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’. Then the cyclic word Λ1 formed by the a-components of the k-steps in P is α...αα’. Since Σ2 is a mos pattern of βb + (n − β)(a~c) and β’a + (n − β’)(a~c), the cyclic word Λ2 = the pattern of β and β’ must be a mos. Similarly, Λ3 = the pattern of γ and γ’ ~~must form~~ a mos.

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’. Then the cyclic word Λ1 formed by the a-components of the k-steps in P is α...αα’. Since Σ2 is a single-period mos pattern of βb + (n − β)(a~c) and β’a + (n − β’)(a~c), the cyclic word Λ2 = the pattern of β and β’ must be a single-period mos. Similarly, Λ3 = the pattern of γ and γ’ is a single-period mos.

Suppose Λ2 is the mos λβ μβ’. Then Λ3 is the mos (λ ± 1)γ (μ ∓ 1)γ’. 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.

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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’. Then the cyclic word Λ<sub>1</sub> formed by the a-components of the k-steps in P is α...αα’. Since Σ<sub>2</sub> is a mos pattern of βb + (n &minus; β)(a~c) and β’a + (n &minus; β’)(a~c), the cyclic word Λ<sub>2</sub> = the pattern of β and β’ must be a mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ’ must form a mos.
+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’. Then the cyclic word Λ<sub>1</sub> formed by the a-components of the k-steps in P is α...αα’. Since Σ<sub>2</sub> is a single-period mos pattern of βb + (n &minus; β)(a~c) and β’a + (n &minus; β’)(a~c), the cyclic word Λ<sub>2</sub> = the pattern of β and β’ must be a single-period mos. Similarly, Λ<sub>3</sub> = the pattern of γ and γ’ is a single-period mos.
 Suppose Λ<sub>2</sub> is the mos λβ μβ’. Then Λ<sub>3</sub> is the mos (λ ± 1)γ (μ ∓ 1)γ’. 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.