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Rank-3 scale theorems: Difference between revisions

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* Every max variety 3 block is a triple Fokker block. (However, not every max-variety 3 scale, in general, need be a Fokker block.)
* Every max variety 3 block is a triple Fokker block. (However, not every max-variety 3 scale, in general, need be a Fokker block.)
* Triple Fokker blocks form a [http://en.wikipedia.org/wiki/Trihexagonal_tiling trihexagonal tiling] on the lattice.
* Triple Fokker blocks form a [http://en.wikipedia.org/wiki/Trihexagonal_tiling trihexagonal tiling] on the lattice.
* A scale imprint is that of a Fokker block if and only if it is the [[product word]] of two DE scale imprints with the same number of notes. See [https://link.springer.com/chapter/10.1007/978-3-642-21590-2_24 Introduction to Scale Theory over Words in Two Dimensions | SpringerLink]
* A scale imprint is that of a Fokker block if and only if it is the [[product word|product]] of two DE scale imprints with the same number of notes. See [https://link.springer.com/chapter/10.1007/978-3-642-21590-2_24 Introduction to Scale Theory over Words in Two Dimensions | SpringerLink]
* If the step sizes for a rank-3 Fokker block are L, m, n, and s, where L > m > n > s, then the following identity must hold: (n-s) + (m-s) = (L-s), hence n+m=L+s
* If the step sizes for a rank-3 Fokker block are L, m, n, and s, where L > m > n > s, then the following identity must hold: (n-s) + (m-s) = (L-s), hence n+m=L+s
* Any convex object on the lattice can be converted into a hexagon.
* Any convex object on the lattice can be converted into a hexagon.
* Any convex scale with 3 step sizes is a hexagon on the lattice, in which each set of parallel lines corresponds to one of the steps.
* Any convex scale with 3 step sizes is a hexagon on the lattice, in which each set of parallel lines corresponds to one of the steps.
 
* An MV3 scale always has two of the step sizes occurring the same number of times, except powers of abacaba. Except multi-period MV3's, such scales are always either pairwise-well-formed, a power of abcba, or a "twisted" word constructed from the mos 2qX rY. A pairwise-well-formed scale has odd size, and is either [[generator-offset]] or of the form abacaba. The PWF scales are exactly the single-period rank-3 [[billiard scales]].
== Unproven Conjectures ==
== Conjectures ==
* Every rank-3 Fokker block has mean-variety < 4, meaning that some interval class will come in less than 4 sizes.
* Every rank-3 Fokker block has mean-variety < 4, meaning that some interval class will come in less than 4 sizes.
== MV3 proofs ==
Under construction
=== Definitions and theorems ===
Throughout, let ''S'' be a scale word in steps ''x'', ''y'', ''z'' (and assume all three of these letters are used).
==== Definition: Unconditionally MV3 ====
''S'' is ''unconditionally MV3'' or ''intrinsically MV3'' if ''S'' is MV3 for all possible choices of step ratio x:y:z.
==== Definition: EMOS ====
''S'' is ''elimination-MOS'' (EMOS) if the result of removing (all instances of) any one of the step sizes is a MOS.
==== Definition: PMOS ====
''S'' is ''pairwise MOS'' (PMOS) if the result of equating any two of the step sizes is a MOS.
==== Definition: AG ====
''S'' satisfies the ''alternating generator property'' (AG) if it satisfies the following equivalent properties:
# ''S'' can be built by stacking a single chain of alternating generators g1 and g2, resulting in a circle of the form  either g1 g2 ... g1 g2 g1 g3 or g1 g2 ... g1 g2 g3.
# ''S'' is generated by two chains of generators separated by a fixed interval; either both chains are of size ''m'', or one chain has size ''m'' and the second has size ''m-1''.
These are equivalent, since the separating interval can be taken to be g1 and the generator of each chain = g1 + g2.
==== Definitions: LQ ====
Let n = a_1 + ... + a_r be the scale size, w a scale word with signature a_1 X_1, ..., a_r X_r, let L be a line of the form L(t) = (a_1, ..., a_r)t + v_0, where v_0 is a constant vector in R^r. We say that L is ''in generic position'' if L contains a point (0, α_1, α_2, ... α_{r-1}) where α_i and α_i/α_j for i != j are irrational.
