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'''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'''. | 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. '''R'''P<sup>2</sup> can be visualized as a sphere with diametrically opposite points | 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 viewed as the same point. | ||
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'' − ''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.--> | 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'' − ''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.--> | ||