Recursive structure of MOS scales: Difference between revisions

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Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that W' is not a mos. Then for some k, W' must have k-steps of the following sizes:

# p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in W', corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's

# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking W₂ to the right ~~if necessary~~, we assume W₂ begins in λ.

# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking W₂ to the right until it begins in λ, which will never decrease the number of λ's, we can assume W₂ begins in λ.

~~where~~ pᵢ + qᵢ = k and p₂ - p₁ ≥ 2

Here, pᵢ + qᵢ = k and p₂ - p₁ ≥ 2.

Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps:

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 Without loss of generality assume r ≥ 1 (otherwise flip the roles of L and s). Let W'(λ, σ) be the reduced word with step sizes σ = r\n and λ = (r + 1)\n, and assume that W' is not a mos. Then for some k, W' must have k-steps of the following sizes:
 # p₁ λ's and q₁ σ's, represented by subword W₁(λ, σ) in W', corresponding to an interval in the mos with (p₁(r + 1) + q₁r) L's and k s's
-# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking W₂ to the right if necessary, we assume W₂ begins in λ.
+# p₂ λ's and q₂ σ's, represented by subword W₂(λ, σ) in W', corresponding to an interval in the mos with (p₂(r + 1) + q₂r) L's and k s's. By slinking W₂ to the right until it begins in λ, which will never decrease the number of λ's, we can assume W₂ begins in λ.
-where pᵢ + qᵢ = k and p₂ - p₁ ≥ 2
+Here, pᵢ + qᵢ = k and p₂ - p₁ ≥ 2.
 Let K = p₁(r + 1) + q₁r + k. Then we have at least 3 different sizes for (K+1)-steps: