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Theorem: "mandolin chords shapes" is isomorphic to "guitar chord shapes (lower 4 strings)"
Proof:
Constructively, let f(c) = reverse(c).
For example,
Let's start with the low E-string:
c[1] is the finger placement on the low-E string
then f(c)[4] plays the same note (under octave equivalence), since the mandolin's 4th string is an E.
Likewise, for all i, c[i] plays the same note on the guitar (under octave equivalence) as f(c)[5-i] plays in the mandolin. This is because while the guitar is starting from the low-E and going up in fourths, the mandolin is starting from the high-E and coming down in fifths. Since fourths and fifths are octave-complements of each other, that means that the difference between the guitar-note and mandolin-note will remain the same after each iteration (under octave equivalence). Since the difference was 0 to start with, then it remains 0. Do note that the absolute difference becomes one octave smaller at each step.
(It kinda bothers me that properly formalizing the math is so much work. I think mathematical language needs to be redesigned by good software engineers.)
I think one important reason to call this an "isomorphism" instead of "bijection" is that there are relations between chords that are preserved. For example, if you know that to go from an E to an E7, you let go of one finger...
you can do the same on the mandolin.
In fact, you have lots of relations between chords like this one, that are all preserved under this simple isomorphism. Another obvious one is transposition (i.e. move the bar in bar chords, NB: all chords are bar chords).
Trying this for all basic guitar chords (i.e. non-bar): C, D, E, G, A... a problem appears: If you try to translate A into the mandolin, it won't sound right. This is because there is no 3rd degree present in the lower 4 strings. One possible fix is to play an F# on the D string, which corresponds to a non-standard A chord on the guitar.
Similarly, when translating a D from the guitar, you should play the low-E string: i.e. you should play D/F#.
Credits to Thorne, who playing mandolin for the first time, said that the mandolin was like a guitar backwards. I thought that he was playing a left-handed mandolin, but then I knew what he meant.
Proof:
Constructively, let f(c) = reverse(c).
For example,
E = (0,2,2,1,0,0) : ---|---|---|--- ---|---|---|--- ---o---|---|--- G# 3 ---|---o---|--- E 1 ---|---o---|--- B 5 ---|---|---|--- E 1 becomes E = (1,2,2,0): ===|===|===|=== E 1 ===|===o===|=== B 5 ===|===o===|=== E 1 ===o===|===|=== G# 3
Let's start with the low E-string:
c[1] is the finger placement on the low-E string
then f(c)[4] plays the same note (under octave equivalence), since the mandolin's 4th string is an E.
Likewise, for all i, c[i] plays the same note on the guitar (under octave equivalence) as f(c)[5-i] plays in the mandolin. This is because while the guitar is starting from the low-E and going up in fourths, the mandolin is starting from the high-E and coming down in fifths. Since fourths and fifths are octave-complements of each other, that means that the difference between the guitar-note and mandolin-note will remain the same after each iteration (under octave equivalence). Since the difference was 0 to start with, then it remains 0. Do note that the absolute difference becomes one octave smaller at each step.
(It kinda bothers me that properly formalizing the math is so much work. I think mathematical language needs to be redesigned by good software engineers.)
I think one important reason to call this an "isomorphism" instead of "bijection" is that there are relations between chords that are preserved. For example, if you know that to go from an E to an E7, you let go of one finger...
E7 = (0,2,0,1,0,0) : ---|---|---|--- ---|---|---|--- ---o---|---|--- G# 3 ---|---|---|--- D 7 ---|---o---|--- B 5 ---|---|---|--- E 1
you can do the same on the mandolin.
E7 = (1,0,2,0): ===|===|===|=== E 1 ===|===o===|=== B 5 ===|===|===|=== D 7 ===o===|===|=== G# 3
In fact, you have lots of relations between chords like this one, that are all preserved under this simple isomorphism. Another obvious one is transposition (i.e. move the bar in bar chords, NB: all chords are bar chords).
Trying this for all basic guitar chords (i.e. non-bar): C, D, E, G, A... a problem appears: If you try to translate A into the mandolin, it won't sound right. This is because there is no 3rd degree present in the lower 4 strings. One possible fix is to play an F# on the D string, which corresponds to a non-standard A chord on the guitar.
Similarly, when translating a D from the guitar, you should play the low-E string: i.e. you should play D/F#.
Credits to Thorne, who playing mandolin for the first time, said that the mandolin was like a guitar backwards. I thought that he was playing a left-handed mandolin, but then I knew what he meant.