> > http://mathworld.wolfram.com/CycleIndex.html
> >
> > > The one expanding with (x^n + 1) I am calling the "second"
> > > polynomial.
> > >
> > > I like to know why the formulas work, not just how to use
> > > them.
> >
> > Yes, that is my way, too. It is a quite fascinating theory,
halfway
> > between group theory and combinatorics.
> >
> > As for D4 X S3, I just tried that by hand (still affordable), and
> got
> >
> > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 + 4*x12 )
> >
> > No warranty...

So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the Cycle
Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do these
combine to give the above?

Paul Hj
> >
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

🔗David Bowen <dmb0317@gmail.com>

Paul,

My copy of the book arrived from Amazon yesterday. I got through a
couple chapters last night and already had some experience with Polya
polynomials from the discussion here. I don't know if there is an
easy formula for the cycle index of a product of groups given the
indexes of the products. I normally use GAP to create the whole group
from a small number of generators and have it give me the cycle
structures for all of the elements. D4 X S3 will be generated by (1
2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8 11)(3 6
9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get you
started on working out the whole group. I'll get back in touch when
I've finished the book.

David Bowen

On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:
> Han Straub wrote:
>
> > > > As for D4 X S3, I just tried that by hand (still affordable), and
> > > got
> > > >
> > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 + 4*x12 )
> > > >
> > > > No warranty...
> >
> > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the Cycle
> > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do these
> > combine to give the above?
>
> I can see its not a simple matter of just combining cycle indices for
> D4 and S3, could someone show me how the cycle index for D4 X S3 is
> extracted? Thanks!
>
> > Paul Hj
> > > >
> > >
> >
>
>
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, David Bowen <dmb0317@...> wrote:
>
> Paul,
>
> My copy of the book arrived from Amazon yesterday. I got through
a
> couple chapters last night and already had some experience with
Polya
> polynomials from the discussion here. I don't know if there is an
> easy formula for the cycle index of a product of groups given the
> indexes of the products. I normally use GAP to create the whole
group
> from a small number of generators and have it give me the cycle
> structures for all of the elements. D4 X S3 will be generated by (1
> 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8 11)(3
6
> 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get you
> started on working out the whole group. I'll get back in touch when
> I've finished the book.
>
> David Bowen

Great! I am actually just studying the generators for D4 X S3, and
trying to figure out how to extract the cycle index. What book did
you get - is it "From Polynomials to Polya"? or something else?

>
> On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@...> wrote:
> > Han Straub wrote:
> >
> > > > > As for D4 X S3, I just tried that by hand (still
affordable), and
> > > > got
> > > > >
> > > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 +
4*x12 )
> > > > >
> > > > > No warranty...
> > >
> > > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the
Cycle
> > > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do these
> > > combine to give the above?
> >
> > I can see its not a simple matter of just combining cycle indices
for
> > D4 and S3, could someone show me how the cycle index for D4 X S3
is
> > extracted? Thanks!
> >
> > > Paul Hj
> > > > >
> > > >
> > >
> >
> >
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
> >
>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗David Bowen <dmb0317@gmail.com>

Hi, Paul,

Yes, I got a copy of "Polychords to Polya". Here's my analysis of
your question with an attempt to provide a general answer for any
product of groups. Sorry if I bore you with things you already know.

If G and H are permutation groups of order g and h operating on
sets M and N of size m and n respectively, then G X H can be
considered a permutation group of order gh operating on the set M X N
of size mn. Let (a,b) \elem of M X N and p \elem G, q \elem H. Then (p
x q) (a,b) = (p(a), q(b)) and so if a is part of a k-cycle and b part
of an l-cycle then (a,b) will be part of an lcm(k,l)-cycle.

Now we can use this to work out the cycle structure of G X H given
the structures of G and H. For your particular case S_3 has cycle
index z_1^3 + 3z_1z_2 + 2z_3 and D4 has cycle index z_1^4 + 3z_2^2 +
2z_1^2z_2 + 2z_4. The z_1^3, z_1^4 combination will produce a z_1^12.
z_1^ and 3 z_2^2 will produce a 3z_2^6. z_1^3 and 2z_1^2z_2 produces
2z_1^6z_2^3 and z_1^3 and z4 produces z_4^3. For 3z_1z_2 we get
3z_1^4z_2^4, 9z_2^6, 6z_1^2z_2^5 and 6z_4^3. And finally, for 2z_3 we
2z_3^4, 6z_6^2, 4z_3^2z_6, and z_12. I've skipped through the details
and will happily answer questions about any of the calculations or
about how the calculations would go in any other direct products you
might be interested in. The important thing to remember is the lcm
rule, when 2-cycles match up with 2-cycles you just get 2-cycles not
4-cycles and when 2-cycles and 4-cycles match up you just get
4-cycles. If you haven't picked up a copy of GAP, think seriously
about doing so. I've been using it for about 15 years and it's great
for computational group theory.

