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8.6 Vector Bundles and regular schemes

Most of the examples of vector space schemes that we have seen so far are vector bundles; these are vector space schemes that are ``locally'' isomorphic to $ \mathbb {V}$n for some fixed n. In other words, E$ \to$X is a vector bundle if there is a collection of open subschemes Ui $ \subset$ X such that $ \cup$ Ui(A) = X(A) for every finite local ring and E×XUi is isomorphic to $ \mathbb {V}$n×Ui as a vector space scheme over Ui for every i. A collection of open sets satisfying the first property is referred to as an open cover of X. The vector bundle $ \mathbb {V}$q×X is called the trivial vector bundles on X. The number n is called the rank of the vector bundle.

Recall, that L was defined as the subscheme of $ \mathbb {V}$p + 1×$ \mathbb {P}$p consisting of pairs of tuples (b0,..., bp;a0 : ... : ap) such that aibj = ajbi for all i and j between 0 and p. An open cover of $ \mathbb {P}$p is given by the open subschemes Ui = V(0;Xi). We see easily that L×$\scriptstyle \mathbb {P}$pUi is given by the equations bj = (aj/ai)bi since ai is a unit. Thus the map from $ \mathbb {G}$a×Ui to L×$\scriptstyle \mathbb {P}$pUi given by

(c;a0 : ... : ap) $\displaystyle \mapsto$ $\displaystyle \left(\vphantom{ (a_0/a_i)c,\dots,(a_p/a_0)c ; a_0:\cdots:a_p
}\right.$(a0/ai)c,...,(ap/a0)c;a0 : ... : ap$\displaystyle \left.\vphantom{ (a_0/a_i)c,\dots,(a_p/a_0)c ; a_0:\cdots:a_p

gives an isomorphism. Thus L is a vector bundle of rank 1 or a line bundle. Recall also that H was defined as the subscheme of $ \mathbb {P}$p + 1 which is the complement of the point (0 : ... : 0 : 1). The morphism H$ \to$$ \mathbb {P}$p is the projection away from this point and the zero-section is V(Xp + 1). For each i between 0 and p we have a natural homomorphism si : $ \mathbb {G}$a×$ \mathbb {P}$p$ \to$H given by

(c;a0 : ... : ap) $\displaystyle \mapsto$ (a : 0 : ... : ap : c . ai)

Note that this is an isomorphism outside the hyperplane V(Xi); in other words this is an isomorphism on Ui = V(0;Xi). Thus H is also a line bundle.

The automorphisms of the vector space $ \mathbb {V}$n are given as the closed subscheme GLn of $ \mathbb {A}$n2 + 1 consisting of ((Xij)i, j = 1n, T) such that det((Xij))T = 1. For any scheme X, any automorphism of the vector space scheme $ \mathbb {V}$n×X corresponds naturally to a morphism g : X$ \to$GLn. Moreover, it is clear that GLn is a group scheme.

Now let E be a vector bundle over a scheme X, {Ui} be an open cover of X and $ \phi_{i}^{}$ be the isomorphism of vector space schemes $ \phi_{i}^{}$ : E×XUi$ \to$$ \mathbb {V}$n×Ui. For any i and j it is clear that we get a morphism $ \phi_{ij}^{}$ : Ui $ \cap$ Uj$ \to$GLn by comparing the two isomorphisms of E×X(Ui $ \cap$ Uj) with $ \mathbb {V}$n×(Ui $ \cap$ Uj). These morphisms satisfy $ \phi_{ij}^{}$ . $ \phi_{jk}^{}$ = $ \phi_{ik}^{}$ on Ui $ \cap$ Uj $ \cap$ Uk. Conversely, it is clear that we can use such a collection of morphisms $ \phi_{ij}^{}$ : Ui $ \cap$ Uj$ \to$GLn to construct a vector bundle on X by patching together the vector bundles $ \mathbb {V}$n×Ui. More generally, we can easily show that for any vector space scheme V on X, the group scheme GLn operates on V$\scriptstyle \oplus$ n. Thus we can use the $ \phi_{ij}^{}$ to patch together V$\scriptstyle \oplus$ n×XUi to obtain a vector space scheme. This vector space scheme is denoted E $ \otimes$ V and is called the tensor product of E with V. It is clear that $ \mathbb {V}$1 $ \otimes$ V = V. One can show that Hn = H$\scriptstyle \otimes$ n and H $ \otimes$ L = $ \mathbb {V}$1×$ \mathbb {P}$p.

As before we define the K-group of vector bundles of a scheme S as the quotient K0(S) of the free abelian group on isomorphism classes of vector bundles by the subgroup generated by relations of the form [V] + [U] - [W] where 0$ \to$V$ \to$W$ \to$U$ \to$ 0 is an exact sequence of vector bundles. Note that any vector bundle is a vector space scheme and an exact sequence of vector bundles is also an exact sequence of vector space schemes. Thus we have a natural homomorphism K0(S)$ \to$G0(S). When S is a regular scheme this is an isomorphism; usually one gives a definition of regular schemes in terms of ring theory and proves the equivalence, but we could equally well use this as a definition. As a particular case we have the ``Jacobian criterion'' which says that a scheme is regular if the Zariski tangent vector space scheme is a vector bundle; note however that this is not in general necessary. For example the subscheme of $ \mathbb {A}$2 defined by XY = p for some prime p is regular but its Zariski tangent space is not a vector bundle.

In fact the tensor product construction makes K0(S) into a ring and G0(S) a module over this ring.

next up previous
Next: 8.7 Action of correspondences Up: 8 Algebraic Schemes for Previous: 8.5 The category of
Kapil Hari Paranjape 2002-10-20