Why is $ \pi^{-1} (U) $ an affine open set in $ E $, with coordinate ring : $ \mathrm{Sym}_A \Gamma ( U , \mathcal{E}^{ \vee } ) $?

Question

Let $ E $ a vector bundle of rank $ r $ on a scheme $ X $. This means that $ E $ is a scheme equipped with a morphism $ \pi : E \to X $, satisfying the following conditions:

There is an open covering $ \{ U_i \} $ of $ X $, and isomorphisms $ \varphi_i $ of $ \pi^{-1} (U) $ with $ U_i \times \mathbb{A}^r_K $, such that: over $ U_i \cap U_j $, the composition $ \varphi_i \circ \varphi_{j}^{-1} $ is linear, i.e.: given by a morphism: $ g_{ij} : U_i \cap U_j \to \mathrm{GL} ( r , K ) $. These transitions functions satisfy : $$ g_{ik} = g_{ij} g_{jk } , \ \ \ \ \ g_{ij}^{-1} = g_{ji}, \ \ \ \ \ g_{ii} = 1 $$ Conversely, any such transition functions determine a vector bundle.

My question is:

For an affine open set: $ U \subset X $, with coordinate ring $ A $, why is $ \pi^{-1} (U) $ an affine open set in $ E $, with coordinate ring the symmetric algebras: $ \mathrm{Sym}_A \Gamma ( U , \mathcal{E}^{ \vee } ) $, such that: $ \mathcal{E} $ is the sheaf of sections of $ E $, and $ \mathcal{E}^{ \vee } = \mathrm{Hom}_{ \mathcal{O}_X } ( \mathcal{E} , \mathcal{O}_X ) $, and $ \Gamma ( U , \mathcal{E}^{ \vee } ) = H^0 ( U , \mathcal{E}^{ \vee } ) $ is the space of sections?

Thank you in advance.

Answer 1

I start quickly from the begin: what is a rank $r$ trivial (vector) bundle $(E,\pi)$ on a scheme $X$? A first possible answer: it is a scheme such that $E=\pi^{-1}(X)\cong X\times_{\mathbb{Z}}\mathbb{K}^r$, where $\mathbb{K}$ is a field.

Remark 1: Every scheme is canonically a scheme over $\operatorname{Spec}\mathbb{Z}$!

But, because the notation $\mathbb{K}^r$ indicates the (affine) scheme $\operatorname{Spec}\mathbb{K}^r$, one has $$ \operatorname{Spec}\mathbb{K}^r=\coprod_1^r\operatorname{Spec}\mathbb{K}=\coprod_1^r\{(0)\} $$ and therefore this answer is unlike!

The second possible (but partially right) answer: it is a scheme such that $E=\pi^{-1}(X)\cong X\times_{\mathbb{Z}}\mathbb{A}^r_{\mathbb{K}}$; but what means one for $\mathbb{A}^r_{\mathbb{K}}$?

The first (and right) answer is the following: $$ \mathbb{A}^r_{\mathbb{K}}=\operatorname{Spec}\mathbb{K}[t_1,\dots,t_r]=\operatorname{Spec}\bigoplus_{n\in\mathbb{N}_{\geq0}}S^n\left(\mathbb{E}^{\vee}\right)=\operatorname{Spec}\operatorname{Sym}(\mathbb{E}) $$ where:

1. $\mathbb{E}$ is the $\mathbb{K}$-vector space generated from $t_1,\dots,t_r$;
2. $S^n(\cdot)$ is the $n$-th symmetric power of "$\cdot$".

Remark 2: Let $R$ be a (commutative) ring (with unit), let $M$ be the free $R$-module of rank $r$; one defines $$ \mathbb{A}^r_R=\operatorname{Spec}\operatorname{Sym}(M). $$

At this point, the question comes: what is the sheaf $\mathcal{E}$ of the sections of $E$?

By definition: a section $s$ of $E$ over an open subset $U$ of $X$ is a morphism $s:U\to E$ of schemes such that $\pi\circ s=Id_U$; let $T$ be a scheme, let: $$ s(T):U(T)\to E(T)\cong X(T)\times\mathbb{A}^r_{\mathbb{K}}(T) $$ one has that $s$ maps the $T$-valued points $u$ of $U$ in a pair of $T$-valued points in $E$ such that the first term is $u$; because this statement holds for any scheme $T$ then $s$ is identifiable as a morphism $$ s\in\operatorname{Hom}_{\mathbf{Sch}}\left(U,\mathbb{A}^r_{\mathbb{K}}\right); $$ in particular \begin{gather*} \operatorname{Hom}_{\mathbf{Sch}}\left(U,\mathbb{A}^1_{\mathbb{K}}\right)^r\cong\operatorname{Hom}_{\mathbf{Sch}}\left(U,\mathbb{A}^r_{\mathbb{K}}\right)\cong\operatorname{Hom}_{\mathbf{Ring}}(\operatorname{Sym}(\mathbb{E}),\mathcal{O}_X(U))\cong\\ \cong\operatorname{Sym}\left(\operatorname{Hom}_{\mathbf{Ring}}\left(\mathbb{E}^{\vee},\mathcal{O}_X(U)\right)\right)=\operatorname{Sym}\operatorname{Hom}_{\mathbf{Ring}}\left(\widetilde{\mathbb{E}^{\vee}}(U),\mathcal{O}_X(U)\right)\cong\\ \cong\operatorname{Sym}\operatorname{Hom}_{\mathcal{O}_X-\mathbf{Mod}}\left(\mathcal{E}^{\vee},\mathcal{O}_X\right)(U)\cong\operatorname{Sym}\Gamma(U,\mathcal{E}) \end{gather*} where $\mathcal{E}=\widetilde{\mathbb{E}}$ is the $\mathcal{O}_X$-module associated to $\mathbb{E}$.

Remark 3: Until this point, never I had use the hypothesis that $\mathbb{E}$ is a $\mathbb{K}$-vector space; indeed, this reasoning holds for any finite rank free module over a ring.

After all, if $U$ is an affine open subset of $X$, then $\pi^{-1}(U)\cong U\times_{\mathbb{Z}}\mathbb{A}^r_{\mathbb{K}}$ is an open affine subscheme of $E$, because it is the fibre product of affine schemes; in particular, let $U\cong\operatorname{Spec}R$ then $\pi^{-1}(U)\cong\mathbb{A}^r_R$ and by previous statement: $$ \pi^{-1}(U)\cong\operatorname{Spec}\operatorname{Sym}\left(R^{\oplus r}\right)\cong\operatorname{Spec}\operatorname{Sym}_R\Gamma(U,\mathcal{E}). $$

This reasoning is generalizable to general vector bundle on a scheme!