Cartesian product of two ET scales

We all know that any ET scale of N elements can be seen as a multiplicative
cyclic group in which, for example, G1=(2)**(1/N) could be taken as the
basic generator. The obvious example is the 12 ET scale, with the
tempered semitone as a generator.

Now, we could take another scale, also ET, such that the basic generator
would be g2=(2)**(1/M) where N and M are coprimes (this is key, otherwise
in reality we do not get a different scale).

The product of both scales or groups would produce another group with NxM
elements and it would also be cyclic but in this case the generator would
be a pair (G1,G2).

I am wondering which, if any of the historical relevant tunnings produces
scales which are cyclic groups of this kind, i.e. with two independent
generators.

Many of you have a better knowledge than me of historical scales and I
would appreciatte a comment about this.

Thanks

Carlos

--- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:

> I am wondering which, if any of the historical relevant tunnings
produces
> scales which are cyclic groups of this kind, i.e. with two
independent
> generators.

Cyclic groups by definition can be expressed in terms of a single
generator. They can also be factored into cyclic groups of prime
power order, and if N and M are relatively prime, we are simply
combining the two factorizations. If you take N, M (not necessarily
coprime) and combine the corresponding equal temperaments then you
will get the cyclic group of order LCM(N, M).

Ok, I think I gave a wrong example. What you say is clear in the case we
take cyclic groups in which both generators are of the form g=2**(1/n)
where n is some number. This is, in the case of assuming the octave
equivalence.

I think what I was having in mind was something different and more general
like, gaving cyclic groups with generators of the form

g1= 3**(1/n) (I am think of the Bohlen-Pierce scale, for example n=13),

g2= 2**(1/m) i.e a regular ET scale (say m=2)

What would be the product group look like? It has to be cyclic of order nxm
but it does not have a single generator. It would seem that the generator
has to be the pair (g1,g2), or not?

If you could provide some clarification I would appreciatte.

Carlos

On Friday 15 August 2003 13:09, Gene Ward Smith wrote:
> --- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:
> > I am wondering which, if any of the historical relevant tunnings
>
> produces
>
> > scales which are cyclic groups of this kind, i.e. with two
>
> independent
>
> > generators.
>
> Cyclic groups by definition can be expressed in terms of a single
> generator. They can also be factored into cyclic groups of prime
> power order, and if N and M are relatively prime, we are simply
> combining the two factorizations. If you take N, M (not necessarily
> coprime) and combine the corresponding equal temperaments then you
> will get the cyclic group of order LCM(N, M).
>
>
>
>
> To unsubscribe from this group, send an email to:
> tuning-math-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/

Carlos wrote:

>Ok, I think I gave a wrong example. What you say is clear in the case we >take cyclic groups in which both generators are of the form g=2**(1/n) >where n is some number. This is, in the case of assuming the octave >equivalence.
>
>I think what I was having in mind was something different and more general >like, gaving cyclic groups with generators of the form
>
>g1= 3**(1/n) (I am think of the Bohlen-Pierce scale, for example n=13), >
>g2= 2**(1/m) i.e a regular ET scale (say m=2)
> >
If the equivalence interval isn the same, I think they should both be considered of infinite order. In which case adding them won't give a cyclic group, but a linear temperament. For which, see

http://x31eq.com/temper/method.html <http://local.microtonal.co.uk/temper/method.html>

>What would be the product group look like? It has to be cyclic of order nxm >but it does not have a single generator. It would seem that the generator >has to be the pair (g1,g2), or not?
> >
If it doesn't have a single generator then it isn't cyclic, by definition. As they do have an interval in common, you could define 12-equal to be 19 notes to a 3:1. Then the resultant group will be cyclic of order 19*13=247.

Graham

--- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:

> I think what I was having in mind was something different and more
general
> like, gaving cyclic groups with generators of the form
>
> g1= 3**(1/n) (I am think of the Bohlen-Pierce scale, for example
n=13),
>
> g2= 2**(1/m) i.e a regular ET scale (say m=2)
>
> What would be the product group look like?

