Double Steiner System

This is from Dr. Noam Elkies, by way of the "This Week in
Mathematical Physics" Column (Week 234) of Dr. John Baez:

http://math.ucr.edu/home/baez/week234.html

{0,1,2,3,4,6},
{0,1,2,3,5,7},
{0,1,2,3,6,7},
{0,1,2,4,5,8},
{0,1,3,4,6,9},
{0,1,3,5,7,9},
{0,1,2,4,5,9},
{0,1,2,4,7,8},
{0,1,2,5,6,8},
{0,1,2,4,7,9},
{0,1,3,5,6,8}.

These are sets 0 through 10 in Dr. Elkies system.

These form a double Steiner System S(5,6,12) so that each hexachord,
along with it's transpositions, inversions, and complements, contain
every pentachord exactly twice. (There are C(12,5) pentachords, which
equals 792). Pentachords are neat because they divide out nicely and
have no modes of limited transposition.

In my system, they are N1, X, O1, B1, H, V5, E, F, F, R5, R5.
The two F's and R5's are inverse complements of each other, so
they may be redundant.

Here's a cool graph:

http://www.math.harvard.edu/~elkies/m12.pdf

Down the middle spine, you have sets 6,3,4,0,2,1 and 5. You have
8 off the left of 2, 7 off the right, and 10 off the left of 1 and
9 off the right. Lines determine a common edge. (pentachord).
I think the crosshatch means you have to invert.

In my system this is E,B1,H, N1, O1, X and V5 down the middle
and F, R5 off each side.

Here's with weights:

E - 3
B1 - 2
H - 1
N1 - 2
01 - 1
X - 0
V5 - 1

F, F - 2
R5,R5 - 1

The bottom four are two Z-related pairs (also inverted). Every
set is also asymmetrical. H, and the F's map to self through
D4-neg X C3. E and X map to self through S3-neg X C4. The others
map to complement through inversion, except the Z-related ones.

Everything hooks up to a weight with different parity, in fact,
always off by one. (3,2,1,2,1,0,1 and 2,1,2 and 1,0,1)

In my I-ching system, which classifies hexachords based on tritone
scrambling or non-tritone scrambling, this is balanced, with the
exception of E having no tritones, B1, N1, X, R5, R5 having one,
and rest having two, you also have this correspondency:

E

B1N1<->O1V5
R5R5<->FF
X<->H

Based on exactly how far apart tritones or non-tritones are
in the I-ching diagrams. (Tritones for 1 tritone sets and
non-Tritones for 2 tritone sets). You scramble non-tritones
in the first group and tritones in the second group, though.

E stands alone, it is special because it is the only asymmetrical
hexachord that maps into its complement, without inversion.

Dr. Elkies also has a single Steiner system which canvasses
all pentachords once, but with only even transpositions. Don't have
that yet...

I am also working to see how the three Z related pentachord
types fit into this, (There are 6 total, of this config 2+2, 2+1, and
1+1) and the three weakly related complices (2 symmetrical and
1 assymmetrical pair making 4 total).

One of these is Z-related (0,1,3,5,6) and in fact is the
only weakly related set complex that cannot be embedded in
its own complement. The other three get to their complement
through a Z-related hexachord.

