by $x$ and add the resultant expression to (2.3) and then prove the equation $P_2=0$, where $$P_2=f_{xt}f^*-f_xf_t^*-{\rm i}\rho^2(g_xg^*-f_xf^*)+\kappa\rho^2(gg^*-ff^*). \eqno(3.18)$$ Substituting (3.8)-(3.11), (3.13) and (3.14)-(3.16) into (3.18) and rearranging, $P_2$ reduces to $$P_2={{\kappa}\over \rho^2}\Bigl\{|D||D({\bf z}^*,{\bf z}_t^*;{\bf z},{\bf z}_t)|-|D({\bf z}^*;{\bf z})||D({\bf z}_t^*;{\bf z}_t)|
+|D({\bf z}^*;{\bf z}_t)||D({\bf z}_t^*;{\bf z})|\Bigr\}$$ $$+{{\rm i}\over \rho^2}\Bigl\{-|D||D({\bf z}^*,{\bf z}_t^*;{\bf z}_x,{\bf z}_t)|+|D({\bf z}^*;{\bf z}_x)||D({\bf z}_t^*;{\bf z}_t)|
-|D({\bf z}^*;{\bf z}_t)||D({\bf z}_t^*;{\bf z}_x)|\Bigr\}$$ $$\!+{1\over \rho^2}\Bigl\{\!-|D({\bf z}^*;{\bf z})||D({\bf z}^*,{\bf z}_t^*;{\bf z}_x,{\bf z}_t)|+|D({\bf z}^*;{\bf z}_x)||D({\bf z}_t^*,{\bf z}^*;{\bf z}_t,{\bf z})|
-|D({\bf z}^*;{\bf z}_t)||D({\bf z}^*,{\bf z}_t^*;{\bf z},{\bf z}_x)|\Bigr\}. \eqno(3.19)$$ The first and second terms on the right-hand side of (3.19) vanish by virtue of Jacobi’s identity. To show that the third term becomes zero as well, we consider the determinantal identity $$\begin{vmatrix}|D({\bf z}^*;{\bf z})| & |D({\bf z}^*;{\bf z})|& |D({\bf z}_t^*;{\bf z})|\\
                 |D({\bf z}^*;{\bf z}_x)| & |D({\bf z}^*;{\bf z}_x)|& |D({\bf z}_t^*;{\bf z}_x)| \\                        
                 |D({\bf z}^*;{\bf z}_t)| & |D({\bf z}^*;{\bf z}_t)|& |D({\bf z}_t^*;{\bf z}_t)| \end{vmatrix}=0.$$ It is obvious that this determinant is zero since the first two columns coincide. The above assertion follows immediately by expanding the determinant with respect to the first column and using Jacobi’s identity. Consequently, $P_2=0$. $\Box$

Before proceeding to the proof of (2.10), we emphasis that the constraints (3.2c) have not been 