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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 n for some fixed n. In other words, EX is a vector bundle if there is a collection of open subschemes Ui X such that Ui(A) = X(A) for every finite local ring and E×XUi is isomorphic to 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 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 p + 1×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 p is given by the open subschemes Ui = V(0;Xi). We see easily that L×pUi is given by the equations bj = (aj/ai)bi since ai is a unit. Thus the map from a×Ui to L×pUi given by

(c;a0 : ... : ap) (a0/ai)c,...,(ap/a0)c;a0 : ... : ap

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 p + 1 which is the complement of the point (0 : ... : 0 : 1). The morphism Hp 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 : a×pH given by

(c;a0 : ... : ap) (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 n are given as the closed subscheme GLn of 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 n×X corresponds naturally to a morphism g : XGLn. 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 be the isomorphism of vector space schemes : E×XUin×Ui. For any i and j it is clear that we get a morphism : Ui UjGLn by comparing the two isomorphisms of E×X(Ui Uj) with n×(Ui Uj). These morphisms satisfy . = on Ui Uj Uk. Conversely, it is clear that we can use such a collection of morphisms : Ui UjGLn to construct a vector bundle on X by patching together the vector bundles n×Ui. More generally, we can easily show that for any vector space scheme V on X, the group scheme GLn operates on V n. Thus we can use the to patch together V n×XUi to obtain a vector space scheme. This vector space scheme is denoted E V and is called the tensor product of E with V. It is clear that 1 V = V. One can show that Hn = H n and H L = 1×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 0VWU 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)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 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: 8.7 Action of correspondences Up: 8 Algebraic Schemes for Previous: 8.5 The category of
Kapil Hari Paranjape 2002-10-20