# Correspondence between line bundles $L$ and $\mathcal{O}_X(d)$

I understand well the notion of a line bundle $L \to X$ where $X$ is the base. Locally at some $U \subset X$ we have that $L = U \times \mathbb{C}$ (if we work over the complex numbers).

I also understand the notion of the structure sheaf $\mathcal{O}_X$ and Serre's twisting sheaf $\mathcal{O}_X(d)$ in terms of regular functions over $U$ and in terms of rational functions of degree $d$ over $U$ respectively.

What I struggle to understand is how to understand the connection between sections of the line bundle with those sheaves though.

• Can you be more precise? What exactly are you not understanding? There is a one to one correspondence between locally free sheaves and vectorbundles but this makes use of the relative spectrum construction, See for example here: math.stackexchange.com/questions/593875/… – Maik Pickl Sep 22 '16 at 10:09
• What you're looking for is as follows. If $L\to X$ is a line bundle, then the sheaf of sections of this map is a line bundle sheaf. If $\mathcal{L}$ is a line bundle sheaf then $\mathrm{Sym}(\mathcal{L}^\vee)$ is a quasi-coherent sheaf of $\mathcal{O}_X$-algebras and its relative spectrum is a line bundle. These operations are inverse. – Alex Youcis Sep 22 '16 at 10:10
• @AlexYoucis I am not familiar with Sym$\mathcal{L^{\vee}}$ is. Would you mind expanding a little bit more? – Marion Sep 22 '16 at 10:25
• @Marion If $\mathcal{M}$ is any coherent sheaf you can get a finite type $\mathcal{O}_X$-algebra defined by sheafifying $U\mapsto \mathrm{Sym}(\mathcal{M}(U))$ where here 'Sym' stands for the usual symmetric algebra over $\mathcal{O}_X(U)$. Of course, one can check that if $\mathcal{M}=\mathcal{O}_X^n$ then $\mathrm{Sym}(\mathcal{M})=\mathcal{O}_X[T_1,\ldots,T_n]$ and thus $\underline{\text{Spec}}(\mathrm{Sym}(\mathcal{M}))=\mathbb{A}^n\times X$. So, for example, if $\mathcal{M}$ is a vector bundle, so locally $\mathcal{O}_X^n$, – Alex Youcis Sep 22 '16 at 10:31
• then the space $\underline{\text{Spec}}(\text{Sym} (\mathcal{M}))\to X$ is locally $X\times\mathbb{A}^n$, so a vector bundle in the traditional sense. – Alex Youcis Sep 22 '16 at 10:32