If $X$ is a scheme and $\mathcal{F}$ is a locally free sheaf of rank $n$, then in Vakil's book the total space of $\mathcal{F}$ is defined to be $Spec(\text{Sym}^\bullet \mathcal{F}^\vee)$, the relative spectrum of the sheaf of algebras. The total space is a rank $n$ vector bundle, which is easy to see, just take any open subset $U$ such that $\mathcal{F}|_U$ is trivial, then we have \begin{equation} Spec(\text{Sym}^\bullet \mathcal{F}^\vee|_U) \cong \mathbb{A}_U^n \end{equation}

In Ex. 17.1.G, it claims that $\mathcal{F}$ is isomorphic to the sheaf of sections of the total space $Spec(\text{Sym}^\bullet \mathcal{F}^\vee)$. It is easy to see that the sections of $Spec(\text{Sym}^\bullet \mathcal{F}^\vee)$ on $U$ is isomorphic to $\mathcal{O}_U^n$, i.e. when $U$ is affine, say $U=\text{Spec}(A)$, then we have $\text{Sym}^\bullet \mathcal{F}^\vee|_U \cong A[x_1,\cdots,x_n]$ and the sections is in one to one bijection with the $A$-map $A[x_1,\cdots,x_n] \rightarrow A$, which is determined by the values of $x_i$, thus it is isomorphic to $A^n$.

However, I could not show the transition functions are actually the same. If the transition functions for $\mathcal{F}$ is $T_{ij}$ with respect to the affine open cover $U_i$, then the transition function for $\mathcal{F}^\vee$ is $(T_{ij}^t)^{-1}$, the inverse of transpose. It seems to me that the transition function for the vector bundle $Spec(\text{Sym}^\bullet \mathcal{F}^\vee)$ is actually $(T_{ij}^t)^{-1}$. How could the transition function for the sheaf of its sections be the same as that of $\mathcal{F}$, i.e. $T_{ij}$?


1 Answer 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 $T_{ij}$.


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