The wedgie - Revision history

Fredg999: Fix double redirect

2025-08-15T04:36:38Z

Fix double redirect

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	#~~redirect~~ [[~~Wedgies and multivals~~]]		#REDIRECT [[Wedgie/Archived version]]

Xenwolf: skip double redirect

2021-06-05T11:23:44Z

skip double redirect

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	#redirect [[Wedgies and ~~Multivals~~]]		#redirect [[Wedgies and multivals]]

Inthar: Redirected page to Wedgies and Multivals

2021-01-03T19:07:50Z

Redirected page to Wedgies and Multivals

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	~~__FORCETOC__~~		#redirect [[Wedgies and Multivals]]
	~~The '''wedgie''' is a way of defining and working with an [[abstract regular temperament]]. If one takes r independent [[vals]] in a p-limit group of n primes, then the wedgie is defined by taking the~~ [[Wedgies and Multivals\|wedge product]] of the vals, and dividing out the greatest common divisior of the coefficients, to produce an r-multival. If the first non-zero coefficient of this multival is negative, it is then scalar multiplied by -1, changing the sign of the first non-zero coefficient to be positive. The result is the wedgie. Wedgies are in a one-to-one relationship with abstract regular temperaments; that is, regular temperaments where no tuning has been decided on.
	~~==How the period and generator falls out of a rank-2 wedgie==~~
	The following is a procedure for finding a period and a generator for a rank-2 regular temperament on the 2.q_1.(...).q_n [[Subgroup temperaments\|JI subgroup]], given the temperament's wedgie. We also give a (hopefully convincing and enlightening) proof of why the procedure always works. We'll assume that the equave (interval of equivalence) is the octave, but non-octave JI equaves can be substituted for the octave if needed, by substituting the appropriate JI ratio for 2/1.

	~~The following assumes that:~~
	* you can think of JI ratios as vectors living in the n-dimensional lattice of the "JI subgroup"
	* you know what a "period" and a "generator" of a rank-2 temperament are
	* you know what a [[val]] is
	* you know what the words "basis", "linear map", and "determinant" mean.

	~~===The procedure===~~

	~~Consider the entries of the wedgie W. The entries of W are W(2, q_1), ..., W(2, q_n), and W(q_i, q_j) for i < j.~~

	~~To find the '''period''': Let d = gcd(W(2, q_1), ..., W(2, q_n)). Then your period is 1\d.~~

	To find the '''generator''': Use the [https://en.wikipedia.org/wiki/Extended_Euclidean_algorithm extended Euclidean algorithm] to find a linear combination g = a_1 q_1 + ... + a_n q_n (i.e. a JI ratio g = q_1^a_1 ... q_n^a_n) such that W(2, g) = a_1 W(2, q_1) + ... a_n W(2,q_n) = d.

	~~===Example===~~

	Consider the wedgie W = <<1 4 4\|\| for 2.3.5 meantone. We have W(2,3) = 1 and W(2,5) = 4, so d = 1, and our period is 1\1. We already have W(2,3) = 1, so we can use 3/1 as our generator. Alternatively, W(2, 3/2) = W(2,3) - W(2, 2) = W(2, 3) = 1, so 3/2 is a valid generator for meantone as well.

	~~===Proof===~~
	The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.

	In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:

	Let K_1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.

	~~By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.~~

	[In geometric terms, given JI ratios u and v, and wedgie W, the number W(u,v) is the signed area of the parallelogram spanned by (tempered versions of) u and v. The entries of the wedgie give the values of the wedgie on the basis elements of the JI subgroup that the temperament is on. By the alternating property [i.e. W(u, v) = -W(v, u)] and bilinearity [W is linear in each argument separately], specifying the values on basis elements of the JI subgroup is enough to define W as an alternating bilinear form on all of the JI subgroup. This is the determinant of the tempered versions of u and v. The musical interpretation of the parallelogram spanned by u and v is: If you want to consider intervals that are multiples of u apart the same note (for example, if you want an octave-equivalent scale), W(u, v) tells you how many generators it take to get to v.]

	The key fact about the determinant we use here is that two integer vectors v_1, v_2 form a basis for the rank-2 integer lattice '''Z'''<sup>2</sup> iff det(v_1, v_2) = ±1. So in order to find a period and generator for our tempearment, we need a pair of vectors {p, g} such that W(p, g) = 1 and p is 1\d for some integer d.