Dave Bowen

On 2/17/06, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:
> --- In tuning-math@yahoogroups.com, David Bowen <dmb0317@...> wrote:
> >
> > Paul,
> >
> > My copy of the book arrived from Amazon yesterday. I got through
> a
> > couple chapters last night and already had some experience with
> Polya
> > polynomials from the discussion here. I don't know if there is an
> > easy formula for the cycle index of a product of groups given the
> > indexes of the products. I normally use GAP to create the whole
> group
> > from a small number of generators and have it give me the cycle
> > structures for all of the elements. D4 X S3 will be generated by (1
> > 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8 11)(3
> 6
> > 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get you
> > started on working out the whole group. I'll get back in touch when
> > I've finished the book.
> >
> > David Bowen
>
> Great! I am actually just studying the generators for D4 X S3, and
> trying to figure out how to extract the cycle index. What book did
> you get - is it "From Polynomials to Polya"? or something else?
>
>
> >
> > On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@...> wrote:
> > > Han Straub wrote:
> > >
> > > > > > As for D4 X S3, I just tried that by hand (still
> affordable), and
> > > > > got
> > > > > >
> > > > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 +
> 4*x12 )
> > > > > >
> > > > > > No warranty...
> > > >
> > > > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the
> Cycle
> > > > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do these
> > > > combine to give the above?
> > >
> > > I can see its not a simple matter of just combining cycle indices
> for
> > > D4 and S3, could someone show me how the cycle index for D4 X S3
> is
> > > extracted? Thanks!
> > >
> > > > Paul Hj
> > > > > >
> > > > >
> > > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Yahoo! Groups Links
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> >
>
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@...> wrote:
>
> Hi, Paul,
>
> Yes, I got a copy of "Polychords to Polya". Here's my analysis of
> your question with an attempt to provide a general answer for any
> product of groups. Sorry if I bore you with things you already know.
>
> If G and H are permutation groups of order g and h operating on
> sets M and N of size m and n respectively, then G X H can be
> considered a permutation group of order gh operating on the set M X
N
> of size mn. Let (a,b) \elem of M X N and p \elem G, q \elem H. Then
(p
> x q) (a,b) = (p(a), q(b)) and so if a is part of a k-cycle and b
part
> of an l-cycle then (a,b) will be part of an lcm(k,l)-cycle.
>
> Now we can use this to work out the cycle structure of G X H
given
> the structures of G and H. For your particular case S_3 has cycle
> index z_1^3 + 3z_1z_2 + 2z_3 and D4 has cycle index z_1^4 + 3z_2^2 +
> 2z_1^2z_2 + 2z_4. The z_1^3, z_1^4 combination will produce a
z_1^12.
> z_1^ and 3 z_2^2 will produce a 3z_2^6. z_1^3 and 2z_1^2z_2 produces
> 2z_1^6z_2^3 and z_1^3 and z4 produces z_4^3. For 3z_1z_2 we get
> 3z_1^4z_2^4, 9z_2^6, 6z_1^2z_2^5 and 6z_4^3. And finally, for 2z_3
we
> 2z_3^4, 6z_6^2, 4z_3^2z_6, and z_12. I've skipped through the
details
> and will happily answer questions about any of the calculations or
> about how the calculations would go in any other direct products you
> might be interested in. The important thing to remember is the lcm
> rule, when 2-cycles match up with 2-cycles you just get 2-cycles not
> 4-cycles and when 2-cycles and 4-cycles match up you just get
> 4-cycles. If you haven't picked up a copy of GAP, think seriously
> about doing so. I've been using it for about 15 years and it's great
> for computational group theory.
>
> Dave Bowen

Thanks, this is great. About the book - I sent Michael Keith an email
saying I enjoyed it. He sent me a reply. I sent a couple more emails
and I am waiting to hear back, regarding creating metanecklaces that
could count interval vectors (the full kind, not just open intervals).
I thought the use of metanecklaces was pretty cool. I have also
combined this book with a paper by Gilbert and Riordan, 1961, to
calculate complementability. (My other Polya post). I am a little
hung up on symmetrical necklaces, they don't pan out quite right
in terms of Tn. (tranposition by n). But the counts are still
right, in general, so I guess that is the point of the formulas.

>
> On 2/17/06, Paul G Hjelmstad <paul_hjelmstad@...> wrote:
> > --- In tuning-math@yahoogroups.com, David Bowen <dmb0317@> wrote:
> > >
> > > Paul,
> > >
> > > My copy of the book arrived from Amazon yesterday. I got
through
> > a
> > > couple chapters last night and already had some experience with
> > Polya
> > > polynomials from the discussion here. I don't know if there is
an
> > > easy formula for the cycle index of a product of groups given
the
> > > indexes of the products. I normally use GAP to create the whole
> > group
> > > from a small number of generators and have it give me the cycle
> > > structures for all of the elements. D4 X S3 will be generated
by (1
> > > 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8
11)(3
> > 6
> > > 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get
you
> > > started on working out the whole group. I'll get back in touch
when
> > > I've finished the book.
> > >
> > > David Bowen
> >
> > Great! I am actually just studying the generators for D4 X S3, and
> > trying to figure out how to extract the cycle index. What book did
> > you get - is it "From Polynomials to Polya"? or something else?
> >
> >
> > >
> > > On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@> wrote:
> > > > Han Straub wrote:
> > > >
> > > > > > > As for D4 X S3, I just tried that by hand (still
> > affordable), and
> > > > > > got
> > > > > > >
> > > > > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 +
> > 4*x12 )
> > > > > > >
> > > > > > > No warranty...
> > > > >
> > > > > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the
> > Cycle
> > > > > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do
these
> > > > > combine to give the above?
> > > >
> > > > I can see its not a simple matter of just combining cycle
indices
> > for
> > > > D4 and S3, could someone show me how the cycle index for D4 X
S3
> > is
> > > > extracted? Thanks!
> > > >
> > > > > Paul Hj
> > > > > > >
> > > > > >
> > > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> >
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@...> wrote:
>

Hi Dave: Please see questions below * * * Thanks

> Hi, Paul,
>
> Yes, I got a copy of "Polychords to Polya". Here's my analysis of
> your question with an attempt to provide a general answer for any
> product of groups. Sorry if I bore you with things you already know.
>
> If G and H are permutation groups of order g and h operating on
> sets M and N of size m and n respectively, then G X H can be
> considered a permutation group of order gh operating on the set M X
N
> of size mn. Let (a,b) \elem of M X N and p \elem G, q \elem H. Then
(p
> x q) (a,b) = (p(a), q(b)) and so if a is part of a k-cycle and b
part
> of an l-cycle then (a,b) will be part of an lcm(k,l)-cycle.
>
> Now we can use this to work out the cycle structure of G X H
given
> the structures of G and H. For your particular case S_3 has cycle
> index z_1^3 + 3z_1z_2 + 2z_3 and D4 has cycle index z_1^4 + 3z_2^2 +
> 2z_1^2z_2 + 2z_4. The z_1^3, z_1^4 combination will produce a
z_1^12.
> z_1^ and 3 z_2^2 will produce a 3z_2^6. z_1^3 and 2z_1^2z_2 produces
> 2z_1^6z_2^3 and z_1^3 and z4 produces z_4^3. For 3z_1z_2 we get
> 3z_1^4z_2^4, 9z_2^6, 6z_1^2z_2^5 and 6z_4^3. And finally, for 2z_3
we
> 2z_3^4, 6z_6^2, 4z_3^2z_6, and z_12. I've skipped through the
details
> and will happily answer questions about any of the calculations or
> about how the calculations would go in any other direct products you
> might be interested in. The important thing to remember is the lcm
> rule, when 2-cycles match up with 2-cycles you just get 2-cycles not
> 4-cycles and when 2-cycles and 4-cycles match up you just get
> 4-cycles. If you haven't picked up a copy of GAP, think seriously
> about doing so. I've been using it for about 15 years and it's great
> for computational group theory.
>
> Dave Bowen