It doesn't make much sense to reduce this to octave pitch classes; if
you don't you have a free group of rank two, generated by g1 and g2.

Graham and Gene,

Thanks. I think that I was trying to think of something which does not make
much sense, you need to have the same equivalence (octave or otherwise) in
both groups to have a meaninfull product.

What I was trying to do is to find an application to the concept of "direct
product" of groups.

Say you have the groups two cyclic groups G1= {0*,1**} and G2={0**,1**,2**}
the direct product would be G1 x G2 =

{ (0*,0**),(0*,1**),(0*,2**), (1*,0**),(1*,1**),(1*,2**) }

which is cyclic of order 6 and which has as a generator (1*,1**)

I was trying to figure out if these could have some sort of application to
scales with more than one generator.

Thanks

Carlos

On Friday 15 August 2003 22:19, Gene Ward Smith wrote:
> --- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:
> > I think what I was having in mind was something different and more
>
> general
>
> > like, gaving cyclic groups with generators of the form
> >
> > g1= 3**(1/n) (I am think of the Bohlen-Pierce scale, for example
>
> n=13),
>
> > g2= 2**(1/m) i.e a regular ET scale (say m=2)
> >
> > What would be the product group look like?
>
> It doesn't make much sense to reduce this to octave pitch classes; if
> you don't you have a free group of rank two, generated by g1 and g2.
>
>
>
>
> To unsubscribe from this group, send an email to:
> tuning-math-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to
> http://docs.yahoo.com/info/terms/

balzano tried to make such application in his beautiful papers
generalizing the 12-equal diatonic scale to any n*(n+1)-equal tuning,
where the generalized diatonic scale has n+n+1 notes. i find his use
of the direct product sneaky and his thesis invalid . . .

--- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:
> Graham and Gene,
>
> Thanks. I think that I was trying to think of something which does
not make
> much sense, you need to have the same equivalence (octave or
otherwise) in
> both groups to have a meaninfull product.
>
> What I was trying to do is to find an application to the concept
of "direct
> product" of groups.
>
> Say you have the groups two cyclic groups G1= {0*,1**} and G2=
{0**,1**,2**}
> the direct product would be G1 x G2 =
>
> { (0*,0**),(0*,1**),(0*,2**), (1*,0**),(1*,1**),(1*,2**) }
>
> which is cyclic of order 6 and which has as a generator (1*,1**)
>
> I was trying to figure out if these could have some sort of
application to
> scales with more than one generator.
>
> Thanks
>
> Carlos
>
>
>
>
> On Friday 15 August 2003 22:19, Gene Ward Smith wrote:
> > --- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...>
wrote:
> > > I think what I was having in mind was something different and
more
> >
> > general
> >
> > > like, gaving cyclic groups with generators of the form
> > >
> > > g1= 3**(1/n) (I am think of the Bohlen-Pierce scale, for
example
> >
> > n=13),
> >
> > > g2= 2**(1/m) i.e a regular ET scale (say m=2)
> > >
> > > What would be the product group look like?
> >
> > It doesn't make much sense to reduce this to octave pitch
classes; if
> > you don't you have a free group of rank two, generated by g1 and
g2.
> >
> >
> >
> >
> > To unsubscribe from this group, send an email to:
> > tuning-math-unsubscribe@yahoogroups.com
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/

--- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
wrote:
> balzano tried to make such application in his beautiful papers
> generalizing the 12-equal diatonic scale to any n*(n+1)-equal
tuning,
> where the generalized diatonic scale has n+n+1 notes. i find his
use
> of the direct product sneaky and his thesis invalid . . .

Are the papers actually beautiful? The thesis seems frankly brain-
damaged; should I read them anyway?