	Let d = gcd(W(2/1, q_1), ..., W(2/1, q_n)). This tells you that for any JI ratio v in your JI subgroup, W(2/1, v) = 2n(v) for some number n(v) [that depends linearly on v]. This equation is also true when we replace 2/1 with any JI ratio u that is equated to 2/1. This tells us that for W(p, g) = 1, we (up to some choices) need p to be an interval such that d*p is equated to 2/1, i.e. p represents 1/d of the octave.

	~~Choose a basis e_1, e_2 for the temperament group and write (the image of) 2/1 as 2/1 = a_1 e_1 + a2 e_2. Then:~~
	*W(2/1, e_1) = W(a_2 e_2, e_1) = -a_2 W(e_1, e_2) = -a_2
	*W(2/1, e_2) = W(a_1 e_1, e_2) = a_1 W(e_1, e_2) = a_1.
	Divisibility (by d = gcd(W(2, q_1), ..., W(2, q_n))) and the fact that e1 and e2 represent JI ratios in the 2.q_1. ... q_n subgroup impliy that a_1 and a_2 are both divisible by d, and hence 2/1 is a dth power in M' (the temperament space). Since gcd(W(2, q_1), ..., W(2, q_n)) = d, we can always find a linear combination g = a_1 q_1 + ... + a_n q_n such that W(2, g) = a_1 W(2, q_1) + ... a_n W(2,q_n) = d using the extended Euclidean algorithm. Then since W(2, g) = W(dp, g) = dW(p, g) = d, we have W(p,g) = 1. Ta-da!

	~~== Truncation of wedgies ==~~
	A useful operation to perform on any multivector, including wedgies, is truncation of the wedgie to a lower prime limit. This in effect sets all the basis vectors of a p-limit wedgie which are greater than q, the prime limit being truncated to, to zero. An algorithm to produce the truncation is to list the r-subsets of the primes to p in alphabetical order, and add the corresponding coefficient to the list of the q-limit truncation if and only if the maximum prime in the r-subet is less than or equal to q. Truncating a wedgie can lead to a non-wedgie if the GCD of the coefficients is greater than one; this means that in the lower limit, [[Wedgies_and_Multivals\|contortion]] has appeared.

	~~== Conditions on being a wedgie ==~~
	If we take any three integers <<a b c\|\| such that GCD(a, b, c) = 1 and a ≥ 1 the result is always a wedgie, the wedgie tempering out the [[The_dual\|dual]] [[monzos\|monzo]] \|c -b a>. Since three such integers chosen at random are unlikely to produce a suitably small comma, the temperament will probably not be worth much, but at least it can be defined.

	However, this is no longer the case in higher limits. There, not everything which looks like a wedgie will be one; for instance the wedgies must also satisfy the condition, for any wedgie W, that W∧W = 0, where the "0" means the multival of rank 2r obtained by wedging W with W. For prime limits 7 and 11 this condition suffices for rank two, but in general we need to check, for every prime q ≤ p and every basis val v sending q to 1 and everything else to 0, that (W∨q)∧W and (W∧v)º∧Wº = 0, where "∨" denotes the [[Interior_product\|interior product]]. These conditions, the complete set along with the basic reduction conditions for being a wedgie, are known as the [http://en.wikipedia.org/wiki/Pl%C3%BCcker_embedding Plücker relations]. Note that the Plücker relations must be satisfied, since for a rank r multival, W∨q is a rank r-1 multival corresponding to tempering out all the commas of W, as well as q.

