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6.10 The Einstein and Ernst Equations  295

          Exercise. The one-variable Hirota operators H x and H xx are defined in
          Section 5.7 and the determinants A n and E n , each of which is a function of
          ρ and z, are defined in (6.10.8) and (6.10.10). Apply Lemma 6.20 to prove
          that
                                             n − 1

            H ρ (A n−1 ,E n ) − ωH z (A n ,E n−1 )=  A n−1 E n ,
                                              ρ
                                              n − 2             2

            H ρ (A n ,E n−1 ) − ωH z (A n−1 ,E n )= −  A n E n−1  (ω = −1).
                                                ρ
          Using the notation
                                          1

                           2
                         K (f, g)=  H ρρ + H ρ + H zz (f, g),
                                          ρ
          where f = f(ρ, z) and g = g(ρ, z), prove also that
                                                     n(n − 2)
                                         2
                                        K (E n ,A n )=      E n A n ,
                                                        ρ 2

                                 2n − 4                1
                             2
                           K +          (A n ,A n−1 )= −  A n A n−1 ,
                                   ρ                  ρ 2
                         /                       0
                      K 2  ρ n(n−2)/2 E n ,ρ n(n−2)/2  A n  =0,
                    /   2                        0
                 K 2  ρ (n −4n+2)/2 A n−1 ,ρ n(n−2)/2  A n  =0,
                       /   2                     0
                    K 2  ρ (n −2)/2 A n+1 ,ρ n(n−2)/2  A n  =0.
                                                         (Sasa and Satsuma)


          6.10.4  Preparatory Theorems

          Define a Vandermondian (Section 4.1.2) V 2n (x) as follows:
                                         j−1
                              V 2n (x)= x

                                        i
                                           2n
                                    = V (x 1 ,x 2 ,...,x 2n ),     (6.10.24)
                                                            (2n)
          and let the (unsigned) minors of V 2n (c) be denoted by M  (c). Also, let
                                                            ij
                             (2n)
                   M i (c)= M   (c)= V (c 1 ,c 2 ,...,c i−1 ,c i+1 ,...,c 2n ),
                             i,2n
                             (2n)
                  M 2n (c)= M    (c)= V 2n−1 (c).                  (6.10.25)
                             2n,2n
                             x j =  z + c j  ,
                                    ρ
                                     3
                                                2
                             ε j = e ωθ j  1+ x 2  (ω = −1)
                                           j
                               =  τ j  ,                           (6.10.26)
                                  ρ
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