A beginner's question of Riemannian Geometry.

Question

In picture below ,I don't know why $\Phi^{-1}(F)=(F(\phi^i,e_j))_{i,j=1}^n$

Answer 1

$\{e_j\}$ is a basis on $V$. $\{\phi_i\}$ the complementary basis on $V^*$. That is, for all $i,j, \phi_i(e_j) = \delta_{ij}$.

Now if $A\in \operatorname{End}(V)$, then $\Phi(A)(\phi_i, e_j) = \phi_i(A(e_j)) = A_{ij}$, that is, the entry in the $i^\text{th}$ row and $j^\text{th}$ column of the matrix representation of $A$ with respect to the basis $\{e_j\}$. Letting $\Phi(A) = F$, we see that $A_{ij} = F(\phi_i, e_j)$.

This is what is meant by $\Phi^{-1}(F)=(F(\phi^i,e_j))_{i,j=1}^n$