	In the 7-limit case, if we wedge a prospective rank two multival W = <<a b c d e f\|\| with itself, we obtain W∧W = 2(af-be+cd). The quantity af-be+cd is the [http://en.wikipedia.org/wiki/Pfaffian Pfaffian] of the wedgie, and that the Pfaffian is zero tells us that in the five-dimensional projective space '''P⁵''' in which wedgies live, the wedgie lies on a (four-dimensional) [http://en.wikipedia.org/wiki/Hypersurface hypersurface], known as the [[Abstract_regular_temperament#The Geometry of Regular Temperaments\|Grassmannian]] '''Gr'''(2, 4). For an 11-limit rank-two wedgie W = <<w1 w2 w3 w4 w5 w6 w7 w8 w9 w10\|\| we have that W∧W = 2<<<<w1w8-w2w6+w3w5, w1w9-w2w7+w4w5, w1w10-w3w7+w4w6, w2w10-w3w9+w4w8, w5w10-w6w9+w7w8\|\|\|\| is zero. These conditions allow us to solve for three of the coefficients in terms of the other seven, and so that '''Gr'''(2, 5), the Grassmannian of rank-two 11-limit temperaments, is a six-dimensional projective [http://en.wikipedia.org/wiki/Algebraic_variety algebraic variety] in nine-dimensional projective space '''P⁹'''. Wedgies correspond to rational points on this variety. For 11-limit rank three temperaments, we have w6w1-w5w2+w4w3 = w9w1-w8w2+w7w3 = w10w1-w8w4+w7w5 = w10w2-w9w4+w7w6 = w10w3-w9w5+w8w6 = 0; again, this leads to a six-dimensional variety, this time '''Gr'''(3, 5).

	~~== Constrained wedgies ==~~
	Most of the wedgies which are legitimate according to the previous section do not represent temperaments which are in any way reasonable. To get temperaments which are, we need to constrain the relevant metrics--complexity should not be too high, error should not be too high, and badness should not be so high that competing temperaments are much better. Let us consider how bounding [[Tenney-Euclidean_temperament_measures#TE simple badness\|relative error]] E, aka simple badness, constrains a 7-limit rank two wedgie W = <<a b c d e f\|\|.

	By definition, E = \|\|J∧Z\|\|, where Z is the weighted version of W; if q3, q5 and q7 are the logarithms base two of 3, 5, and 7, then Z = <<a/q3 b/q5 c/q7 d/(q3q5) e/(q3q7) f/(q5q7)\|\|. We now have

	~~<math>\displaystyle (\frac{d}{q_3q_5}-\frac{b}{q_5}+\frac{a}{q_3})^2+(\frac{e}{q_3q_7}-\frac{c}{q_7}+\frac{a}{q_3})^2+(\frac{f}{q_5q_7}-\frac{c}{q_7}+\frac{b}{q_5})^2 \\~~
	~~+(\frac{f}{q_5q_7}-\frac{e}{q_3q_7}+\frac{d}{q_3q_5})^2 = 4 E^2</math>~~

	From this we can conclude that d, e and f satisfy \|d - q3b + q5a\| ≤ 2Eq3q5, \|e - q3c + q7a\| ≤ 2Eq3q7 and \|f - q5c + q7d\| ≤ 2Eq5q7. This has an interesting interpretation: since <1 q3 q5 q7\|∧<0 a b c\| = <<a b c q3b-q5a q3c-q7a q5c-q7b\|\|, if E ≤ 1/(4q5q7), then the full wedgie can be recovered from the octave equivalent (OE) portion of the wedgie simply by wedging it with <1 q3 q5 q7\| and rounding to the nearest integer. This is not a very serious constraint to place on relative error; it seems unlikely anyone would be interested in a temperament which did not fall well under this low standard. Hence we may compile lists of reasonable temperaments by presuming "reasonable" requires this bound to be met, searching through triples <<a b c ...\|\| (note that if all of these are zero, 2 is being tempered out) up to some complexity bound, wedging with <1 q3 q5 q7\| and rounding, then checking if the GCD is one and the Pfaffian af-be+cd is zero. Then we may toss everthing which does not meet the bound on relative error; however, for a reasonable list we will want a tighter bound.

	If C = \|\|W\|\| is the TE complexity, then the formula for the [[Tenney-Euclidean_metrics#Logflat TE badness\|logflat badness]] B in the 7-limit rank-two case is particularly simple: B = CE. If complexity is bounded by, for example, 20 (which allows for some quite complex temperaments) then since E ≤ 1/(4q5q7), B ≤ 20/(4q5q7) = 0.767. This badness figure is easily met. While simply bounding complexity will lead to a finite list, the list would be enormous. An alternative is also to bound badness; for instance, we might produce a list of 7-limit rank-two temperaments with complexity less than 20 and a more reasonable badness limit, such as 0.05 or 0.06.

