Equivalence continuum: Difference between revisions

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This has a particularly simple description when ''r'' = 1 (i.e. when ''T'' is an edo), ''n'' = 3 (for example, when ''S'' is the [[5-limit]], 2.3.7 or 2.5.7) and ''k'' = 2 (so that we're considering the equivalence continua of rank-2 temperaments associated with an edo), as then '''G''' = '''Gr'''(1, 2) = '''R'''P1 (the real projective line), which can be viewed as a circle. Then the continuum corresponds to the set of lines with rational slope passing through the origin on the Cartesian plane '''R'''2 where the lattice of ker(''T'') lives. The lattice of ker(''T'') is generated by a [[basis]] of some choice of two commas '''u''' and '''v''' in ''S'' tempered out by the edo; view the plane as having two perpendicular axes corresponding to '''u''' and '''v''' directions. A rational point, i.e. a temperament on the continuum, then corresponds to a rational or infinite ratio ''t'' = ''p''/''q'', where the temperament is defined by the identification ''p'''''u''' ~ ''q'''''v''' (written additively). When ''t'' = 0, this corresponds to the temperament tempering out '''v'''. When ''t'' = (unsigned) infinity, this corresponds to the temperament tempering out '''u'''.

A higher-dimensional example: Say that ''r'' = 1, ''n'' = 4 (e.g. when ''S'' is the [[7-limit]]), and ''k'' = 2, for example the set of rank-2 [[7-limit]] temperaments supported by [[31edo]]. Then our Grassmannian '''G''' becomes '''Gr'''(2, 3). Define a coordinate system (''x'', ''y'', ''z'') for ker(T) using some fixed comma basis '''u'''''x'', '''u'''''y'', '''u'''''z'' for ker(T). Then our Grassmannian can be identified with '''R'''P2 (the real projective plane, the space of lines through the origin in 3-dimensional space) by taking the unique line '''Rv''' perpendicular (according to the dot product given by the given coordinates) to the plane of commas tempered out for each temperament. Say that the vector '''v''' (which depends on ''T'') defining ~~the~~ unique line has components (''v''1, ''v''2, ''v''3), so that the plane associated with the rank-2 temperament has equation ''v''1''x'' + ''v''2''y'' + ''v''3''z'' = 0. [We may further assume that ''v''1, ''v''2, ''v''3 are integers with gcd 1, since the condition of being perpendicular to two integer vectors is defined by a system of linear equations with integer coefficients, thus has a unique rational solution up to scaling.] One coordinate ''v''i is always guaranteed to be nonzero, for any temperament. Assuming ''v''1 ≠ 0, we can scale '''v''' by 1/''v''1, then the resulting vector '''v'''/''v''1 = (1, ''v''2/''v''1, v3/''v''1) = (1, ''s'', ''t'') points in the same direction as '''v''' and describes two rational (or infinite) parameters ''s'' and ''t'' which defines any temperament with ''v''1 ≠ 0 on 31edo's 7-limit rank-2 continuum uniquely. Two-dimensional coordinates can similarly be assigned for the set of all temperaments such that ''v''2 ≠ 0 and the set of all temperaments such that ''v''3 ≠ 0.

A higher-dimensional example: Say that ''r'' = 1, ''n'' = 4 (e.g. when ''S'' is the [[7-limit]]), and ''k'' = 2, for example the set of rank-2 [[7-limit]] temperaments supported by [[31edo]]. Then our Grassmannian '''G''' becomes '''Gr'''(2, 3). Define a coordinate system (''x'', ''y'', ''z'') for ker(T) using some fixed comma basis '''u'''''x'', '''u'''''y'', '''u'''''z'' for ker(T). Then our Grassmannian can be identified with '''R'''P2 (the real projective plane, the space of lines through the origin in 3-dimensional space) by taking the unique line '''Rv''' perpendicular (according to the dot product given by the given coordinates) to the plane of commas tempered out for each temperament. '''R'''P2 can be visualized as a sphere with diametrically opposite points identified.

