Paul,

I've changed the routine, hoping the results are right.

Do you have a means to check them?

I have to change the file consist_limits.txt too.

For 12 to 24-tET:

1200.0000 cents divided by 12, step = 100.0000 cents

Highest harmonic represented consistently : 10

Highest harmonic represented uniquely : 6

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 13, step = 92.3077 cents

Highest harmonic represented consistently : 4

Highest harmonic represented uniquely : 6

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 14, step = 85.7143 cents

Highest harmonic represented consistently : 4

Highest harmonic represented uniquely : 8

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 15, step = 80.0000 cents

Highest harmonic represented consistently : 8

Highest harmonic represented uniquely : 8

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 16, step = 75.0000 cents

Highest harmonic represented consistently : 8

Highest harmonic represented uniquely : 6

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 17, step = 70.5882 cents

Highest harmonic represented consistently : 4

Highest harmonic represented uniquely : 6

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 18, step = 66.6667 cents

Highest harmonic represented consistently : 8

Highest harmonic represented uniquely : 8

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 19, step = 63.1579 cents

Highest harmonic represented consistently : 10

Highest harmonic represented uniquely : 9

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 20, step = 60.0000 cents

Highest harmonic represented consistently : 4

Highest harmonic represented uniquely : 9

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 21, step = 57.1429 cents

Highest harmonic represented consistently : 4

Highest harmonic represented uniquely : 10

Highest harm. represented uniquely inv. equiv.: 8

1200.0000 cents divided by 22, step = 54.5455 cents

Highest harmonic represented consistently : 12

Highest harmonic represented uniquely : 9

Highest harm. represented uniquely inv. equiv.: 6

1200.0000 cents divided by 23, step = 52.1739 cents

Highest harmonic represented consistently : 6

Highest harmonic represented uniquely : 10

Highest harm. represented uniquely inv. equiv.: 8

1200.0000 cents divided by 24, step = 50.0000 cents

Highest harmonic represented consistently : 6

Highest harmonic represented uniquely : 9

Highest harm. represented uniquely inv. equiv.: 6

Manuel

> I've changed the routine, hoping the results are right.

> Do you have a means to check them?

> I have to change the file consist_limits.txt too.

What prompted the change? Did you find a bug?

-Carl

--- In tuning-math@y..., <manuel.op.de.coul@e...> wrote:

>

> Paul,

>

> I've changed the routine, hoping the results are right.

I thought your results were right before. I never said the results

were wrong, only that the wording was wrong.

> I have to change the file consist_limits.txt too.

I felt confident that the results in that file were correct the way

they stood for several years, since I referred to that table many,

many times over the last few years. It's just the wording that I was

objecting to.

My own wording is "19-tET uniquely articulates ratios through integer-

limit 7, and through odd-limit 5."

Carl wrote:

>What prompted the change? Did you find a bug?

Yes or at least I think I did. First I discovered

the wrong assumption of octave equivalence, so the

value for Bohlen-Pierce for example wouldn't be right.

Try for me the 19-tET case, I think it should be

9, not 7 (without inversional equivalence).

You can do this by hand by doing "eq 19" and then

repeatedly replace:

rep/near 3/2

rep/near 4/3

rep/near 5/3

rep/near 5/4

..etc until

rep/near 9/7

rep/near 9/8

and checking each time if the pitch replaced

wasn't a just value before.

Manuel

--- In tuning-math@y..., <manuel.op.de.coul@e...> wrote:

>

> Carl wrote:

> >What prompted the change? Did you find a bug?

>

> Yes or at least I think I did. First I discovered

> the wrong assumption of octave equivalence, so the

> value for Bohlen-Pierce for example wouldn't be right.

> Try for me the 19-tET case, I think it should be

> 9, not 7 (without inversional equivalence).

In 19-tET, 6:7 and 7:8 are approximated by the same interval. So 9 is too high.

>In 19-tET, 6:7 and 7:8 are approximated by the same interval. So 9 is too

high.

Oops, replaced one bug with another one. After moving the misplaced

if-statement I get the same old values again. So false alarm.

Manuel

Regular temperaments enforce consistency. I'm struggling once

again for Gene's way of saying this... if the consonant intervals

are a group closed under multiplication, then there's a group

homomorphism to elements of the temperament that preserves the

closure. How's that?

Anyway, what about uniqueness... can the cardinality (do groups

have cardinality, or is that only sets?) of the group decrease

after the mapping to the temperament?

If, after figuring out what I'm asking and answering, I can go

back and phrase it correctly, I'd be tremendously pleased.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

>

> Anyway, what about uniqueness... can the cardinality (do groups

> have cardinality, or is that only sets?) of the group decrease

> after the mapping to the temperament?

>

Huh? Groups _are_ sets, and therefore have cardinality, of course.