	~~== Reconstituting wedgies in general ==~~
	Essentially the same situation obtains for rank two temperaments in higher limits. The rule then is that if E ≤ 1/(C(n, 3)lb(q)lb(p)) then wedging K = <1 lb(3) lb(5) ... lb(p)\| with the val consisting of 0 followed by the first n-1 coefficients of the wedgie and rounding will give the wedgie. Here p and q are the largest and second largest primes in the prime limit, lb(x) is log base two, and C(n, 3) is n choose three, n(n-1)(n-2)/6.

	More generally, we can reconstitute W by rounding Y = (W∨2)∧K to the nearest integer coefficients, where K is the JI point <1 lb(3) lb(5) ... lb(p)\| in unweighted coordinates. Then we have \|\|(W-Y)+Y\|\| ≤ \|\|W-Y\|\| + \|\|Y\|\| by the triangle inequality, and since \|\|W-Y\|\| is bounded by the fact that W has been obtained by rounding, complexity, which is \|\|(W-Y)+Y\|\|=\|\|W\|\|, can be bounded by \|\|Y\|\|; which means it can be bounded by the coefficients of Y, which are those coefficients of W which can be found in W∨2 and over which we could be conducting a search. Moreover, we have from Y∧K = ((W∨2)∧K)∧K = 0 that relative error, which is \|\|W∧K\|\|, is \|\|((W-Y) + Y)∧K\|\|=\|\|(W-Y)∧K\|\|, hence relative error is also bounded by the fact that \|\|W-Y\|\| is bounded. This means that unless relative error is large, W can be recovered by rounding Y, and hence all wedgies within such a bound, which we may call ''recoverable'', can be found by a search on only some prospective coefficients. Temperaments which are not recoverable seem of little interest and may be ruled out of consideration. Search spaces for complexity measures such as [[Tenney-Euclidean_temperament_measures#TE Complexity\|TE complexity]] which are defined in terms of the wedgie can be obtained by assuming all wedgie coefficients which are not being used to recover a wedgie are zero, which gives a minimum value for the complexity. In the case of rank two temperaments, an especially efficient complexity measure for such searches, and one with some other desirable properties, is [[Generator_complexity\|generator complexity]].

	In the particular case of the 11-limit in rank three, we have that (W∨2)∧K gives the full wedgie, which has ten coefficents, in terms of the first six upon rounding off. Using this for a search is less difficult than it sounds, since the complexity numbers for rank three are so much lower. If the relative error E satisifes E ≤ 1/(2√5 q5q7q11), then the rounding off is guaranteed to lead to the correct result. This amount, 0.0099, is again easily met.

Inthar: /* How the period and generator falls out of a rank-2 wedgie */

2021-01-03T00:49:10Z

How the period and generator falls out of a rank-2 wedgie

← Older revision		Revision as of 00:49, 3 January 2021
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	* you can think of JI ratios as vectors living in the n-dimensional lattice of the "JI subgroup"		* you can think of JI ratios as vectors living in the n-dimensional lattice of the "JI subgroup"
	* you know what a "period" and a "generator" of a rank-2 temperament are		* you know what a "period" and a "generator" of a rank-2 temperament are
			* you know what a [[val]] is
	* you know what the words "basis", "linear map", and "determinant" mean.		* you know what the words "basis", "linear map", and "determinant" mean.

