Jake Huryn's scratchpad

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I'm trying to learn tuning theory. Please point out anything I might be missing or have wrong! My apologies for any difficult-to-understand descriptions or weird notation, this is just a personal page meant for getting my thoughts on paper. If there's anything you don't understand I'll do my best to explain better. Thanks!

Suppose we start with the JI subgroup 2.3.5 and temper out a comma, say 81/80. In terms of abstract algebra, we can say that the result is the quotient group G{2, 3, 5}/G{81/80}, where G{a, b, c, …} represents the group generated by a, b, c, etc, under multiplication. We have thus produced a group homomorphic to 2.3.5 but in which 81/80 (and all of its powers) are the kernel, meaning that they are mapped to the identity (unison, or 1/1). The elements of this quotient group are the cosets G{81/80}p/q, or the sets produced by multiplying every element of G{81/80} by p/q. Given that a/b and c/d do not differ by an integer multiple of 81/80 (excluding 1/1), G{81/80}a/b and G{81/80}c/d will be distinct cosets. These cosets may be multiplied: G{81/80}a/b*G{81/80}c/d = G{81/80}(ac)/(bd). This works and produces a group (G{2, 3, 5}/G{81/80}, as above) because multiplication is commutative, so any subgroup of 2.3.5 is commutative and thus a normal subgroup.

(Although this makes everything appear very complicated, I will now notate intervals as monzos as they are easier to work with.)

As it turns out, this quotient group can be generated by only two elements; in tuning theory terms, tempering one comma from a three dimensional system produces a two dimensional system. Furthermore, for each two generators we chose, say p and q, we can produce a mapping of the form |<a b c|, <d e f|>, where G{|-4 4 -1>}|1 0 0> = (G{|-4 4 -1>}p)^a*(G{|-4 4 -1>}q)^d. In other words, our tempered 2/1 (its corresponding coset in the quotient group) can be written as a product of the tempered interval p to the a-th power times the tempered interval q to the d-th power. This can of course be done for |0 1 0> and |0 0 1>, corresponding to the second and third elements of each val, respectively.

Using this same quotient group, let's chose G{|-4 4 -1>}|1 0 0> and G{|-4 4 -1>}|0 1 0> as our generators. Clearly we can already fill out the first two elements of each val: |<1 0|, <0 1|>. However, we still need to find a mapping for the fifth harmonic. Starting with G{|-4 4 -1>}|0 0 1>, we can multiply |0 0 1> by |-4 4 -1> to produce the same coset: G{|-4 4 -1>}|-4 4 0>. Now it is clear how we may write this in terms of our generators: G{|-4 4 -1>}|-4 4 0> = (G{|-4 4 -1>}|1 0 0>)^-4 * (G{|-4 4 -1>}|0 1 0>)^4, so our completed mapping is |<1 0 -4|, <0 1 4|>.

This is not always so easy, however. Suppose we wanted to find the mapping when we use G{|-4 4 -1>}|-3 2 0> and G{|-4 4 -1>}|4 -1 -1> (9/8 and 16/15, respectively) as our generators. Even for G{|-4 4 -1>}|1 0 0> it is not clear how we may write this in terms of these generators. So, let's write what we're trying to figure out: G{|-4 4 -1>}|1 0 0> = (G{|-4 4 -1>}|-3 2 0>)^a * (G{|-4 4 -1>}|4 -1 -1>)^d, and expanding everything out, G{|-4 4 -1>}|1 0 0> = G{|-4 4 -1>}|-3a+4d 2a-d -d>. It almost appears that we could set up a system of linear equations, equating each element of the monzos on each side, but if you do this you'll find that there are three equations but only two variables and they do not produce a consistent system, so we need to introduce another variable. If you recall from above, you may realize the solution—multiply one side by k|-4 4 -1>, i.e. the some multiple of our vanishing comma, which does not change a coset's identity. Doing this to the left side, we have G{|-4 4 -1>}|1-4k 4k -k> = G{|-4 4 -1>}|-3a+4d 2a-d -d>, which can now be written as a system of three linear equations (now in three variables):

-3a + 4d = 1 - 4k

2a - d = 4k

-d = -k.

Solving for a and d (k is irrelevant here), we find that our mapping begins |<5|, <2|>—that is, an octave can be reached by stacking five whole steps and two half steps! Repeating this process for |0 1 0> and |0 0 1> we can complete this mapping.