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In a 2-val Fokker group built around a common 1-val, <val| \/ |2/1>
tells you how many notes are in the scale.

In a 3-val Fokker group built around a common 2-val, <<bival|| \/
|2/1> doesn't tell you how many notes are in the scale. You need to
input a bimonzo to get that, meaning <<bival|| \/ ||bimonzo>> will
give you what you want.

One interpretation of this is that if you do <<bival|| \/ |2/1>, you
end up with a single <val| remaining. Therefore, the "number of notes"
in a bival with respect to 2/1 as period is actually a linear
functional.

For instance, <<1 4 4|| \/ |1 0 0> = <0 1 4|. This is the "number of
notes per octave" in meantone temperament. This is strange, but makes
sense if you consider that there really is no answer to the question
"how many notes are in meantone, per octave?" This is because it
depends on what MOS you're using (if you're using an MOS at all), or
how many generators you're using, or what have you.

This is what <0 1 4| represents: a linear functional telling you how
many notes are in meantone per octave if you input an additional
monzo, which you can interpret as a "chroma" if you'd like. Therefore,
the "number of notes in meantone per 2/1" is a linear functional
requiring another monzo to give you an answer.

You can also consider the result that you get if you put a bimonzo in
directly. For instance <<1 4 4|| \/ ||1 0 0> = 1, which tells you all
of these things
1) if 2/1 is a period, and 3/1 is the "chroma," you get 1 note per octave.
2) if 2/1 is a chroma, and 3/1 is the "period," you get 1 note per octave.
3) 2/1 and 3/1 are a pair of generators for meantone.
4) if you go into the tempered meantone lattice, || |tempered 2/1> ^
|tempered 3/1> || = 1.
5) if you go into val space, and you take any pair of vals such that
v1^v2 = <<1 4 4||, and you draw out the parallelogram with the origin
and those two vals as vertices, and then you "project" this
parallelogram onto the plane formed by the vals <1 0 0| and <0 1 0|,
the area of the resulting parallelogram will be 1.

etc, but they're all subtly saying the same thing.

-Mike