Douglas Blumeyer's RTT How-To: Difference between revisions

Line 1,452:

Note that multicommas could just as well serve the purpose of a unique identifier for temperaments. There’s no particular reason to prefer the canonical multimap to the canonical multicomma. By convention, however, the canonical multimap is the one that’s used.

If you need to do this process for a higher dimensionality than 6, then you'll need to understand how I found the symbols for each cell of Figure 6b. Here's how:

If you need to do this process for a higher dimensionality than 6, then you'll need to understand how I found the symbols for each cell of Figure 6b. Here's how<ref>

You may wonder: but why are these sign patterns the way they are? Well, to deeply understand this, you'd probably have to hit some math books. I can shed a little more light on it, but it will still be fairly hand-wavy. The basic gist of it is this. The sets of prime numbers we've looked at, such as (2,3) or (3,5,11) are really like combinations of basis vectors, or in other words, atomic elements that are incomparable, in different terms, orthogonal, however you want to call it. In most exterior algebra texts, you'll see them in abstract, variable form, like this: e₁, e₂, e₃, etc. and for this demonstration it will be easier to use them that way, so that's what we'll do. To be clear, 2 is like e₁, 3 is like e₂, 5 is like e₃, etc. So as you've seen, when we take the dual, the resulting dual's terms correspond to the complementary set of these basis vectors; i.e. if there are 5 of them, the corresponding term in the dual for the term with basis vector combination e₁e₄ will be e₂e₃e₅. We can write this like:

∗(e₁∧e₄) = (e₂∧e₃∧e₅)

Let's look at a simple case: dimensionality 3. Here's all such pairs of dual basis vector combinations:

e₁ ↔ e₂e₃

e₂ ↔ e₁e₃

e₃ ↔ e₁e₂

We can rewrite this like

1|23

2|13

3|12

Now, for each line, we need to swap elements until they are in order. Note that this is the kind of swapping where you imagine the objects are in boxes and we pick up two and a time and swap which box they're in; not the kind of swapping where you slide them around a table to change order and afterwards shift things around to fill in gaps.

1|23 (already done)

2|13, 1|23 (done, 1 swap necessary)

3|12, 1|32, 1|23 (done, 2 swaps necessary)

Finally, you count the number of swaps required to get things in order. If the count is odd, then this term will be negative in the dual. If it is positive, this term will be positive. This is the derivation of the sign change pattern for d=3 g=1 being +-+. But we still haven't really explained why this is. That's because a property of exterior algebra is that it's not exactly commutative. e₁e₂ ≠ e₂e₁. Instead, e₁e₂ = -e₂e₁! So every time you need to swap these elements, it introduces another sign change. let's try one more example: d=5 r=3.

e₁e₂e₃ ↔ e₄e₅

e₁e₂e₄ ↔ e₃e₅

e₁e₂e₅ ↔ e₃e₄

e₁e₃e₄ ↔ e₂e₅

e₁e₃e₅ ↔ e₂e₄

e₁e₄e₅ ↔ e₂e₃

e₂e₃e₄ ↔ e₁e₅

e₂e₃e₅ ↔ e₁e₄

e₂e₄e₅ ↔ e₁e₃

e₃e₄e₅ ↔ e₁e₂

Rewrite it:

123|45

124|35

125|34

134|25

135|24

145|23

234|15

235|14

245|13

345|12

Swap until in order:

123|45

124|35, 123|45

125|34, 123|54, 123|45

134|25, 124|35, 123|45

135|24, 125|34, 123|54, 123|45

145|23, 125|43, 123|45

234|15, 134|25, 124|35, 123|45

235|14, 135|24, 125|34, 123|54, 123|45

245|13, 145|23, 125|43, 123|45

345|12, 145|32, 125|34, 123|54, 123|45

Counting swaps, we see 0,1,2,2,3,2,3,4,3,4 and that gives even,odd,even,even,odd,even,odd,even,odd,even, so that gives +-++-+-+-+. Great!

</ref>:

# Take the rank, halved, rounded up. In our case, <math>\lceil \frac{r}{2} \rceil = \lceil \frac{2}{2} \rceil = \lceil 1 \rceil = 1</math>. Save that result for later. Let’s call it <math>x</math>.

