Journal of Mathematics and Music Vol. 01, No. 03, December 2007, 1–2

David Lewin and Maximally Even Sets Emmanuel AMIOT 1 rue du Centre, F 66570 St NAZAIRE, France (v1.0.0 released june 2007)

Supplementary II: pictures and proofs

Figure 1. These two hexachords share intervallic content

On figure 1 with the two complementary hexachords, the fifths have been signaled with arrows. Each hexachord has the same number of fifths, three in this example.

Figure 2. Etude N◦ 5 opus 10, Fr´ ed´ eric Chopin

Proof of prop. ??: Proof From Thm. ?? we have A is τ −periodic ⇐⇒ ∀t ∈ Zc

FA (t) = e−2iπτ t/c FA (t) ⇐⇒ ∀t ∈ Zc

(1 − e−2iπτ t/c )FA (t) = 0

Professor in Class Preps, Perpignan, France. Email: [email protected]

Journal of Mathematics and Music c 2007 Taylor & Francis Ltd. ISSN 1745-9737 print / ISSN 1745-9745 online http://www.tandf.co.uk/journals DOI: 10.1080/17459730xxxxxxxxx

2

title on some pages

Unless e−2iπτ t/c = 1, this compels FA (t) to be 0. Now the condition e−2iπτ t/c = 1 is equivalent to c | τ t, i.e. t multiple of m = c/ gcd(c, τ ) – this makes sense for any representative of the residue classes τ and t. This is compatible with reduction modulo c, and means t ∈ m Zc ⊂ Zc . Conversely, if FA is nil except on a subgroup, say m Zc with 0 < m | c in Z (we recall all subgroups of Zc are cyclic) then, by inverse Fourier Transform ∀k ∈ Zc

1A (k) =

1X 1 X 1 0 FA (t)e2iπ k t/c = FA (t0 )e2iπ k t /c = c c 0 c t ∈m Zc

t∈Zc

00

X t00 =1...

FA (m t00 )e2iπ k t

m/c

c m

c c as a period, as each term in the sum is and this is obviously periodic with (the residue class of ) m m periodic.  Proof of prop ??: Proof For transposition and inversion it is theorem ??. For complementation we see that |FZc \A (c − d)| = |FZc \A (−d)| = | − FA (d)| = |FA (d)| holds for any subset A. So the one value is maximal whenever the other is, e.g. A is a ME set iff Zc \ A is maximally even.  Proof of remark ??, linking he Fourier coefficients of A and its reduction B mod c0 : FA (d) =

X k∈A

−2iπdk/c

e

=

X m−1 X k00 ∈B `=0

−2iπd(k00 +` c0 )/c

e

=

X k00 ∈B

−2iπd k00 /c

e

m−1 X `=0

e−2iπ` = m

X k00 ∈B

0

e−2iπd

k00 /c0

= m FB (d0 )