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).

8000
8050
8100
8150
8200
8250
8300
8350
**8400**
8450
8500
8550
8600
8650
8700
8750
8800
8850
8900
8950

**8400 -**
8425 -

Message: 8400 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:01:46 Subject: Re: Vals? From: Graham Breed Dave Keenan wrote:> "m" here is the grade of the object, i.e. the number of nested > brakets. A more intuitive (for me) alternative to m(m+1)/2 is > Ceiling(m/2). If the sum of the indices plus this quantity is even > then you negate it when complementing.That's an interesting short cut. The way I've been doing it is: Join the old basis on the left and the new basis on the right to get a list of all primitive bases (or whatever they are). If this is an odd permuatation, negate the coefficient. This assumes all bases are being stored in numerical order of their components (what's all this talk about alphabetical order?) You can test for an odd permutation by swapping adjacent pairs of primitive bases until their in the right order, and it's an odd permutation if you did an odd number of swaps. This follows from the antisymmetry of the wedge product, which is the main thing you need to know about it. You can also compare every pair of numbers in the basis, and it's an odd permutation iff an odd number of them are the wrong way round. Graham

Message: 8401 - Contents - Hide Contents Date: Wed, 19 Nov 2003 04:39:07 Subject: Re: Vals? From: Dave Keenan OK. With lexicographic ordering of the indices, it isn't as simple as negating every second coefficient. There's sometimes a hiccup in the middle. It's explained in Section 5.4 of http://www.ses.swin.edu.au/homes/browne/grassmannalgebra/book/bookpdf/TheComplement.pdf - Type Ok * [with cont.] (Wayb.)

Message: 8402 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:04:18 Subject: Re: Vals? From: Graham Breed Paul Erlich wrote:> yes, as you know i (and especially graham) like that idea very much -- > BUT 88cET has no octaves!You can make 88cET equivalent with respect to some other interval -- like 88 cents for example. Graham

Message: 8403 - Contents - Hide Contents Date: Wed, 19 Nov 2003 04:43:25 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> So the 3D wedge product is not quite the same as the cross-product.Nor is the 2D wedge product exactly multiplication of complex numbers. You need to take a compliment, since a 3D wedge is a 2- vector, not a vector.> The cross product is actually the wedge-product followed by a > complementation. What should we use for the complement operator? Tilde?I've been using asterisk, but if tilde is clearer I will change what I've written.> E.g. With Pauls's example of the syntonic comma and diaschisma and > 12-ET, wedging the two comma monzos gives > > [-4 4 -1> ^ [-11 4 2> = [[28 19 12>> a bimonzoGoing strictly by alphabetical ordering, this would be [-4 4 -1> ^ [-11 4 2> = [[28 -19 12>> after which [[28 -19 12>>* = [12 19 28> and we have the cross product.> So why no problems with minus signs in 3D?You used a basis without them, because you didn't alphabetically order the basis wedges.

Message: 8404 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:09:45 Subject: Re: Vals? From: Graham Breed Dave Keenan wrote:> Well, in ASCII * is of course most commonly infix for multiplication. > As postfix I'm used to it being the complex-conjugate operator which > doesn't seem as analogous as prefix ~ for the logical complement.I prefer prefix ~ because I can get it to work in Python. But there certainly is a connection with complex conjugates as it's one thing Dirac's bras and kets distinguish. Quantum mechanics works in a Hilbert space where |x| = xx* = 1. Hermitian conjugates if I remember rightly. Logical complements are often done with a bar above the symbol. The usual - sign is a more obvious analogy. Graham

Message: 8405 - Contents - Hide Contents Date: Wed, 19 Nov 2003 04:44:56 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> I suspect if you just put in "don't cares" for some of the > coefficients, they will propagate sensibly. e.g. Use NaNs > (Not-a-number) in IEEE Floating point, and display them as "X"s.What does a don't care do to dimensionality? If we are not worried about that, we can use 0.

Message: 8406 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:11:26 Subject: Re: Vals? From: Graham Breed Dave Keenan wrote:>> I'm still confused here. >> >> So the complement operation keeps the braket pointing in the same >> direction? >> >> So <12 19 28] is not the complement of [[28 -19 12>> but is simply >> _equal_ to it (because it has a reciprocal basis)?The bra and ket notation should make complements explicit, although Gene seems to like them specified anyway. Graham

Message: 8407 - Contents - Hide Contents Date: Wed, 19 Nov 2003 04:46:50 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> Further to my question of whether the cross-product is the 3D wedge > product followed by complementation, is it the case that the > dot-product is complementation of the second argument followed by the > wedge-product followed by complementation?This sounds a little backwards--it's the inner product for n-vals with n-monzos that lets you define the compliment.