* * * *

The two terms I don't understand:

The group direct product (is that right?) of 3z_1z_2 and 3z_2^2
combine to form 9z_2^6 (where does ^6 come from?) Also,
3z_1z_2 and 2z_1^2z_2 combine to form 6z_1^2z_2^5 (where does ^5 come
from?). Otherwise, seems to make sense, even though I expected
z_4^3 term to have 2 in front of it and z^12 term to have 4 in front
of it. Thanks!

> On 2/17/06, Paul G Hjelmstad <paul_hjelmstad@...> wrote:
> > --- In tuning-math@yahoogroups.com, David Bowen <dmb0317@> wrote:
> > >
> > > Paul,
> > >
> > > My copy of the book arrived from Amazon yesterday. I got
through
> > a
> > > couple chapters last night and already had some experience with
> > Polya
> > > polynomials from the discussion here. I don't know if there is
an
> > > easy formula for the cycle index of a product of groups given
the
> > > indexes of the products. I normally use GAP to create the whole
> > group
> > > from a small number of generators and have it give me the cycle
> > > structures for all of the elements. D4 X S3 will be generated
by (1
> > > 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8
11)(3
> > 6
> > > 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get
you
> > > started on working out the whole group. I'll get back in touch
when
> > > I've finished the book.
> > >
> > > David Bowen
> >
> > Great! I am actually just studying the generators for D4 X S3, and
> > trying to figure out how to extract the cycle index. What book did
> > you get - is it "From Polynomials to Polya"? or something else?
> >
> >
> > >
> > > On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@> wrote:
> > > > Han Straub wrote:
> > > >
> > > > > > > As for D4 X S3, I just tried that by hand (still
> > affordable), and
> > > > > > got
> > > > > > >
> > > > > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 +
> > 4*x12 )
> > > > > > >
> > > > > > > No warranty...
> > > > >
> > > > > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the
> > Cycle
> > > > > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do
these
> > > > > combine to give the above?
> > > >
> > > > I can see its not a simple matter of just combining cycle
indices
> > for
> > > > D4 and S3, could someone show me how the cycle index for D4 X
S3
> > is
> > > > extracted? Thanks!
> > > >
> > > > > Paul Hj
> > > > > > >
> > > > > >
> > > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > > Yahoo! Groups Links
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > >
> >
> >
> >
> >
> >
> >
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
> >
> >
> >
>

🔗David Bowen <dmb0317@gmail.com>

Paul,

The real short explanation is each of the 12 elements of M X N will
be part of a 2-cycle so there must be 6 2-cycles total, hence the ^6.
If you want to go into more detail, look at the various possibilities
for a and b. a can be part of a 1-cycle or a 2-cycle while b must
always be part of a 2-cycle. In either case (a,b) will be part of a
2-cycle.

For 3z_1z_2 and 2z_1^2z_2, we have two cases for a and two for b.
There are two cases where both a and b are part of a 1-cycle resulting
in the z_1^2. There is two cases where a is a 1-cycle and b is part of
a 2-cycle. That gives us one 2-cycle. There are four cases where a is
part of a two cycle and b is part of a 1-cycle (two choices for a and
two for b) which gives us two more 2-cycles. Finally there are four
cases where both a and b are part of a 2-cycle which create two more
2-cycles for a total of five 2-cycles and the z_2^5.

I believe you are correct about the missing coefficients for z_4^3
and z_12. I was working a little to quickly.

Dave Bowen

On 2/21/06, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:
>
> * * * *
>
> The two terms I don't understand:
>
> The group direct product (is that right?) of 3z_1z_2 and 3z_2^2
> combine to form 9z_2^6 (where does ^6 come from?) Also,
> 3z_1z_2 and 2z_1^2z_2 combine to form 6z_1^2z_2^5 (where does ^5 come
> from?). Otherwise, seems to make sense, even though I expected
> z_4^3 term to have 2 in front of it and z^12 term to have 4 in front
> of it. Thanks!
>
>
> > On 2/17/06, Paul G Hjelmstad <paul_hjelmstad@...> wrote:
> > > --- In tuning-math@yahoogroups.com, David Bowen <dmb0317@> wrote:
> > > >
> > > > Paul,
> > > >
> > > > My copy of the book arrived from Amazon yesterday. I got
> through
> > > a
> > > > couple chapters last night and already had some experience with
> > > Polya
> > > > polynomials from the discussion here. I don't know if there is
> an
> > > > easy formula for the cycle index of a product of groups given
> the
> > > > indexes of the products. I normally use GAP to create the whole
> > > group
> > > > from a small number of generators and have it give me the cycle
> > > > structures for all of the elements. D4 X S3 will be generated
> by (1
> > > > 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8
> 11)(3
> > > 6
> > > > 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get
> you
> > > > started on working out the whole group. I'll get back in touch
> when
> > > > I've finished the book.
> > > >
> > > > David Bowen
> > >
> > > Great! I am actually just studying the generators for D4 X S3, and
> > > trying to figure out how to extract the cycle index. What book did
> > > you get - is it "From Polynomials to Polya"? or something else?
> > >
> > >
> > > >
> > > > On 2/16/06, Paul G Hjelmstad <paul_hjelmstad@> wrote:
> > > > > Han Straub wrote:
> > > > >
> > > > > > > > As for D4 X S3, I just tried that by hand (still
> > > affordable), and
> > > > > > > got
> > > > > > > >
> > > > > > > > 1/48*( x1^12 + 2*x1^6*x2^3 + 3*x1^4*x2^4 + 6*x1^2*x2^5 +
> > > > > > > > 12*x2^6 + 2*x3^4 + 4*x3^2*x6 + 8*x4^4 + 6*x6^2 +
> > > 4*x12 )
> > > > > > > >
> > > > > > > > No warranty...
> > > > > >
> > > > > > So, the Cycle Index for S3 is 1/6[x1^3+3x1x2 + 2x3] and the
> > > Cycle
> > > > > > Index for D4 is 1/8[x1^4 + 1/2x2x1^2+3x2^2+1/2x4]. How do
> these
> > > > > > combine to give the above?
> > > > >
> > > > > I can see its not a simple matter of just combining cycle
> indices
> > > for
> > > > > D4 and S3, could someone show me how the cycle index for D4 X
> S3
> > > is
> > > > > extracted? Thanks!
> > > > >
> > > > > > Paul Hj
> > > > > > > >
> > > > > > >
> > > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > > Yahoo! Groups Links
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > > >
> > > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > Yahoo! Groups Links
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> >
>
>
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@...> wrote:
>
> Paul,
>
> The real short explanation is each of the 12 elements of M X N
will
> be part of a 2-cycle so there must be 6 2-cycles total, hence the
^6.
> If you want to go into more detail, look at the various
possibilities
> for a and b. a can be part of a 1-cycle or a 2-cycle while b must
> always be part of a 2-cycle. In either case (a,b) will be part of a
> 2-cycle.
>
> For 3z_1z_2 and 2z_1^2z_2, we have two cases for a and two for b.
> There are two cases where both a and b are part of a 1-cycle
resulting
> in the z_1^2. There is two cases where a is a 1-cycle and b is part
of
> a 2-cycle. That gives us one 2-cycle. There are four cases where a
is
> part of a two cycle and b is part of a 1-cycle (two choices for a
and
> two for b) which gives us two more 2-cycles. Finally there are four
> cases where both a and b are part of a 2-cycle which create two more
> 2-cycles for a total of five 2-cycles and the z_2^5.
>
> I believe you are correct about the missing coefficients for
z_4^3
> and z_12. I was working a little to quickly.
>
> Dave Bowen

I stared at this for quite awhile before figuring it out. (Confusing!)
I see you are defining "cycle" based on z_2 for example, not the
power, (I'm not disagreeing, I just got used to "cycle" being based
on the power, like (x^2+1)^6 which is defined as "6 cycles of 2
notes") Thanks for giving me a toe-hold in this!

🔗David Bowen <dmb0317@gmail.com>

Paul,

Well, I said that the answer in the general case was likely to be
messy and S3 X D4 is actually a pretty easy case. Hopefully, it
covered enough of the tricky details so that you can work out more
complicated examples that might interest you. If you get stuck, drop
me a line and I'll try to answer your questions.

Dave

P.S. With your last name, are you residing somewhere in Minnesota?

On 2/21/06, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:
> --- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@...> wrote:
> >
> > Paul,
> >
> > The real short explanation is each of the 12 elements of M X N
> will
> > be part of a 2-cycle so there must be 6 2-cycles total, hence the
> ^6.
> > If you want to go into more detail, look at the various
> possibilities
> > for a and b. a can be part of a 1-cycle or a 2-cycle while b must
> > always be part of a 2-cycle. In either case (a,b) will be part of a
> > 2-cycle.
> >
> > For 3z_1z_2 and 2z_1^2z_2, we have two cases for a and two for b.
> > There are two cases where both a and b are part of a 1-cycle
> resulting
> > in the z_1^2. There is two cases where a is a 1-cycle and b is part
> of
> > a 2-cycle. That gives us one 2-cycle. There are four cases where a
> is
> > part of a two cycle and b is part of a 1-cycle (two choices for a
> and
> > two for b) which gives us two more 2-cycles. Finally there are four
> > cases where both a and b are part of a 2-cycle which create two more
> > 2-cycles for a total of five 2-cycles and the z_2^5.
> >
> > I believe you are correct about the missing coefficients for
> z_4^3
> > and z_12. I was working a little to quickly.
> >
> > Dave Bowen
>
> I stared at this for quite awhile before figuring it out. (Confusing!)
> I see you are defining "cycle" based on z_2 for example, not the
> power, (I'm not disagreeing, I just got used to "cycle" being based
> on the power, like (x^2+1)^6 which is defined as "6 cycles of 2
> notes") Thanks for giving me a toe-hold in this!
>
>
>
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, David Bowen <dmb0317@> wrote:
> >
> > Paul,
> >
> > My copy of the book arrived from Amazon yesterday. I got
through
> a
> > couple chapters last night and already had some experience with
> Polya
> > polynomials from the discussion here. I don't know if there is an
> > easy formula for the cycle index of a product of groups given the
> > indexes of the products. I normally use GAP to create the whole
> group
> > from a small number of generators and have it give me the cycle
> > structures for all of the elements. D4 X S3 will be generated by
(1
> > 2)(4 5)(7 8)(10 11), (2 3)(5 6)(8 9)(11 12), (1 4 7 10)(2 5 8 11)
(3
> 6
> > 9 12), and (1 4)(2 5)(3 6)(7 10)(8 11)(9 12). That should get you
> > started on working out the whole group. I'll get back in touch
when
> > I've finished the book.
> >
> > David Bowen

Interesting. 2 of the 3 octotonic scales, triple-diminished, and
triple-split-diminished...

Actually, do have a quick question.

Gene gives these generators:

D(4) x S(3)
Order 48

a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
b := [[1, 5], [2, 10], [4, 8], [7, 11]]
e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
k := [[1, 7], [3, 9], [5, 11]]

I imagine that D(4) X S(3) might be expressed by different sets of
generators?

