This is an Opt In Archive . We would like to hear from you if you want your posts included. For the contact address see About this archive. All posts are copyright (c).
- Contents - Hide Contents - Home - Section 21000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850 1900 1950
1350 - 1375 -
Message: 1350 - Contents - Hide Contents Date: Tue, 21 Aug 2001 20:27:27 Subject: Re: Generators and unison vectors From: Paul Erlich --- In tuning-math@y..., graham@m... wrote:> I was thinking about what n "unison vectors" in n-dimensional space could > mean. In fact, it's what I've already called a "basis". See > <404 Not Found * [with cont.] Search for http://www.microtonal.co.uk/matrix.htm in Wayback Machine>. I assume that's analogous to > Gene's "kernel". Let's revisit the example there: > > (t) ( 1 -2 1) > (s) = ( 4 -1 -1)H > (p) (-4 4 -1) > > Which can be used to define this scale > > C D E F G A B C > t+p t s t+p t p+t sHow did you get this, without appealing to a parallelogram or some such construction?> > The inverse of the matrix is > > ( 5 2 3) > ( 8 3 5) > (12 4 7) > > It defines H in terms of t, s and p. That I already knew. Now, the > interesting thing is, taking p out of the scale > > C D E F G A B C > t t s t t t s > > is the same as making p a commatic unison vector. By definition. > And we can define the > approximate H in terms of t and s > > ( 5 2)(t) > H' = ( 8 3)(s) > (12 4) > > I already knew this, but I didn't think of it as "designating a commatic > unison vector" or whatever. Before we've been thinking about starting > with commatic unison vectors and making them chromatic. It actually makes > more sense the other way around.If you mean, start with the unison vectors as non-zero intervals and then selectively bring some of them to zero, yes, that's how I think of it in the lattice/PB regime.> > Making (4 -1 -1)H or 16:15 another commatic unison vector gives us > 5-equal: > > ( 5)(t) > H' = ( 8) > (12) > > Gene's already explained this in algebraic terms, but I think the example > makes it clearer. It means I can now interpret > > ( 1 -2 1) > ( 4 -1 -1) > (-4 4 -1) > > as the octave-specific equivalent of a periodicity block. Unfortunately > it only has 1 note, How so? > so the analogy breaks down. But whatever, I'll call > all the intervals unison vectors anyway._All_ the intervals?> Now, we can also define meantone as > > ( 1 0 0) (1 0) > ( 0 1 0)C = (0 1) > (-4 4 -1) (0 0) > > Here, the octave and twelfth are taking the place of the chromatic unison > vectors. So, can we call them unison vectors?A unison vector has to come out to a musical _unison_! An augmented unison maybe, but still a unison.> I think not, because C > evaluates to > > ( 1 0) > ( 0 1) > (-4 4) > > That has negative numbers in it, which means the generators (for so they > are) don't add up to give all the primary consonances. Don't follow. > So now we have a > definition for what interval qualifies as a "unison vector". > Interestingly, a fifth would work as a generator here, but not for a > diatonic scale.You've completely lost me.> > An alternative to "unison vector" would be "melodic generator".Ouch! Are you sure? The steps in the scale can't be unison vectors, nor can the generator of the scale be a unison vector.> > Usually we call the octave the period and a fifth (which is therefore > equivalent to the twelfth) the generator. It isn't much of a stretch to > think of two generators.Gene has referred to that concept, I believe.> > In octave-equivalent terms, a unison is: > > v v v v v v v v v v v w > > where v is s fifth and w is a wolf. The same algebra applies, but it's > harder to visualise. I think the fifth here does count as a unison > vector.Please, please, please don't refer to a fifth as a unison vector. Musical intervals go: unison, second, third, fourth, fifth. There's a long way between a unison and a fifth.> Defining an MOS in octave-equivalent terms is easy, once you realize that > the generators that pop out needn't be unison vectors in octave- specific > space.They can't be unison vectors in any space! If the generator is a unison vector, you never get beyond the first note of the scale and its chromatic alterations.> > That's the way I see it now. I've also realized that I've been saying > "octave invariant" when "octave equivalent" would probably be more to the > point.I suppose -- but you got me doing it too :)
Message: 1351 - Contents - Hide Contents Date: Tue, 21 Aug 2001 20:57:27 Subject: Re: The hypothesis From: genewardsmith@xxxx.xxx --- In tuning-math@y..., graham@m... wrote:>> Considering scales is another level of generality altogether-- first >> we have approximations (kernels, unison vectors, and so forth) then >> we have tuning, and finally we select a subset and have scales.> Oh, right, but whatever they are, they're octave invariant. CLAMPITT.PDF > is a relevant place for definitions, as it's already been referenced: "By > /scale/ we refer to a set of pitches ordered according to ascending > frequencies (pitch height) bounded by an interval of periodicity."This definition of a scale does not assume it repeats octaves, so it does not assume octave invariance.> I don't agree with this anyway. You can have a scale without a tuning. > For example, a C major scale can be tuned to either 12- or 19-equal.OK, we can consider a conceptual scale to be something defined on the level of the homomorphic image (including the identity map.) Then C major (or mean tone diatonic) is a conceptual scale in the mean-tone linear temperament, which becomes a tuned scale if you pick a tuning. How's that?> It looks like a commatic UV would be in the kernel, and a chromatic UV > would not.So far as I can see, they are both in the kernel and refer to the same thing.>> If we consider equivalence classes modulo octaves, the 5-limit is >> free of rank two, but I don't know what you mean by "cyclic around >> the octave".