Generator-offset property: Difference between revisions

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We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c).

Suppose for sake of contradiction that ~~P has~~ only one ~~preimage~~ (α, β, γ) in S. Then projecting to S2 shows that S2's generator is the k-step (α + γ)*(a~c) + βb, and Σ2's imperfect generator is located at index n, like Σ1's imperfect generator is. Then S1 and S2 are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.

Suppose for sake of contradiction that only one size (α, β, γ) in S projects to P in S1. Then projecting to S2 shows that S2's generator is the k-step (α + γ)*(a~c) + βb, and Σ2's imperfect generator is located at index n, like Σ1's imperfect generator is. Then S1 and S2 are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S1 (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S1 come from a. Then the steps of S2 corresponding to the L of S1 must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S1 and S2 are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.

Only two sizes of 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.

Only two sizes of k-steps of S can project to P in S1, for if there are three sizes of k-steps (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S that project to P, 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 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:

So 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. By way of illustration, the chain of k-steps might look like this in S:

1 2 3 4 5 6 7 8 9

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Λ3 = γ γ γ γ’ γ γ γ γ γ

We ~~don’t know which P’s have preimage Q and which P’s have preimage R, but we~~ now do know that Σ2 has one more R than Σ3, and thus that:

We now do know that Σ2 has one more R than Σ3, and thus that:

Λ2 is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).

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We need only consider stacks up to f-many k-steps. Either:

# the stack has only ~~preimages of P’s~~ and ~~it either contains~~ R ~~or not~~; or

# the stack has only Q's and R's; or

# the stack has one T and does not contain any R (since it’s more than f − 1 generators away).

These give exactly three distinct sizes for every interval class. Hence S is SV3.

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 We can write sizes of intervals in S as vectors (p, q, r) using the basis (a, b, c).
-Suppose for sake of contradiction that P has only one preimage (α, β, γ) in S. Then projecting to S<sub>2</sub> shows that S<sub>2</sub>'s generator is the k-step (α + γ)*(a~c) + βb, and Σ<sub>2</sub>'s imperfect generator is located at index n, like Σ<sub>1</sub>'s imperfect generator is. Then S<sub>1</sub> and S<sub>2</sub> are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.
+Suppose for sake of contradiction that only one size (α, β, γ) in S projects to P in S<sub>1</sub>. Then projecting to S<sub>2</sub> shows that S<sub>2</sub>'s generator is the k-step (α + γ)*(a~c) + βb, and Σ<sub>2</sub>'s imperfect generator is located at index n, like Σ<sub>1</sub>'s imperfect generator is. Then S<sub>1</sub> and S<sub>2</sub> are the same mode of the same mos pattern (up to knowing which step size is the bigger one). Assume the L of S<sub>1</sub> (it could be s, but it doesn’t matter) is the result of identifying b and c, and all instances of s in S<sub>1</sub> come from a. Then the steps of S<sub>2</sub> corresponding to the L of S<sub>1</sub> must be either all b’s or all a~c’s, thus these steps are all b’s in S (otherwise they would be identified with the a, against the assumption that S<sub>1</sub> and S<sub>2</sub> are the same mos pattern and mode). So S has only two step sizes (a and b), contradicting the assumption that S has exactly three step sizes.
-Only two sizes of 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>.
+Only two sizes of k-steps of S can project to P in S<sub>1</sub>, for if there are three sizes of k-steps (α, β, γ), (α, β’, γ’), (α, β’’, γ’’) in S that project to P, 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 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:
+So 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. By way of illustration, the chain of k-steps might look like this in S:
 2 3 4  5 6 7 8 9
@@ Line 139: / Line 139: @@
   Λ<sub>3</sub> = γ γ γ γ’ γ γ γ γ γ
-We don’t know which P’s have preimage Q and which P’s have preimage R, but we now do know that Σ<sub>2</sub> has one more R than Σ<sub>3</sub>, and thus that:
+We now do know that Σ<sub>2</sub> has one more R than Σ<sub>3</sub>, and thus that:
 Λ<sub>2</sub> is λβ μβ’, 2 ≤ μ ≤ ceil(n/2).
@@ Line 169: / Line 169: @@
 We need only consider stacks up to f-many k-steps. Either:
-# the stack has only preimages of P’s and it either contains R or not; or
+# the stack has only Q's and R's; or
 # the stack has one T and does not contain any R (since it’s more than f &minus; 1 generators away).
 These give exactly three distinct sizes for every interval class. Hence S is SV3.