# 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 $\mathcal{O}_X^n|_U \to \mathcal{F}|_U$....

• You probably already know this, but Ravi Vakil covers this (motivating locally free sheaves by studying vector bundles) in Ch. 13 of his notes: math.stanford.edu/~vakil/216blog/FOAGnov2815public.pdf – Richard D. James Dec 30 '15 at 16:26
• The correspondence between vector bundles and locally free sheaves is not exactly obvious. There's an exercise in Hartshorne outlining a proof. – Matt Samuel Dec 30 '15 at 22:41
• could you mention the exercise number? – Patricio Dec 30 '15 at 22:45
• I just remember this because I did the exercise. My copy of Hartshorne is buried somewhere in my house and would take some time to find. I'd be willing to bet you can find locally free sheaves in the index, and it will probably have a page number for the exercise. – Matt Samuel Dec 30 '15 at 22:48
• okey thank you very much! i will have a look! – Patricio Dec 30 '15 at 22:49

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.

• I have always seen the vector bundle associated to a locally free sheaf with some symmetric algebra involved, but you don't mention it in your answer. Is the symmetric algebra of the sheaf only there to turn the locally free sheaf into a sheaf of algebras and being able to aply relative Spec? – Pedro Jan 10 '18 at 1:06
• @Pedro: Yes, that's correct. – Daniel McLaury Jan 10 '18 at 2:41
• @Pedro could you please elaborate this last part, on why the symmetric algebra is used? I feel like it should be more obvious to me. Is this analogous to affine n-space (which would correspond to a locally free sheaf in this analogy) and the coordinate ring - the polynomial ring over n variables (which would correspond to the symmetric algebra over our locally free sheaf in this analogy)? – ZxJx Sep 13 '19 at 10:21
• @ZxJx yes, that's right. – Daniel McLaury Sep 24 '19 at 6:35