* Assume ''S'' is a 2-step scale. Then ''S'' is ''slope-LQ'' if the slope between any two pair of points (representing a ''k''-mosstep) is one of the two nearest possible slopes (in the set {k/0,...,0/k}) to b/a.
* Say that a 2-step scale ''S'' is ''floor-LQ'' if some mode ''M'' of ''S'' satisfies that γ(''M'') = the graph of floor(b/a*x).
* Say that an r-step scale ''S'' is ''LQ'' if any appropriate line in generic position, (a_1, ..., a_r)t + v_0, has intersections with coordinate level planes x_i = k that spell out the scale as you move in the positive t direction.
<!--===== MV2 is equivalent to floor-LQ in 2-step scales (WIP) =====
Assume wlog there are more L's than s's.
Take the graph of the brightest mode of the mos, M_b(x) (right = L, up = s). We claim that this is the required graph of F(x) = floor(b/a*x).
M_b <= F: Prove that F(x) describes a mos.
Say F has #s s's and #L L's across interval [m, m']. Say there is #s+t small steps and #L-t large steps on some k step [r, r'], t >= 2. This implies that the slope of the line b/a* x itself satisfies
(F(m')-F(m)-1)/(m'-m) <= b/a <= (F(m')-F(m)+1)/(m'-m).
(bounded by "floor minus ceiling" and "ceiling minus floor" slopes; this is because x-x' <= x-floor(x') <= floor(x)+1-floor(x').)
Rearranging,
F(m') - F(m) - 1 <= b/a(m'-m) <= F(m') - F(m) + 1
But F(m') -F(m) = #s and m'-m = #L. So #s -1 <= b/a*#L <= #s+1. Do the same thing for the "bad" interval [r', r] and you get #s+t-1 <= b/a(#L-t) <= #s+t+1.
Thus b/a#L <= b/a(#L-t), a contradiction.
M_b >= F: (bc it's a mos) Suppose there is an x-value n_0 where M_b(n_0) <= F(n_0) - 1. n_0 > 1 since otherwise, M_b(1) < 0. Let k = min(n_0, n-n_0), n = scale size. Then find three different k-mossteps/average slopes by taking the interval [n_0-k, n_0] before n_0, one interval containing n_0 and one interval after n_0. (We already know that mosses are slope-LQ.)
Since M_b is a mos mode, there is a k-step within [0, n_0] that has the slope which is just smaller than (F(n_0)-1)/n_0 (1). Similarly, there is a k-step within [n_0, n] that has the slope which is just bigger than (F(n_0)+1)/(n-n_0). These slopes are "two or more steps away" from each other, which is a contradiction. (State this more formally)
<!--==== MV3 Theorem 1 (WIP) ====
''The following are equivalent for a non-multiperiod scale word S with steps x, y, z:''
# ''S is unconditionally MV3.''
# ''(a) S is PMOS, or (b) S is of the form x'y'z'y'x', or (c) S has signature nx ny nz, n ≥ 2.''ppp
# ''(a) S is AG *and* is of the form ax by bz, or (b) S is of the form x'y'x'z'x'y'x', or (c) S is one of the exceptions to PMOS in statement 2.''
====== MV3 implies LQ except in the case "xyzyx" (WIP) ======
====== MV3 + LQ implies EMOS (WIP) ======
Proof sketch:
Let L = L(t) = (a, b, c)t + (0, α, β) be a line in generic position corresponding to the signature aX bY cZ. The projection matrices
<math>
P_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \
P_2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \
P_3 = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},
</math>
(which map (x, y, z) to (x, y), (y, z) and (z, x), respectively) map L to lines in R^2 that are in generic position (i.e. they intersect the x- and y-axes at irrational points). The projections record intersections with two of the planes to intersections with x- and y- axes, and these intersections must spell out the result of removing one of the step sizes; hence the resulting scales must be mosses.
(MV3 has not been used yet)
This in particular implies that xyzyx is not LQ.
====== MV3 + EMOS implies PMOS (WIP) ======
Suppose w_{YZ} is a word made by identifying Y and Z into one letter, say Q.
Suppose w_{YZ} has a k-step, 1 < k <= n/2, which comes in three sizes
* v_1 = A_1 X + B_1 Q,
* v_2 = A_2 X + B_2 Q,
* v_3 = A_3 X + B_3 Q.
