Generator-offset property: Difference between revisions

Line 117:

==== Proof ====

===== Preliminaries =====

Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S1 (via identifying b ~ c), S2 (via identifying a ~ c) and S3 (via identifying a ~ b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" words, which we call Σ2 and Σ3, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S1 and S2, which come in at most two sizes. Since S1 is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, km is ''not'' divisible by n and km-steps come in ''exactly'' two sizes; hence both Σ2 and Σ3 are single-period mosses.

Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S1 (via identifying b with c), S2 (via identifying a with c) and S3 (via identifying a with b). Suppose S1's generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S2 and S3, respectively. These two-"step-size" words, which we call Σ2 and Σ3, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S1 and S2, which come in at most two sizes. Since S1 is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, km is ''not'' divisible by n and km-steps come in ''exactly'' two sizes; hence both Σ2 and Σ3 are single-period mosses.

index: 1 2 3 4 ... n

@@ Line 117: / Line 117: @@
 ==== Proof ====
 ===== Preliminaries =====
-Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S<sub>1</sub> (via identifying b ~ c), S<sub>2</sub> (via identifying a ~ c) and S<sub>3</sub> (via identifying a ~ b). Suppose S<sub>1</sub>'s generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S<sub>2</sub> and S<sub>3</sub>, respectively. These two-"step-size" words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S<sub>1</sub> and S<sub>2</sub>, which come in at most two sizes. Since S<sub>1</sub> is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, km is ''not'' divisible by n and km-steps come in ''exactly'' two sizes; hence both Σ<sub>2</sub> and Σ<sub>3</sub> are single-period mosses.
+Suppose S has n notes (after dealing with small cases, we may assume n ≥ 7) and S projects to single-period mosses S<sub>1</sub> (via identifying b with c), S<sub>2</sub> (via identifying a with c) and S<sub>3</sub> (via identifying a with b). Suppose S<sub>1</sub>'s generator is a k-step, which comes in two sizes: P, the perfect k-step, and I, the imperfect k-step. By stacking k-steps, we get two words of length n of k-steps of S<sub>2</sub> and S<sub>3</sub>, respectively. These two-"step-size" words, which we call Σ<sub>2</sub> and Σ<sub>3</sub>, must be mosses, since m-steps in the new words correspond to mk-steps in the mos words S<sub>1</sub> and S<sub>2</sub>, which come in at most two sizes. Since S<sub>1</sub> is a single-period mos, gcd(k, n) = 1. Hence when 0 < m < n, km is ''not'' divisible by n and km-steps come in ''exactly'' two sizes; hence both Σ<sub>2</sub> and Σ<sub>3</sub> are single-period mosses.
   index: 1 2 3 4 ...   n