Rank-3 scale theorems: Difference between revisions

Inthar (talk | contribs)
Inthar (talk | contribs)
Line 32: Line 32:
==== Definitions: Billiard scale ====
==== Definitions: Billiard scale ====
Let  
Let  
* w a scale word with signature a_1 X_1, ..., a_r X_r;
* w be a scale word with signature a<sub>1</sub>X<sub>1</sub>, ..., a<sub>r</sub>X<sub>r</sub>;
* n = len(w) = a_1 + ... + a_r;
* n = a<sub>1</sub> + ... + a<sub>r</sub> be the length of w;
* 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.
* L be a line of the form L(t) = (a<sub>1</sub>, ..., a<sub>r</sub>)t + v<sub>0</sub>, where v<sub>0</sub> is a constant vector in R<sup>r</sup>.
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.
We say that L is ''in generic position'' if L intersects the hyperplane x<sub>1</sub> = 0 at a point (0, α<sub>1</sub>, α<sub>2</sub>, ... α<sub>r-1</sub>) where α<sub>i</sub> and α<sub>i</sub>/α<sub>j</sub> for i j are irrational.


Say that an r-step scale ''S'' is a ''billiard scale'' 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.
We say that ''w'' is a ''billiard scale'' if any appropriate line in generic position, (a<sub>1</sub>, ..., a<sub>r</sub>)t + v<sub>0</sub>, has intersections with coordinate level planes x<sup>i</sup> = 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) ======
-->


[[Category:Fokker block]]
[[Category:Fokker block]]