Inthar: /* Proof / the extended* Euclidean algorithm finds the coefficients a_i

2021-01-03T00:47:11Z

Proof: the *extended* Euclidean algorithm finds the coefficients a_i

← Older revision		Revision as of 00:47, 3 January 2021
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	*W(2/1, e_1) = W(a_2 e_2, e_1) = -a_2 W(e_1, e_2) = -a_2		*W(2/1, e_1) = W(a_2 e_2, e_1) = -a_2 W(e_1, e_2) = -a_2
	*W(2/1, e_2) = W(a_1 e_1, e_2) = a_1 W(e_1, e_2) = a_1.		*W(2/1, e_2) = W(a_1 e_1, e_2) = a_1 W(e_1, e_2) = a_1.
	Divisibility (by d = gcd(W(2, q_1), ..., W(2, q_n))) and the fact that e1 and e2 represent JI ratios in the 2.q_1. ... q_n subgroup impliy that a_1 and a_2 are both divisible by d, and hence 2/1 is a dth power in M' (the temperament space). Since gcd(W(2, q_1), ..., W(2, q_n)) = d, we can always find a linear combination g = a_1 q_1 + ... + a_n q_n such that W(2, g) = a_1 W(2, q_1) + ... a_n W(2,q_n) = d using the Euclidean algorithm. Then since W(2, g) = W(dp, g) = dW(p, g) = d, we have W(p,g) = 1. Ta-da!		Divisibility (by d = gcd(W(2, q_1), ..., W(2, q_n))) and the fact that e1 and e2 represent JI ratios in the 2.q_1. ... q_n subgroup impliy that a_1 and a_2 are both divisible by d, and hence 2/1 is a dth power in M' (the temperament space). Since gcd(W(2, q_1), ..., W(2, q_n)) = d, we can always find a linear combination g = a_1 q_1 + ... + a_n q_n such that W(2, g) = a_1 W(2, q_1) + ... a_n W(2,q_n) = d using the extended Euclidean algorithm. Then since W(2, g) = W(dp, g) = dW(p, g) = d, we have W(p,g) = 1. Ta-da!

	== Truncation of wedgies ==		== Truncation of wedgies ==

Inthar: /* Proof */ typo

2021-01-03T00:45:27Z

Proof: typo

← Older revision		Revision as of 00:45, 3 January 2021
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	In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:		In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:

	Let K_1 = the kernel of ~~_the~~ temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.		Let K_1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.

	By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.		By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.

Inthar: /* Proof */ typo

2021-01-03T00:45:00Z

Proof: typo

← Older revision		Revision as of 00:45, 3 January 2021
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	The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.		The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.

	In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the ~~biliear~~ form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:		In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:

	Let K_1 = the kernel of _the temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.		Let K_1 = the kernel of _the temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.

Inthar: /* Proof */

2021-01-03T00:40:31Z

Proof

← Older revision		Revision as of 00:40, 3 January 2021
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	By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.		By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.

	[In geometric terms, given JI ratios u and v, and wedgie W, the number W(u,v) is the signed area of the parallelogram spanned by (tempered versions of) u and v. The entries of the wedgie give the values of the wedgie on the basis elements of the JI subgroup that the temperament is on. By the alternating property [i.e. W(u, v) = -W(v, u)] and bilinearity [W is linear in each argument separately], specifying the values on basis elements of the JI subgroup is enough to define W as an alternating bilinear form on all of the JI subgroup. This is the determinant of the tempered versions of u and v.]		[In geometric terms, given JI ratios u and v, and wedgie W, the number W(u,v) is the signed area of the parallelogram spanned by (tempered versions of) u and v. The entries of the wedgie give the values of the wedgie on the basis elements of the JI subgroup that the temperament is on. By the alternating property [i.e. W(u, v) = -W(v, u)] and bilinearity [W is linear in each argument separately], specifying the values on basis elements of the JI subgroup is enough to define W as an alternating bilinear form on all of the JI subgroup. This is the determinant of the tempered versions of u and v. The musical interpretation of the parallelogram spanned by u and v is: If you want to consider intervals that are multiples of u apart the same note (for example, if you want an octave-equivalent scale), W(u, v) tells you how many generators it take to get to v.]

	[The musical interpretation of the parallelogram spanned by u and v is: If you want to consider intervals that are multiples of u apart the same note (for example, if you want an octave-equivalent scale), W(u, v) tells you how many generators it take to get to v.]

	The key fact about the determinant we use here is that two integer vectors v_1, v_2 form a basis for the rank-2 integer lattice '''Z'''<sup>2</sup> iff det(v_1, v_2) = ±1. So in order to find a period and generator for our tempearment, we need a pair of vectors {p, g} such that W(p, g) = 1 and p is 1\d for some integer d.		The key fact about the determinant we use here is that two integer vectors v_1, v_2 form a basis for the rank-2 integer lattice '''Z'''<sup>2</sup> iff det(v_1, v_2) = ±1. So in order to find a period and generator for our tempearment, we need a pair of vectors {p, g} such that W(p, g) = 1 and p is 1\d for some integer d.

Inthar: /* Proof */

2021-01-03T00:40:15Z

Proof

← Older revision		Revision as of 00:40, 3 January 2021
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	===Proof===		===Proof===

	The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.		The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.

