Wedgie/Archived version: Difference between revisions

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In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group ''M''; this means that (ignoring sign) it acts like the operation of finding the determinant of two vectors on the appropriate quotient group ''M' '' = ''M''/''K'' of ''M'', where ''K'' is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament, as follows. (Hence ''M''/''K' '' is a rank-2 lattice on which W is an alternating non-degenerate bilinear form, which justifies the intuition of viewing W as a determinant-like function.)

Let ''K''1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and K2 = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''1, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''2. Conversely, if '''v''' ∈ ''K''2, then W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0 for all w, which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all w. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent in ''M*'' (the dual '''Z'''-module of M); so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''1 and ''K''1 = ''K''2; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".

Let ''K''1 = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and ''K''2 = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''1, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''2. Conversely, if '''v''' ∈ ''K''2, then W('''v''', '''w''') = a('''v''')b('''w''') − a('''w''')b('''v''') = 0 for all w, which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all w. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent in ''M*'' (the dual '''Z'''-module of M); so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''1 and ''K''1 = ''K''2; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".

By the First Isomorphism Theorem it follows that ''M' '' is the group of intervals in the rank-2 temperament in question.

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 In the language of linear algebra, the wedgie is an "alternating bilinear form" on the appropriate JI group ''M''; this means that (ignoring sign) it acts like the operation of finding the determinant of two vectors on the appropriate quotient group ''M' '' = ''M''/''K'' of ''M'', where ''K'' is the kernel of the bilinear form W. Using the fact that W = a&b where a and b are two edos (properly, rank-1 [[val]]s), you can verify that K is exactly the kernel of the rank-2 temperament, as follows. (Hence ''M''/''K' '' is a rank-2 lattice on which W is an alternating non-degenerate bilinear form, which justifies the intuition of viewing W as a determinant-like function.)
-Let ''K''<sub>1</sub> = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and K<sub>2</sub> = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''<sub>1</sub>, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''<sub>2</sub>. Conversely, if '''v''' ∈ ''K''<sub>2</sub>, then W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0 for all w, which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all w. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent in ''M*'' (the dual '''Z'''-module of M); so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''<sub>1</sub> and ''K''<sub>1</sub> = ''K''<sub>2</sub>; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".
+Let ''K''<sub>1</sub> = the kernel of the temperament (i.e. the set of commas tempered out by the temperament), and ''K''<sub>2</sub> = ker W = {'''v''' ∈ ''M'' : W('''v''', '''w''') = 0 ∀'''w''' ∈ ''M''}. If '''v''' ∈ ''K''<sub>1</sub>, then '''v''' is tempered out by both a and b, so W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0, and '''v''' ∈ ''K''<sub>2</sub>. Conversely, if '''v''' ∈ ''K''<sub>2</sub>, then W('''v''', '''w''') = a('''v''')b('''w''') &minus; a('''w''')b('''v''') = 0 for all w, which implies a('''v''')b('''w''') = a('''w''')b('''v''') (*) for all w. Since a and b both have rank 1 but a&b has rank 2, a and b are linearly independent in ''M*'' (the dual '''Z'''-module of M); so we can choose '''w''' such that a('''w''') = 0 but b('''w''') ≠ 0. Then (*) shows a('''v''') = 0. By the same argument, b('''v''') = 0. So '''v''' is in ''K''<sub>1</sub> and ''K''<sub>1</sub> = ''K''<sub>2</sub>; the kernel of the temperament is exactly the intervals that the wedgie "treats as zero".
 By the First Isomorphism Theorem it follows that ''M' '' is the group of intervals in the rank-2 temperament in question.