Four disjoint 7-limit temperaments

Griffiths and Harris (pg 206) give as the simplest example of the
Schubert calculus in action the statement that four lines in general
position in three-dimensional projective space have two lines which
meet all four of the lines. I tried this out for meantone,
ennealimmal, porcupine and rodan for the four lines, and got two lines
involving complex numbers. This isn't surprising, but it is known
that for theorems of this kind you can always find real examples. We
need rational examples; it would be interesting to find an example of
four 7-limit linear temperaments, each of which has no common comma
with any of the others, such that there is a pair of linear
temperaments which have a common comma with each of these four.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> Griffiths and Harris (pg 206) give as the simplest example of the
> Schubert calculus in action the statement that four lines in general
> position in three-dimensional projective space have two lines which
> meet all four of the lines. I tried this out for meantone,
> ennealimmal, porcupine and rodan for the four lines, and got two
lines
> involving complex numbers. This isn't surprising, but it is known
> that for theorems of this kind you can always find real examples. We
> need rational examples; it would be interesting to find an example
of
> four 7-limit linear temperaments, each of which has no common comma
> with any of the others, such that there is a pair of linear
> temperaments which have a common comma with each of these four.

So, do 7-limit linear temperaments have two commas apiece? Could you
give an example of this last statement?

Paul

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul.hjelmstad@m...> wrote:

> So, do 7-limit linear temperaments have two commas apiece? Could you
> give an example of this last statement?

All 7-limit linear temperaments are generated by two commas, but the
two commas are not uniquely determined. If the two commas are A and B,
only the kernel of the temperament, which is the set of all commas A^i
B^j for integral i and j, is intrinsic. Writing the two commas as
monzos, taking the wedge product, and then the complement and reducing
to a standard form (or equivalently, regarding the result as a
projective point) gives the wedgie, and this is also intrinsic and
uniquely attached to the temperament, but not to the two commas.

An example would be 81/80 and 126/125, which determine septimal
meantone. So, however, do 81/80 and 225/224, or 126/125 and 225/224.