User:Frostburn/Theory From First Principles: Difference between revisions

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Usually the basis is obvious from context e.g. <math>a = 2, b = 3, c = 5</math>. In these cases we use a left-facing arrow e.g.

<math>\overleftarrow{12} := \mathrm{val}(12; 2, 3, 5) = 12 ~~e_2~~ + 19 ~~e_3~~ + 28 ~~e_5~~ =: \langle 12, 19, 28 \rbrack</math>

<math>\overleftarrow{12} := \mathrm{val}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>

The superscripts are simply the geometric reciprocals of our basis. See unit considerations below.

We can use these new objects to calculate how many steps of 12edo a tempered interval spans e.g.

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The geometric inverses are mainly relevant for subgroup temperaments. Consider [[The_Archipelago#Subgroup_temperaments|Barbados]]:

<math>\overleftarrow{5} := \mathrm{val}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 ~~e_2~~ + 8 ~~e_3~~ - \frac{7}{2}~~e_5~~ + \frac{7}{2}e_{13}</math>

<math>\overleftarrow{5} := \mathrm{val}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>

We can verify that the comma 676/675 indeed vanishes using this val:

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i.e. we have <math>e^2 = e_2 / w_2^2 = \hat{i} / w_2</math>. A reciprocal cent can now be expressed as <math>¢^{-1} = 1200 e^2 = 100 d^{-1} \hat{i}</math> or 100 reciprocal demitones in the <math>\hat{i}</math> direction.

== ~~Exterior~~ algebra nonsense ==

== Clifford algebra nonsense ==

Both 12edo and 7edo temper out the syntonic comma:

<math>\overleftarrow{12} \cdot \overrightarrow{81/80} = 0 = \overleftarrow{7} \cdot \overrightarrow{81/80}</math> .

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We can identify the plane spanned by <math>\overleftarrow{12}</math> and <math>\overleftarrow{7}</math> as the (5-limit) Meantone temperament. We can use wedges to represent it symbolically:

<math>\overleftarrow{12} \wedge \overleftarrow{7} = -4 ~~e_3~~ \wedge ~~e_5~~ + 4 ~~e_5~~ \wedge ~~e_2~~ - e_2 \wedge e_3</math> ,

<math>\overleftarrow{12} \wedge \overleftarrow{7} = -4 e^3 \wedge e^5 + 4 e^5 \wedge e^2 - e^2 \wedge e^3</math> ,

where the components are basis planes. E.g. <math>e^3 \wedge e^5</math> is the plane where octaves are tempered out. The wedge of any vector with itself is zero i.e. you can't span a plane with only one direction. The wedge product is also antisymmetric and the planes come with signed weights but we mostly care about the orientation they represent. Note that <math>e^2</math> represents the "line" where tritaves and pentaves are tempered out.

The largest possible wedge combines all of the basis vectors and represents just intonation i.e. no tempering whatsoever: <math>e^2 \wedge e^3 \wedge e^5</math>.

We can define the dual operator:

<math>

\begin{align}

\overline{e_2} &= e^3 \wedge e^5 \\

\overline{e_3} &= e^5 \wedge e^2 \\

\overline{e_5} &= e^2 \wedge e^3 , \\

\end{align}

</math>

which is is extended linearly to all vectors. It is obvious from the definitions that the Meantone temperament is represented by <math>\overline{\overrightarrow{81/80}}</math>.

Defining the vee operator <math>a \vee b := \overline{\overline{b} \wedge \overline{a}}</math>, where the inner overlines are the inverses of the dual (always obvious from context). Wedges combine vals into temperaments that are closer to just intonation while vees do progressive damage to just intonation. As an exercise let's calculate the damage done by combining Meantone with Augmented:

~~where the components are basis planes. E.g.~~ <math>e_3 \wedge e_5</math> is the plane where octaves are tempered out. The wedge of any vector with itself is zero i.e. you can't span a plane with only one direction. The wedge product is also antisymmetric and the planes come with signed weights but we mostly care about the orientation they represent.

<math>

\begin{align}

& \overline{\overrightarrow{81/80}} \vee \overline{\overrightarrow{128/125}} \\

& = \overline{-4 e_2 + 4 e_3 - e_5 } \vee \overline{7 e_2 - 3 e_5} \\

& = \overline{(7 e_2 - 3 e_5) \wedge (-4 e_2 + 4 e_3 - e_5)} \\

& = \overline{28 e_2 \wedge e_3 - 7 e_2 \wedge e_5 + 12 e_5 \wedge e_2 - 12 e_5 \wedge e_3} \\

& = \overline{12 e_3 \wedge e_5 + 19 e_5 \wedge e_2 + 28 e_2 \wedge e_3} \\

& = 12 e^2 + 19 e^3 + 28 e^5 \\

& = \overleftarrow{12}

\end{align}

</math>

~~The largest possible wedge combines all of the basis vectors and represents just intonation i.e. no tempering whatsoever: <math>e_2 \wedge e_3 \wedge e_5</math>.~~

Neat!

Exterior algebras do have a sense of ~~orthogonality~~ but we need a metric to do projection and tuning which we already ~~defined implicitly by giving numerical values to geometric inverses and dot products~~. As far as data structures go, full Clifford algebras are memory-hungry. Not worth the complication in a general purpose tool like Scale Workshop.

Exterior algebras do have a sense of lying-in-the-plane-of but we need a metric to do projection and tuning which we already touched upon in the units section. As far as data structures go, full Clifford algebras are memory-hungry. Not worth the complication in a general purpose tool like Scale Workshop.