All for now,

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<phjelmstad@...> wrote:
>
> This is from Dr. Noam Elkies, by way of the "This Week in
> Mathematical Physics" Column (Week 234) of Dr. John Baez:
>
> http://math.ucr.edu/home/baez/week234.html
>
> {0,1,2,3,4,6},
> {0,1,2,3,5,7},
> {0,1,2,3,6,7},
> {0,1,2,4,5,8},
> {0,1,3,4,6,9},
> {0,1,3,5,7,9},
> {0,1,2,4,5,9},
> {0,1,2,4,7,8},
> {0,1,2,5,6,8},
> {0,1,2,4,7,9},
> {0,1,3,5,6,8}.
>
> These are sets 0 through 10 in Dr. Elkies system.
>
> These form a double Steiner System S(5,6,12) so that each hexachord,
> along with it's transpositions, inversions, and complements, contain
> every pentachord exactly twice. (There are C(12,5) pentachords,
which
> equals 792). Pentachords are neat because they divide out nicely and
> have no modes of limited transposition.
>
> In my system, they are N1, X, O1, B1, H, V5, E, F, F, R5, R5.
> The two F's and R5's are inverse complements of each other, so
> they may be redundant.
>
> Here's a cool graph:
>
> http://www.math.harvard.edu/~elkies/m12.pdf
>
> Down the middle spine, you have sets 6,3,4,0,2,1 and 5. You have
> 8 off the left of 2, 7 off the right, and 10 off the left of 1 and
> 9 off the right. Lines determine a common edge. (pentachord).
> I think the crosshatch means you have to invert.
>
> In my system this is E,B1,H, N1, O1, X and V5 down the middle
> and F, R5 off each side.
>
> Here's with weights:
>
> E - 3
> B1 - 2
> H - 1
> N1 - 2
> 01 - 1
> X - 0
> V5 - 1
>
> F, F - 2
> R5,R5 - 1
>
> The bottom four are two Z-related pairs (also inverted). Every
> set is also asymmetrical. H, and the F's map to self through
> D4-neg X C3. E and X map to self through S3-neg X C4. The others
> map to complement through inversion, except the Z-related ones.
>
> Everything hooks up to a weight with different parity, in fact,
> always off by one. (3,2,1,2,1,0,1 and 2,1,2 and 1,0,1)
>
> In my I-ching system, which classifies hexachords based on tritone
> scrambling or non-tritone scrambling, this is balanced, with the
> exception of E having no tritones, B1, N1, X, R5, R5 having one,
> and rest having two, you also have this correspondency:
>
> E
>
> B1N1<->O1V5
> R5R5<->FF
> X<->H
>
> Based on exactly how far apart tritones or non-tritones are
> in the I-ching diagrams. (Tritones for 1 tritone sets and
> non-Tritones for 2 tritone sets). You scramble non-tritones
> in the first group and tritones in the second group, though.
>
> E stands alone, it is special because it is the only asymmetrical
> hexachord that maps into its complement, without inversion.
>
> Dr. Elkies also has a single Steiner system which canvasses
> all pentachords once, but with only even transpositions. Don't have
> that yet...
>
> I am also working to see how the three Z related pentachord
> types fit into this, (There are 6 total, of this config 2+2, 2+1,
and
> 1+1) and the three weakly related complices (2 symmetrical and
> 1 assymmetrical pair making 4 total).
>
> One of these is Z-related (0,1,3,5,6) and in fact is the
> only weakly related set complex that cannot be embedded in
> its own complement. The other three get to their complement
> through a Z-related hexachord.
>
> All for now,
>
> Paul

Review:

In 12-Et there are 35 pentachord/septachords, after reducing for
direction (reversibility of necklaces) and Z-relation (There are three
Z-related pentachord/septachords (Three pairs of pentachords where
each pair shares a common interval vector). The Z-related pentachords
are:

(0,1,2,5,9)---Z---(0,1,3,4,8)
(0,1,2,5,8)---Z---(0,1,4,5,7)
(0,1,3,5,6)---Z---(0,1,2,4,7)

The first two are also related by "M5" symmetry: D4XS3 Group symmetry.
(0,1,3,5,6) is also an "impassible" weakly-related 7/5 set complex:
The only way to go from the septachord to the pentachord is through a
Z-related hexachord (All the Z-related hexachords are complements of
each other) (0,1,3,5,6) is also special as a WR7/5SC in that the
pentachord does not fit into the complementary septachord. The other
two WR7/5SC complexes are:

(0,1,4,7,8) and
(0,1,2,4,7)

The last one is also a Z-relation to the first (0,1,3,5,6).

* * * * *

Now, forgetting about asymmetry, which affects the second Z-related
pair, and the right side of the third, I find that:

(0,1,2,5,9) embeds in set 6 and it's inverse, and so does it's
Z-relation (0,1,4,5,7). This Z relation has the M5 symmetry

(0,1,2,5,8) embeds into set 3 and set 8. It's Z-relation (0,1,4,5,7)
embeds into set 2 and set 8.

The last pair, which are also each a WRSC:
(0,1,3,5,6) (impassible) embeds into set 10 and its inverse, while
(01,2,4,7) embeds into sets 9 and 7, and finally

The remaining WRSC (0,1,4,7,8) embeds into set 7 and it's inverse

* * * * *

So what of this? Will I ever tie this into tuning? Or at least some
kind of lattice? (Well, lattices first, the 7-limit that is).

Here's a start: 7 set types are represented. 4 are symmetrical.
3 are not. Two Z-related pairs have M5 symmetry. 1 pair doesn't
(except unto self). All WRSC have M5 symmetry unto themselves.

Steiner sets per Z-related pentachords:

A) 6, -6 <-> 6, -6
B) 3, 8 <-> 2, 8
C) 10, -10 <-> 9, 7

Steiner sets per WRSC:

A) 10, -10
B) 9, 7
C) 7, -7

My lettering: 6=E, 3=B1, 8=F, 2=O1, 10=R5 (V7/I), 9=Z(R5) (Inverse
Blues-scale), 7=Z(F). WRSC only use Z-related hexachords. Z-related
pentachords make use of a mixture of hexachords. 7,8,9,10,-10 get
used, but also 6, -6, 3 and 2. I think Dr. Elkies included Z-relations
of 7, 9 (8, 10) because they are not obvious complements, but I am
not sure. Otherwise, you can whittle this list down to 9 hexachord
(types).

Z relations use only E, B1, O1, F, and R5
WRSC use only F and R5.

F and R5 balance in my I-ching system
B1 and 01 also balance

E stands alone

N1,V5; X,H are not used here (which also balance)

Paul Hj