Remark 2

Let $(e_\lambda)_{\lambda \in L}$, $(\bar{e}_\lambda)_{\lambda \in L}$ be two bases of M and P the matrix of passage from $(e_\lambda)$ to $(\bar{e}_\lambda)$. On account of what was seen in Example 1, the element of P appearing in the row of index $\lambda$ and the column of index $\mu$ is denoted by $\alpha_\mu^\lambda$ and the element of the contragredient ${}^tP^{-1}$ appearing in the row of index $\lambda$ and the column of index $\mu$ is denoted by $\beta_\lambda^\mu$. Then (in the notation introduced above)

for all mappings $f': I \to L$ and $g': J \to L$. The coordinates $\zeta_g^f$ of a tensor $z \in T_J^I(M)$ with respect to the basis $(e_\lambda)$ can therefore be expressed in terms of the coordinates $\bar{\zeta}_{g'}^{f'}$ of z with respect to the basis $(\bar{e}_\lambda)$ using the formulae The matrix $P^{-1}$ of passage from the basis $(\bar{e}_\lambda)$ to the basis $(e_\lambda)$ is the transpose of ${}^tP^{-1}$, so that $\beta_\lambda^\mu$ is the element which appears in the column of index $\lambda$ and the row of index $\mu$ of $P^{-1}$. The calculation of $e_f^g$ in terms of the $\bar{e}_{f'}^{g'}$ and that of the $\bar{\zeta}_{g'}^{f'}$ in terms of the $\zeta_g^f$ are therefore made by replacing $\alpha_\mu^\lambda$ by $\beta_\lambda^\mu$ and $\beta_\lambda^\mu$ by $\alpha_\mu^\lambda$ in the above calculations and exchanging the roles of $f$ and $f'$ and those of $g$ and $g'$. Then

(23)

The above formulae are such that the summation is over indices which appear once as a lower index and once as an upper index. Certain authors allow themselves because of this circumstance to suppress the summation signs.