> You describe an ET as cyclic below.I said an ET modulo ocatves was cyclic. In the case above, where we are considering equivalence classes of notes with representative elements 3^a * 5^b, we have't modded out and we have something which is free, not cyclic. By setting h(3^a*5^b) = 7*a+2*b (mod 12) we do get a circle of note equivalence classes.>> An ET would be free of rank 1, or "cyclic of infinite order". If we >> mod out by octaves, it would no longer be free but would (still) be >> cyclic, which implies one generator.> But if we "mod out" by one scale step, it would still have rank 1?It would have one generator, but the rank is the rank of the group modulo torsion, so in this case the rank is 0. There is a structure theorem for finitely generated abelian groups (which is what we have been considering) which gives a complete set of invariants specifying the group; the rank being one of them and perhaps the most important. In the case of C(12), it is in terms of groups of prime power order C(3) x C(4), which specifies it.> You certainly can tell a small interval if 2 is taken out. Of course, you > need a way of calculating interval size. As I see it, that lies outside > of the group theory we've been discussing so far. There are two ways of > doing it, corresponding to the two interpretations above.I think your two ways in effect put the 2 back in by calculating what it must be.>If > it doesn't work out that way perhaps you, as the mathematician, can tell > us the size constraints on the unison vectors.So far as I can see, there aren't any. A unison vector is simply anything you've decided to regard as a unison, and so lies in (or generates?) the kernel.>> You can generate by fifths and octaves if you want, but you don't >> need to.> You do need to if you want to enforce octave equivalence.I don't see why. Why not by octaves and major thirds instead?>> However, there is nothing in the nature of the problem to suggest I >> need to make any interval exact. One obvious way to decide would be >> to pick a set of intervals {t1, ... , tn} which I want to be well >> approximated, and a corresponding set of weights {w1, ... , wn} >> defining how important I think it is to have that interval >> approximate nicely. Perhaps I could do this using harmonic entropy? >> In any case, having done this I now have an optimization problem >> which I can decide using the method of least squares. If I have two >> generators, which I have in the case of the 5-limit with 81/80 a >> unison vector, then solving this will give me tunings for the >> generators and hence tunings for the entire system. There is no >> special treatment given to the octave in this method, but I see no >> reason in terms of psychological acoustics why there needs to be.> Hold on, you need more than that.Some one else said people around here have done such computations many times, so there seems to be some confusion on that score. I could certainly do one using the method above, at any rate.> So the nature of this problem is such that I do need to make two intervals > exact. There are plenty of other problems that have different natures, > but I'm puzzled as to why you insist on bringing them up when we're > discussing octave-invariant systems.I was discussing, among other things, tuning and temperament, where the question of octave tuning is relevant. As I pointed out, tuning an octave as a 2 is a choice of tuning.
Message: 1353 - Contents - Hide Contents Date: Tue, 21 Aug 2001 21:50:01 Subject: Re: Chromatic = commatic? From: genewardsmith@xxxx.xxx --- In tuning-math@y..., graham@m... wrote:>You seem to have a low view of algebra.Actually, I'm an algebraist and number theorist, and algebraists are notorious for wanting to solve a problem using only algebraic methods when possible, which is what I was doing. If you calculate the "7" using determinants, you can get the rest of the homomorphism by using floating point calculations: 7 log_2(3) = 11.09... and 7 log_2(5) = 16.25... As an algebraist, I naturally want to do the computation using only algebra. As an algebraist, I also want to say "Now hold on! What if I want to say a third has 3 steps, instead of two?" Therefore I distinguish the systems [7, 11, 16] and [7, 11, 17].> For example, here's that pair of octave-invariant unison vectors: > > [ 4 -1] > [ 0 -3] > > Invert it to get > > [-3 1] > [ 0 4] > > Each column is a mapping of generators.Each column is a mapping of everything in the 5-limit modulo octaves, but I don't see how you draw the conclusions you do about them.> That's all true, but missing the point. Commas and chromatic semitones > are certainly treated differently in diatonic music.If you are looking *only* at diatonic music and not allowing any accidentals, then neither one is a scale step; they seem therefore to be treated in the same way. The comma, being smaller, lends itself better to approximations, of course.>> Now I translate this to saying that if the rank of the kernel is n, >> then we get a linear temperament. Since the rank of the set of notes >> is n+1, this means the codimension is 1 and hence the rank of the >> homomorphic image is 1, meaning we have an et--which is precisely >> what we did get in the case where we had the 7-et. Why do you say >> linear temperament, which we've just determined means rank 2?> Because the rest of the world says "linear temperament" and has done so > for longer than we've been alive.You're missing my point--the world does not call something we get from codimension 1 a linear temperament, it calls codimension 2 (with 2 generators) a linear temperament.> How about you stop telling us what we should be doing, and start listening > to what we're trying to say?In a mathematical subject, mathematics is your friend; which means potentially so is a friendly mathematician. Why not view me as an ally in this endeavor?