If some A_i and A_j differed by more than 2, a contradiction would result, as all intermediate combinations must be attained (since scooting over by one step changes the numbers of Q's and X's by <= 1). So we assume we have sizes
* v_1 = v(s_1) = AX + BQ,
* v_2 = v(s_2) = (A-1)X + (B+1)Q,
* v_3 = v(s_3) = (A+1)X + (B-1)Q.
Plan: v_2 and v_3 are sizes that are problematic when they occur together.
Let EX(w), EY(w), EZ(w) be mosses that result from eliminating X, Y and Z. MV3 implies that for any possible choices of s_i, EX(s_1), EX(s_2), EX(s_3) each only comes in one possible size as B-, (B+1)- and (B-1)-steps in EX(w). [EX means eliminate X]
====== PMOS implies AG (except in the case xyxzxyx) (WIP) ======
-->
==== AG + unconditionally MV3 implies "ax by bz" and that the scale's cardinality is odd or 4 ====
'''Assuming both AG and unconditional MV3''', we have two chains of generator g0 (going right). The two cases are:
O-O-...-O (m notes)
O-O-...-O (m notes)
and
O-O-O-...-O (m notes)
O-O-...-O (m-1 notes).
Label the notes (1,k) and (2,k), 1 ≤ k ≤ m or m-1, for notes in the upper and lower chain respectively.
In case 1 (even scale size), let g1 = (2,1) - (1,1) and g2 = (1,2) - (2,1). We have the chain g1 g2 g1 g2... g1 g3. Consider the sizes of the n/2-step (which is an odd number of generator steps):
# from g1 ... g1, get a1 = (n/2-1)*g0 + g1 = n/2 g1 + (n/2-1) g2
# from g2 ... g2, get a2 =(n/2-1)*g0 + g2 = (n/2-1) g1 + n/2 g2
# from g2 (even) g1 g3 g1 (even) g2, get a3 = (n/2-1) g1 + (n/2-1) g2 + g3
# from g1 (odd) g1 g3 g1 (odd) g1, get a4 = n/2 g1 + (n/2-2) g2 + g3.
Choose a tuning where g1 and g2 are both very close to but not exactly 1/2*g0 (i.e. they differ from 1/2*g0 by ε, a quantity much smaller than the chroma of the n/2-note mos generated by g0, which is |g3 - g2|). Assuming n > 4, We have 4 distinct sizes for n/2-steps, a contradiction to unconditional-MV3:
# a1, a2 and a3 are clearly distinct.
# a4 - a3 = g1 - g2 != 0, since the scale is a non-trivial AG.
# a4 - a1 = g3 - g2 = (g3 + g1) - (g2 + g1) != 0. This is exactly the chroma of the mos generated by g0.
# a4 - a2 = g1 - 2 g2 + g3 = (g3 - g2) + (g1 - g2) = (chroma ± ε) != 0 by choice of tuning.
In case 2, let (2,1)-(1,1) = g1, (1,2)-(2,1) = g2 be the two alternating generators. Let g3 be the leftover generator after stacking alternating g1 and g2. Then the generator circle looks like g1 g2 g1 g2 ... g1 g2 g3. Then the generators corresponding to a step are:
# k g1 + (k-1) g2
# (k-1) g1 + k g2
# (k-1) g1 + (k-1) g2 + g3,
if a step is an odd number of generators (since the scale size is odd, we can always ensure this by taking octave complements of all the generators). The first two sizes must occur the same number of times.
This proof shows that AG and unconditionally-MV3 scales must have odd size or size 4.
==== An AG scale is unconditionally MV3 iff its cardinality is odd ====
We only need to see that AG + odd cardinality => MV3. But the argument in case 2 above works for any interval class (MV3 wasn't used), hence any interval class comes in at most 3 sizes regardless of tuning.
==== An even-cardinality MV3 is of the form W(x,y,z)W(y,x,z) (WIP) ====
==== 3-DE implies MV3 (WIP) ====
We prove that 3-DE + not abcba implies PMOS, which is known to imply MV3.


[[Category:Fokker block]]
[[Category:Fokker block]]
[[Category:Math]]
[[Category:Math]]
[[Category:Rank 3]]
[[Category:Rank 3]]
[[Category:Scales]]
[[Category:Scale]]
[[Category:Theory]]
[[Category:Pages with open problems]]
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