Detempered torsion

Here's a scale obtained by taking (135/128)^i (45/32)^j for
-5 <= i <= 6, 0 <= j <= 1, and reducing via 648/625 and 2048/2025.
Garibaldi seems to be a good way of adding some 7-limit harmony; that
gives, in terms of fifths, [-18, -17, -16, -11, -10, -9, -8, -7, -6,
-3, -2, -1, 0, 1, 2, 3, 7, 8, 9, 10, 11, 12, 16, 17].

! torb24.scl
detempering C2 x C12 {648/625, 2048/2025} with generators 45/32 and
135/128
24
!
81/80
25/24
16/15
10/9
9/8
32/27
6/5
5/4
32/25
4/3
27/20
25/18
45/32
40/27
3/2
25/16
8/5
5/3
27/16
16/9
9/5
15/8
48/25
2

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> ! torb24.scl
> detempering C2 x C12 {648/625, 2048/2025} with generators 45/32 and

This torb24 scale isn't epimorphic, of course, but it has a related
property of abelian group structure nonetheless. It shows in the
following way: if we take every other scale step, we get a twelve note
scale
1, 25/24, 10/9, 32/27, 5/4, 4/3, 25/18, 40/27, 25/16, 5/3, 16/9, 15/8

This scale is epimorphic by <12 19 28|. Moreover, if we take every
other scale step, starting at the second note 81/80, and divide by
81/80, making the second step the 1/1, we get
1, 256/243, 10/9, 32/27, 512/405, 4/3, 25/18, 40/27, 128/81, 5/3,
16/9, 256/135

This scale is also epimorphic via <12 19 28|. The full scale is A +
(81/80)B, where A is the first epimorphic scale, and B is the second.
Both A and B are epimorphic via the same val, <12 19 28|, which sends
the 81/80 we multiply B by to zero. Such scales with noncyclic abelian
group structures generalizes the notion of epimorphic/CS, and it would
be interesting to test for it. In any case, it is possible to create
such scales, which is interesting in itself. Simply gluing together
the appropriate number of epimorphic scales with the same val and the
appropriate generator is the more obvious way to do it.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> Simply gluing together
> the appropriate number of epimorphic scales with the same val and the
> appropriate generator is the more obvious way to do it.

Not easier, but perhaps more elegant--a Fokker block. An example
Fokker block with C2 x C12 structure, obtained from commas 648/625 and
2048/2025, is

1, 128/125, 25/24, 16/15, 10/9, 9/8, 32/27, 6/5, 5/4, 32/25, 4/3,
27/20, 64/45, 36/25, 40/27, 3/2, 25/16, 8/5, 5/3, 128/75, 16/9, 9/5,
15/8, 48/25

Presumably this is typical of the breed. The even and odd scale steps,
taken independently, are both <12 19 28| epimorphic, and in terms of
this val the steps go 0,0,1,1,2,2 etc. This is the C2 x C12 group
structure.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
>
> > Simply gluing together
> > the appropriate number of epimorphic scales with the same val and
the
> > appropriate generator is the more obvious way to do it.
>
> Not easier, but perhaps more elegant--a Fokker block. An example
> Fokker block with C2 x C12 structure, obtained from commas 648/625
and
> 2048/2025, is
>
> 1, 128/125, 25/24, 16/15, 10/9, 9/8, 32/27, 6/5, 5/4, 32/25, 4/3,
> 27/20, 64/45, 36/25, 40/27, 3/2, 25/16, 8/5, 5/3, 128/75, 16/9, 9/5,
> 15/8, 48/25

>
> Presumably this is typical of the breed. The even and odd scale
steps,
> taken independently, are both <12 19 28| epimorphic, and in terms of
> this val the steps go 0,0,1,1,2,2 etc. This is the C2 x C12 group
> structure.

Could you explain epimorphic again? I looked it up in Tonalsoft, but
don't quite get it. Is h=<12 19 28). How does h(qn)=n?

Paul

same as superparticular. the top number is one higher than the lower.as in 3/2, 7/6 49,48 etc.
If you construct a lambdoma with any series where one uses epimoric ratios, the final result will also be epimorphic

Paul G Hjelmstad wrote:

>--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...> >wrote:
> >
>>--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
>>wrote:
>>
>> >>
>>>Simply gluing together
>>>the appropriate number of epimorphic scales with the same val and >>> >>>
>the
> >
>>>appropriate generator is the more obvious way to do it.
>>> >>>
>>Not easier, but perhaps more elegant--a Fokker block. An example
>>Fokker block with C2 x C12 structure, obtained from commas 648/625 >> >>
>and
> >
>>2048/2025, is
>>
>>1, 128/125, 25/24, 16/15, 10/9, 9/8, 32/27, 6/5, 5/4, 32/25, 4/3, >>27/20, 64/45, 36/25, 40/27, 3/2, 25/16, 8/5, 5/3, 128/75, 16/9, 9/5,
>>15/8, 48/25
>> >>
>
> >
>>Presumably this is typical of the breed. The even and odd scale >> >>
>steps,
> >
>>taken independently, are both <12 19 28| epimorphic, and in terms of
>>this val the steps go 0,0,1,1,2,2 etc. This is the C2 x C12 group
>>structure.
>> >>
>
>Could you explain epimorphic again? I looked it up in Tonalsoft, but >don't quite get it. Is h=<12 19 28). How does h(qn)=n?
>
>Paul
>
>
>
>
> >Yahoo! Groups Links
>
>
>
> >
>
>
> >

