Total space of a finite rank locally free sheaf, Vakil's 17.1.4 & 17.1.G

Solution 1:

First of all, pick a basis $\{v_k^i\}$ for $\mathcal{F}$ on each trivializing open set $U_i$, and let $\{x_k^i\}$ be its dual basis, which trivializes $\mathcal{F}^\vee$. Over some open affine subset of an intersection $U_{ij}$, the total space is $\DeclareMathOperator{spec}{Spec} \spec(A[x_1^i,\ldots ,x_n^i])\cong \spec(A[x_1^j,\ldots ,x_n^j])$, where the isomorphism is the transition function $\phi_{ij}$ we are looking for. It is induced by the map of rings in the opposite direction $$ \phi_{ij}^*:A[x_1^j,\ldots ,x_n^j]\to A[x_1^i,\ldots ,x_n^i] $$ Which maps a generator $x_k^j$ to $T_{ij}^t(x_k^j)$. If we let $v$ be a point in the total space, then $\phi_{ij}(v)$ is such that $x_k^j(\phi_{ij}(v))=\phi_{ij}^*(x_k^j)(v)$, just by how a ring homomorphism induces a map of schemes. Now the fact that for any basis element $x_k^j\circ \phi_{ij}=\phi_{ij}^*(x_{k}^j)$ implies that $\phi_{ij}$ is the dual map to $\phi_{ij}^*$, that is, its matrix with respect to the bases $\{v_k^i\}$ and $\{v_k^j\}$ will be the transpose of $T_{ij}^t$, which gives us $T_{ij}$ back.

Summing up: the $x_i$'s form a basis for the dual of the fiber of the total space, and if the transition function acts like $T_{ji}^t$ on the dual, it must act on the original space as $T_{ij}$.