How do either of these sets of generators lead to the cycle
structures? (As opposed to deriving the cycle index from D(4) X S(3)
as you have shown...) Or is it really the same thing... Thanks!

🔗David Bowen <dmb0317@gmail.com>

Paul,

Remember that the cycle index depends on the action of the group on
the set. For example, the group of rotations of the cube has natural
actions on a six element set (the faces), an eight element set (the
verteces) and a twelve element set (the edges) and each will have a
different cycle index. You can even have different actions on sets of
the same size. The easiest example is S3 which has two different
actions on a six point set; one generated by (12)(36)(45) and
(14)(23)(56) and the other generated by (12)(45) and (23)(56). The
first is transitive while the second is two copies of the three point
representation of S3, so the actions are different and they will have
different cycle indexes.

But the case you asked about isn't like that, at least from my
quick analysis. Both Gene and I took reasonable generators for S3 and
D4 and then mapped them onto 3 X 4. We differed in our choices for
those generators. I used (12) and (23) for S3 and (12)(34) and (24)
for D4. Gene went with (123) and (12) for S3 and (1234) and (12)(34)
for D4. But we should have the isomorphic actions and the same cycle
indexes.

Now to change the topic, how are you defining a symmetric set? I
guess for S3 X D4 you can think of a necklace with three strands of
four beads. S3 interchanges strands while the D4 represents the
rotational and inversional aspects of a four bead necklace, so there
may be a geometric notion to fall back on here. Do you have a
definition that will work even in the absence of a geometric
intuition.?

I remember the name Dan Mack from my Cray days, but I can't put a
face to it or say which group he worked in. I'm still living in
Lakeville, so if e-mail explanations get too messy maybe we can switch
to phone or even a face-to-face meeting.

Dave Bowen

On 2/22/06, Paul G Hjelmstad <paul_hjelmstad@allianzlife.com> wrote:

>
> D(4) x S(3)
> Order 48
>
> a := [[12, 4, 8], [1, 5, 9], [2, 6, 10], [3, 7, 11]]
> b := [[1, 5], [2, 10], [4, 8], [7, 11]]
> e := [[12, 3, 6, 9], [1, 4, 7, 10], [2, 5, 8, 11]]
> k := [[1, 7], [3, 9], [5, 11]]
>
> I imagine that D(4) X S(3) might be expressed by different sets of
> generators?
>
> How do either of these sets of generators lead to the cycle
> structures? (As opposed to deriving the cycle index from D(4) X S(3)
> as you have shown...) Or is it really the same thing... Thanks!
>
>
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
>
>
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@...> wrote:
>
> Paul,
>
> Remember that the cycle index depends on the action of the group
on
> the set. For example, the group of rotations of the cube has natural
> actions on a six element set (the faces), an eight element set (the
> verteces) and a twelve element set (the edges) and each will have a
> different cycle index. You can even have different actions on sets
of
> the same size. The easiest example is S3 which has two different
> actions on a six point set; one generated by (12)(36)(45) and
> (14)(23)(56) and the other generated by (12)(45) and (23)(56). The
> first is transitive while the second is two copies of the three
point
> representation of S3, so the actions are different and they will
have
> different cycle indexes.
>
> But the case you asked about isn't like that, at least from my
> quick analysis. Both Gene and I took reasonable generators for S3
and
> D4 and then mapped them onto 3 X 4. We differed in our choices for
> those generators. I used (12) and (23) for S3 and (12)(34) and (24)
> for D4. Gene went with (123) and (12) for S3 and (1234) and (12)(34)
> for D4. But we should have the isomorphic actions and the same cycle
> indexes.
>
> Now to change the topic, how are you defining a symmetric set? I
> guess for S3 X D4 you can think of a necklace with three strands of
> four beads. S3 interchanges strands while the D4 represents the
> rotational and inversional aspects of a four bead necklace, so there
> may be a geometric notion to fall back on here. Do you have a
> definition that will work even in the absence of a geometric
> intuition.?
>
> I remember the name Dan Mack from my Cray days, but I can't put a
> face to it or say which group he worked in. I'm still living in
> Lakeville, so if e-mail explanations get too messy maybe we can
switch
> to phone or even a face-to-face meeting.
>
> Dave Bowen

Dan said he worked in the Sun System Admin group of the Software
division there. Thanks for your answer. The necklace illustration
is especially helpful. I have also studied S2-complementability and
S3-complementability (three color necklaces) or higher...in a paper
by Gilbert and Riordan (1961 Ill. Journ. of Math.) which is
a slightly different application of Sn groups which one can
combine into these group definitions (I am not so sure of the
notation, but hexachords reduced for both D4 X S3 and S2
complementability come to 26, which is my system of hexachords
which uses the letters of the alphabet....)

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@> wrote:
> >
> > Paul,
> >
> > Remember that the cycle index depends on the action of the
group
> on
> > the set. For example, the group of rotations of the cube has
natural
> > actions on a six element set (the faces), an eight element set
(the
> > verteces) and a twelve element set (the edges) and each will have
a
> > different cycle index. You can even have different actions on
sets
> of
> > the same size. The easiest example is S3 which has two different
> > actions on a six point set; one generated by (12)(36)(45) and
> > (14)(23)(56) and the other generated by (12)(45) and (23)(56).
The
> > first is transitive while the second is two copies of the three
> point
> > representation of S3, so the actions are different and they will
> have
> > different cycle indexes.
> >
> > But the case you asked about isn't like that, at least from my
> > quick analysis. Both Gene and I took reasonable generators for S3
> and
> > D4 and then mapped them onto 3 X 4. We differed in our choices for
> > those generators. I used (12) and (23) for S3 and (12)(34) and
(24)
> > for D4. Gene went with (123) and (12) for S3 and (1234) and (12)
(34)
> > for D4. But we should have the isomorphic actions and the same
cycle
> > indexes.
> >
> > Now to change the topic, how are you defining a symmetric set?
I
> > guess for S3 X D4 you can think of a necklace with three strands
of
> > four beads. S3 interchanges strands while the D4 represents the
> > rotational and inversional aspects of a four bead necklace, so
there
> > may be a geometric notion to fall back on here. Do you have a
> > definition that will work even in the absence of a geometric
> > intuition.?
> >
> > I remember the name Dan Mack from my Cray days, but I can't
put a
> > face to it or say which group he worked in. I'm still living in
> > Lakeville, so if e-mail explanations get too messy maybe we can
> switch
> > to phone or even a face-to-face meeting.
> >
> > Dave Bowen
>
> Dan said he worked in the Sun System Admin group of the Software
> division there. Thanks for your answer. The necklace illustration
> is especially helpful. I have also studied S2-complementability and
> S3-complementability (three color necklaces) or higher...in a paper
> by Gilbert and Riordan (1961 Ill. Journ. of Math.) which is
> a slightly different application of Sn groups which one can
> combine into these group definitions (I am not so sure of the
> notation, but hexachords reduced for both D4 X S3 and S2
> complementability come to 26, which is my system of hexachords
> which uses the letters of the alphabet....)
>

