The hexachordal theorem

by math_explorer, Jan 25, 2020, 3:55 AM

Let $n$ be a positive integer. Let $R$ be a subset of $\mathbb{Z}/2n\mathbb{Z}$ with size $n$. We denote the complement of $R$ with respect to $\mathbb{Z}/2n\mathbb{Z}$ as $\overline{R}$.

Let $d$ be an element of $\mathbb{Z}/2n\mathbb{Z}$.

Let $H(R, d)$ be the number of ordered pairs $(a, b) \in R^2$ such that $b - a \equiv d \bmod{2n}$.

Theorem. For all $d$, $H(R, d) = H(\overline{R}, d)$.

Proof. Let $A = \{ (a, a + d) \mid a \in R \}$. Let $B = \{ (b - d, b) \mid b \in \overline{R} \}$. They are the same size: $|A| = |R| = n = |\overline{R}| = |B|$. So their differences $A \setminus B$ and $B \setminus A$ are the same size too. But $|A \setminus B|$ is precisely $H(R, d)$ and $|B \setminus A|$ is precisely $H(\overline{R}, d)$ (why?). $\square$

Remark. This is more or less what's known as the "Hexachordal theorem" in "musical set theory". The usual interpretation is when $n = 6$, $\mathbb{Z}/2n\mathbb{Z}$ is identified with the pitch classes in twelve-tone equal temperament, and $R$ is a "hexachord", a set of six pitch classes. Apparently there is also a crystallography interpretation.

This is a paper on a generalization and I love how many citations they managed to come up with for proofs, some of which are apparently more concise than others: http://cs.smith.edu/~jorourke/Papers/CHexa-MCM.pdf

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