Say that the vector '''v''' (which depends on ''T'') defining this unique line has components (''v''1, ''v''2, ''v''3), so that the plane associated with the rank-2 temperament has equation ''v''1''x'' + ''v''2''y'' + ''v''3''z'' = 0. [We may further assume that ''v''1, ''v''2, ''v''3 are integers with gcd 1, since the condition of being perpendicular to two integer vectors is defined by a system of linear equations with integer coefficients, thus has a unique rational solution up to scaling.] One coordinate ''v''i is always guaranteed to be nonzero, for any temperament. Assuming ''v''1 ≠ 0, we can scale '''v''' by 1/''v''1, then the resulting vector '''v'''/''v''1 = (1, ''v''2/''v''1, v3/''v''1) = (1, ''s'', ''t'') points in the same direction as '''v''' and describes two rational (or infinite) parameters ''s'' and ''t'' which defines any temperament with ''v''1 ≠ 0 on 31edo's 7-limit rank-2 continuum uniquely. Two-dimensional coordinates can similarly be assigned for the set of all temperaments such that ''v''2 ≠ 0 and the set of all temperaments such that ''v''3 ≠ 0.

NB: Not all equivalence continua are projective spaces, because not all Grassmannians are; for example, '''Gr'''(2, 4) is not a projective space.

