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## Compositions - op.1 to op.6

Both musics and mathematics deal with things and the laws of their relations to each other. In mathematics, these are numbers and other, more abstract objects. In musics these are tones, chords and rhythms.

As for opus one to six, we have a balance act between free musical idea and mathematical structures. More precisely, I transfer the concepts of GROUPS, FIELDS and VECTOR SPACES to the world of composition.

### The Works

#### Opus 1 - Multiplikative Gruppe I

Multiplication Group I

The work is based on the GROUP [{1,-1},⊗] with laws 1⊗1=1, 1⊗-1=-1, -1⊗1=-1, -1⊗-1=1 over the tone set bB - C - G - D - E - B - #F. It has several layers: Layer I contains the multipliers / chords:

  [1,1,1,1,1,1,1,1,1]	≙ bB-F-C-G-D-A-E-b-#F
[1,1,·,·,1,·,·,·,·] 	≙ bB-D-F
[·,1,1,·,·,1,·,·,·] 	≙ F-A-C
[·,·,1,1,·,·,1,·,·] 	≙ C-E-G
[·,·,·,1,1,·,·,1,·] 	≙ G-B-D
[·,·,·,·,1,1,·,·,1] 	≙ D-#F-A
[·,·,·,·,1,1,·,·,1] 	≙ D-#F-A
[·,·,·,1,1,·,·,1,·] 	≙ G-B-D
[·,·,1,1,·,·,1,·,·] 	≙ C-E-G
[·,1,1,·,·,1,·,·,·] 	≙ F-A-C
[1,1,·,·,1,·,·,·,·] 	≙ bB-D-F


For simplicity I write "·" für "-1", and those mean "not sounding". The first chord is the identity, the ten others make for a bouncing rondo. Applied all of them one after the other, they neutralize. Layer II contains the multipliers

  [1,1,1,1,1,1,1,1,1]	≙ bB-F-C-G-D-A-E-B-#F
[1,1,·,1,·,1,·,1,1] 	≙ bB-F-G-A-B-#F
[·,·,1,·,1,1,1,·,1] 	≙ C-D-A-E-#F
[1,1,·,1,·,1,·,1,1] 	≙ bB-F-G-A-B-#F
[·,·,1,·,1,1,1,·,1] 	≙ C-D-A-E-#F


Also combining all those leads to a neutralization. Layer III contains

  [1,1,1,1,1,1,1,1,1]	≙ bB-F-C-G-D-A-E-B-#F
[1,·,1,·,1,·,1,·,1] 	≙ bB-C-D-E-#F
[·,1,·,1,·,1,·,1,·] 	≙ F-G-A-B
[·,1,·,1,·,1,·,1,·] 	≙ F-G-A-B
[1,·,1,·,1,·,1,·,1] 	≙ bB-C-D-E-#F


The composition starts with the chord: A - C - #F - E - G - B. It gets multiplied with the multiplicators of all levels, in a hierarchical manner:

K · E11 · E21 · E31
K · E11 · E21 · E32
K · E11 · E21 · E33
...
K · E11 · E22 · E31
K · E11 · E22 · E32
...
K · E11 · E23 · E31
...
K · E11 · E25 · E31
...
K · E12 · E21 · E31
...

However, a descending to level III happens only if the chord contains both C and B at that point. This in the end will give us 120 chords, which get spread uniformly over 120 bars. Empty chords get replaced by the off-system tone bE. Other than that, melody and rhythm are free.

The score is here, in Youtube you'll find a video here.

#### Opus 2 - Multiplikative Gruppe II

Multiplication Group II

In this work we start with 17 chords of a tonal tune. Then we combine each neighbor pair by the GROUP [{1,-1},⊗] with laws 1⊗1=1, 1⊗-1=-1, -1⊗1=-1, -1⊗-1=1 over all twelve tones, resulting in 16 chords. We do the same again, giving 15 chords, and so one down to one chord. The 1 + 2 + ... + 17 chords are getting played in reverse order, that is first the one fully reduced chord, then the two one step before the last reduction, then three, and so on. Other than that the chords get used in a freely manner.

The score is here, an Youtube Video here.

#### Opus 3 - Lineare Transformation I

Lineary Transformation I

The chords of this work find themselves in a vector space over the FIELD [{0,-1,1},+;·] with laws: 1⊗1=1, 1⊗-1=-1, -1⊗1=-1, -1⊗-1=1, and -1⊕-1=1, -1⊕0=-1, -1⊕1=0, 0⊕-1=-1, 0⊕0=0, 0⊕1=1, 1⊕-1=0, 1⊕0=1, 1⊕1=-1. Here we use two bases:

BAS1 =

C – G – D – A – E – B – #F
C – E – G
D – #F – A
E – #G – B
#F – #A – #C
bA – C – bE
bB – D – F
bD – F – bA
bE – G – bB
F – A – C
G – B – D
A – #C – E

und BAS2 =

#F – #C – #G – #D – bB – F – C
C – E – G
bE – G – bB
#F – #A – #C
A – #C – E
bD – F – bA
E – #G – B
G – B – D
bB – D – F
D – #F – A
F – A – C
bA – C – bE

The transformation matrix from BAS1 to BAS2 gets reinterpreted as a linear mapping and calculates to: T = BAS2 · BAS1-1, with the chords of each basis as column vectors and a matrix multiplication.