Message: 1354 - Contents - Hide Contents Date: Wed, 22 Aug 2001 12:56 +0 Subject: Re: A counterexample to the hypothesis? From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9lvm8o+9v8r@xxxxxxx.xxx> Gene wrote:> This scale has intervals of sizes 81/80, 50/49, 25/24, 21/20, > 16/15 and 200/189. Since six intervals of widely differing sizes will > surely drive us nuts, we decide to temper out the two unison vectors, > namely 64/63 and 126/125. One simple way to do this is to use the 27- > et, which has both of them in its kernel (like the 15-et.) When we do > this, we find that 81/80 and 50/49 go to intervals of one scale step, > while the rest go to intervals of two scale steps. We now have the > pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a > third of an octave. Isn't this a counterexample?No, it's an MOS with a period of a third of an octave for the purposes of the hypothesis. The generator is the interval L. Graham
Message: 1355 - Contents - Hide Contents Date: Wed, 22 Aug 2001 23:50:00 Subject: Re: About CS From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Not often enough, sadly. How's your French?Even worse than his English. :(
Message: 1356 - Contents - Hide Contents Date: Wed, 22 Aug 2001 12:56 +0 Subject: Re: Interpreting Graham's matrix From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9lv41h+ag6a@xxxxxxx.xxx> Gene wrote:> The inverse of this is > > [ 7 0 0] > M^(-1) = 1/7 [11 1 2] > [16 4 1]. > > The question is how to interpret M^(-1). One use for it can be found > when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1) > and select all notes [a b c] such that > > 0 <= i < 1 > -1/2 <= j < 1/2 > -1/2 <= k < 1/2 > > we obtain the seven-note scale > > 1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)That is clever! The list of [a b c] is [0 0 0] [1 -2 1] [1 1 -1] [2 -1 0] [-1 1 0] [0 -1 1] [0 2 -1] [1 0 0] So you multiply that by the inverse (*not* the adjoint) and get [0 0 0] [0.142857143 0.285714286 -0.428571429] [0.285714286 -0.428571429 0.142857143] [0.428571429 -0.142857143 -0.285714286] [0.571428571 0.142857143 0.285714286] [0.714285714 0.428571429 -0.142857143] [0.857142857 -0.285714286 0.428571429] [1 0 0] Where, indeed, the first column lies between 0 and 1, and the others between -1/2 and 1/2. Furthermore, the list is already ordered by pitch, before we even worked out any pitches! That's because the left hand column, being "the thing that's left over when all the unison vectors are taken out", defines approximate pitch order. Over to Paul: is this a general method for finding a periodicity block?> with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone > system which equates 9/8 and 10/9 (which means we are using the fact > that 81/80 is a commatic vector) we get a PB scale with just two > sizes of intervals, LsLLLsL, ready to delight the fans of modal > melodies. Does this have anything to do with Graham's comments on > this matrix?I'll bring the inverse down again, and lose the 1/7 to make it the adjoint. [ 7 0 0] [11 1 2] [16 4 1] The first row defines the octave (which also happens to be the interval of equivalence) the second defines the 3:1 approximation and the third the 5:4 approximation. The interval 9:8 is [-3 2 0]. So multiply [-3 2 0] by adj(M) to get [1 2 4] The first column tells you the number of steps in the scale LsLLLsL the interval spans. 9:8 is L so spans 1 step. The second column tells you the number of generators give the interval in the system where 81:80 is tempered out (ie meantone). The generator is the fifth (or twelfth, etc), so 9:8 is two fifths (octave reduced). The last column defines the interval in the system where 25:24 is tempered out. I'll guess that's correct. 10:9 is [1 -2 1]. Multiply by Adj(M) and we get [1 2 3]. The first two columns are the same as for 9:8 which is no surprise because 9:8 and 10:9 are identical in meantone. That the third column differs by 1 could well tell you that 9:8 and 10:9 differ by a comma. 16:15 is [4 -1 -1]. This multiplies out to [1 -5 -3]. So it's also one scale step (s this time). And it's generated by -5 fifths, or 5 fourths. That is C-F-Bb-Eb-Ab-Db. The same process should work for all intervals. Now, the clever part is that if we take m, the octave equivalent of M, [-1 2] [ 4 -1] the adjoint is [-1 -2] [-4 -1] sign isn't important, so make that [1 2] [4 1] This is identical to a subset of adj(M). This will always be the case for correct unison vectors, but I don't have the criteria for those unison vectors being "correct". It tells us the octave-equivalent mapping in terms of fifths for all 5-limit intervals. The determinant tells us the number of notes in the periodicity block, but nothing I can see tells us what order they should be in. So that's all we lose by being octave equivalent. Graham
Message: 1357 - Contents - Hide Contents Date: Wed, 22 Aug 2001 23:59:26 Subject: Re: A counterexample to the hypothesis? From: genewardsmith@xxxx.xxx --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:> Why would this be a counterexample if LssssLssss wasn't?I had somehow gotten the idea such a thing would be a counterexample. I didn't look at 10 or 12 because they would have already been picked over, so 15 looked like a good composite to try. I think by the way that 15 notes out of a 27 scale would be a fine system to try for anyone who likes sharp systems and who wants to wash the 12-et out of their head for a while. I have the idea that what we are talking about is some sort of equidistribution property, but I can't seem to pin down the property. Suppose I define a two-value scale structure to be a periodic sequence of L and s values, containing both values. I could try the following: (1) Every contiguous set of n values has the same number of L's and s's as every other such set. The trouble with this is that nothing passes this condition. (2) Contiguous sets of n values have on average the same number of L's and s's. The trouble with this is that everything passes this condition.
Message: 1359 - Contents - Hide Contents Date: Wed, 22 Aug 2001 15:05 +0 Subject: Re: Generators and unison vectors From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9lug7f+8e65@xxxxxxx.xxx> Paul wrote:>> C D E F G A B C >> t+p t s t+p t p+t s >> How did you get this, without appealing to a parallelogram or some > such construction?It's an arbitrary scale, not generated from anything more fundamental.>> Making (4 -1 -1)H or 16:15 another commatic unison vector gives us >> 5-equal: >> >> ( 5)(t) >> H' = ( 8) >> (12) >> >> Gene's already explained this in algebraic terms, but I think the > example>> makes it clearer. It means I can now interpret >> >> ( 1 -2 1) >> ( 4 -1 -1) >> (-4 4 -1) >> >> as the octave-specific equivalent of a periodicity block. > Unfortunately>> it only has 1 note, > > How so?Everything's a unison!>> so the analogy breaks down. But whatever, I'll call >> all the intervals unison vectors anyway. >> _All_ the intervals? Yep!>> Now, we can also define meantone as >> >> ( 1 0 0) (1 0) >> ( 0 1 0)C = (0 1) >> (-4 4 -1) (0 0) >> >> Here, the octave and twelfth are taking the place of the chromatic > unison>> vectors. So, can we call them unison vectors? >> A unison vector has to come out to a musical _unison_! An augmented > unison maybe, but still a unison.Fokker says, in "Unison Vectors and Periodicity Blocks" <A.D. Fokker: Unison Vectors and Periodicity Bl... * [with cont.] (Wayb.)> "All pairs of notes differing by octaves only are considered unisons." That's not far from "an octave is a unison vector".>> I think not, because C >> evaluates to >> >> ( 1 0) >> ( 0 1) >> (-4 4) >> >> That has negative numbers in it, which means the generators (for so > they>> are) don't add up to give all the primary consonances. > > Don't follow.You can't define the 5:4 approximation by adding octaves and twelfths. This gives us a way of defining "small" intervals without needing interval sizes. We need to construct a scale somehow or other, and say that scale has to be constructed by adding together a set of linearly independent small intervals. The only problem is that they can be smaller than a unison. So adding a major and minor tone can be replaced by a major third and a descending minor tone. So the next constraint is that the scale has to have monotonically increasing pitch, which you need the metric for.>> So now we have a >> definition for what interval qualifies as a "unison vector". >> Interestingly, a fifth would work as a generator here, but not for > a >> diatonic scale. >> You've completely lost me.Four fifths add up to a 5:1 approximation. But you can't define a diatonic scale by adding fifths and octaves.>> An alternative to "unison vector" would be "melodic generator". >> Ouch! Are you sure? The steps in the scale can't be unison vectors, > nor can the generator of the scale be a unison vector.I don't know what to call them.>> Usually we call the octave the period and a fifth (which is > therefore>> equivalent to the twelfth) the generator. It isn't much of a > stretch to>> think of two generators. >> Gene has referred to that concept, I believe. Yes.>> In octave-equivalent terms, a unison is: >> >> v v v v v v v v v v v w >> >> where v is s fifth and w is a wolf. The same algebra applies, but > it's>> harder to visualise. I think the fifth here does count as a unison >> vector. >> Please, please, please don't refer to a fifth as a unison vector. > Musical intervals go: unison, second, third, fourth, fifth. There's a > long way between a unison and a fifth.Probably it'd be better written as v v v v v v v v v v v v p where v is a fifth and p is a Pythagorean comma. That means p is the (chromatic) unison vector. That leaves v as "the thing you have left when you take out all the unison vectors". That's what we need a name for.>> Defining an MOS in octave-equivalent terms is easy, once you > realize that>> the generators that pop out needn't be unison vectors in octave- > specific >> space. >> They can't be unison vectors in any space! If the generator is a > unison vector, you never get beyond the first note of the scale and > its chromatic alterations.But the generator does have a close relationship to the chromatic unison vector, in an octave equivalent system. You're taking out everything except the chromatic unison vector, and describing what you have left in terms of the generator. Graham
Message: 1360 - Contents - Hide Contents Date: Wed, 22 Aug 2001 00:02:20 Subject: Re: The hypothesis From: Dave Keenan --- In tuning-math@y..., graham@m... wrote:> In this case, it happens that the > simplest tuning will always have one interval in the set under > consideration just. As a mathematician, perhaps you can prove this.I'm pretty sure I once found a counterexample (but the margin was too small to contain it :-) where minimising the max absolute error did not produce any rational n-limit interval because the two equal maximum errors were in intervals that had no common factor e.g. 2:3 and 5:7. -- Dave Keenan
Message: 1361 - Contents - Hide Contents Date: Wed, 22 Aug 2001 15:05 +0 Subject: Re: The hypothesis From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9luhvn+n2lb@xxxxxxx.xxx> Gene wrote:>> Oh, right, but whatever they are, they're octave invariant. > CLAMPITT.PDF>> is a relevant place for definitions, as it's already been > referenced: "By>> /scale/ we refer to a set of pitches ordered according to ascending >> frequencies (pitch height) bounded by an interval of periodicity." >> This definition of a scale does not assume it repeats octaves, so it > does not assume octave invariance.It assumes it repeats, so it may as well repeat at the octave.> OK, we can consider a conceptual scale to be something defined on the > level of the homomorphic image (including the identity map.) Then C > major (or mean tone diatonic) is a conceptual scale in the mean-tone > linear temperament, which becomes a tuned scale if you pick a tuning. > How's that? Sounds okay.>> It looks like a commatic UV would be in the kernel, and a chromatic > UV >> would not. >> So far as I can see, they are both in the kernel and refer to the > same thing.They aren't the same thing. I'm not sure if they'd be in the kernel. I'd need to understand what "kernel" means a little better.