@@ Line 1,452: / Line 1,452: @@
 Note that multicommas could just as well serve the purpose of a unique identifier for temperaments. There’s no particular reason to prefer the canonical multimap to the canonical multicomma. By convention, however, the canonical multimap is the one that’s used.
-If you need to do this process for a higher dimensionality than 6, then you'll need to understand how I found the symbols for each cell of Figure 6b. Here's how:
+If you need to do this process for a higher dimensionality than 6, then you'll need to understand how I found the symbols for each cell of Figure 6b. Here's how<ref>
+You may wonder: but why are these sign patterns the way they are? Well, to deeply understand this, you'd probably have to hit some math books. I can shed a little more light on it, but it will still be fairly hand-wavy. The basic gist of it is this. The sets of prime numbers we've looked at, such as (2,3) or (3,5,11) are really like combinations of basis vectors, or in other words, atomic elements that are incomparable, in different terms, orthogonal, however you want to call it. In most exterior algebra texts, you'll see them in abstract, variable form, like this: e₁, e₂, e₃, etc. and for this demonstration it will be easier to use them that way, so that's what we'll do. To be clear, 2 is like e₁, 3 is like e₂, 5 is like e₃, etc. So as you've seen, when we take the dual, the resulting dual's terms correspond to the complementary set of these basis vectors; i.e. if there are 5 of them, the corresponding term in the dual for the term with basis vector combination e₁e₄ will be e₂e₃e₅. We can write this like:<br>
+<br>
+∗(e₁∧e₄) = (e₂∧e₃∧e₅)<br>
+<br>
+Let's look at a simple case: dimensionality 3. Here's all such pairs of dual basis vector combinations:<br>
+<br>
+e₁ ↔ e₂e₃<br>
+e₂ ↔ e₁e₃<br>
+e₃ ↔ e₁e₂<br>
+<br>
+We can rewrite this like<br>
+<br>
+|23<br>
+|13<br>
+|12<br>
+<br>
+Now, for each line, we need to swap elements until they are in order. Note that this is the kind of swapping where you imagine the objects are in boxes and we pick up two and a time and swap which box they're in; not the kind of swapping where you slide them around a table to change order and afterwards shift things around to fill in gaps.<br>
+<br>
+|23 (already done)<br>
+|13, 1|23 (done, 1 swap necessary)<br>
+|12, 1|32, 1|23 (done, 2 swaps necessary)<br>
+<br>
+Finally, you count the number of swaps required to get things in order. If the count is odd, then this term will be negative in the dual. If it is positive, this term will be positive. This is the derivation of the sign change pattern for d=3 g=1 being +-+. But we still haven't really explained why this is. That's because a property of exterior algebra is that it's not exactly commutative. e₁e₂ ≠ e₂e₁. Instead, e₁e₂ = -e₂e₁! So every time you need to swap these elements, it introduces another sign change. let's try one more example: d=5 r=3.<br>
+<br>
+e₁e₂e₃ ↔ e₄e₅<br>
+e₁e₂e₄ ↔ e₃e₅<br>
+e₁e₂e₅ ↔ e₃e₄<br>
+e₁e₃e₄ ↔ e₂e₅<br>
+e₁e₃e₅ ↔ e₂e₄<br>
+e₁e₄e₅ ↔ e₂e₃<br>
+e₂e₃e₄ ↔ e₁e₅<br>
+e₂e₃e₅ ↔ e₁e₄<br>
+e₂e₄e₅ ↔ e₁e₃<br>
+e₃e₄e₅ ↔ e₁e₂<br>
+<br>
+Rewrite it:<br>
+<br>
+|45<br>
+|35<br>
+|34<br>
+|25<br>
+|24<br>
+|23<br>
+|15<br>
+|14<br>
+|13<br>
+|12<br>
+<br>
+Swap until in order:<br>
+<br>
+|45<br>
+|35, 123|45<br>
+|34, 123|54, 123|45<br>
+|25, 124|35, 123|45<br>
+|24, 125|34, 123|54, 123|45<br>
+|23, 125|43, 123|45<br>
+|15, 134|25, 124|35, 123|45<br>
+|14, 135|24, 125|34, 123|54, 123|45<br>
+|13, 145|23, 125|43, 123|45<br>
+|12, 145|32, 125|34, 123|54, 123|45<br>
+<br>
+Counting swaps, we see 0,1,2,2,3,2,3,4,3,4 and that gives even,odd,even,even,odd,even,odd,even,odd,even, so that gives +-++-+-+-+. Great!
+</ref>:
 # Take the rank, halved, rounded up. In our case, <span><math>\lceil \frac{r}{2} \rceil = \lceil \frac{2}{2} \rceil = \lceil 1 \rceil = 1</math></span>. Save that result for later. Let’s call it <span><math>x</math></span>.