In numbers:

$T = \left( \begin{array}{cccccccccccc} -1 & 0 & 1 & 0 & -1 & -1 & 0 & 0 & 1 & -1 & 0 & 0 & \\ -1 & 1 & 1 & -1 & 0 & -1 & 0 & 1 & 0 & -1 & 0 & 1 & \\ 1 & -1 & 1 & -1 & 0 & -1 & 1 & -1 & 0 & 1 & 0 & 0 & \\ 1 & 1 & 0 & 1 & -1 & 1 & 0 & 0 & 1 & 1 & -1 & 0 & \\ 1 & 0 & -1 & -1 & 1 & 0 & -1 & -1 & 0 & -1 & -1 & -1 & \\ 0 & -1 & 1 & 0 & 1 & 0 & -1 & -1 & 1 & 0 & -1 & 1 & \\ 0 & 1 & 0 & 1 & 1 & 0 & -1 & -1 & -1 & 1 & 0 & 1 & \\ 0 & 1 & 1 & 1 & 0 & 1 & 1 & 1 & -1 & -1 & 1 & 1 & \\ 0 & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 0 & 1 & 1 & 1 & \\ 0 & 1 & 1 & 0 & 0 & -1 & 0 & 0 & 0 & -1 & 0 & 0 & \\ -1 & -1 & 1 & -1 & -1 & -1 & 1 & -1 & -1 & -1 & 0 & 0 & \\ 1 & 1 & 0 & 1 & -1 & -1 & 0 & 0 & 1 & 0 & -1 & 0 & \end{array} \right)$

Op. 3 is the application of T once or several times. Other than that, motivs, rhythms and metrics are free.

The score is here, a Youtube video here.

#### Opus 4 - Lineare Transformation II

Linear Transformation II

Opus 4 moves like opus 3 in the vector space over FIELD [{0,-1,1},+,·]. The mapping from opus 3 gets used again. Motives and rhythm are different, however. Another distinction is, that mapping are getting applied to both chords and chord parts, for example induced by distributing over the two instruments.

The score is here, an Youtube video herer.

#### Opus 5 - Lineare Transformation III

Linear Transformation III

Opus 5 like opus 3 and 4 moves in the vector space over the FIELD [{0,-1,1},+,·]. Also the mapping used there gets used here again. In contrary to opus 3 and 4, where we started with accord chains and mapped them once or twice, in opus 5 we start with only two accords and apply the mapping often.

The chords used are: D - #F - #A - #C - F und bB - bE - bA - bD.

Hier is tje score and an Youtube video.

#### Opus 6 - Kontinuum I

Continuum I

Opus 6 deals with chords using real numbered components. Here we use a mixture of distributing fractions among the instruments and among the stress layers inside the metrics. The chords used come from a Gram-Schmidt orthogonalization:

  K1 = (1;0;0;  0;1;0;  0;1;0;  0;1;0)
K2 = (0;1;0;  0;0;1;  0;0;1;  0;0;1)
K3 = (0.75;0;1;  0;-0.25;0;  1;-0.25;0;  1;-0.25;0)
K4 = (-0.4;0.75;0.13;  1.0;-0.53;-0.25;  0.13;0.47;-0.25;  0.13;0.47;-0.25)
K5 = (-0.63;-0.29;1;  0.4;0.74;-0.69;  -0.18;-0.05;0.49;  -0.18;-0.05;0.49)
K6 = (0.27;-0.51;-0.15;  1;0.23;0.58;  -0.58;-0.25;-0.04;  0.47;-0.25;-0.04)
K7 = (-0.42;0.41;-0.41;  0.17;1;0.18;  0.76;-0.95;-0.3;  0.07;0.38;-0.3)
K8 = (-0.6;-0.34;0.54;  -0.16;0.1;0.98;  0.28;0.54;-1;  -0.22;-0.04;0.37)
K9 = (0.23;-0.56;-0.28;  0.62;-0.18;0.11;  1;0.2;0.49;  -0.96;-0.26;-0.03)
K10 = (-0.3;0.26;-0.41;  -0.04;0.52;-0.14;  0.22;0.78;0.12;  0.49;-1;-0.23)
K11 = (-0.45;-0.26;0.32;  -0.24;-0.06;0.53;  -0.04;0.15;0.73;  0.17;0.36;-1)
K12 = (-0.65;-0.88;-0.8;  -0.1;-0.33;-0.25;  0.45;0.22;0.29;  1;0.76;0.84)

with orthogonal meaning that we have for all pairs (Ki;Ki): ∑mKim·Kjm = 0.

For handling the fraction parts of the numbers we split the interval [0; 1] into five parts

  [0;p]    ]p;p+λ]    ]p+λ; p+2λ]    ]p+2λ; p+3λ]    ]p+3λ; 1],  λ = (1-p)/4

with the first part standing for a break and the other parts distribute among the four instruments. The parameter p, which determines the ratio of breaks, gets different values for different work parts. Each instrument interval gets split into another three parts, telling whether a tone should go to a half, a crotchet or a quaver.

Here is the score and an Youtube video.