--
Kraig Grady
North American Embassy of Anaphoria Island <http://anaphoria.com/>
The Wandering Medicine Show
KXLU <http://www.kxlu.com/main.html> 88.9 FM Wed 8-9 pm Los Angeles

At 10:28 AM 7/1/2005, you wrote:
>same as superparticular. the top number is one higher than the lower.as
>in 3/2, 7/6 49,48 etc.
> If you construct a lambdoma with any series where one uses epimoric
>ratios, the final result will also be epimorphic

Heya Kraig,

You're thinking of epimoric

http://tonalsoft.com/enc/epimore.htm

Gene's got something else called "epimorphic". Paul E. possibly
understands it, and Manuel (Scala can do it).

-Carl

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

> Could you explain epimorphic again? I looked it up in Tonalsoft, but
> don't quite get it. Is h=<12 19 28). How does h(qn)=n?

"h" is whatever it needs to be to make the epimorphic condition true;
you can find it by sticking the scale in Scala and using the show data
command.

Let's take an example of a scale where h = <12 19 28|. Suppose our
scale is 1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8. Extending past the octave
in both directions, we have a function s[n] such that s[n] corresponds
to the nth scale degree. Hence, s[0] is 1, s[7] is 2, s[-7] is 1/2,
s[4] is 3/2 and so forth; to every integer n we've associated a scale
degree s[n]. It now happens to be true that

h(s[n]) = n

This means the scale s is epimorphic, with the defining val being h.

The situation I was talking about is a good deal more complicated, but
similar. Now we have a scale s with S notes to the octave, and a val
h which is an n-edo val--that is, h(2)=n. Here n divides s, so that
g = S/n is also an integer. Little g is the order of a finite abelian
group big G, and instead of a group ismorphic to the integers Z being
what a homomorphic map would map to, we have G x Z. That is, a rank
one free part, like the integers Z, and a finite torsion part G. The
val h only maps to the Z part of this group, and everything with the
same image by h is equivalent, with a structure of G. We have scale
steps which go 0,0,0 ... g times, then 1,1,1, ... g times, and so
forth. If we take the ratios between any set of g successive related
notes mapped to k,k,k ... g times, there is a set of commas for the
val h such that modulo these commas, the g ratios map to g different
values, each of which is sent to 0 by h, and some power of which will
reduce to 1 by the commas--torsion elements for these commas.

This is, unfortunately, a hell of a lot more complicated than just an
ordinary epimorpic scale. Maybe I can think of a way to make it
simpler. However, such scales seem interesting. They show up as Fokker
blocks, as we learned to our sorrow, and they have a structure which
makes them something like constant strucure and epimorphic, even
though they are neither if the group G is not trivial.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Could you explain epimorphic again? I looked it up in Tonalsoft,
but
> > don't quite get it. Is h=<12 19 28). How does h(qn)=n?
>
> "h" is whatever it needs to be to make the epimorphic condition
true;
> you can find it by sticking the scale in Scala and using the show
data
> command.
>
> Let's take an example of a scale where h = <12 19 28|. Suppose our
> scale is 1, 9/8, 5/4, 4/3, 3/2, 5/3, 15/8. Extending past the octave
> in both directions, we have a function s[n] such that s[n]
corresponds
> to the nth scale degree. Hence, s[0] is 1, s[7] is 2, s[-7] is 1/2,
> s[4] is 3/2 and so forth; to every integer n we've associated a
scale
> degree s[n]. It now happens to be true that
>
> h(s[n]) = n

*** How does h(3/2)=4 when h=<12 19 28| ? For example...
>
> This means the scale s is epimorphic, with the defining val being h.
>
> The situation I was talking about is a good deal more complicated,
but
> similar. Now we have a scale s with S notes to the octave, and a val
> h which is an n-edo val--that is, h(2)=n. Here n divides s, so that
> g = S/n is also an integer. Little g is the order of a finite
abelian
> group big G, and instead of a group ismorphic to the integers Z
being
> what a homomorphic map would map to, we have G x Z. That is, a rank

I take it S=24, G=2 and Z=12?

one free part, like the integers Z, and a finite torsion part G. The
> val h only maps to the Z part of this group, and everything with the
> same image by h is equivalent, with a structure of G. We have scale
> steps which go 0,0,0 ... g times, then 1,1,1, ... g times, and so
> forth. If we take the ratios between any set of g successive related
> notes mapped to k,k,k ... g times, there is a set of commas for the
> val h such that modulo these commas, the g ratios map to g different
> values, each of which is sent to 0 by h, and some power of which
will
> reduce to 1 by the commas--torsion elements for these commas.
>
> This is, unfortunately, a hell of a lot more complicated than just
an
> ordinary epimorpic scale. Maybe I can think of a way to make it
> simpler. However, such scales seem interesting. They show up as
Fokker
> blocks, as we learned to our sorrow, and they have a structure which
> makes them something like constant strucure and epimorphic, even
> though they are neither if the group G is not trivial.

