Generator-offset property: Difference between revisions

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===== Some implications of the above =====

Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’~~. Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ~~. The cyclic words Λ1 = the pattern of α and α’, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ’ must form mosses, and Λ1 = α...αα’. ~~By the assumption β’’= β, Σ2 has one more R than Σ3, and thus:~~

Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. The cyclic words Λ1 = the pattern of α and α’, Λ2 = the pattern of β and β’, and Λ3 = the pattern of γ and γ’ must form mosses, and Λ1 = α...αα’.

* Λ2 is λβ μβ’~~, 2 ≤ μ ≤ ceil(n/2)~~.

Suppose Λ2 is λβ μβ’. Then Λ3 is a (λ &pm; 1)γ (μ &mp; 1)γ’ mos. Since neither Λ2 nor Λ3 are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.

* Λ3 is a (λ ± 1)γ (μ ∓ 1)γ’ mos.

~~Since neither Λ2 nor~~ Λ3 ~~are multimosses, and at least one of μ and~~ (μ ∓ 1) ~~are even, it is now immediate that n is odd~~.

Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ, and Λ3 is a (λ + 1)γ (μ − 1)γ’ mos. Also assume that the first k-step in Σ is Q:

~~We can~~ assume that the first k-step in Σ is Q:

1 … n

Σ = Q W(Q, R) T

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 ===== Some implications of the above =====
-Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ. The cyclic words Λ<sub>1</sub> = the pattern of α and α’, Λ<sub>2</sub> = the pattern of β and β’, and Λ<sub>3</sub> = the pattern of γ and γ’ must form mosses, and Λ<sub>1</sub> = α...αα’. By the assumption β’’= β, Σ<sub>2</sub> has one more R than Σ<sub>3</sub>, and thus:
+Suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two k-steps in S that project to P. Then T = (α’, β’’, γ’’) projects to I. Here the values in each component differ by at most 1, and α ≠ α’. The cyclic words Λ<sub>1</sub> = the pattern of α and α’, Λ<sub>2</sub> = the pattern of β and β’, and Λ<sub>3</sub> = the pattern of γ and γ’ must form mosses, and Λ<sub>1</sub> = α...αα’.
-* Λ<sub>2</sub> is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).
+Suppose Λ<sub>2</sub> is λβ μβ’. Then Λ<sub>3</sub> is a (λ &pm; 1)γ (μ &mp; 1)γ’ mos. Since neither Λ<sub>2</sub> nor Λ<sub>3</sub> are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.
-* Λ<sub>3</sub> is a (λ ± 1)γ (μ ∓ 1)γ’ mos.
-Since neither Λ<sub>2</sub> nor Λ<sub>3</sub> are multimosses, and at least one of μ and (μ ∓ 1) are even, it is now immediate that n is odd.
+Either β’’ = β or β’’ = β’. Assume β’’ = β’. Then γ’’ = γ, and Λ<sub>3</sub> is a (λ + 1)γ (μ &minus; 1)γ’ mos. Also assume that the first k-step in Σ is Q:
-We can assume that the first k-step in Σ is Q:
 …        n
   Σ  = Q W(Q, R)  T