User:Frostburn/Geometric algebra for regular temperaments: Difference between revisions

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 An expression such as <math>\mathbf{T} = \overleftarrow{12} \wedge \overleftarrow{19}</math> is nice for getting the feel of a temperament: Meantone feels like what [[12edo]] and [[19edo]] have in common. However it's hard  to figure out how a given monzo should be mapped to periods and generators.
-We can invert <math>\overrightarrow{p}</math> and <math>\overrightarrow{g}</math> ''(TODO: How?)'' to produce a [[mapping]] for the temperament. Let's choose <math>\overrightarrow{p} = \overrightarrow{2/1} = [1, 0, 0></math> and <math>\overrightarrow{g} = \overrightarrow{3/2} = [-1, 1, 0> </math> as the period and generator for Meantone. The mapping vals will be:
+With a temperament of rank 2 the period <math>\overrightarrow{p}</math> and the generator <math>\overrightarrow{g}</math> together with n - 2 commas <math>\overrightarrow{c}_i</math> that are tempered out form a basis for the whole n-dimensional JI subgroup and can thus be inverted to produce a [[mapping]] for the temperament.
+Let's choose <math>\overrightarrow{p} = \overrightarrow{2/1} = [1, 0, 0></math> and <math>\overrightarrow{g} = \overrightarrow{3/2} = [-1, 1, 0> </math> as the period and generator for Meantone. In this case there's only a single comma <math>\overrightarrow{c} = \overrightarrow{81/80} = [-4, 4, 1 ></math>. Inverting the matrix <math>[\overrightarrow{p}, \overrightarrow{g}, \overrightarrow{c}]</math> gives us the mapping:
 :<math>\begin{align}
 \overleftarrow{p} &= < 1, 1, 0 ] \\
-\overleftarrow{g} &= < 0, 1, 4 ]
+\overleftarrow{g} &= < 0, 1, 4 ] \\
+\overleftarrow{c} &= < 0, 0, -1]
 \end{align}</math>
-Now <math>\mathbf{T} = \overleftarrow{p} \wedge \overleftarrow{g}</math> and for any monzo <math>\overrightarrow{m}</math> we can get the number of periods as <math>\overleftarrow{p} \cdot \overrightarrow{m}</math> and the number of generators as <math>\overleftarrow{g} \cdot \overrightarrow{m}</math>.
+Now <math>\mathbf{T} = \overleftarrow{p} \wedge \overleftarrow{g}</math> and for any monzo <math>\overrightarrow{m}</math> we can get the number of periods as <math>\overleftarrow{p} \cdot \overrightarrow{m}</math> and the number of generators as <math>\overleftarrow{g} \cdot \overrightarrow{m}</math>. (We don't care about how many commas are needed because they're tempered out anyway.)
-For example <math>\overleftarrow{p} \cdot \overrightarrow{5/3} = -1</math> and <math>\overleftarrow{g} \cdot \overrightarrow{5/3} = 3</math>. Indeed <math>\frac{5}{3} \sim (\frac{2}{1})^{-1}(\frac{3}{2})^3 = \frac{27}{16}</math> in Meantone because <math>\frac{5}{3}\frac{81}{80} = \frac{27}{16}</math> and the comma <math>\frac{81}{80}</math> vanishes in Meantone. Musically this means that the major sixth is constructed by stacking three perfect fifths and reducing by an octave.
+For example <math>\overleftarrow{p} \cdot \overrightarrow{5/3} = -1</math> and <math>\overleftarrow{g} \cdot \overrightarrow{5/3} = 3</math>. Indeed <math>\frac{5}{3} \sim (\frac{2}{1})^{-1}(\frac{3}{2})^3 = \frac{27}{16}</math> in Meantone. Musically this means that the major sixth is constructed by stacking three perfect fifths and reducing by an octave. If we want to be pedantic we can calculate <math>\overleftarrow{c} \cdot \overrightarrow{5/3} = -1</math> for the final numerically correct expression <math>\frac{5}{3} = \frac{27}{16}(\frac{81}{80})^{-1}</math>.
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