Generator-offset property: Difference between revisions

Line 125:

(Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)

~~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 pattern (up to knowing which step size is the bigger one) and even the same mode:

Suppose 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 pattern (up to knowing which step size is the bigger one) and even the same mode:

S1: L ... L s

Line 134:

We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only 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, 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 we denote T. 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. By way of illustration, the chain of k-steps might look like this in S:

So suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which we denote T. 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. By way of illustration, the chain of k-steps might look like this in S:

1 2 3 4 5 6 7 8 9

@@ Line 125: / Line 125: @@
 (Note: We say “assume” as shorthand for either “assume without loss of generality” or “assume after excluding other possibilities”.)
-Assume that S<sub>2</sub>'s generator is also a k-step, and that Σ<sub>2</sub>'s I is located at index n. Then Σ<sub>1</sub> and Σ<sub>2</sub> are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode:
+Suppose that S<sub>2</sub>'s generator is also a k-step, and that Σ<sub>2</sub>'s I is located at index n. Then Σ<sub>1</sub> and Σ<sub>2</sub> are the same mos pattern (up to knowing which step size is the bigger one) and even the same mode:
   S<sub>1</sub>:    L  ... L  s
@@ Line 134: / Line 134: @@
 We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c). Only two k-steps of S can project to P in S<sub>1</sub>, for if P has three preimages (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S, then β, β’ and β’’ are three distinct values. Thus these would project to three different k-steps in S<sub>3</sub>, contradicting the mos property of S<sub>3</sub>.
-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 we denote T. 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. By way of illustration, the chain of k-steps might look like this in S:
+So suppose Q = (α, β, γ) ≠ R = (α, β’, γ’) are the two preimages of P in S. Then I has preimage (α’, β’’, γ’’), which we denote T. 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. By way of illustration, the chain of k-steps might look like this in S:
 2 3 4  5 6 7 8 9