>> You certainly can tell a small interval if 2 is taken out. Of > course, you>> need a way of calculating interval size. As I see it, that lies > outside>> of the group theory we've been discussing so far. There are two > ways of>> doing it, corresponding to the two interpretations above. >> I think your two ways in effect put the 2 back in by calculating what > it must be.Oh. Well, we can take factors of 2 out of the algebra.>> If >> it doesn't work out that way perhaps you, as the mathematician, can > tell>> us the size constraints on the unison vectors. >> So far as I can see, there aren't any. A unison vector is simply > anything you've decided to regard as a unison, and so lies in (or > generates?) the kernel.When working on my MOS finding program, I found some unison vectors didn't work. For example, take the meantone matrix [ 1 0 0] [-3 -1 2] [-4 4 -1] The (negative) adjoint is [ 7 0 0] [11 1 2] [16 4 1] The left hand column defines 7-equal. Now, if you replace 81:80 by 160:81, [ 1 0 0] [-3 -1 2] [ 5 -4 1] you get [ 7 0 0] [13 1 -2] [17 4 -1] so it doesn't work. Which means the size of the unison vectors must be important. However, for octave-equivalent matrices it doesn't seem to matter at all. Which is good, because "interval size" is harder to define anyway. The adjoint of [-1 2] [ 4 -1] is [-1 -2] [-4 -1] And the adjoint of [-1 2] [-4 1] is [1 -2] [4 -1] both define the same linear temperaments.>>> You can generate by fifths and octaves if you want, but you don't >>> need to. >>> You do need to if you want to enforce octave equivalence. >> I don't see why. Why not by octaves and major thirds instead?It won't work for meantone temperament. The temperament it does work for is excellent. I have an example in it at <http://www.microtonal.co.uk/magicpump.mp3 - Type Ok * [with cont.] (Wayb.)>.>> So the nature of this problem is such that I do need to make two > intervals>> exact. There are plenty of other problems that have different > natures,>> but I'm puzzled as to why you insist on bringing them up when we're >> discussing octave-invariant systems. >> I was discussing, among other things, tuning and temperament, where > the question of octave tuning is relevant. As I pointed out, tuning > an octave as a 2 is a choice of tuning.The subject line says "The hypothesis" and Paul's hypothesis most definitely concerns an octave-equivalent system. Graham
Message: 1362 - Contents - Hide Contents Date: Wed, 22 Aug 2001 00:23:55 Subject: Re: Chromatic = commatic? From: Dave Keenan --- In tuning-math@y..., graham@m... wrote:> In-Reply-To: <9lt3q7+gtig@e...> > In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:>> Why do you say >> linear temperament, which we've just determined means rank 2? >> Because the rest of the world says "linear temperament" and has done so > for longer than we've been alive.Sorry if I'm responsible for confusing you, Gene. Meantone remains a 5-limit linear temperament whether you are including 2s or ignoring them. When including 2s it appears 2 dimensional, when ignoring 2s it appears 1 dimensional. I really like your approach of treating 2s just like the other primes. We then need to switch our thinking to integer-limit rather than odd-limit when deciding which errors we care about. Or how about an a*b complexity limit for ratios a:b? -- Dave Keenan
Message: 1363 - Contents - Hide Contents Date: Wed, 22 Aug 2001 15:05 +0 Subject: Re: Chromatic = commatic? From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9lufbl+8f7m@xxxxxxx.xxx> Paul wrote:> --- In tuning-math@y..., graham@m... wrote: >>> For example, here's that pair of octave-invariant unison vectors: >> >> [ 4 -1] >> [ 0 -3] >> >> Invert it to get >> >> [-3 1] >> [ 0 4] >> >> The inverse times the determinant. Yes.>> Equal temperaments are tricky when they're octave invariant. > Effectively,>> it means the new period is a scale step. Hence all notes are > identical. >> Here you're going with the WF-related view that the interval of > repetition must be equal to the interval of equivalence. I suggest > instead the alternate view that they be allowed to differ, but that > the interval of equivalence is always an integer number of intervals > of repetition.I don't think MOS is any more clearly defined than WF in this respect. So let's assume they are identical. In which case for the things my program spits out to be MOSes, and therefore your hypothesis to be true, they must be the same. Or, at least, we need to redefine both to use the "period" instead of "interval of equivalence". The octave-equivalent algebra doesn't distinguish the different repeating blocks. The determinant tells you the number of steps to the octave, but to get the number of steps to any given interval you need to go octave-specific. The linear temperament result tells you the number of times the period goes into the interval of equivalence (octave) and the mapping in terms of generators. Again, you need the octave-specific algebra, or the metric, to get the mapping by generators *and* periods. Effectively, the octave-equivalent algebra collapses so that the interval of equivalence becomes the period in any given context.>> How about you stop telling us what we should be doing, and start > listening>> to what we're trying to say? >> Graham, I agree with you but maybe we should be willing to meet Gene > halfway? Dave Keenan expressed an opinion on this but I'm not sure > what it was.Gene's a better mathematician than I am, so he should be able to understand the octave-equivalent case. From his comments, it looks like he doesn't, so I'll keep trying to explain it until he does. If I can't manage that, what hope will I have with people who don't know their non-Abelian ring from their anomalous symmetric suspension? Graham
Message: 1364 - Contents - Hide Contents
Date: Wed, 22 Aug 2001 02:05:37
Subject: Interpreting Graham's matrix
From: genewardsmith@xxxx.xxx
Let's consider the situation where 81/80 is the commatic unison
vector, 25/24 is the chromatic unison vector, and 2 is what you might
call the repetition vector, defining the unit which is repeated
(since of course that doesn't actually need to be octaves.) Then if
each of these defines the row of a matrix, we get Graham's matrix
[ 1 0 0]
M = [-3 -1 2]
[-4 4 -1]
The inverse of this is
[ 7 0 0]
M^(-1) = 1/7 [11 1 2]
[16 4 1].