Message: 8408 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:19:34 Subject: Re: Vals? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Dave Keenan wrote: >>> Well, in ASCII * is of course most commonly infix for multiplication. >> As postfix I'm used to it being the complex-conjugate operator which >> doesn't seem as analogous as prefix ~ for the logical complement. >> I prefer prefix ~ because I can get it to work in Python. But there > certainly is a connection with complex conjugates as it's one thing > Dirac's bras and kets distinguish. Quantum mechanics works in a Hilbert > space where |x| = xx* = 1. Hermitian conjugates if I remember rightly. > > Logical complements are often done with a bar above the symbol. The > usual - sign is a more obvious analogy.Grassman himself apparently used a prefix vertical bar. John Browne uses a horizontal bar above the symbol (or above a whole expression) exactly as you describe for logical complements. But this is usually translated to a prefix tilde ~ in ASCII, and it has the advantage of looking similar to a minus sign - which you say is more analogous, but is different from a prefix minus sign which would have the more obvious interpretation of negating _all_ the coefficients (and not reversing their order or the brakets).

Message: 8409 - Contents - Hide Contents Date: Wed, 19 Nov 2003 05:05:09 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> Could a minus sign or two be required in the bimonzo, which then > disappear when you take the complement?The most straightforward way to do it would be that. 5-limit [a1,a2,a3>* = <<a3,-a2,a1]] [[a1,a2,a3>>* = <a3,-a2,a1] 7-limit [[a1,a2,a3,a4,a5,a6>>* = <<a6,-a5,a4,a3,-a2,a1]] 11-limit [[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10>>* = <<<a10,-a9,a8,a7,-a6,a5,-a4,a3,-a2,a1]]] [[[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10>>>* = <<-a10,a9,-a8,a7,-a6,a5,-a4,a3,a2,-a1]] 13-limit [[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15>>* = <<<<a15,-a14,a13,-a12,a11,a10,-a9,a8,-a7,a6,-a5,a4,a3,-a2,a1]]]] [[[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18,a19, a20>>>* = <<<-a20,a19,-a18,a17,-a16,a15,-a14,-a13,a12,-a11,a10,-a9,a8,a7,- a6,a5,-a4,a3,-a2,a1]]] [[[[a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15>>>>* = <<a15,-a14,a13,-a12,a11,a10,-a9,a8,-a7,a6,-a5,a4,a3,-a2,a1]] Unless I've made a mistake; I'll check this.> The basis for the monzos is > lg(2) lg(3) lg(5) > where lg() is a logarithm function of arbitrary base.This strikes me as a weird notion. It is true that <1 log2(3) log2(5)] (not a val, but a perfectly acceptable bra vector) is important in the 5-limit, as mapping monzos to log2 of the corrsponding rational number--or we could use <1200 1200*log2(3) 1200*log2(5)] instead.> According to John Browne, the above bimonzo is correct if its basis is > > lg(2)^lg(3) lg(5)^lg(2) lg(3)^lg(5) (indices 12 31 23)Browne doesn't use alphabetical ordering of the basis wedges? I thought alphabetical was what you had settled on.> And it starts to look like the general complement (for any grade and > dimension) should not only reverse the order of coefficients, but > negate every second one. But I'm really not sure. Gene, help!It isn't even that easy; the coefficients get reversed, but the sign has to be figured out from whether the permutation you get is even or odd.

Message: 8410 - Contents - Hide Contents Date: Wed, 19 Nov 2003 22:02:24 Subject: Finding the compliment From: Gene Ward Smith This is what Browne calls the Euclidean compliment, and there is a somewhat nice and standard formula for this. I'm going to apply it to the 13-limit case for some examples. The basic idea is that we find the signs, for which we can use the formula, apply them, reverse the ordering, change the m-monzo to an (n-m)-val, and we have the compliment of a monzo. The same applies the other way around for vals. If we have a 13-limit 2-monzo, it will have dimension 6*5/2 = 15 Let [[c1 c2 c3 c4 c5 c6 c7 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>> be that monzo. Now take the ordered combinations of six things, taken two at a time, in alphabetical order: [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5], [2, 6], [3, 4], [3, 5], [3, 6], [4, 5], [4, 6], [5, 6]] Sum the numbers in each and put down a 1 if the sum is even, and a -1 if it is odd, giving [-1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1] Now note that m=2, so m(m+1)/2 is 2*3/2 = 3, an odd number. Hence we must multiply the above by -1; at the same time we will reverse the ordering, giving us now [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1] We now use these signs, together with the coefficients in descending order, to get the complimentary 4-val: <<<<c15 -c14 c13 c12 -c11 c10 -c9 c8 -c7 c6 c5 -c4 c3 -c2 c1]]]] This isn't what I got before, where I presumably goofed. A 13-limit 3-monzo has dimension 6*5*4/6 = 20, and may be written [[[c1 c2 c3 c4 c5 c6 c6 c8 c9 c10 c11 c12 c13 c14 c15 c16 c17 c18 c19 c20>>> The ordered combinations of [1,2,3,4,5,6] taken three at a time are [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5], [1,3, 6], [1, 4, 5], [1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6], [3, 4, 5], [3, 4, 6], [3, 5, 6], [4, 5, 6]] Taking the sums of the coefficients of these and putting 1 for even, -1 for odd gives us [1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1] We are taking the compliment of a 3-monzo, so m=3 and m(m+1)/2 = 6, an even number; hence we don't change sign. However, we do as always reverse the ordering and take the coefficients in reverse order: [-1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1] <<<-c20 c19 -c18 c17 -c16 c15 -c14 -c13 c12 -c11 c10 -c9 c8 c7 -c6 c5 -c4 c3 -c2 c1]]] Now for the compliment of a 4-monzo, which is a 6*5*4*3/24 = 15 dimensional object again [[[[c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>>>> The combinations of [1,2,3,4,5,6] taken four at a time in order are [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 3, 6], [1, 2, 4, 5], [1, 2, 4, 6], [1, 2, 5, 6], [1, 3, 4, 5], [1, 3, 4, 6], [1, 3, 5, 6], [1, 4, 5, 6], [2, 3, 4, 5], [2, 3, 4, 6], [2, 3, 5, 6], [2, 4, 5, 6], [3, 4, 5, 6]] Even and odd for this gives [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1] m=4, and 4*5/2 = 10, an even number, so we don't change sign but as always do reverse the order [1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1] Taking these signs with descending coefficients gives <<c15 -c14 c13 -c12 c11 c10 -c9 c8 -c7 c6 -c5 c4 c3 -c2 c1]] as the compliment.

Message: 8411 - Contents - Hide Contents Date: Thu, 20 Nov 2003 17:33:23 Subject: Re: "does not work in the 11-limit" From: Manuel Op de Coul Carl wrote:>If you can do MTS, I don't know what it does to the track->chan >mapping. Maybe Manuel can chime in.Then it's just a one-to-one mapping from tracks to channels. But if no editing of the seq files is needed, then it's of course easier to transform the Cakewalk midi file directly to another midi file also without channel swapping if MTS is used.>Another reason for >assigning channels this way is that I experienced that having two >different patches assigned to the same channel results tends to >corrupt the quality of the sound.With an external instrument that will always be the case. Maybe a good quality soundcard doesn't have this problem? Manuel

Message: 8412 - Contents - Hide Contents Date: Thu, 20 Nov 2003 18:59:41 Subject: Re: Definition of val etc. From: monz hi Dave, thanks *very* much for these suggestions. just a couple of things before i actually do incorporate them ... in general, the shortest and most compact terminology is the one that gets the most use anyway, and the one i prefer to promote. thus, i think that the nice new definition you gave below should be that for "map", with all the other terms pointing to *it*. and likewise for "monzo". ... of course i also have my own reasons for promoting that particular term. ;-) also, i believe that "vector" should retain its general defintion. there's another term "interval vector" which i haven't yet put in the Dictionary but which is common currency in atonal music-theory. -monz --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> I suggest that the definition of val stay as Gene had it, as a > definition of a pure math term. But that we add something like the > following text at the start of it. > > -------------------------------------------------------------------- -- > "val" is a term coined by Gene Ward Smith for the mathematical object > described below. When vals are applied to tuning theory they are > usually interpreted as "prime exponent mappings". > -------------------------------------------------------------------- -- > > Then a link should take you to a definition of "prime exponent mapping": > > -------------------------------------------------------------------- -- > A "prime exponent mapping", sometimes shortened to "prime mapping", > "exponent mapping", "mapping" or simply "map", is a list of numbers > (integers) enclosed in < ... ] that tell you how a particular > temperament maps each prime number (up to some limit) to numbers of a > particular "generator" in that temperament. The prime numbers here > represent frequency ratios. > > The simplest case is an equal temperament where the generator is the > step interval. For example, the 5-limit map for 12-equal is <12 19 28] > which means it takes 12 steps to make an octave (1:2), 19 steps to > make a twelfth (1:3), and 28 steps to make a 1:5 interval. > > When an interval is represented in the complementary form, as a > prime-exponent-vector, we can find the number of generators > corresponding to it in some temperament by multiplying each number in > the temperament's map by the corresponding number in the vector, and > adding up the results. In mathematical terms this is called the > dot-product, scalar-product or inner-product of the map with the vector. > > For example the interval 3:5 (a major sixth), has the 5-limit exponent > vector [0 -1 1>. To find how many steps of 12-equal it maps to, we write > > <12 19 28].[0 -1 1> > = 12*0 + 19*-1 + 28*1 > = 28 - 19 > = 9 > -------------------------------------------------------------------- -- > > We should also have a definition for "prime exponent vector" and the > definition for "monzo" should simply say "See "prime exponent > vector"", and have a link to it. > > Links should also take you from > "map", > "mapping", > "prime mapping" and > "exponent mapping" > to > "prime exponent mapping", > > and from > > "monzo", > "vector", > "prime vector" and > "exponent vector" > to > "prime exponent vector". > > The definition for "prime exponent vector" would be pretty much what > Monz has now for "vector", with the addition of some stuff about the [ > ... > notation and the relationship to "prime exponent mapping". > > -- Dave Keenan

Message: 8413 - Contents - Hide Contents Date: Thu, 20 Nov 2003 20:25:38 Subject: Re: Definition of val etc. From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote: {{A "prime exponent mapping", sometimes shortened to "prime mapping", "exponent mapping", "mapping" or simply "map", is a list of numbers (integers) enclosed in < ... ] that tell you how a particular temperament maps each prime number (up to some limit) to numbers of a particular "generator" in that temperament.}} This assumes that all such mappings are (equal, and you need to say that) temperaments, which is not true. Also, simply calling it a "map" won't work as a specific shorthand, since that already has a well-established meaning, as another word for "function" which is more often used in some contexts ("homomorphic map" being one example.) I find "prime exponent mapping" too clumsy, too confusing, and too verbose, and have no plans to use the term. {{The prime numbers here represent frequency ratios.}} This is at best confusing; the prime numbers are prime numbers and don't represent anything else. Tuning is another matter. {{When an interval is represented in the complementary form...}} "Complimentary form" is not a good phrase to use here. {{...as a prime-exponent-vector, we can find the number of generators corresponding to it in some temperament by multiplying each number in the temperament's map by the corresponding number in the vector, and adding up the results.}} This is assuming the mapping in question defines an equal temperament (and again leaves out the word equal), which is hardly always the case. {{In mathematical terms this is called the dot-product, scalar- product or inner-product of the map with the vector.}} This is unfortunately the case--unfortunate, in that these in most contexts mean a product defined on a single vector space, not on a vector space with its dual. As if that were not enough, "interior product" in the context of exterior algebra has a specialized meaning that we probably don't want to mess with. What about simply calling it the bracket and leaving it at that?

Message: 8414 - Contents - Hide Contents Date: Thu, 20 Nov 2003 20:26:22 Subject: Re: Finding the wedge product? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> So Gene, you've given us a nice description of how to compute the > complement for arbitrary dimension and grade. Now how about the wedge > product between pairs of multivectors of arbitrary dimension (and > possibly different) grades? Graham, feel free to chime in here too. :-)I suggest simply using the definition.

Message: 8415 - Contents - Hide Contents Date: Thu, 20 Nov 2003 20:30:52 Subject: Re: Definition of val etc. From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote:> hi Dave, > > > thanks *very* much for these suggestions. > just a couple of things before i actually do > incorporate them ...I hope you take into account my comments before rushing out to do this.> in general, the shortest and most compact terminology > is the one that gets the most use anyway, and the one > i prefer to promote. I concur. > thus, i think that the nice new definition you gave > below should be that for "map", with all the other > terms pointing to *it*. and likewise for "monzo"."Map" is a synonym for "function", let's leave it that way. I came up with the word "val" because we needed a term, and the reason is still a good one.> also, i believe that "vector" should retain its general > defintion.I concur. Just to make matters worse, vectors of integers define free abelian groups rather than vector spaces.

