Generator-offset property: Difference between revisions

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Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the corresponding steps of S2 must be either all a’s or all b~c’s. Thus these steps are all a’s or all b’s. This contradicts the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.

At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S2, contradicting the mos property of S2.

At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.

So assume Q = (α, β, γ) != R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which by abuse we also denote I. Here the values in each component differ by at most 1, and α != α’. Recall that I has preimage (α’, β’’, γ’’). 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. By way of illustration, the chain of k-steps might look like this in s:

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 Assume that S2's generator is also a k-step, and that Σ2's I is located at index n. Then Σ1 and Σ2 are the same mos and even the same mode. Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying a and b, and all instances of s in S1 come from c. Then the corresponding steps of S2 must be either all a’s or all b~c’s. Thus these steps are all a’s or all b’s. This contradicts the assumption that S has exactly three step sizes. This shows that the maximum variety of S must be at least 3, and that P has at least two preimages in S.
-At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S2, contradicting the mos property of S2.
+At most two k-steps of S can project to P in S1, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S (written in the basis a, b, c), then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S3, contradicting the mos property of S3.
 So assume Q = (α, β, γ) != R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which by abuse we also denote I. Here the values in each component differ by at most 1, and α != α’. Recall that I has preimage (α’, β’’, γ’’). 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. By way of illustration, the chain of k-steps might look like this in s: