Geometric invariant theory

Transcription

1 Geometric invariant theory Shuai Wang October Introduction Some very basic knowledge and examples about GIT(Geometric Invariant Theory) and maybe also Equivariant Intersection Theory. 2 Finite flat group schemes by John Tate 3 Linearizations of line bundles Let, G be a connected linear algebraic group, X a normal G-variety, and L a line bundle over it, then L n admits a G-linearization for some n. Note first that this doesn t mean large enough powers of L admit G-linearization, just some of them. The second thing is that X being normal is crucial, see the following example Example 3.1 (L n admits no linearization for any n). Consider the nodal curve in P 2 C : y 2 z = x 3 x 2 z Then we can get C by identifying and 0 in P 1. Namely t (t 2 + 1, t 3 ) Identify the fibres over and 0 of O P 1(k), (k 0), we can get a line bundle L over C. Let G m act on P 1 by [x, y] [tx, t 1 y], 0 are fixed, so we have a natural G m action on C, if L admits a G m linearization, then by pulling it back, we can get a G m linearization of O P 1(k), and G m acts on fibres over, 0 with the same character. However, this is impossible for and G m linearization of O P 1(k). To see this, we have a SES 0 X(G) P ic G (X) P ic(x) P ic(g). 1

2 In our particular situation, it is 0 Z P ic G (P 1 ) Z 0. Which simply says that if O P 1(k) is linearizable, then we have different linearizations parametrized by Z, we can construct all these linearizations explicitly T ot(o P 1(k)) = P(1, 1, k) [0, 0, 1] P(1, 1, k) [0, 0, 1] [x, y, z] [tx, ty, t j z]; j Z Then the characters of G m action on firbres over and 0 differs by t k. This contradiction tells us that L is not G m linearizable, same obstruction holds for L n. Remark. The obstruction for a line bundle to be linearizable is given by the Schur multiplier H 2 (G, k ), i.e we have a SES 0 Hom(G, k ) = X(G) P ic G (X) P ic(x) G H 2 (G, k ) Note that if G is a connected linear algebraic group, then the action of G on P ic(x) must be trivial, in other words P ic(x) G = P ic(x), and for a connected affine algebraic group acting on a normal variety, we actually have 0 X(G) P ic G (X) P ic(x) P ic(g) Remark (Intuitively, why O P n(k) is not P GL(n + 1)-linearizable, if n + 1 k?). My intuition comes from the following sequence is true for X affine or proper and connected over k, specially, for projective smooth varieties, everyting works fine. 0 X(G) P ic G (X) P ic(x) P ic(g) Now, it s intuitively easy to understand the following statements every O P n(k) admits Z different GL(n + 1)-linearization given by det n. Let me just write down the group action on the total space GL(n + 1) g : P(1, 1,..., k) [0, 0,..., 1] P(1, 1,..., k) [0, 0,..., 1] [x 0,..., x n ; z] [g(x 0, x 1,..., x n ); det n z]. every O P n(k) admits a unique SL(n + 1)-linearization SL(n + 1) g : P(1, 1,..., k) [0, 0,..., 1] P(1, 1,..., k) [0, 0,..., 1] [x 0,..., x n ; z] [g(x 0, x 1,..., x n ); z]. L P ic(p n ) admits a P GL(n + 1)-linearization if and only if its image in P ic(g) is zero. Note that the last map in the sequence is given by δ : P ic(x) P ic(g); L pr 2L σ L 1 G x0. 2

3 x 0 is a choose point in X, so actually δ is not canonical. And we know pr 2O P n(k) σ O P n( k) G x0 = O P N ( k) G. And the last group is trivial if and only if n + 1 k. Haha, you might ask where does the last equality comes from? It s comes from several facts P GL(n + 1) = P N, where means the determinate variety. From the decomposition of P N above, we get P ic(p GL(n + 1)) = Z/(n + 1)Z, with the restriction of P N (1) as a generator. since the complement has codimension 2, we actually have P ic(p N P n ) = P ic(p GL(n + 1) P n ). In other words, every line bundle on P GL(n+1) P n is the restriction of a line bundle on P N P n = Z Z, this is how we actually write down σ L 1. the group action is the restriction of the birational map P N P n P n n n [(a ij ), (x 0, x 1,..., x n )] [ a 1j x j,..., a nj x j ]. Computation of σ is easy since the group action is just a bilinear function on every component, the pull-back just means separate this linearity. Actually, we can really write down the P GL(n+1)-linearization explicitly, say for the cannonical bundle O P n( (n + 1)) g : P(1, 1,..., (n + 1)) [0, 0,..., 1] P(1, 1,..., (n + 1)) [0, 0,..., 1] [x 0,..., x n ; z] [g(x 0, x 1,..., x n ); z det 1 (g)]; g P GL(n + 1). Remark (fixed-point and character). Remark (induced action on cohomology groups H i (X, L )). Instead of the the chain of linear maps j=0 j=0 Γ(X, F ) Γ(G X, σ I F ) Γ(G X, pr2f ) = C[G] Γ(X, F ). On higher cohomology groups, we have H i (X, F ) H i (G X, σ I F ) H i (G X, pr2f ) = C[G] H i (X, F ). 3

4 The identity H i (G X, pr 2F ) = C[G] H i (X, F ) comes form the fact pr 2 is a flat morphism(fibrewise Hilbert polynomial criterion, or actually it s just the trivial family), thus from the general fact if you have a flat morphism f : X X between noetherian schemes H i (X, F ) A A = H i (X, f F ). This simply because if A is flat over A, then A A commutes with taking cohomology groups of the Cech complex. I like a more explicit way of constructing the group action on cohomology groups, let {U i } be an open covering of X, then so is {gu i }, if α Γ(X, F ) is given by (α Ui ), we simply define gα to be the global section defined by the image of I 1 gu i : F (U i ) F (gu i ). Namely let β = gα, we have β gui = I 1 gu i (α Ui ). Same construction works for H i. The equivariant structure of F makes everything fit perfectly. 4 Picard group: special push-forward and pullback If π : X Y is an morphism between varieties, we have π O Y = O X π 1 O Y π 1 O Y = O X. This follows from the construction. This is more than just view f O Y as an element f π O X. For push-forward, we don t have similar property in general, but in some important situations, we do have, and also by constriction, namely the global Proj. For example, of we have a projective bundle p : PE X, we do have π O PE = O X. For global Spec, α : Spec(A ) X, we have similar property Remember A is a sheaf of O X -algebra. α O SpecA = A. Theorem 4.1 (P ic(pe ) = P ic(x) Z). If X is a noetherian regualr scheme and E is a vector bundle on X with rank at least 2, then we have P ic(pe ) = P ic(x) Z. Proof. We prove the following map is an isomorphims P ic(x) Z P ic(pe ) (L, m) O PE (m) p L. 4

5 injectivity. If we have p (L (m)) = O PE, then O X = p O PE = p (O PE (m) p L ) = L p O PE (m). The last step is by the push-full formula. Note by the cohomology and base change theorem, we know p O PE (m) = Sym m E. Since rank(e ) 2, the identity above is possible only if m = 0, consequently, we get L = O X. surjectivity. By tensoring a suitble O PE (m), we may assume M P ic(pe ) is trivial on p 1 U i, where {U i } is a suitable open covering of X, then M is given by some cocycles p ij : p O X Ui p O X Uj, and since p p O X Ui = O X Ui, this naturally defines a line bundle L on X, and by construction, we have p L = M. Figure 1: The Universe References 5