Message: 8416 - Contents - Hide Contents Date: Thu, 20 Nov 2003 21:05:10 Subject: Re: Finding the wedge product? From: Graham Breed Dave Keenan wrote:> So Gene, you've given us a nice description of how to compute the > complement for arbitrary dimension and grade. Now how about the wedge > product between pairs of multivectors of arbitrary dimension (and > possibly different) grades? Graham, feel free to chime in here too. :-)Certainly, to find the wedge product of two wedgies x an y: Initialize the result to the empty wedgie. Take every distinct basis element of x in turn (by distinct, I mean don't take both . For each of them, take every distinct basis element of y in turn. If both basis elements have a primitive basis in common (I still don't know the proper word for this) ignore the term. Otherwise, form a new basis by concatenating the original two, and add the product of the coefficients of x and y for their respective basis elements to the current coefficient for the resultant basis in the result. Here's some abbreviated Python code in case it helps: def wedgeProduct(x, y): result = Wedgie() for base1, value1 in x.items(): for base2, value2 in y.items(): for element in base1: if element in base2: break else: base = base1+base2 result[base] = result[base] + value1*value2 return result All this assumes you know how to reduce each basis element to a consistent form, which I think was covered in the complement discussion. For an example, I'll try |1 2 3> ^ |4 5 6> They can be written as (e1 + 2*e2 + 3*e3) ^ (4*e1 + 5*e2 + 6*e3) So the first elements of each are e1 and 4*e1. We see that there's an e1 in both of these, so ignore the term. The next element of y is the 5*e2. So the new basis element is e1e2 and the coefficient is 5. The result's value for e1e2 is 1+5=5. Then from y we have 6*e3, so the result is now (5*e1e2 + 6*e1e3) Now we take the next element of x, 2*e2. (2*e2) ^ (4*e1) is 8*e2e1. The current value of e2e1 is -5 (because e1e2 is 5) so e2e1 is now 8-3=3 (or e1e2 is -3) and result is (-3*e1e2 + 6*e1e3). Then we have two terms with an e2, so move on. 2*e2 ^ 6*e3 = 12*e2e3. The result doesn't already have a term in e2e3, so it's now (-3*e1e2 + 6*e1e3 + 12*e2e3). Now we move on to the last element of x, 3*e3. 3*e3 ^ 4*e1 is 12*e3e1 or -12*e1e3. Result is now (-3*e1e2 - 6*e1e3 + 12*e2e3). 3*e3 ^ 5*e2 is 15*e3e2 or -15*e2e3 giving a result of (-3*e1e2 - 6*e1e3 - 3*e2e3). And that's the final result, because we can throw away the term in e3^e3. Hence (e1 + 2*e2 + 3*e3) ^ (4*e1 + 5*e2 + 6*e3) = (-3*e1e2 - 6*e1e3 - 3*e2e3) or, the way I happen to do the flattening [1 2 3> ^ [4 5 6> = [[-3 -6 -3>> Graham