In answer to your question, I tend to study these symmetries using
a 3 X 4 grid: [0 3 6 9], [4 7 10 1], [8 11 2 5]. I don't know
if that is the best way to tackle this, but different flips
and rotations seem to correspond to what is going on with the
generators. It might be nice if we could chat on the phone. Please
email me and we can exchange phone numbers. Thanks.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

This is so obscure and half-formed that I am not sure why I am
posting it, but I found this weekend, working on creating my "D4 X S3
transposition grid", that adding 80 + 20 + 16 + 20 /4 gives 34
hexachords, the correct result for D4 X S3. 80 is the total number
of hexachords (after reducing for transposes) 20 is those sets that
map into themselves, TnI. 16 is those sets that map into their
M5 counterpart through TnI. 20 is those sets that map into their
M5 counterpart through Tn. There is overlap between the two, so
there are 28 sets that map into their M5 counterpart through Tn
and/or TnI. One can find the M5 counterpart by holding 0,2,4,6,8,10
fixed and swaping (1&7,3&9,5&11).

What I haven't figured out is that fact that adding 80 + 20 /2 gives
50, which is D4 X C3, and strangely, adding 80 + 16 /2 gives 48, which
is C4 X S3. This might be coincidental - because the S3 function
is NOT just inversion, even though the D4 function is a perfect M5
swap. If you take a 3 X 4 grid [0 3 6 9], [4 7 10 1], [8 11 2 5],
then holding 0, 2 columns fixed and swapping 1,3 IS D4, swapping
any two rows is S3, and turning the whole thing 180 degrees is
inversion. I think the C4 X S3 thing works because S3 actually
does do the D4 thing, and inverts it too!

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> This is so obscure and half-formed that I am not sure why I am
> posting it, but I found this weekend, working on creating my "D4 X
S3
> transposition grid", that adding 80 + 20 + 16 + 20 /4 gives 34
> hexachords, the correct result for D4 X S3. 80 is the total number
> of hexachords (after reducing for transposes) 20 is those sets that
> map into themselves, TnI. 16 is those sets that map into their
> M5 counterpart through TnI. 20 is those sets that map into their
> M5 counterpart through Tn. There is overlap between the two, so
> there are 28 sets that map into their M5 counterpart through Tn
> and/or TnI. One can find the M5 counterpart by holding 0,2,4,6,8,10
> fixed and swaping (1&7,3&9,5&11).
>
> What I haven't figured out is that fact that adding 80 + 20 /2 gives
> 50, which is D4 X C3, and strangely, adding 80 + 16 /2 gives 48,
which
> is C4 X S3. This might be coincidental - because the S3 function
> is NOT just inversion, even though the D4 function is a perfect M5
> swap. If you take a 3 X 4 grid [0 3 6 9], [4 7 10 1], [8 11 2 5],
> then holding 0, 2 columns fixed and swapping 1,3 IS D4, swapping
> any two rows is S3, and turning the whole thing 180 degrees is
> inversion. I think the C4 X S3 thing works because S3 actually
> does do the D4 thing, and inverts it too!

Correction - I should have said S3 does the M5 thing, and inverts it
too. Sorry.
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

Dave:

I see this creates 12 terms, and the coefficients add up to 48,
which of course is divided by 48. I was just wondering, (and I can
prove this by hand - but it would take awhile). Does each one
of these terms correspond to a transposition? (It does with
the Dihedral Group D12 for example). In other words, choosing
3z_2^6, this expands out to 3(x^2+1)^6 = 3*
(x^12+6x^10+15x^8+20x^6+15x^4+6x^2+1)

Could this be one of the tranpositions (T0-T11) of the D4 X S3 group?

