We introduce one final significant concept: the dual mode. The dual of a mode simply has the notes that are not in the original mode. This means dual of the mode with mode number m will have mode number 4095-m. For example, the dual of the major mode m:1010 1101 0101 is m:0101 0010 1010. Clearly, if a mode has n notes, then its dual with have 12-n notes. We will denote the dual of a mode m by m̄
(At the risk of being confusing, m:0101 0010 1010 is generally named “the pentatonic scale”, although the meaning of “scale” used in these posts, and “pentatonic” can mean any mode or scale with five notes. If you play the major scale on piano starting on C, you will use all the white keys. The dual is then the pentatonic scale that uses all the black keys).
The concept of a dual mode can be used to explain a number of symmetries we have noticed between things involving the numbers x and 12-x in some way.
In the first post, we grouped together related modes using their gap patterns. We assert that if two modes m1 and m2 have the same canonical gap patterns, then their respective dual modes, m̄1 and m̄2 will also have the same canonical gap patterns. This is left as an exercise for the reader.
This leads onto one of the important results for these posts: a mode m and its dual m̄ will have same number of imperfections of degree d, for all d.
This means, for example, that the (seven-note) major scale and it dual, “the” pentatonic scale, have the same number of notes without (for example) a minor third above them in the scale! This result is more surprising when you consider a mode and its dual generally have different numbers of notes.
To explain this result, we will introduce the concept of a chain. Consider a mode with at least one imperfection of degree d. That means there is an imperfect note of degree d – i.e. a note in the mode that doesn’t have another note in the mode d semitones above it. We will denote this by x1. However, we can see if there is another note x2 d semitones below x1, and this note becomes a note in the chain. We can then repeat the process until we eventually find a note without another note in mode d semitones below it, and this completes the chain. We then have a sequence of notes like the following::
- xL → … → x3 → x2 → x1
(In some cases, x1 won’t have any notes in the mode d semitones below it. This means L=1, and the chain is simply the note x1. This doesn’t affect the rest of this argument). (L has a maximum value of 12).
We can repeat the process for every imperfect note of degree d, resulting in a chain for each imperfect note of degree d.
Now consider what happen when you up d semitones from x1. This note cannot be in the mode m (because x1 is an imperfect note of degree d). This means it must in the dual mode m̄, because m̄ contains precisely the notes that aren’t in m. We will denote note by y1. We can then form another chain starting with y1, as follows:
- xL → … → x3 → x2 → x1…→; y1 → y2 → … → yK
This chain must have an end note, yK. (If it didn’t, then the chain must have rejoined itself, which would imply that one note in the chain had two notes feeding into it. This is impossible, as it would mean that note had two different note d semitones below it.) (In general L <> K. Also, K can equal to 1, which again does not affect this argument.) That end note will be an imperfect note of degree d.
We can repeat this process for every imperfect note x of degree d in m: finding the note y in m̄ that is d semitones above x, building a chain starting on y, and then finding an imperfect note of degree d in m̄. This implies the following:
- The (non-zero) number of imperfections of degree d in a mode m cannot be higher than the number of imperfections of degree d in its dual mode m̄
However, we can reverse the process and apply the same logic, because the dual of the mode m̄ is simply the original mode m: for each imperfect note of degree d in m̄, we can find the note in m that is d semitones higher, build a chain in m starting on that note and ending with an imperfect note of degree d in m. This implies the following:
- The (non-zero) number of imperfections of degree d in a dual mode m̄ cannot be higher than the number of imperfections of degree d in the original mode m.
Together, these two statements imply that the number of imperfections of degree d in a mode and its dual will the same, for all d, assuming that number is not zero.
But what if it is zero? Assume (for a contradiction) that the mode m has zero imperfections of degree d, and its dual mode m̄ has a non-zero number of imperfections of degree d. However, the latter statement implies (by the above) that m will have the same non-zero number of imperfections of degree d. This contradicts our original statement. Hence m̄ must zero imperfections of degree d also.
This argument can also be applied using the chains developed above. If a mode m has zero imperfections of degree d, then this means the chain starting with a note x1 must eventually wind up back at that note:
- x1 → x2 → x3 → … → xL → x1
We say this is a closed chain. The length of the closed chain (L) could be same as the number of notes in the mode, or less. However, it cannot be equal to 1. (L=1 implies you could take a note, go up d<12 semitones, and end up back at the original note.) For a given mode with zero imperfections of degree d, there will a set of one or more closed chains that together contain all notes in the chain. Because the chains are closed, this means every note in the mode has another note that d semitones below it in the mode.
Now, consider the dual mode m̄. If there was an imperfect note of degree d, then the note d semitones above it would be in the original mode m, or conversely, the imperfect note in m̄ is d semitones below some note in m. However, we know that descending d semitones from any note in m produces another note in m. Hence there can be no imperfect notes of degree d in m̄.