Ternary scale theorems: Difference between revisions

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# Either such a stack has '''i''' as one of its ''k''-steps, or it does not.

Depending on point (2), ''m'' may be even or odd. If ''m'' is odd, then in any stack in ''S'' that does not have '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. On the other hand, '''i''' has an even number of non-'''X''' steps, thus the numbers of '''Y''' and '''Z''' are the same in '''v''' and equal to min(, |'''u'''1|'''Y''', |'''u'''2|'''Y''').

Depending on point (2), ''m'' may be even or odd. If ''m'' is odd, then in any stack in ''S'' that does not have '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. On the other hand, '''i''' has an even number of non-'''X''' steps, thus the numbers of '''Y''' and '''Z''' are the same in '''v''' and equal to min(|'''u'''1|'''Y''', |'''u'''2|'''Y''').

In summary: Let '''u'''1, '''u'''2 be the two sizes that include '''i''', and '''v''' be the size that does not. Say that '''u'''1 has one more '''g'''1 than '''g'''2. Then

* |'''u'''1|'''X''' = |'''u'''2|'''X''' = |v|'''X''' − 1

* |'''u'''1|'''Y''' = |'''u'''2|'''Y'''</sub + 1

* |'''u'''1|'''Y''' = |'''u'''2|'''Y''' + 1

* |'''u'''1|'''Z''' = |'''u'''2|'''Z'''</sub − 1

* |'''u'''1|'''Z''' = |'''u'''2|'''Z''' − 1

* |'''v'''|'''Y''' = |'''v'''|'''Z''' = |'''u'''1|'''Y'''.

This proves that the set of ''j''-steps is balanced. The argument for when ''m'' is ~~odd~~ is similar. {{qed}}

This proves that the set of ''j''-steps is balanced. The argument for when ''m'' is even is similar. {{qed}}

== Theorem 6 (Classification of MV3 scales) ==

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 # Either such a stack has '''i''' as one of its ''k''-steps, or it does not.
-Depending on point (2), ''m'' may be even or odd. If ''m'' is odd, then in any stack in ''S'' that does not have '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. On the other hand, '''i''' has an even number of non-'''X''' steps, thus the numbers of '''Y''' and '''Z''' are the same in '''v''' and equal to min(, |'''u'''<sub>1</sub>|<sub>'''Y'''</sub>, |'''u'''<sub>2</sub>|<sub>'''Y'''</sub>).
+Depending on point (2), ''m'' may be even or odd. If ''m'' is odd, then in any stack in ''S'' that does not have '''i''', the number of stacked '''g''' generators in the projection is odd, hence the number of non-'''X''' letters in the corresponding ''km''-step word in ''S'' is odd. Each incremental shift of the boundary that does not result in including '''i''' results in one '''Y''' being swapped for a '''Z''', or vice versa, while the number of '''X''' steps remains ''t''. On the other hand, '''i''' has an even number of non-'''X''' steps, thus the numbers of '''Y''' and '''Z''' are the same in '''v''' and equal to min(|'''u'''<sub>1</sub>|<sub>'''Y'''</sub>, |'''u'''<sub>2</sub>|<sub>'''Y'''</sub>).
 In summary: Let '''u'''<sub>1</sub>, '''u'''<sub>2</sub> be the two sizes that include '''i''', and '''v''' be the size that does not. Say that '''u'''<sub>1</sub> has one more '''g'''<sub>1</sub> than '''g'''<sub>2</sub>. Then
 * |'''u'''<sub>1</sub>|<sub>'''X'''</sub> = |'''u'''<sub>2</sub>|<sub>'''X'''</sub> = |v|<sub>'''X'''</sub> &minus; 1
-* |'''u'''<sub>1</sub>|<sub>'''Y'''</sub> = |'''u'''<sub>2</sub>|<sub>'''Y'''</sub + 1
+* |'''u'''<sub>1</sub>|<sub>'''Y'''</sub> = |'''u'''<sub>2</sub>|<sub>'''Y'''</sub> + 1
-* |'''u'''<sub>1</sub>|<sub>'''Z'''</sub> = |'''u'''<sub>2</sub>|<sub>'''Z'''</sub &minus; 1
+* |'''u'''<sub>1</sub>|<sub>'''Z'''</sub> = |'''u'''<sub>2</sub>|<sub>'''Z'''</sub> &minus; 1
 * |'''v'''|<sub>'''Y'''</sub> = |'''v'''|<sub>'''Z'''</sub> = |'''u'''<sub>1</sub>|<sub>'''Y'''</sub>.
-This proves that the set of ''j''-steps is balanced. The argument for when ''m'' is odd is similar. {{qed}}
+This proves that the set of ''j''-steps is balanced. The argument for when ''m'' is even is similar. {{qed}}
 == Theorem 6 (Classification of MV3 scales) ==