Thanks

Paul Hj

Thanks

Paul Hj

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "David Bowen" <dmb0317@> wrote:
> >
> > Hi, Paul,
> >
> > Yes, I got a copy of "Polychords to Polya". Here's my analysis
of
> > your question with an attempt to provide a general answer for any
> > product of groups. Sorry if I bore you with things you already
know.
> >
> > If G and H are permutation groups of order g and h operating on
> > sets M and N of size m and n respectively, then G X H can be
> > considered a permutation group of order gh operating on the set M
X
> N
> > of size mn. Let (a,b) \elem of M X N and p \elem G, q \elem H.
Then
> (p
> > x q) (a,b) = (p(a), q(b)) and so if a is part of a k-cycle and b
> part
> > of an l-cycle then (a,b) will be part of an lcm(k,l)-cycle.
> >
> > Now we can use this to work out the cycle structure of G X H
> given
> > the structures of G and H. For your particular case S_3 has cycle
> > index z_1^3 + 3z_1z_2 + 2z_3 and D4 has cycle index z_1^4 +
3z_2^2 +
> > 2z_1^2z_2 + 2z_4. The z_1^3, z_1^4 combination will produce a
> z_1^12.
> > z_1^ and 3 z_2^2 will produce a 3z_2^6. z_1^3 and 2z_1^2z_2
produces
> > 2z_1^6z_2^3 and z_1^3 and z4 produces z_4^3. For 3z_1z_2 we get
> > 3z_1^4z_2^4, 9z_2^6, 6z_1^2z_2^5 and 6z_4^3. And finally, for
2z_3
> we
> > 2z_3^4, 6z_6^2, 4z_3^2z_6, and z_12. I've skipped through the
> details
> > and will happily answer questions about any of the calculations or
> > about how the calculations would go in any other direct products
you
> > might be interested in. The important thing to remember is the lcm
> > rule, when 2-cycles match up with 2-cycles you just get 2-cycles
not
> > 4-cycles and when 2-cycles and 4-cycles match up you just get
> > 4-cycles. If you haven't picked up a copy of GAP, think seriously
> > about doing so. I've been using it for about 15 years and it's
great
> > for computational group theory.
> >
> > Dave Bowen
>
> Dave:
>
> I see this creates 12 terms, and the coefficients add up to 48,
> which of course is divided by 48. I was just wondering, (and I can
> prove this by hand - but it would take awhile). Does each one
> of these terms correspond to a transposition? (It does with
> the Dihedral Group D12 for example). In other words, choosing
> 3z_2^6, this expands out to 3(x^2+1)^6 = 3*
> (x^12+6x^10+15x^8+20x^6+15x^4+6x^2+1)
>
> Could this be one of the tranpositions (T0-T11) of the D4 X S3
group?

Actually, I see that it can be split into 48 terms. I believe these
correspond to 4 rows of 12 transpositions, giving Tn, TnI, "M5->Tn"
and "M5->TnI" but I have no proof. Also it seems like it would be
difficult to determine which term fits into each slot. Of course,
each of the 48 slots would themselves contain cardinalities 0-12.
So a matrix of 12 x 12 could be constructed (the 4 rows would
be collapsed, otherwise you get 4 x 12 x 12 which would be confusing.)

I have the 12 x 12 grid for plain C3 X C4 which I will post to my
files section.

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

> > I see this creates 12 terms, and the coefficients add up to 48,
> > which of course is divided by 48. I was just wondering, (and I
can
> > prove this by hand - but it would take awhile). Does each one
> > of these terms correspond to a transposition? (It does with
> > the Dihedral Group D12 for example). In other words, choosing
> > 3z_2^6, this expands out to 3(x^2+1)^6 = 3*
> > (x^12+6x^10+15x^8+20x^6+15x^4+6x^2+1)
> >
> > Could this be one of the tranpositions (T0-T11) of the D4 X S3
> group?
>
> Actually, I see that it can be split into 48 terms. I believe these
> correspond to 4 rows of 12 transpositions, giving Tn, TnI, "M5->Tn"
> and "M5->TnI" but I have no proof. Also it seems like it would be
> difficult to determine which term fits into each slot. Of course,
> each of the 48 slots would themselves contain cardinalities 0-12.
> So a matrix of 12 x 12 could be constructed (the 4 rows would
> be collapsed, otherwise you get 4 x 12 x 12 which would be
confusing.)
>
> I have the 12 x 12 grid for plain C3 X C4 which I will post to my
> files section.

More necklace stuff:

Actually, Inverse(D4) is S3, right? Here is an interesting result
taking "hexachord partitions" (meaning all the 6 white 6 black
necklaces, with complementability, where colors can be swapped).
These are reduced for transposition, which all divide out evenly with
Polya's methods. (Chords of limited transposition multiply out to be
the same as primitive ones, that is, 12 transpositions)

80 hexachords -> 44 partitions
There are 26 hexachord-partitions that map to the same partition
under TnI: This is 20 + 32 /2 using the necklace formula.

Also,

18 hexachord-partitions map to the same partition under S3
16 hexachord-partitions map to the same partition under D4
12 hexachord-partitions map under both S3 and D4
6 hexachord-partitions map under S3 only
4 hexachords-partitions map under D4 only

So, 80 + 26 + 18 + 16 = 104 /4 = 26, the letters of My Hexachord
System which uses the letters of the alphabet

One could also study reduction for S3 and D4 only, (both normal
hexachords and hexachord-partitions) to obtain a kind of
C4 X S3 X S2C measure and a D4 X C3 X S2C measure. (Where SC2
stands for S2-complementability group. These answers are left
as an exercise to the reader.

Just kidding. Even if I posted this on sci.math I doubt anyone in
the world would want to calculate this. (BTW, I got answers
to a math question in 20 minutes from Portugal and Germany both on
sci.math - Cool It was a question on Discrete Logarithms).

Incidentally, does anyone think Discrete Logarithms might come
in handy for the work on this newsgroup???

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> > > I see this creates 12 terms, and the coefficients add up to 48,
> > > which of course is divided by 48. I was just wondering, (and I
> can
> > > prove this by hand - but it would take awhile). Does each one
> > > of these terms correspond to a transposition? (It does with
> > > the Dihedral Group D12 for example). In other words, choosing
> > > 3z_2^6, this expands out to 3(x^2+1)^6 = 3*
> > > (x^12+6x^10+15x^8+20x^6+15x^4+6x^2+1)
> > >
> > > Could this be one of the tranpositions (T0-T11) of the D4 X S3
> > group?
> >
> > Actually, I see that it can be split into 48 terms. I believe
these
> > correspond to 4 rows of 12 transpositions, giving Tn, TnI, "M5-
>Tn"
> > and "M5->TnI" but I have no proof. Also it seems like it would be
> > difficult to determine which term fits into each slot. Of course,
> > each of the 48 slots would themselves contain cardinalities 0-12.
> > So a matrix of 12 x 12 could be constructed (the 4 rows would
> > be collapsed, otherwise you get 4 x 12 x 12 which would be
> confusing.)
> >
> > I have the 12 x 12 grid for plain C3 X C4 which I will post to my
> > files section.
>
> More necklace stuff:
>
> Actually, Inverse(D4) is S3, right? Here is an interesting result
> taking "hexachord partitions" (meaning all the 6 white 6 black
> necklaces, with complementability, where colors can be swapped).
> These are reduced for transposition, which all divide out evenly
with
> Polya's methods. (Chords of limited transposition multiply out to
be
> the same as primitive ones, that is, 12 transpositions)
>
> 80 hexachords -> 44 partitions
> There are 26 hexachord-partitions that map to the same partition
> under TnI: This is 20 + 32 /2 using the necklace formula.
>
> Also,
>
> 18 hexachord-partitions map to the same partition under S3
> 16 hexachord-partitions map to the same partition under D4
> 12 hexachord-partitions map under both S3 and D4
> 6 hexachord-partitions map under S3 only
> 4 hexachords-partitions map under D4 only
>
> So, 80 + 26 + 18 + 16 = 104 /4 = 26, the letters of My Hexachord
> System which uses the letters of the alphabet

Oops - 44 + 26 + 18 + 16 = 104 /4 = 26

This is also (((80 + 8) + (20 + 32)/2) + 18 + 16)/4 for what its
worth...

>
> One could also study reduction for S3 and D4 only, (both normal
> hexachords and hexachord-partitions) to obtain a kind of
> C4 X S3 X S2C measure and a D4 X C3 X S2C measure. (Where SC2
> stands for S2-complementability group. These answers are left
> as an exercise to the reader.
>
> Just kidding. Even if I posted this on sci.math I doubt anyone in
> the world would want to calculate this. (BTW, I got answers
> to a math question in 20 minutes from Portugal and Germany both on
> sci.math - Cool It was a question on Discrete Logarithms).
>
> Incidentally, does anyone think Discrete Logarithms might come
> in handy for the work on this newsgroup???
>

🔗Paul G Hjelmstad <paul_hjelmstad@allianzlife.com>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > > > I see this creates 12 terms, and the coefficients add up to
48,
> > > > which of course is divided by 48. I was just wondering, (and
I
> > can
> > > > prove this by hand - but it would take awhile). Does each one
> > > > of these terms correspond to a transposition? (It does with
> > > > the Dihedral Group D12 for example). In other words, choosing
> > > > 3z_2^6, this expands out to 3(x^2+1)^6 = 3*
> > > > (x^12+6x^10+15x^8+20x^6+15x^4+6x^2+1)
> > > >
> > > > Could this be one of the tranpositions (T0-T11) of the D4 X
S3
> > > group?
> > >
> > > Actually, I see that it can be split into 48 terms. I believe
> these
> > > correspond to 4 rows of 12 transpositions, giving Tn, TnI, "M5-
> >Tn"
> > > and "M5->TnI" but I have no proof. Also it seems like it would
be
> > > difficult to determine which term fits into each slot. Of
course,
> > > each of the 48 slots would themselves contain cardinalities 0-
12.
> > > So a matrix of 12 x 12 could be constructed (the 4 rows would
> > > be collapsed, otherwise you get 4 x 12 x 12 which would be
> > confusing.)
> > >
> > > I have the 12 x 12 grid for plain C3 X C4 which I will post to
my
> > > files section.
> >
> > More necklace stuff:
> >
> > Actually, Inverse(D4) is S3, right? Here is an interesting result
> > taking "hexachord partitions" (meaning all the 6 white 6 black
> > necklaces, with complementability, where colors can be swapped).
> > These are reduced for transposition, which all divide out evenly
> with
> > Polya's methods. (Chords of limited transposition multiply out to
> be
> > the same as primitive ones, that is, 12 transpositions)
> >
> > 80 hexachords -> 44 partitions
> > There are 26 hexachord-partitions that map to the same partition
> > under TnI: This is 20 + 32 /2 using the necklace formula.
> >
> > Also,
> >
> > 18 hexachord-partitions map to the same partition under S3
> > 16 hexachord-partitions map to the same partition under D4
> > 12 hexachord-partitions map under both S3 and D4
> > 6 hexachord-partitions map under S3 only
> > 4 hexachords-partitions map under D4 only
> >
> > So, 80 + 26 + 18 + 16 = 104 /4 = 26, the letters of My Hexachord
> > System which uses the letters of the alphabet
>
> Oops - 44 + 26 + 18 + 16 = 104 /4 = 26
>
> This is also (((80 + 8) + (20 + 32)/2) + 18 + 16)/4 for what its
> worth...
>
> >
> > One could also study reduction for S3 and D4 only, (both normal
> > hexachords and hexachord-partitions) to obtain a kind of
> > C4 X S3 X S2C measure and a D4 X C3 X S2C measure. (Where SC2
> > stands for S2-complementability group. These answers are left
> > as an exercise to the reader.
> >
> > Just kidding. Even if I posted this on sci.math I doubt anyone in
> > the world would want to calculate this. (BTW, I got answers
> > to a math question in 20 minutes from Portugal and Germany both
on
> > sci.math - Cool It was a question on Discrete Logarithms).
> >
> > Incidentally, does anyone think Discrete Logarithms might come
> > in handy for the work on this newsgroup???

Triads:

Triads that map into themselves under Tn: 19
Triads that map into themselves under TnI: 5
Triads that map into themselves under D4: 7
Triads that map into themselves under S3: 5

Therefore you obtain (19 + 5 + 7 + 5)/4 = 9 Triads for S3 X D4
TnI + S3 /2 = 5, but I don't know what this would be...

Hexachords:

D4 X C3 = (80 + 20)/2 = 50
C4 X S3 = (80 + 16)/2 = 48
(D4 X C3) X SC2 (44 + 16)/2 = 30
(C4 X S3) X SC2 (44 + 18)/2 = 31
(D4 X S3) X SC2 (44 + 26 + 16 + 18)/4 = 26

Just like 44 and 26 can be broken down into (80 + 8)/2 and (32 + 20)/2
so can 16 and 18. Also, interesting to note that 6 + 14 = 20 in cells
3 and 6 is actually (32 + 8)/2 (Sets that map into themselves + Sets
that map into their complements under TnI /2)

7 7 6
8 8 14
8 8 14