Message: 8417 - Contents - Hide Contents Date: Thu, 20 Nov 2003 01:11:28 Subject: Re: Finding the complement From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> This is what Browne calls the Euclidean compliment, and there is a > somewhat nice and standard formula for this. I'm going to apply it to > the 13-limit case for some examples.Thanks very much for these. Please don't be offended by my attempts to translate some parts in a way that I imagine might be more easily comprehended by Monz and others.> The basic idea is that we find > the signs, for which we can use the formula, apply them, reverse the > ordering, change the m-monzo to an (n-m)-val, and we have the > compliment of a monzo. The same applies the other way around for vals. > > If we have a 13-limit 2-monzo, it will have dimension 6*5/2 = 15It think it will cause less confusion if we call them bi-monzos and tri-monzos (with or without hyphens). There are enough numbers flying around as it is. But by all means start using the numbers once we get beyond grade 3. And why not use "g" for a variable grade instead of "m"? And call them multi-monzos when the grade is unspecified and doesn't need to be referred to in any formula. This is the sort of stuff that doesn't matter to someone who already understands it, but for someone trying to break into it, every little bit helps. Now if the "dimension" of a multi-monzo is the number of coefficients in it, then what do you call the thing that I've been calling the dimension, which in our application is the index of the highest prime that's represented? That seems more like "dimension" to me, since it is the dimension of the underlying simple space. The other I have been calling simply "number of coefficients". To say that in 4D-space a plane has dimension 6 wouldn't make a lot of sense to most people. To say that the bi-vector representing it has 6 coefficients, is more likely to be understood. Translation: A 13-limit monzo (prime-exponent-vector) has 6 coefficients so a 13-limit bi-monzo, will have 6*5/2 = 15 coefficients. In general, if the monzo has n coefficients, the corresponding bi-monzo will have n*(n-1)/2 coefficients. The same goes of course for maps and bi-maps (vals and bi-vals).> Let > > [[c1 c2 c3 c4 c5 c6 c7 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>> > > be that monzo.You mean "bi-monzo".> Now take the ordered combinations of six things, taken > two at a time, in alphabetical order: > > [[1, 2], [1, 3], [1, 4], [1, 5], [1, 6], [2, 3], [2, 4], [2, 5], > [2, 6], [3, 4], [3, 5], [3, 6], [4, 5], [4, 6], [5, 6]]It might be clearer to use compound indices in the bimonzo as follows (this trick will work up to the 23-limit, beyond which we have more than 9 coefficients in the monzo or map, but we hardly ever go there). Graham raised the question, why talk about alphabetical order when it's numerical order. Well, up to simple indices of 9, they are the same thing, but beyond that if you want it to still be numerical you have to use a number base greater than the maximum simple index. The monzo [c1 c2 c3 c4 c5 c6> represents the 13-limit ratio 2^c1 * 3^c2 * 5^c3 * 7^c4 * 1^c5 * 13^c6. The bimonzo [[c12 c13 c14 c15 c16 c23 c24 c25 c26 c34 c35 c36 c45 c46 c56>> represents a two 13-limit ratios simultaneously, but in such a way that the bimonzo for any other pair which are linear combinations of this pair would give the same bimonzo. For example, if these are two commas which vanish in some temperament, then any other pair of commas whose vanishing would define the same temperament will have the same bimonzo. There's no redundancy as there would be in a matrix representation of this (which would have 6x6 = 36 coefficients). Notice that _within_ each compound index for a single coefficient, the digits are in numerical/alphabetical order, and the same digit does not occur more than once. Notice also that the coefficients themselves are listed in numerical/alphabetical order of their compound index.> Sum the numbers in each and put down a 1 if the sum is even, and a -1 > if it is odd, giving > > [-1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1] > > Now note that m=2,The grade g is 2 since it's a _bi_monzo. This is what the number of brackets shows us.> so m(m+1)/2 is 2*3/2 = 3, an odd number.so Ceiling(g/2) = 1, an odd number.