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 This has a particularly simple description when ''r'' = 1 (i.e. when ''T'' is an edo), ''n'' = 3 (for example, when ''S'' is the [[5-limit]], 2.3.7 or 2.5.7) and ''k'' = 2 (so that we're considering the equivalence continua of rank-2 temperaments associated with an edo), as then '''G''' = '''Gr'''(1, 2) = '''R'''P<sup>1</sup> (the real projective line), which can be viewed as a circle. Then the continuum corresponds to the set of lines with rational slope passing through the origin on the Cartesian plane '''R'''<sup>2</sup> where the lattice of ker(''T'') lives. The lattice of ker(''T'') is generated by a [[basis]] of some choice of two commas '''u''' and '''v''' in ''S'' tempered out by the edo; view the plane as having two perpendicular axes corresponding to '''u''' and '''v''' directions. A rational point, i.e. a temperament on the continuum, then corresponds to a rational or infinite ratio ''t'' = ''p''/''q'', where the temperament is defined by the identification ''p'''''u''' ~ ''q'''''v''' (written additively). When ''t'' = 0, this corresponds to the temperament tempering out '''v'''. When ''t'' = (unsigned) infinity, this corresponds to the temperament tempering out '''u'''.
-A higher-dimensional example: Say that ''r'' = 1, ''n'' = 4 (e.g. when ''S'' is the [[7-limit]]), and ''k'' = 2, for example the set of rank-2 [[7-limit]] temperaments supported by [[31edo]].  Then our Grassmannian '''G''' becomes '''Gr'''(2, 3). Define a coordinate system (''x'', ''y'', ''z'') for ker(T) using some fixed comma basis '''u'''<sub>''x''</sub>, '''u'''<sub>''y''</sub>, '''u'''<sub>''z''</sub> for ker(T). Then our Grassmannian can be identified with '''R'''P<sup>2</sup> (the real projective plane, the space of lines through the origin in 3-dimensional space) by taking the unique line '''Rv''' perpendicular (according to the dot product given by the given coordinates) to the plane of commas tempered out for each temperament. Say that the vector '''v''' (which depends on ''T'') defining the unique line has components (''v''<sub>1</sub>, ''v''<sub>2</sub>, ''v''<sub>3</sub>), so that the plane associated with the rank-2 temperament has equation  ''v''<sub>1</sub>''x'' + ''v''<sub>2</sub>''y'' + ''v''<sub>3</sub>''z'' = 0. [We may further assume that ''v''<sub>1</sub>, ''v''<sub>2</sub>, ''v''<sub>3</sub> are integers with gcd 1, since the condition of being perpendicular to two integer vectors is defined by a system of linear equations with integer coefficients, thus has a unique rational solution up to scaling.] One coordinate ''v''<sub>i</sub> is always guaranteed to be nonzero, for any temperament. Assuming ''v''<sub>1</sub> ≠ 0, we can scale '''v''' by 1/''v''<sub>1</sub>, then the resulting vector '''v'''/''v''<sub>1</sub> = (1, ''v''<sub>2</sub>/''v''<sub>1</sub>, v<sub>3</sub>/''v''<sub>1</sub>) = (1, ''s'', ''t'') points in the same direction as '''v''' and describes two rational (or infinite) parameters ''s'' and ''t'' which defines any temperament with ''v''<sub>1</sub> ≠ 0 on 31edo's 7-limit rank-2 continuum uniquely. Two-dimensional coordinates can similarly be assigned for the set of all temperaments such that ''v''<sub>2</sub> ≠ 0 and the set of all temperaments such that ''v''<sub>3</sub> ≠ 0.<!-- Note that this continuum is actually part of a mathematical manifold with a more complicated topology and needs to be described using more than one local chart (coordinate system) constructed like this; unlike for the ''k'' &minus; ''r'' = 1 case, a single circle won't define every point on this 2-dimensional continuum, just like a single circle won't define every point on a 2-dimensional sphere.-->
+A higher-dimensional example: Say that ''r'' = 1, ''n'' = 4 (e.g. when ''S'' is the [[7-limit]]), and ''k'' = 2, for example the set of rank-2 [[7-limit]] temperaments supported by [[31edo]].  Then our Grassmannian '''G''' becomes '''Gr'''(2, 3). Define a coordinate system (''x'', ''y'', ''z'') for ker(T) using some fixed comma basis '''u'''<sub>''x''</sub>, '''u'''<sub>''y''</sub>, '''u'''<sub>''z''</sub> for ker(T). Then our Grassmannian can be identified with '''R'''P<sup>2</sup> (the real projective plane, the space of lines through the origin in 3-dimensional space) by taking the unique line '''Rv''' perpendicular (according to the dot product given by the given coordinates) to the plane of commas tempered out for each temperament. '''R'''P<sup>2</sup> can be visualized as a sphere with diametrically opposite points identified.
+Say that the vector '''v''' (which depends on ''T'') defining this unique line has components (''v''<sub>1</sub>, ''v''<sub>2</sub>, ''v''<sub>3</sub>), so that the plane associated with the rank-2 temperament has equation  ''v''<sub>1</sub>''x'' + ''v''<sub>2</sub>''y'' + ''v''<sub>3</sub>''z'' = 0. [We may further assume that ''v''<sub>1</sub>, ''v''<sub>2</sub>, ''v''<sub>3</sub> are integers with gcd 1, since the condition of being perpendicular to two integer vectors is defined by a system of linear equations with integer coefficients, thus has a unique rational solution up to scaling.] One coordinate ''v''<sub>i</sub> is always guaranteed to be nonzero, for any temperament. Assuming ''v''<sub>1</sub> ≠ 0, we can scale '''v''' by 1/''v''<sub>1</sub>, then the resulting vector '''v'''/''v''<sub>1</sub> = (1, ''v''<sub>2</sub>/''v''<sub>1</sub>, v<sub>3</sub>/''v''<sub>1</sub>) = (1, ''s'', ''t'') points in the same direction as '''v''' and describes two rational (or infinite) parameters ''s'' and ''t'' which defines any temperament with ''v''<sub>1</sub> ≠ 0 on 31edo's 7-limit rank-2 continuum uniquely. Two-dimensional coordinates can similarly be assigned for the set of all temperaments such that ''v''<sub>2</sub> ≠ 0 and the set of all temperaments such that ''v''<sub>3</sub> ≠ 0.<!-- Note that this continuum is actually part of a mathematical manifold with a more complicated topology and needs to be described using more than one local chart (coordinate system) constructed like this; unlike for the ''k'' &minus; ''r'' = 1 case, a single circle won't define every point on this 2-dimensional continuum, just like a single circle won't define every point on a 2-dimensional sphere.-->
 NB: Not all equivalence continua are projective spaces, because not all Grassmannians are; for example, '''Gr'''(2, 4) is not a projective space.