The question is how to interpret M^(-1). One use for it can be found
when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(-1)
and select all notes [a b c] such that
0 <= i < 1
-1/2 <= j < 1/2
-1/2 <= k < 1/2
we obtain the seven-note scale
1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)
with step sizes 9/8, 10/9 and 27/25. If we now temper by a meantone
system which equates 9/8 and 10/9 (which means we are using the fact
that 81/80 is a commatic vector) we get a PB scale with just two
sizes of intervals, LsLLLsL, ready to delight the fans of modal
melodies. Does this have anything to do with Graham's comments on
this matrix?
Message: 1365 - Contents - Hide Contents Date: Wed, 22 Aug 2001 15:05 +0 Subject: Re: Chromatic = commatic? From: graham@xxxxxxxxxx.xx.xx In-Reply-To: <9lul29+57gl@xxxxxxx.xxx> Gene wrote:> Actually, I'm an algebraist and number theorist, and algebraists are > notorious for wanting to solve a problem using only algebraic methods > when possible, which is what I was doing. If you calculate the "7" > using determinants, you can get the rest of the homomorphism by using > floating point calculations: 7 log_2(3) = 11.09... and 7 log_2(5) = > 16.25... As an algebraist, I naturally want to do the computation > using only algebra. As an algebraist, I also want to say "Now hold > on! What if I want to say a third has 3 steps, instead of two?" > Therefore I distinguish the systems [7, 11, 16] and [7, 11, 17].This is because you're expecting to get an octave-specific result. The octave-equivalent equivalent of an equal temperament is an MOS. (Or family of MOS scales, there's nothing special about the particular numbers of generators that give two interval sizes.) Hence the system gets defined a different way, but *can* be defined by pure algebra. For example, you could have 5-limit scales generated by fifths, with the octave as the period. The most common of these is meantone, where a third is four fifths. You can write that as [1 4]. An alternative is schismic, explained by either Ellis or Helmholtz, where a third is eight fourths. That's written [1 -8]. Both of these are fifth scales, but where a third is mapped differently. That's the same as the examples of 7 scales you gave, where a third is also mapped differently. Both cases submit to exactly the same algebra. The only difference is that there's a complication when you convert octave-equivalent vectors into pitch sizes. Another example of a fifth scale is the diaschismic family, which has already come up. There, a third is 2 half-octave reduced fourths. Crazily enough, that's written [2 -4]. It means a homomorphism of [1 -2] where the interval of equivalence is a half-octave.>> For example, here's that pair of octave-invariant unison vectors: >> >> [ 4 -1] >> [ 0 -3] >> >> Invert it to get >> >> [-3 1] >> [ 0 4] >> >> Each column is a mapping of generators. >> Each column is a mapping of everything in the 5-limit modulo octaves, > but I don't see how you draw the conclusions you do about them.Ah, well, that'll be why you think algebra doesn't apply. Usually, you think of 12-equal as being ... Bb B C C# D Eb E F F# G G# A Bb B C C# D ... Where it continues infinitely in either direction. In which case 5 steps is always a fourth, 7 is a fifth, 4 is a major third, etc. Truncate it to within an octave and you have C C# D Eb E F F# G G# A Bb B Now, the interval from B to C isn't a scale step any more. And the interval from F to C isn't a fifth. This is a problem. It means equal temperaments can't be expressed in terms of scale steps in an octave equivalent system, unless you explicitly define the system to be cyclic. That is, say "the note after B is C again". However, those notes can also be written like this: Eb Bb F C G D A E B F# C# G# That's the circle of fifths. You'll have seen it before. Now, the interval from F to C is a fifth, and C to F is a fourth. B to C is the same as E to F, and so that can be called a "semitone". However, the interval from G# to Eb should be a fifth, but isn't. So the system still needs to be made cyclic: ... Db Ab Eb Bb F C G D A E B F# C# G# D# ... By making G#=Ab, C#=Db, and so on, we have another definition of 12-equal. However, by taking the sequence at face value, turning the circle into a spiral we've defined meantone. All the notes are in a line, so it's a linear temperament. It's this temperament that setting 81:80 to be a commatic unison vector gives. Its homomorphism is given by the right hand column of [-3 1] [ 0 4] Or [1 4] as a row vector. It tells you that a fifth is 1 step and a major third is four steps. Check with the line above, you'll see it's correct. All the algebra doesn't tell us is how to tune that "generator" step ...> If you are looking *only* at diatonic music and not allowing any > accidentals, then neither one is a scale step; they seem therefore to > be treated in the same way. The comma, being smaller, lends itself > better to approximations, of course.Thanks! I think I understand the term "chromatic unison vector" better now!> You're missing my point--the world does not call something we get > from codimension 1 a linear temperament, it calls codimension 2 (with > 2 generators) a linear temperament.I'm fully aware that the world has this wrong, but I go along with the accepted usage. Still, as I show above, you can describe a linear temperament using a codimension of 1.> In a mathematical subject, mathematics is your friend; which means > potentially so is a friendly mathematician. Why not view me as an > ally in this endeavor?But I do! Why else would I spend so much time with you? Graham
Message: 1366 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:19:22 Subject: Re: Chromatic = commatic? From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> > If you are looking *only* at diatonic music and not allowing any > accidentals,We _are_ allowing accidentals, and they play a very specific role in diatonic music (as opposed to, say, atonal serial music).