> Hence we > must multiply the above by -1; at the same time we will reverse the > ordering, giving us now > > [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1] > > We now use these signs, together with the coefficients in descending > order, to get the complimentary 4-val: > > <<<<c15 -c14 c13 c12 -c11 c10 -c9 c8 -c7 c6 c5 -c4 c3 -c2 c1]]]]That's "compl_e_mentary", although it's true Gene didn't charge any money for it. ;-) Because the 13-prime-limit is a 6-dimensional space (2 3, 5 7 11, 13), the complement of a bi-monzo is a 6-2 = 4-map (4-val) and where the bi-monzo has 2 brackets pointing to the right, its complement has 4 brackets pointing to the left. <<<<c56 -c46 c45 c36 -c35 c34 -c26 c25 -c24 c23 c16 -c15 c14 -c13 c12]]]] So you don't actually have to write out the 1's and -1's. Just calculate Ceiling(g/2) and add it to the sum of the digits making up each compound index and put a minus sign if the result is odd. You ask, how can I remember whether I put the minus sign when it's odd or even? Easy. You're actually multiplying them by -1^(sum_of_indices+Ceiling(g/2)) and minus-one to an even power is 1, while to an odd power it is -1. For example, consider c13 in our bimonzo. Ceiling(g/2) = 1 and 1 + 1 + 3 = 5, an odd number so it becomes -c13 in the complement.> This isn't what I got before, where I presumably goofed. > > A 13-limit 3-monzo has dimension 6*5*4/6 = 20, and may be written > > [[[c1 c2 c3 c4 c5 c6 c6 c8 c9 c10 > c11 c12 c13 c14 c15 c16 c17 c18 c19 c20>>> > > The ordered combinations of [1,2,3,4,5,6] taken three at a time are > > [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5], > [1,3, 6], [1, 4, 5], [1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5], > [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6], [3, 4, 5], [3, 4, 6], > [3, 5, 6], [4, 5, 6]] > > Taking the sums of the coefficients of these and putting 1 for even, > -1 for odd gives us > > [1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, -1] > > We are taking the compliment of a 3-monzo, so m=3 and m(m+1)/2 = 6, an > even number; hence we don't change sign. However, we do as always > reverse the ordering and take the coefficients in reverse order: > > [-1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1] > > <<<-c20 c19 -c18 c17 -c16 c15 -c14 -c13 c12 -c11 > c10 -c9 c8 c7 -c6 c5 -c4 c3 -c2 c1]]]In other words the tri-monzo is: [[[c123 c124 c125 c126 c134 c135 c136 c145 c146 c156 c234 c235 c236 c245 c246 c256 c345 c346 c356 c456>>> and we find Ceiling(g/2) = Ceiling(3/2) = Ceiling(1.5) = 2. So we add 2 to the sum of the digits in each case, which of course is the same as adding zero, since all we care about is whether the result is odd or even. In fact we don't even have to add them up. We can just count how many are odd, and if this count is itself odd, then we negate. So we negate all those coefficients with an odd number of odd digits (in their compound index), and we reverse their order, to get the tri-map: [[[-c456 c356 -c346 c345 -c256 c246 -c245 -c236 c235 -c234 c156 -c146 c145 c136 -c135 c134 -c126 c125 -c124 c123>>>> > Now for the compliment of a 4-monzo, which is a 6*5*4*3/24 = 15 > dimensional object again > > [[[[c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 c13 c14 c15>>>> > > The combinations of [1,2,3,4,5,6] taken four at a time in order are > > [[1, 2, 3, 4], [1, 2, 3, 5], [1, 2, 3, 6], [1, 2, 4, 5], [1, 2, 4, 6], > [1, 2, 5, 6], [1, 3, 4, 5], [1, 3, 4, 6], [1, 3, 5, 6], [1, 4, 5, 6], > [2, 3, 4, 5], [2, 3, 4, 6], [2, 3, 5, 6], [2, 4, 5, 6], [3, 4, 5, 6]] > > Even and odd for this gives > > [1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, 1] > > m=4, and 4*5/2 = 10, an even number, so we don't change sign but as > always do reverse the order > > [1, -1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1] > > Taking these signs with descending coefficients gives > > <<c15 -c14 c13 -c12 c11 c10 -c9 c8 -c7 c6 -c5 c4 c3 -c2 c1]] > > as the compliment.Take the 4-monzo [[[[c1234 c1235 c1236 c1245 c1246 c1256 c1345 c1346 c1356 c1456 c2345 c2346 c2356 c2456 c3456>>>> Ceiling(g/2) = Ceiling(4/2) = 2 which is even, so we don't need to add anything more to our count of the odd digits in each index, and the complement of the above 4-monzo is the following bi-map: <<c3456 -c2456 c2356 -c2346 c2345 c1456 -c1356 c1346 -c1345 c1256 -c1246 c1245 c1236 -c1235 c1234]] It should be mentioned that taking the complement of the complement doesn't always give you back what you started with, sometimes it's the negative of what you started with. So in those cases it's analogous to multiplying by i (the square root of -1). This depends on the dimension and the grade. But taking the complement four-times always gives you back exactly what you started with.

