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If $X$ is an irreducible algebraic variety (over $\mathbb C$), an algebraic vector bundle of rank $r$ over $X$ is a couple $(E,\pi)$ where $E$ is an algebraic variety and $\pi: E\longrightarrow X$ is a surjective morphism, with the following properties:

1) $\pi^{-1}(x)$ is a vector space isomorphic to $\mathbb C^r$ for every $x\in X$

2) For every $x\in X$ there are an open neighborhood $U$ and an isomorphism of varieties $\phi_U:\pi^{-1}(U)\longrightarrow U\times\mathbb C^r$ such that:

2a) $ \pi_1\circ\phi_U=\pi$ where $\pi_1:U\times\mathbb C^r\rightarrow U$ is the canonical projection on $U$.

2b) $\phi_U|_{\pi^{-1}(x)}:\pi^{-1}(x)\longrightarrow \{x\}\times\mathbb C^r$ is an isomorphism of vector spaces.


One can also construct an algebraic vector bundle by an open cover $\{U_\alpha\}$ of $X$ and some functions $g_{\alpha\beta}:U_\alpha\cap U_\beta\longrightarrow GL_r(\mathbb C)$ which satisfy some cocicle conditions. Now my question is the following:

Are the functions $g_{\alpha\beta}$ morphisms between varieties? (remember that $GL_r(\mathbb C)$ is an open affine subvariety of $\mathbb C^{r^2}$)

For example Cox in his book on Toric varieties says only that $g_{\alpha\beta}$ is a function. But for smooth manifolds, $g_{\alpha\beta}$ must be a smooth function, so by analogy I think that in the algebraic case $g_{\alpha\beta}$ should be a morphism.

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Yes.

First of all, this is not quite the correct defintion. The vector space structure of the fibers (more generally, on the sections) is not a property. It is an additional structure which belongs to the vector bundle. The modern definition of a vector bundle (which works for manifolds, analytic spaces and schemes at once, in fact arbitrary ringed spaces) is that of a locally free sheaf of modules. Actually this definition simplifies lots of constructions for vector bundles. And it also makes clear what is part of the structure (the sheaf of modules) and what is a property (local triviality). $(\star)$

If $X$ is a ringed space and $F$ is a vector bundle of rank $n$ on $X$, then the definition tells us that there is an open covering $X = \cup_i X_i$ such that $F|_{X_i}$ is free of rank $n$, i.e. isomorphic to $\mathcal{O}_{X_i}^n$. On the intersections we get an automorphism of the sheaf of modules $\mathcal {O}_{X_i \cap X_j}^n$. This can be desribed as an element of $\mathrm{GL}_n(\mathcal{O}(X_i \cap X_j))$. Just for the sake of completeness (skip this if you don't know cohomology yet), this cocycle description may be refined and summarized as $\mathrm{Vect}_n(X) \cong \check{H}^1(X,\mathrm{GL}_n)$.

If $X$ is a scheme (or just a variety), then there is an isomorphism $$\mathrm{GL}_n(\mathcal{O}(X)) \cong \mathrm{Hom}_{\mathrm{Sch}}(X,\mathrm{GL}_n).$$ Actually this could serve as a functorial defintition of the (group) scheme $\mathrm{GL}_n$. More concretely, $\mathrm{GL}_n$ is the spectrum of $k[\{x_{ij}\}][\mathrm{det}^{-1}]$ if $k$ is our base ring which I have omitted so far from the notation.

Of course, $\mathrm{Hom}_{\mathrm{Sch}}(X,\mathrm{GL}_n)$ differs a lot from the (huge and uninteresting) set of all maps of the underlying sets $|X| \to |\mathrm{GL}_n|$. Hence, cocycles $X_i \cap X_j \to \mathrm{GL}_n$ should be morphisms of schemes (or varieties if $X$ is a variety).

You can also specialize to topological vector bundles: here $X$ is a topological space equipped with the sheaf of continuous functions, say with values in $\mathbb{C}$. The cocycles should be continuous maps $X_i \cap X_j \to \mathrm{GL}_n(\mathbb{C})$, not just arbitrary maps. For holomorphic vector bundles, take holomorphic cocycles, etc.

$(\star)$ The connection to the classical / geometric vector bundles is given as follows: If $F$ is a sheaf of modules on a scheme $X$, then $V(F) := \mathrm{Spec}(\mathrm{Sym}(F^*))$ is a scheme which is affine over $X$. For $F=\mathcal{O}_X^n$ we have $V(F)=\mathbb{A}^n_X$. Thus, if $F$ is locally free, we have that $V(F)$ is locally trivial in the usual sense.

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    $\begingroup$ Beautiful! But I have a trouble; a vector bundle should be a variety, but a sheaf of modules on $X$ is not a variety. $\endgroup$ – Dubious Nov 5 '13 at 11:40
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    $\begingroup$ See $(\star)$ ... $\endgroup$ – Martin Brandenburg Nov 5 '13 at 17:38

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