And in a mapping that preserves the group operations (homomorphism),

the cardinality of the image can be lower if the mapping is not

injective. You could call that "descreasing cardinality".

Looks like elemetary math stuff - or did I again miss something?

Hans Straub

hstraub64 wrote:

>Huh? Groups _are_ sets, and therefore have cardinality, of course.

>And in a mapping that preserves the group operations (homomorphism), >the cardinality of the image can be lower if the mapping is not >injective. You could call that "descreasing cardinality".

>Looks like elemetary math stuff - or did I again miss something?

> >

I've been reading up on group theory lately. I don't think it makes sense for a temperament to be an injective mapping. Meaning that at least one interval in the tempered scale should represent more than one interval in the ideal scale. Otherwise, all you have is a subset. Like the 5-limit is a subset of the 7-limit, but not a temperament of it (unless you map the 7:4 in some way).

Furthermore, I suggest a temperament should strictly be a surjective homomorphism, or an epimorphism. That means each interval in the tempered scale represents at least one ideal interval. This definition excludes things like 24-equal in the 5-limit where every other tempered interval isn't a temperament of anything.

As for the cardinality issue, that's the number of elements in the set, isn't it? Only octave-specific equal temperaments have a finite number of elements. Every other regular scale is a countable infinity, which I think means the cardinality is the same (aleph 1). This is assuming they're fully fledged groups -- models that take into account the limits of human perception will have a finite number of elements, but they won't be groups. I think this is what Lindley and Turner Smith were getting at in the paper Gene doesn't like.

Graham

> Huh? Groups _are_ sets, and therefore have cardinality,

> of course.

Yes, I should have looked up "groups" first.

> And in a mapping that preserves the group operations

> (homomorphism), the cardinality of the image can be lower

> if the mapping is not injective. You could call that

> "descreasing cardinality". Looks like elemetary math

> stuff - or did I again miss something?

My question is, can a "regular temperament" be injective.

-Carl

>I don't think it makes sense for a temperament to be an injective

>mapping. Meaning that at least one interval in the tempered scale

>should represent more than one interval in the ideal scale.

What about things like dicot?

>Otherwise, all you have is a subset. Like the 5-limit is a subset

>of the 7-limit, but not a temperament of it (unless you map the 7:4

>in some way).

I don't think there's necessarily a homomorphism from a group

to all its subsets. You have to keep closure. For example,

the integers are closed under addition but the subset (1 2 3)

is not. Am I getting this right?

>Furthermore, I suggest a temperament should strictly be a

>surjective homomorphism, or an epimorphism. That means each

>interval in the tempered scale represents at least one ideal

>interval.

I don't think that's possible. Unless you allow compound

intervals to be ideal. But then you're always going to be

injective, since any comma will confound two compound

intervals (wheras only commas like 25:24 confound two

primary intervals).

>As for the cardinality issue, that's the number of elements in

>the set, isn't it?

I wasn't sure if card. was defined on groups. Turns out that

groups are special types of sets.

>Only octave-specific equal temperaments have a finite number

>of elements. Every other regular scale is a countable infinity,

>which I think means the cardinality is the same (aleph 1).

Wait, are temperaments groups of intervals or pitches? I assumed

intervals.

-Carl

Me:

> >I don't think it makes sense for a temperament to be an injective

> >mapping. Meaning that at least one interval in the tempered scale

> >should represent more than one interval in the ideal scale.

Carl:

> What about things like dicot?

I don't know, what's dicot? It's not in the catalog. A search shows

it's something to do with neutral thirds. Well, I don't see what

stops it being injective. If it's only 5-limit then it isn't

surjective, and so not strictly a temperament by the definition below.

Me:

> >Otherwise, all you have is a subset. Like the 5-limit is a subset

> >of the 7-limit, but not a temperament of it (unless you map the 7:4

> >in some way).

Carl:

> I don't think there's necessarily a homomorphism from a group

> to all its subsets. You have to keep closure. For example,

> the integers are closed under addition but the subset (1 2 3)

> is not. Am I getting this right?

Yes, I should have said "subgroup". And I got it the wrong way round.

The injective homomorphism would have to be from 5-limit to 7-limit,

so 7-limit would be the temperament. That sounds bizarre which

supports the idea that temperaments should be surjective but not

injective.

Me:

> >Furthermore, I suggest a temperament should strictly be a

> >surjective homomorphism, or an epimorphism. That means each

> >interval in the tempered scale represents at least one ideal

> >interval.

Carl:

> I don't think that's possible. Unless you allow compound

> intervals to be ideal. But then you're always going to be

> injective, since any comma will confound two compound

> intervals (wheras only commas like 25:24 confound two

> primary intervals).