Message: 8418 - Contents - Hide Contents Date: Thu, 20 Nov 2003 21:06:47 Subject: Re: Finding the complement From: Graham Breed Dave Keenan wrote:> It should be mentioned that taking the complement of the complement > doesn't always give you back what you started with, sometimes it's the > negative of what you started with. So in those cases it's analogous to > multiplying by i (the square root of -1). This depends on the > dimension and the grade. But taking the complement four-times always > gives you back exactly what you started with.Are you sure? Do you have an example? Graham

Message: 8419 - Contents - Hide Contents Date: Thu, 20 Nov 2003 03:02:37 Subject: Definition of val etc. From: Dave Keenan I suggest that the definition of val stay as Gene had it, as a definition of a pure math term. But that we add something like the following text at the start of it. ---------------------------------------------------------------------- "val" is a term coined by Gene Ward Smith for the mathematical object described below. When vals are applied to tuning theory they are usually interpreted as "prime exponent mappings". ---------------------------------------------------------------------- Then a link should take you to a definition of "prime exponent mapping": ---------------------------------------------------------------------- A "prime exponent mapping", sometimes shortened to "prime mapping", "exponent mapping", "mapping" or simply "map", is a list of numbers (integers) enclosed in < ... ] that tell you how a particular temperament maps each prime number (up to some limit) to numbers of a particular "generator" in that temperament. The prime numbers here represent frequency ratios. The simplest case is an equal temperament where the generator is the step interval. For example, the 5-limit map for 12-equal is <12 19 28] which means it takes 12 steps to make an octave (1:2), 19 steps to make a twelfth (1:3), and 28 steps to make a 1:5 interval. When an interval is represented in the complementary form, as a prime-exponent-vector, we can find the number of generators corresponding to it in some temperament by multiplying each number in the temperament's map by the corresponding number in the vector, and adding up the results. In mathematical terms this is called the dot-product, scalar-product or inner-product of the map with the vector. For example the interval 3:5 (a major sixth), has the 5-limit exponent vector [0 -1 1>. To find how many steps of 12-equal it maps to, we write <12 19 28].[0 -1 1> = 12*0 + 19*-1 + 28*1 = 28 - 19 = 9 ---------------------------------------------------------------------- We should also have a definition for "prime exponent vector" and the definition for "monzo" should simply say "See "prime exponent vector"", and have a link to it. Links should also take you from "map", "mapping", "prime mapping" and "exponent mapping" to "prime exponent mapping", and from "monzo", "vector", "prime vector" and "exponent vector" to "prime exponent vector". The definition for "prime exponent vector" would be pretty much what Monz has now for "vector", with the addition of some stuff about the [ ... > notation and the relationship to "prime exponent mapping". -- Dave Keenan

Message: 8421 - Contents - Hide Contents Date: Thu, 20 Nov 2003 03:19:28 Subject: Finding the wedge product? From: Dave Keenan So Gene, you've given us a nice description of how to compute the complement for arbitrary dimension and grade. Now how about the wedge product between pairs of multivectors of arbitrary dimension (and possibly different) grades? Graham, feel free to chime in here too. :-)

Message: 8422 - Contents - Hide Contents Date: Thu, 20 Nov 2003 22:28:01 Subject: Re: "does not work in the 11-limit" (was:: Vals?) From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> wrote:> But yes, I would also include *both* 108:125 and 1024:1215 as ratios > (or *roles*, as Dave and I call them when a Sagittal symbol is used > to represent multiple ratios) that would be covered by the interval > function of an augmented 2nd in traditional diatonic harmony, just as > I would include both 8:9 and 9:10 as roles in which the interval of a > major 2nd functions. My only requirement is that a single ratio > *not* be represented by multiple vectors (or in this case multiple > scalars) in the scale construct.If you add the condition that *every* ratio (within the prime limit) be represented by one 'scalar in the scale construct', then I believe you *are* speaking of epimorphism -- and equivalently, periodicity, as presented in this paper of mine: http://lumma.org/tuning/erlich/erlich-tFoT.pdf - Type Ok * [with cont.] (Wayb.) Please read it if you haven't yet -- it should be a breeze for you.> I mentioned vectors in the previous paragraph, because in the case of > the 11-limit hexatonic otonal scale that we've been using as our > other example, the tones occur in a 4-dimensional structure. And > there will be only one ratio (or role) for each tone, since the scale > is JI.Just because a scale is in JI doesn't mean there's only one role for each tone, in my opinion. But that may be a separate discussion from 'functional disorientation' . . .

Message: 8423 - Contents - Hide Contents Date: Thu, 20 Nov 2003 07:01:25 Subject: John Browne is back online From: Dave Keenan Grassmann Algebra Book * [with cont.] (Wayb.) But I'm unable to extract from this, a method for calculating the wedge product for anything beyond 3D or arguments beyond grade 1.

Message: 8424 - Contents - Hide Contents Date: Thu, 20 Nov 2003 22:30:53 Subject: Re: Vals? From: Paul Erlich --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:>> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> >> wrote:>>> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> >>> wrote: >>>>>>> what was your original basis choice, and what do the patterns of >>>> signs for duals look like under it? >>>>>> I'd suggest we forget about that. Alphabetical is the most usual >>> approach, and we are already using it. >>>> OK -- but it's interesting to note that the cross product immediately >> gives you the quantity of interest in 3D, regardless of indexing >> conventions. >> Paul. You must have missed where I explained that the cross-product > stays the same no matter what the indexing conventions, because the > wedge-product and the complement change in "complementary" ways when > you change the indexing and A(x)B = ~(A^B).I don't know why you think I missed that, because (even if I did) it's perfectly clear to me, and I was never confused about that. It was the remark that followed that one which was my main point.

8000
8050
8100
8150
8200
8250
8300
8350
**8400**
8450
8500
8550
8600
8650
8700
8750
8800
8850
8900
8950

**8400 -**
8425 -