Vector bundle as locally free coherent sheaves

I am studying coherent sheaves and was looking for a geometric motivation. Hence, in wikipedia and although here is stated that it can be seen as a generalization of vector bundles, which is quite satisfactorical, since this yields a better understanding of what tangent bundle, cotangent bundle or differential forms in sheaf theory and algebraic geometry might be. So, I tryed to see a vector bundle as a locally free coherent sheaf, but I got lost. So here are the two definitions and my first observations:

A (real) vector bundle consists of:

(i) topological spaces $X$ (base space) and $E$ (total space)

(ii) a continuous surjection $\pi:E\mapsto X$ (bundle projection)

(iii) for every $x$ in $X$, the structure of a finite-dimensional real vector space on the fiber $\pi^{-1}(\lbrace x\rbrace)$

where the following compatibility condition is satisfied: for every point in $X$, there is an open neighborhood $U$, a natural number $k$, and a homeomorphism

\begin{align} \varphi :U\times \mathbf {R} ^{k}\to \pi ^{-1}(U) \end{align}

such that for all $x \in U$,

(a) $(\pi \circ \varphi )(x,v)=x$ for all vectors v in $R^k$, and

(b) the map $v\mapsto \varphi (x,v)$ is a linear isomorphism between the vector spaces $R^k$ and $\pi^{−1}(x)$.

and the definition of coherent sheaves is the following:

A sheaf $\mathcal{F}$ of $\mathcal{O}_X$-Modules is coherent if :

1)$ \mathcal{F} $ is of finite type over $ \mathcal{O}_X $, i.e., for any point $ x\in X $ there is an open neighbourhood $ U\subset X $ such that the restriction $ \mathcal{F}|_U $ of $ \mathcal{F} $ to U is generated by a finite number of sections (in other words, there is a surjective morphism $ \mathcal{O}_X^n|_U \to \mathcal{F}|_U $ for some $ n\in\mathbb{N} $);

2) and for any open set $ U\subset X $, any $ n\in\mathbb{N} $ and any morphism $ \varphi\colon \mathcal{O}_X^n|_U \to \mathcal{F}|_U $ of $ \mathcal{O}_X $-modules, the kernel of $ \varphi $ is finitely generated.

Thus, we can see that they both have a topological space $X$ and we can identify $E=\mathcal{O}_X$. Furthermore there is a surjection $ \mathcal{O}_X^n|_U \to \mathcal{F}|_U $....

Solution 1:

I think the comments may be (inadvertently) giving the impression that this is more complicated than it really is. So let's just say how everything fits together:

Given a vector bundle, its sheaf of sections is locally free. Conversely, if we have a locally free sheaf then it's the sheaf of sections of a vector bundle which we can build by taking sheaf Spec of the symmetric algebra of the locally free sheaf.

Why is this true? Well:

• A vector bundle is, by definition, something that's locally isomorphic to a trivial vector bundle.
• A locally free sheaf (which we should really call a "locally free $\mathcal{O}_X$-module") is, by definition, something that's locally isomorphic to a free $\mathcal{O}_X$-module.
• The sheaf of sections of a trivial vector bundle is a free $\mathcal{O}_X$-module.

There is a bit to check here, but the picture itself is pretty clear.