Message: 1367 - Contents - Hide Contents Date: Wed, 22 Aug 2001 07:16:40 Subject: A counterexample to the hypothesis? From: genewardsmith@xxxx.xxx Let's see if I understand the hypothesis well enough to have found a counterexample. Suppose I decide to construct a scale with 15 notes to the octave in 7-limit harmony. I therefore turn my attention to the homomorphic map [15, 24, 35, 42] which in spite of the way I have written it I may consider to be a column vector. The kernel of this homomorphism is generated by 28/27, 64/63 and 126/125, of which the latter two seem to be good candidates for commatic unisons and the first for our chromatic unison. Setting up a Graham-type matrix and inverting it, I find 15 notes within an octave forming a scale: 1 - 21/20 - 10/9 - 9/8 - 6/5 - 5/4 - 4/3 - 7/5 - 10/7 - 3/2 - 8/5 - 5/3 - 16/9 - 9/5 - 40/21 - (2) This scale has intervals of sizes 81/80, 50/49, 25/24, 21/20, 16/15 and 200/189. Since six intervals of widely differing sizes will surely drive us nuts, we decide to temper out the two unison vectors, namely 64/63 and 126/125. One simple way to do this is to use the 27- et, which has both of them in its kernel (like the 15-et.) When we do this, we find that 81/80 and 50/49 go to intervals of one scale step, while the rest go to intervals of two scale steps. We now have the pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a third of an octave. Isn't this a counterexample?
Message: 1368 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:24:53 Subject: Re: Scales From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> Let's see if I understand the situation. In a p-limit system of > harmony G, containing n primes, we select n-1 unison vectors and the > prime 2, which is a sort of special unison vector. This defines a > kernel K and so congruence classes on G, and a homomorphic image H = > G/K which under the best circumstances is cyclic but which in any > case is a finite abelian group of order h.What differentiates the "best circumstances" and just "any case" here?> We now select a set of > coset representatives in G for H by taking elements in one period of > the period lattice defined by the n generators of K. > > Now we select out a certain number m+1 of the unison vectors, > including 2, and call them (aside from the 2) chromatic. The > remaining unison vectors now define a temperament whose codimension > is m+1, and hence is an m-dimensional temperament. We now select a > tuning for that temperament, and we have a scale with h tones in an > octave. If m = 0 we have a just scale;You mean an equal temperament?> if m = 1 we have the scales > with one chromatic vector which people have been talking about. Sounds right.
Message: 1369 - Contents - Hide Contents Date: Wed, 22 Aug 2001 08:14:07 Subject: Re: About CS From: genewardsmith@xxxx.xxx --- In tuning-math@y..., Pierre Lamothe <plamothe@a...> wrote:> The CS property corresponds to the main axiom in the gammoid theory (the > _congruity_ condition) and subtends the _periodicity_ concept in Z- module. > If you flush CS you loss the mathematical sense of _interval class_ (or do > you have a _fuzzy class_ definition to replace it?) > > I can't see how you could define mathematically such terms as step, degree, > class, mode . . . in a system S without the existence of a primordial > epimorphism S --> Z which gives its consistence.Hmmm. A Z-module is the same as an abelian group, which I've been raving about, and an epimorphism to Z sounds like an ET to me. Other than that I don't know what this is about, but I wonder if Pierre shows up here from time to time?
