恒等tensorの成分表示とパラメタ同士の偏微分の関係

$ [\bm{I}]^{\sf E\bar F}_{ij}=[\bm{I}]^{\sf\bar FE}_{ji}=\frac{\partial\bar f_j}{\partial\bar e_i}

$ \bar{\bm{e}}_i=\frac{\partial\bm{r}}{\partial e_i}=\sum_j\frac{\partial\bm{r}}{\partial f_j}\frac{\partial f_j}{\partial e_i}=\sum_j\bar{\bm{f}}_j\frac{\partial f_j}{\partial e_i}=\sum_j\frac{\partial f_j}{\partial e_i}\bar{\bm{f}}_j

$ \underline{[\bm{I}]^\mathsf{\bar{E}F}_{ij}=[\bm{I}]^\mathsf{F\bar{E}}_{ji}=\frac{\partial f_j}{\partial e_i}}

dot積で示す

$ [\bm I]_{ij}^{\sf EF}=\bm e_i\cdot\bm f_j= \sum_k\frac{\partial f_k}{\partial \bar e_i}\frac{\partial\bm r}{\partial f_k}\cdot\bm f_j=\sum_k\frac{\partial f_k}{\partial \bar e_i}\bar{\bm f}_k\cdot\bm f_j=\frac{\partial f_j}{\partial \bar e_i}

$ \because\bar{\bm f}_k\cdot\bm f_i=\llbracket k=i\rrbracket

何通りか示す