6.6 The Matsukidaira–Satsuma Equations 259
Hence applying the Jacobi identity (Section 3.6),
∣
∣
∣
∣
τr+2 τr+1
τr+1 τr
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
A
(n+1)
11
(r)(−1)
n
A
(n+1)
1 ,n+1
(r)
(−1)
n
A
(n+1)
n+1, 1 (r) A
(n+1)
n+1,n+1(r)
∣
∣
∣
∣
∣
=A
(n+1)
(r)A
(n−1)
(r+2).
Replacingrbyr−1,
∣
∣
∣
∣
τr+1 τr
τr τr− 1
∣
∣
∣
∣
=A
(n+1)
(r−1)A
(n−1)
(r+ 1) (6.6.3)
τ
′
r=−A
(n+1)
n,n+1(r)
=−A
(n+1)
n+1,n
(r)
τ
′′
r=A
(n+1)
nn (r).
Hence,
∣
∣
∣
∣
τ
′′
r τ
′
r
τ
′
r τr
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
A
(n+1)
nn (r) A
(n+1)
n,n+1
(r)
A
(n+1)
n+1,n
(r) A
(n+1)
n+1,n+1
(r)
∣
∣
∣
∣
∣
=A
(n+1)
(r)A
(n+1)
n,n+1;n,n+1(r)
=A
(n+1)
(r)A
(n−1)
(r). (6.6.4)
Similarly,
τr+1=−A
(n+1)
1 ,n+1
(r)
=−A
(n+1)
n+1, 1 (r),
τ
′
r+1
=(−1)
n+1
A
(n+1)
1 n
(r),
∣
∣
∣
∣
τ
′
r+1
τr+1
τ
′
r
τr
∣
∣
∣
∣
=A
(n+1)
(r)A
(n−1)
(r+1). (6.6.5)
Replacingrbyr−1,
∣
∣
∣
∣
τ
′
r τr
τ
′
r− 1
τr− 1
∣
∣
∣
∣
=A
(n+1)
(r−1)A
(n−1)
(r). (6.6.6)
Theorem 6.7 follows from (6.6.3)–(6.6.6).
Theorem 6.8.
∣ ∣ ∣ ∣ ∣ ∣
τr τr+1 τ
′
r+1
τr− 1 τr τ
′
r
τ
′
r− 1
τ
′
r
τ
′′
r
∣ ∣ ∣ ∣ ∣ ∣
=0.
Proof. Denote the determinant by F. Then, Theorem 6.7 can be
expressed in the form