@@ Line 42: / Line 42: @@
 Usually the basis is obvious from context e.g. <math>a = 2, b = 3, c = 5</math>. In these cases we use a left-facing arrow e.g.
-<math>\overleftarrow{12} := \mathrm{val}(12; 2, 3, 5) = 12 e_2 + 19 e_3 + 28 e_5 =: \langle 12, 19, 28 \rbrack</math>
+<math>\overleftarrow{12} := \mathrm{val}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>
+The superscripts are simply the geometric reciprocals of our basis. See unit considerations below.
 We can use these new objects to calculate how many steps of 12edo a tempered interval spans e.g.
@@ Line 54: / Line 56: @@
 The geometric inverses are mainly relevant for subgroup temperaments. Consider [[The_Archipelago#Subgroup_temperaments|Barbados]]:
-<math>\overleftarrow{5} := \mathrm{val}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e_2 + 8 e_3 - \frac{7}{2}e_5 + \frac{7}{2}e_{13}</math>
+<math>\overleftarrow{5} := \mathrm{val}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>
 We can verify that the comma 676/675 indeed vanishes using this val:
@@ Line 74: / Line 76: @@
 i.e. we have <math>e^2 = e_2 / w_2^2 = \hat{i} / w_2</math>. A reciprocal cent can now be expressed as <math>¢^{-1} = 1200 e^2 = 100 d^{-1} \hat{i}</math> or 100 reciprocal demitones in the <math>\hat{i}</math> direction.
-== Exterior algebra nonsense ==
+== Clifford algebra nonsense ==
 Both 12edo and 7edo temper out the syntonic comma:
 <math>\overleftarrow{12} \cdot \overrightarrow{81/80} = 0 = \overleftarrow{7} \cdot \overrightarrow{81/80}</math> .
@@ Line 82: / Line 84: @@
 We can identify the plane spanned by <math>\overleftarrow{12}</math> and <math>\overleftarrow{7}</math> as the (5-limit) Meantone temperament. We can use wedges to represent it symbolically:
-<math>\overleftarrow{12} \wedge \overleftarrow{7} = -4 e_3 \wedge e_5 + 4 e_5 \wedge e_2 - e_2 \wedge e_3</math> ,
+<math>\overleftarrow{12} \wedge \overleftarrow{7} = -4 e^3 \wedge e^5 + 4 e^5 \wedge e^2 - e^2 \wedge e^3</math> ,
+where the components are basis planes. E.g. <math>e^3 \wedge e^5</math> is the plane where octaves are tempered out. The wedge of any vector with itself is zero i.e. you can't span a plane with only one direction. The wedge product is also antisymmetric and the planes come with signed weights but we mostly care about the orientation they represent. Note that <math>e^2</math> represents the "line" where tritaves and pentaves are tempered out.
+The largest possible wedge combines all of the basis vectors and represents just intonation i.e. no tempering whatsoever: <math>e^2 \wedge e^3 \wedge e^5</math>.
+We can define the dual operator:
+<math>
+\begin{align}
+ \overline{e_2} &= e^3 \wedge e^5 \\
+ \overline{e_3} &= e^5 \wedge e^2 \\
+ \overline{e_5} &= e^2 \wedge e^3 , \\
+\end{align}
+</math>
+which is is extended linearly to all vectors. It is obvious from the definitions that the Meantone temperament is represented by <math>\overline{\overrightarrow{81/80}}</math>.
+Defining the vee operator <math>a \vee b := \overline{\overline{b} \wedge \overline{a}}</math>, where the inner overlines are the inverses of the dual (always obvious from context). Wedges combine vals into temperaments that are closer to just intonation while vees do progressive damage to just intonation. As an exercise let's calculate the damage done by combining Meantone with Augmented:
-where the components are basis planes. E.g. <math>e_3 \wedge e_5</math> is the plane where octaves are tempered out. The wedge of any vector with itself is zero i.e. you can't span a plane with only one direction. The wedge product is also antisymmetric and the planes come with signed weights but we mostly care about the orientation they represent.
+<math>
+\begin{align}
+& \overline{\overrightarrow{81/80}} \vee \overline{\overrightarrow{128/125}} \\
+& = \overline{-4 e_2 + 4 e_3 - e_5 } \vee \overline{7 e_2 - 3 e_5} \\
+& = \overline{(7 e_2 - 3 e_5) \wedge (-4 e_2 + 4 e_3 - e_5)} \\
+& = \overline{28 e_2 \wedge e_3 - 7 e_2 \wedge e_5 + 12 e_5 \wedge e_2 - 12 e_5 \wedge e_3} \\
+& = \overline{12 e_3 \wedge e_5 + 19 e_5 \wedge e_2 + 28 e_2 \wedge e_3} \\
+& = 12 e^2 + 19 e^3 + 28 e^5 \\
+& = \overleftarrow{12}
+\end{align}
+</math>
-The largest possible wedge combines all of the basis vectors and represents just intonation i.e. no tempering whatsoever: <math>e_2 \wedge e_3 \wedge e_5</math>.
+Neat!
-Exterior algebras do have a sense of orthogonality but we need a metric to do projection and tuning which we already defined implicitly by giving numerical values to geometric inverses and dot products.  As far as data structures go, full Clifford algebras are memory-hungry. Not worth the complication in a general purpose tool like Scale Workshop.
+Exterior algebras do have a sense of lying-in-the-plane-of but we need a metric to do projection and tuning which we already touched upon in the units section. As far as data structures go, full Clifford algebras are memory-hungry. Not worth the complication in a general purpose tool like Scale Workshop.