验证 (3. 6) 式, 即证明 $$\bex \cfrac{\rd J}{\rd t}=J\Div_y {\bf v}. \eex$$
证明: $$\beex \bea \cfrac{\rd J}{\rd t} &=\cfrac{\rd }{\rd t}|{\bf F}|\\ &=\cfrac{\rd }{\rd t} \sum_{j_1\cdots j_n}(-1)^{\tau(j_1\cdots j_n)} f_{1j_1}\cdots f_{nj_n}\\ &=\sum_{j_1\cdots j_n}(-1)^{\tau(j_1\cdots j_n)} \sum_k f_{1j_1}\cdot \cfrac{\rd f_{kj_k}}{\rd t}\cdots f_{nj_n}\\ &=\sum_k \sev{\ba{ccc} f_{11}&\cdots&f_{1n}\\ \vdots&&\vdots\\ \cfrac{\rd f_{k1}}{\rd t}&\cdots&\cfrac{\rd f_{kn}}{\rd t}\\ \vdots&&\vdots\\ f_{n1}&\cdots&f_{nn} \ea}\\ &=\sum_{k,j}\cfrac{\rd f_{kj}}{\rd t}A_{kj}\quad\sex{ A_{kj}:\ f_{kj}\mbox{ 在 }{\bf F}\mbox{ 中的代数余子式} }\\ &=\sum_{k,j} \cfrac{\p v_k}{\p x_j} A_{kj}\\ &\quad\sex{ \cfrac{\rd f_{kj}}{\rd t} =\cfrac{\p }{\p t}\cfrac{\p y_k}{\p x_j} =\cfrac{\p }{\p x_j}\cfrac{\p y_k}{\p t} =\cfrac{\p v_k}{\p x_j}: \mbox{看成 }t,x\mbox{ 的函数} }\\ &=\sum_{k,j,l} \cfrac{\p v_k}{\p y_l}\cfrac{\p y_l}{\p x_j}A_{kj}\quad\sex{\mbox{看成 }t,y\mbox{ 的函数}}\\ &=\sum_{k,l}\cfrac{\p v_k}{\p y_l}\sum_j f_{lj}A_{kj}\\ &=\sum_k \cfrac{\p v_k}{\p y_k}J\\ &=J\Div_y{\bf v}. \eea \eeex$$
