Non permutational serialism (3): What about Schoenberg ?

According to Milton Babbitt [1], Schoenberg’s twelve tone musical system is permutational. However Babbitt’s maths are so hard to understand for me that I seem condemned to believe this claim because of my ignorance. Even in religion this wouldn’t be the right way to believe, so we are going to ask the question about the permutational nature of Schoenberg’s musical theory in a way that makes sense to us (non mathematicians).

The basic schoenbergian material is quite narrow, it is made of 4 expressions, or instances, of the same twelve tone row. If we call S one series of notes chosen among the 479001600 rows made of 12 distinct digits, then Schoenberg admits also the same row read from right to left and he calls this instance of the row its retrograde form, or sometimes also its recurrence. We are going to choose the term ‘reverse’ to name this instance of S, and choose the letter R to design it. The third form used by Schoenberg was called a reversal but consisted in taking the inversion of each interval in the row, so we are going to call it ‘inverse’ and abbreviate it ‘I’. The last traditional form in twelve tone music was sometimes called recurrence of the reversal, but it will be clearer to describe it reverse of the inverse, according to the previous naming since it is only the inverse read from right to left, and we will call it ‘opposite’ or simply ‘O’.

Let’s take a random example:

```S : 87925361A04B
R : B40A16352978
I : 453A796B2081
O : 1802B697A354```

We can verify that R is S read from right to left, as well as O for I. And this reverse form is indubitably permutational because it is obtained easily  by composing S (or I) with the last element of the well ordered permutational set which is BA9876543210. [2]

Now if we look at I, things are not so clear, we can obtain it by taking the inverse value of each note, for example if the first note is 8 (G#) then we take 8 steps down starting at 0 (C) and that is 4 (E), etc. for each other note. But this way of doing is not permutational, it’s arithmetic. In fact for each possible S, there is one precise composed row that will get I, and many different rows will get their inverse by being composed with the same row so there is a whole subset of the total group of permutations that play for I the role BA9876543210 was playing for R .

This set of 10395  inverting rows in base 12 separates the total set of permutations in subsets of 46080 rows. All the 10395 share a common type [3], which is the type of their most remarkable – and last – element : 0BA987654321. [4] We see that having 0 on its index has the result of setting the axis on this note. I don’t know if Schoenberg always chose that note as axis, but if he did then it was an undue priviledge; in Csgrouper one can choose any index as axis by setting the ‘x’ field.

Which further permutational means can make us select the correct row to compose with S in order to obtain I, that’s what I’d like to know; it is perhaps also something Babbitt had explained to the graduates, who knows?

Notes:

[1] Milton Babbitt, 2003, Collected Essays, Princeton University Press.

[2] See the definition of a composition of rows in the ‘code’ section.

[3] See the definition of the type of a row in the ‘code’ section.

[4] There are for instance 46080 rows whose Inverse can be produced by composing them with  0BA987654321, and 10394 other inverting rows like 0BA987654321 with their 0 in indexical position. If we agree to have each note in turn in indexical position and produce this way 12 different inverses for each row, we obtain a set of 124740  inverting rows, among 479 millions.

Update 130607:

We don’t have to check 479 millions permutations to catch those values; as many results about permutations they come from induction:

a)- note that in many bases (B+1) many inverses are produced by the inverting row 0B..1 and call this row T[B+1];
b)- check the small bases first to see how many inverses are produced by T[B] and call this number N[B];
c)- remark that B!/N[B] = the total number of inverting rows, call it R[B];
d)- for the same value of R, bases go by pairs (even then odd), the other mentionned values always differ in a higher base;
e)- observe that B*R[B] = R[B+1] but only when B is odd;
f)- so  N[B+1] = (B+1)!/(B*R[B]) =  (B+1)!/(B*R[B-1]); when B is odd;
g)- and R[B] = (B-1)*R[B-1] = (B-1)*R[B-2]; when B is even;

For example:

```e) 3*1 = 3 = R[4];
f) 4!/3 = 8 = N[4];
e) 5*3 = 15 = R[6];
f) 6!/15 = 48 = N[6];
e) 7*15 = 105 = R[8];
f) 8!/105 = 384 = N[8];
e) 9*105 = 945 = R[10];
f) 10!/945 = 3840 = N[10];
e) 11*945 = 10395 = R[12];
f) 12!/10395 = 46080 = N[12];
..```