Message: 1370 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:27:00 Subject: Re: Chromatic = commatic? From: Paul Erlich --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:> --- In tuning-math@y..., graham@m... wrote: >> In-Reply-To: <9lt3q7+gtig@e...>>> In article <9lt3q7+gtig@e...>, genewardsmith@j... () wrote:>>> Why do you say >>> linear temperament, which we've just determined means rank 2? >>>> Because the rest of the world says "linear temperament" and has done > so>> for longer than we've been alive. >> Sorry if I'm responsible for confusing you, Gene. > > Meantone remains a 5-limit linear temperament whether you are > including 2s or ignoring them. When including 2s it appears 2 > dimensional, when ignoring 2s it appears 1 dimensional. > > I really like your approach of treating 2s just like the other primes. > We then need to switch our thinking to integer-limit rather than > odd-limit when deciding which errors we care about. Or how about an > a*b complexity limit for ratios a:b?When discussing _tuning_, then these can all be useful considerations. But when discussing _scales which use octave- equivalence_, then a*b turns into odd-limit, as I've demonstrated on harmonic_entropy.
Message: 1371 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:28:51 Subject: Re: Interpreting Graham's matrix From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> Let's consider the situation where 81/80 is the commatic unison > vector, 25/24 is the chromatic unison vector, and 2 is what you might > call the repetition vector, defining the unit which is repeated > (since of course that doesn't actually need to be octaves.) Then if > each of these defines the row of a matrix, we get Graham's matrix > > [ 1 0 0] > M = [-3 -1 2] > [-4 4 -1] > > The inverse of this is > > [ 7 0 0] > M^(-1) = 1/7 [11 1 2] > [16 4 1]. > > The question is how to interpret M^(-1). One use for it can be found > when calculating PB's--if we define [i j k] by [i j k] = [a b c]M^(- 1) > and select all notes [a b c] such that > > 0 <= i < 1 > -1/2 <= j < 1/2 > -1/2 <= k < 1/2 > > we obtain the seven-note scale > > 1 - 10/9 - 6/5 - 4/3 - 3/2 - 5/3 - 9/5 - (2)This is exactly how my PB program works, as explained in part 3 of the _Gentle Introduction_.
Message: 1372 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:39:35 Subject: Re: Tetrachordal alterations From: Paul Erlich --- In tuning-math@y..., David C Keenan <D.KEENAN@U...> wrote:> I've ignore tetrachordality for too long, it's time I figured out how to > make it happen in any linear temperament. Paul, I had hoped you had more of > a handle on it than I did. It seems not. > > --- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:>> --- In tuning-math@y..., "Dave Keenan" <D.KEENAN@U...> wrote:>>> Are "tetrachordal alterations" only possible when the interval of >>> repetition is some whole-number fraction of octave? >>>> In an MOS, the interval of repetition is _always_ some whole- number >> fraction of the interval of equivalence. >> I meant "fraction" in the popular sense, as not including the whole. But > no, I now believe tetrachordal (not necc. omnitetrachordal) alterations are > always possible, it's just a question of finding the minimal alteration > that makes it tetrachordal and just how big that alteration turnsout to be> in each case. > > It's not clear to me why a tetrachordal scale in some linear temperament, > must have the same number of notes as a MOS in that temperament, but I'll > assume it for now. Know any good reason(s)?Because the unison vectors define a set of N equivalence classes, which we typically map to letters of the alphabet, etc. Melodies, and even harmonies, will be understood in terms of patterns in this set of N. If you add or eliminate notes, you'll be ruining these patterns. In other words, the tetrachordal scale should be a periodicity block (not necessarily of the Fokker parallelogram type).> > Yes. But I wasn't considering _omni_tetrachordality. That looks too hard > for now. Do any popular or historical scales have it?The most popular ones do -- the diatonic and pentatonic.> > Here's what they look like melodically (in steps of 72-tET). Vertical bar > "|" is used to show tetrachords. > > 7 7 7 9|5 7|7 7 7 9 > 7 7 9 7|5 7|7 7 9 7 7 7 7 9|7 5|7 7 7 9 > 7 9 7 7|5 7|7 9 7 7 7 7 9 7|7 5|7 7 9 7 > 9 7 7 7|5 7|9 7 7 7 7 9 7 7|7 5|7 9 7 7 > 9 7 7 7|7 5|9 7 7 7I might not call these tetrachordal. According to my paper, the disjunction is supposed to contain some pattern of scale steps found with the tetrachords themselves. But in this case, the 5/72 interval isn't found in the tetrachords. Perhaps we can call these "weakly tetrachordal".> But I don't see any sense in which any of these tetrachordal scales are an > "alteration" of the 10 note MOS.Each note should either agree with the corresponding note in the MOS or be a chromatic unison vector different.> It seems to me that tetrachordal scale > creation (in a given temperament) is not related to MOS scale creation (in > the same temperament) in any way.I see the periodicity block concept as underlying both.
Message: 1373 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:41:01 Subject: Re: A counterexample to the hypothesis? From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> We now have the > pattern LLsLLLLsLLLLsLL, which has three repeating divisions of a > third of an octave. Isn't this a counterexample?Why would this be a counterexample if LssssLssss wasn't?
Message: 1374 - Contents - Hide Contents Date: Wed, 22 Aug 2001 19:41:59 Subject: Re: About CS From: Paul Erlich --- In tuning-math@y..., genewardsmith@j... wrote:> but I wonder if Pierre > shows up here from time to time?Not often enough, sadly. How's your French?
1000 1050 1100 1150 1200 1250 1300 1350 1400 1450 1500 1550 1600 1650 1700 1750 1800 1850 1900 1950
1350 - 1375 -