Wedgie/Archived version: Difference between revisions

Inthar (talk | contribs)
mNo edit summary
Inthar (talk | contribs)
mNo edit summary
Line 2: Line 2:
The '''wedgie''' is a way of defining and working with an [[abstract regular temperament]] (that is, a regular temperament where no tuning has been decided on). Wedgies are in a one-to-one relationship with abstract regular temperaments. If one takes r independent [[vals]] V1, ..., Vr in a p-limit group of n primes, then the wedgie for the rank-r temperament V1& ...&Vr is defined by taking the [https://en.wikipedia.org/wiki/Wedge_product wedge product] of the vals (called a '''multival'''), and dividing out the greatest common divisior of the coefficients, to produce an r-multival. If the first non-zero coefficient of this multival is negative, the multival is then scalar multiplied by -1, changing the sign of the first non-zero coefficient to be positive. The result is the wedgie.
The '''wedgie''' is a way of defining and working with an [[abstract regular temperament]] (that is, a regular temperament where no tuning has been decided on). Wedgies are in a one-to-one relationship with abstract regular temperaments. If one takes r independent [[vals]] V1, ..., Vr in a p-limit group of n primes, then the wedgie for the rank-r temperament V1& ...&Vr is defined by taking the [https://en.wikipedia.org/wiki/Wedge_product wedge product] of the vals (called a '''multival'''), and dividing out the greatest common divisior of the coefficients, to produce an r-multival. If the first non-zero coefficient of this multival is negative, the multival is then scalar multiplied by -1, changing the sign of the first non-zero coefficient to be positive. The result is the wedgie.


The entries of a wedgie W for the temperament a&b on a JI subgroup p_1.[...].p_n are  
The entries of a wedgie W for the temperament a&b on a JI subgroup p_1.[...].p_n are (if neither a nor b are [[contorted]] vals):


<math>W(p_i, p_j) = a(p_i)b(p_j) - a(p_j)b(p_i) \text{ for } i < j.</math>  
<math>W(p_i, p_j) = a(p_i)b(p_j) - a(p_j)b(p_i) \text{ for } i < j.</math>