When I say "scale" I mean the whole infinity of notes in the tuning

system. What you describe is "surjective" not "injective". Another

term for "injective" is "one to one" which I find easier to

understand, as long as I remember it isn't the same as "bijective" or

"one to one onto".

Mathworld has a diagram that explains it:

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

> I wasn't sure if card. was defined on groups. Turns out that

> groups are special types of sets.

Yes, a group is a set with a binary operation that fulfils a number of

criteria.

> >Only octave-specific equal temperaments have a finite number

> >of elements. Every other regular scale is a countable infinity,

> >which I think means the cardinality is the same (aleph 1).

>

> Wait, are temperaments groups of intervals or pitches? I assumed

> intervals.

Ah! I meant "octave-equivalent" there. If they're groups, they have

to be sets of intervals, because an operation on elements of the group

has to give another element of the same group. More specifically they

have to be directed intervals. An equal temperament with n notes the

octave can be thought of as a set of n notes in octave equivalent

terms. But it also has n distinct directed intervals, making it a

cyclic group of order n. So for 12-equal, the 12 ascending intervals

are:

0 unison

1 minor second

2 major second

3 minor third

4 major third

5 perfect fourth

6 tritone/augmented fourth/diminished fifth

7 perfect fifth

8 minor sixth

9 major sixth

10 minor seventh

11 major seventh

And octave is the same as a unison in octave-equivalent terms. If you

add an ascending perfect fifth to an ascending minor sixth, you get in

interval of +15 steps. That's the same modulo 12 to a 3 steps, so the

result is an ascending minor third.

The inverse operation relates each ascending interval to a descending

one. A descending minor second has -2 steps, and in octave equivalent

terms that's the same as 10 steps, and so a descending minor second

and an ascending minor seventh are the same octave equivalent

interval.

Graham

>I don't know, what's dicot? It's not in the catalog.

It's in Paul's database. IIRC it's the new name for

neutral thirds.

>Well, I don't see what stops it being injective. If it's

>only 5-limit then it isn't surjective, and so not strictly

>a temperament by the definition below.

//

>Mathworld has a diagram that explains it:

>

>http://mathworld.wolfram.com/Injection.html

Gene's linking to surjection on his regular temperaments

page. But I don't see any definition of "rank" on the

abelian group mathworld page, so I can't follow the bit

about icons -- it looks like Gene's spiced up his def.

of "regular temperament". Actually, I don't see a def.

on this page. And it doesn't seem to be in monz's site

like I thought it was.

-Carl

Carl Lumma wrote:

>It's in Paul's database. IIRC it's the new name for

>neutral thirds.

> >

It's a 5-limit temperament that maps both 6:5 and 5:4 to the neutral third generator. It looks like an epimorphism as good as any other.

>Gene's linking to surjection on his regular temperaments

>page. But I don't see any definition of "rank" on the

>abelian group mathworld page, so I can't follow the bit

>about icons -- it looks like Gene's spiced up his def.

>of "regular temperament". Actually, I don't see a def.

>on this page. And it doesn't seem to be in monz's site

>like I thought it was.

> >

Oh, yes, there are quite a few new things there. He's been quiet round here lately. He also says the icon is an epimorphism, which is the same as a surjection, at least for groups. (Sets can have surjections as well, which might be what I was thinking of before. Well temperaments can be thought of as sets but not groups, and they have an epimorphism of intervals onto an equal temperament. I terms of pitches, there's an epimorphism from JI into a well temperament.) I can't follow the chain of definitions to see if a regular temperament is an epimorphism from JI to something simpler, but it's probably something like that.

Rank is a property of a free abelian group

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

It's the number of (linearly independent) generators -- 2 for 5-limit JI, 3 for 7-limit, 4 for 11-limit, 1 for an equal temperament, 2 for a linear temperament, 3 for a planar temperament.

Graham

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Gene's linking to surjection on his regular temperaments

> page. But I don't see any definition of "rank" on the

> abelian group mathworld page, so I can't follow the bit

> about icons -- it looks like Gene's spiced up his def.

> of "regular temperament". Actually, I don't see a def.

> on this page. And it doesn't seem to be in monz's site

> like I thought it was.

I think I defined rank somewhere or other. The MathWorld definition

seems to be written for mathematicians, but here it is:

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

The tensor product Q tensor G makes the torsion part go away and

turns the free part into a vector space, giving us a vector space of

dimension equal to the number of copies of Z in G. This number is the

rank of an abelian group G; when G is free it doesn't need to be

tensored with Q in order to exterminate the torsion part.

Concretely, if your group is free and group elements are written as

row vectors of integers (always possible for free groups) then the

dimension of these vectors is the rank.

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@l...> wrote:

> Anyway, what about uniqueness...

uniqueness, unlike consistency, is well-defined for temperaments --

it is quite possible for temperaments to violate uniqueness, for

example dicot in the 5-limit . . .