This is very exciting to me to see you explaining tuning in terms
of Group Theory. Ultimately I would like to see how Groups tie
together "musical" set theory (Like C4 X S3) and tuning theory!

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

> *** How does h(3/2)=4 when h=<12 19 28| ? For example...

Sorry, wrong h. I forgot to edit the above so that h = <7 11 16|
after deciding to use a seven note scale example.

If we use the right h, h(3/2) = <7 11 16|-1 1 0> = 11-7 = 4.

> I take it S=24, G=2 and Z=12?

S = 24 and g = 2, G = C2, the cyclic group of order 2, and Z = the
"cyclic group of order infinity", or the additive group of the integers.

> This is very exciting to me to see you explaining tuning in terms
> of Group Theory. Ultimately I would like to see how Groups tie
> together "musical" set theory (Like C4 X S3) and tuning theory!

Well, unfortunately noncyclic group structures seem to be of limited
utility, because when you try to temper to something with such a
structure, the torsion part doesn't seem to correspond to anything
sensible.

Cyclic groups do not decompose into nontrivial elementary factors;
however, there is another standard way of decomposing finitely
generated abelian groups, which is the Kronecker decomposition. This
decomposes the group as a product of "primary" cyclic groups, which
are cyclic groups of prime power order. If you do this for C12, the
cyclic group of order 12, you get C3 x C4 = C12 as its Kronecker
decomposition.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > *** How does h(3/2)=4 when h=<12 19 28| ? For example...
>
> Sorry, wrong h. I forgot to edit the above so that h = <7 11 16|
> after deciding to use a seven note scale example.
>
> If we use the right h, h(3/2) = <7 11 16|-1 1 0> = 11-7 = 4.
>
> > I take it S=24, G=2 and Z=12?
>
> S = 24 and g = 2, G = C2, the cyclic group of order 2, and Z = the
> "cyclic group of order infinity", or the additive group of the
integers.
>
> > This is very exciting to me to see you explaining tuning in terms
> > of Group Theory. Ultimately I would like to see how Groups tie
> > together "musical" set theory (Like C4 X S3) and tuning theory!
>
> Well, unfortunately noncyclic group structures seem to be of limited
> utility, because when you try to temper to something with such a
> structure, the torsion part doesn't seem to correspond to anything
> sensible.
>
> Cyclic groups do not decompose into nontrivial elementary factors;
> however, there is another standard way of decomposing finitely
> generated abelian groups, which is the Kronecker decomposition. This
> decomposes the group as a product of "primary" cyclic groups, which
> are cyclic groups of prime power order. If you do this for C12, the
> cyclic group of order 12, you get C3 x C4 = C12 as its Kronecker
> decomposition.

Does C3 X C4 ever come into play, when detempering scales in C12?

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

> Does C3 X C4 ever come into play, when detempering scales in C12?

It can. We can detemper keeping 128/125 or 648/625, which the C3 (reps
C4.)

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Does C3 X C4 ever come into play, when detempering scales in C12?
>
> It can. We can detemper keeping 128/125 or 648/625, which the C3 (reps
> C4.)

Yes, I thought it was those commas. Interesting. Another question: What
is the value of epimorphic scales? What special properties do they
have?

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

> Yes, I thought it was those commas. Interesting. Another question: What
> is the value of epimorphic scales? What special properties do they
> have?

Epimorphic scales, and the very closely related property of constant
stucture scales, give the scale a systematic, orderly structure. If a
fifth consists of four scale steps at one point, at any other point it
will also consist of four scale steps, for example.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > Yes, I thought it was those commas. Interesting. Another question:
What
> > is the value of epimorphic scales? What special properties do they
> > have?
>
> Epimorphic scales, and the very closely related property of constant
> stucture scales, give the scale a systematic, orderly structure. If a
> fifth consists of four scale steps at one point, at any other point it
> will also consist of four scale steps, for example.

What happens if you go, say, from B to F#? It's four steps, but with an
accidental. Is there some sort of provision for accidentals?

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

> What happens if you go, say, from B to F#? It's four steps, but with an
> accidental. Is there some sort of provision for accidentals?

F# wasn't a part of the scale, so it's not clear what your question means.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul.hjelmstad@m...> wrote:
>
> > What happens if you go, say, from B to F#? It's four steps, but
with an
> > accidental. Is there some sort of provision for accidentals?
>
> F# wasn't a part of the scale, so it's not clear what your question
means.

It is kind of interesting that 45/32 is three steps, so going from
3 steps to 6 steps (B, 15/8) is three steps, which is a fourth.