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6.6 The Matsukidaira–Satsuma Equations  259

          Hence applying the Jacobi identity (Section 3.6),

                                      (n+1)              (n+1)
                    τ                A     (r)    (−1) A
                                                      n
                     r+2  τ r+1       11                 1,n+1  (r)

                                         (n+1)       (n+1)
                               =
                    τ r+1         (−1) A     (r)   A       (r)

                                      n
                                         n+1,1       n+1,n+1
                           τ r
                               = A (n+1) (r)A (n−1) (r +2).
          Replacing r by r − 1,

                        τ
                         r+1   τ r    = A (n+1) (r − 1)A (n−1) (r + 1)  (6.6.3)

                              τ r−1

                         τ r
                                        (n+1)
                                 τ = −A      (r)

                                  r     n,n+1
                                        (n+1)
                                   = −A      (r)
                                        n+1,n

                                 τ = A (n+1) (r).
                                 r     nn
          Hence,

                                      (n+1)       (n+1)
                          τ      τ          (r)  A
                                                  n,n+1
                                      A nn            (r)
                           r   r
                          τ       =    (n+1)     (n+1)

                                       A   (r) A       (r)
                           r  τ r                n+1,n+1
                                      n+1,n
                                  = A (n+1) (r)A (n+1)  (r)
                                             n,n+1;n,n+1
                                  = A (n+1) (r)A (n−1) (r).          (6.6.4)
          Similarly,
                                          (n+1)
                                τ r+1 = −A    (r)
                                          1,n+1
                                          (n+1)
                                     = −A     (r),
                                          n+1,1

                                τ r+1  =(−1) n+1 A (n+1) (r),
                                               1n
                         τ
                                        (n+1)    (n−1)
                          r+1  τ r+1    = A  (r)A    (r +1).         (6.6.5)
                          τ

                           r    τ r
          Replacing r by r − 1,

                            τ           (n+1)       (n−1)
                            r   τ r
                                     = A     (r − 1)A   (r).         (6.6.6)

                            τ r−1  τ r−1
          Theorem 6.7 follows from (6.6.3)–(6.6.6).
          Theorem 6.8.

                                      τ r+1
                                            τ
                                  τ r        r+1
                                τ
                                 r−1   τ r   τ  r    =0.


                                τ      τ     τ
                                 r−1    r    r

          Proof. Denote the determinant by F. Then, Theorem 6.7 can be
          expressed in the form
                                  F 33 F 11 = F 31 F 13 .
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