	In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the biliear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:		In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the biliear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:

	Let K1 = the kernel of ~~the~~ temperament (i.e. the set of commas tempered out by the temperament), and K2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K2. Conversely, if v ∈ K2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have ~~a kernel of dimension 3~~ but a&b ~~(the intersection of the kernels)~~ has ~~dimension~~ 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K1 and K1 = K2, as claimed.		Let K_1 = the kernel of _the temperament (i.e. the set of commas tempered out by the temperament), and K_2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K_2. Conversely, if v ∈ K_2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have rank 1 but a&b has rank 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K_1 and K_1 = K_2, as claimed.

	By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.		By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.

Inthar: /* Proof */

2021-01-03T00:38:09Z

Proof

← Older revision		Revision as of 00:38, 3 January 2021
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	The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.		The period ''p'' (fraction of octave) and generator ''g'' form a basis for all the intervals of a rank-2 temperament. For example, p = 2/1 and g = 3/2 form a basis for meantone. But from a linear algebra perspective, there's nothing special about the basis {p, g}; I could have chosen the basis p' = 3/1 and g' = 2/1. What makes the wedgie a unique identifier for a temperament is that rather than specify a basis directly, the wedgie acts more like a set of constraints that any basis for the temperament must satisfy.

	In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the biliear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament, and ~~thus~~ that M' is the group of intervals in the rank-2 temperament in question. In geometric terms, given JI ratios u and v, and wedgie W, the number W(u,v) is the signed area of the parallelogram spanned by (tempered versions of) u and v. The entries of the wedgie give the values of the wedgie on the basis elements of the JI subgroup that the temperament is on. By the alternating property [i.e. W(u, v) = -W(v, u)] and bilinearity [W is linear in each argument separately], specifying the values on basis elements of the JI subgroup is enough to define W as an alternating bilinear form on all of the JI subgroup. This is the determinant of the tempered versions of u and v.		In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group M; this means that it acts like the operation of finding the determinant of two vectors on the appropriate quotient group M' = M/K of M, where K is the kernel of the biliear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament:

			Let K1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and K2 = ker W = {v ∈ M : W(v, w) = 0 ∀w ∈ M}. If v ∈ K1, then v is tempered out by both a and b, so W(v, w) = a(v)b(w)-a(w)b(v) = 0, and v ∈ K2. Conversely, if v ∈ K2, then W(v, w) = a(v)b(w)-a(w)b(v) = 0 for all w, which implies a(v)b(w) = a(w)b(v) () for all w. Since a and b both have a kernel of dimension 3 but a&b (the intersection of the kernels) has dimension 2, we can choose w such that a(w) = 0 but b(w) ≠ 0. Then () shows a(v) = 0. By the same argument, b(v) = 0. So v is in K1 and K1 = K2, as claimed.

			By the First Isomorphism Theorem it follows that M' is the group of intervals in the rank-2 temperament in question.

			[In geometric terms, given JI ratios u and v, and wedgie W, the number W(u,v) is the signed area of the parallelogram spanned by (tempered versions of) u and v. The entries of the wedgie give the values of the wedgie on the basis elements of the JI subgroup that the temperament is on. By the alternating property [i.e. W(u, v) = -W(v, u)] and bilinearity [W is linear in each argument separately], specifying the values on basis elements of the JI subgroup is enough to define W as an alternating bilinear form on all of the JI subgroup. This is the determinant of the tempered versions of u and v.]

	[The musical interpretation of the parallelogram spanned by u and v is: If you want to consider intervals that are multiples of u apart the same note (for example, if you want an octave-equivalent scale), W(u, v) tells you how many generators it take to get to v.]		[The musical interpretation of the parallelogram spanned by u and v is: If you want to consider intervals that are multiples of u apart the same note (for example, if you want an octave-equivalent scale), W(u, v) tells you how many generators it take to get to v.]

The wedgie - Revision history

Fredg999: Fix double redirect

Xenwolf: skip double redirect

Inthar: Redirected page to Wedgies and Multivals

Inthar: /* How the period and generator falls out of a rank-2 wedgie */

Inthar: /* Proof */ the *extended* Euclidean algorithm finds the coefficients a_i

Inthar: /* Proof */ typo

Inthar: /* Proof */ typo

Inthar: /* Proof */

Inthar: /* Proof */

Inthar: /* Proof */

Inthar: /* Proof / the extended* Euclidean algorithm finds the coefficients a_i