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul Erlich" <perlich@a...>
> wrote:
> > balzano tried to make such application in his beautiful papers
> > generalizing the 12-equal diatonic scale to any n*(n+1)-equal
> tuning,
> > where the generalized diatonic scale has n+n+1 notes. i find his
> use
> > of the direct product sneaky and his thesis invalid . . .
>
> Are the papers actually beautiful? The thesis seems frankly brain-
> damaged; should I read them anyway?

they're beautiful in their trickery. if you were a mathematician who
knew little of music and tuning theory and history, i wouldn't blame
you for being taken in by it.

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

> they're beautiful in their trickery. if you were a mathematician
who
> knew little of music and tuning theory and history, i wouldn't
blame
> you for being taken in by it.

I'm a mathematician who got disgusted and quit doing music theory
when Balzono turned down his attempt to get published, so I'm
prejudiced against.

--- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:
> Graham and Gene,
>
> Thanks. I think that I was trying to think of something which does
not make
> much sense, you need to have the same equivalence (octave or
otherwise) in
> both groups to have a meaninfull product.
>
> What I was trying to do is to find an application to the concept
> of "direct product" of groups.
>
> Say you have the groups two cyclic groups G1= {0*,1**} and G2=
> {0**,1**,2**} the direct product would be G1 x G2 =
>
> { (0*,0**),(0*,1**),(0*,2**), (1*,0**),(1*,1**),(1*,2**) }
>
> which is cyclic of order 6 and which has as a generator (1*,1**)
>
> I was trying to figure out if these could have some sort of
> application to scales with more than one generator.
>

Maybe one point worth to notice is that ET 12 is itself a direct
product of two cyclic groups, Z3 and Z4. Applications might be found
starting from there.

Hans Straub

hstraub64 wrote:

>Maybe one point worth to notice is that ET 12 is itself a direct >product of two cyclic groups, Z3 and Z4. Applications might be found >starting from there.
> >
That's what Balzano did.

I understand and thanks for the observation. After the discussion, is clear
for me that this is a mere coincidence. For example ET31 is prime and has not
such a decomposition, yet the scale is a useful one.

It would seem that Balazano found some aspects wich are specific of 12Et and
too quickly he assumed, without a really good reason, that this had to be
generalized. This is now clear in the introductory statement to his article,
in which he seeks to generalized diatonic scales without regard to any
ratios. What for me is a mistake, based on several aspects, including the
history of the diatonic scales.

Thanks

Carlos

On Wednesday 20 August 2003 13:40, Graham Breed wrote:
> hstraub64 wrote:
> >Maybe one point worth to notice is that ET 12 is itself a direct
> >product of two cyclic groups, Z3 and Z4. Applications might be found
> >starting from there.
>
> That's what Balzano did.
>
>
>
>
>
> To unsubscribe from this group, send an email to:
> tuning-math-unsubscribe@yahoogroups.com
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/

excellent, carlos. not to mention the fact that balzano sneaks the
2:1 ratio in as the octave, contradicting his own purported
independence from ratios.

--- In tuning-math@yahoogroups.com, Carlos <garciasuarez@y...> wrote:
> I understand and thanks for the observation. After the discussion,
is clear
> for me that this is a mere coincidence. For example ET31 is prime
and has not
> such a decomposition, yet the scale is a useful one.
>
> It would seem that Balazano found some aspects wich are specific of
12Et and
> too quickly he assumed, without a really good reason, that this had
to be
> generalized. This is now clear in the introductory statement to his
article,
> in which he seeks to generalized diatonic scales without regard to
any
> ratios. What for me is a mistake, based on several aspects,
including the
> history of the diatonic scales.
>
> Thanks
>
> Carlos
>
>
>
>
>
>
> On Wednesday 20 August 2003 13:40, Graham Breed wrote:
> > hstraub64 wrote:
> > >Maybe one point worth to notice is that ET 12 is itself a direct
> > >product of two cyclic groups, Z3 and Z4. Applications might be
found
> > >starting from there.
> >
> > That's what Balzano did.
> >
> >
> >
> >
> >
> > To unsubscribe from this group, send an email to:
> > tuning-math-unsubscribe@yahoogroups.com
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/