LECTURE NOTES IN EQUIVARIANT ALGEBRAIC GEOMETRY. Spec k = (G G) G G (G G) G G G G i 1 G e


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1 LECTURE NOTES IN EQUIVARIANT ALEBRAIC EOMETRY 8/4/5 Let k be field, not necessaril algebraicall closed. Definition: An algebraic group is a kscheme together with morphisms (µ, i, e), k µ, i, Spec k, which are subject to the following relations.. Associativit propert: the following diagram commutes. ( ) ( ) µ µ () () µ µ. Inverse propert: the following diagram commutes. i µ i e Spec k 3. Identit propert: the following diagrams commutes. e µ Spec k
2 e µ Spec k The above definition can be generalized to define an Sgroup, where S is a scheme. An Sscheme is called an Sgroup if we replace Spec k b the scheme S in the above definition. Suppose that X T are Sschemes. We let X(T ) Hom S (T, X) we refer to the elements of X(T ) as T valued points. Observe that if X(T ), then the two morphisms i T : T T : T X induce the following commutative diagram. X S T X T If T is an Sscheme an Sgroup, then there is a natural multiplication map µ(t ) : (T ) (T ) (T ) defined b if µ(t )(g, g ) µ (g, g ) where (g, g ) : T S is the map induced b the Smorphisms T g T g. We can define an identit element in (T ) b letting e(t ) be the morphism T S i. There is a natural wa to define inverse elements in (T ) b i(t ) : (T ) (T ) given b g i g. It can be checked that (T ) together with µ(t ), i(t ), e(t ) make (T ) a group in the categor of sets. iven a morphism T T of Sschemes, there is an induced morphism of groups t : (T ) ( T ). So algebraic groups over a scheme S defines a contravariant functor from the categor of Sschemes to the categor of groups b sending an Sscheme T to the group (T ). A morphism of algebraic groups over S is a morphism φ : H of Sschemes such that the multiplication map on H is compatible with the multiplication map on, i.e., the following diagram commutes. H H φ φ S µ H µ H φ If φ is an immersion, i.e. locall the map on stalks is surjective, we sa that H is a subgroup of. If φ is a closed immersion then we sa that H is a closed subgroup of. Basic Eamples of Algebraic roups:. Suppose that is a finite group, S is an base scheme. Then we can define a group scheme structure on b Spec( g O S) in the obvious wa. Then (S).
3 ( ). Take Spec C[], a four point scheme over Spec C. Let µ : be the ( 4 ) morphism corresponding to the ring map of global sections µ # C[] : C[] C[] ( 4 ) ( 4 ) ( 4 ) sending ( )( ). Let e : Spec C be the morphism induced b e # : C[]/( 4 ) C sending, i.e. the morphism e picks out the maimal ideal ( ). Finall, we let i : be the morphism induced b i # : C[]/( 4 ) C[]/( 4 ) which sends 3. One can verif that (, µ, e, i) is an algebraic group over Spec C we can compute the group (Spec C). ( ) Observe that the four points of Spec C[] correspond to the four roots of 4 ( 4 ), {,, i, i}. If a, b then µ()(a, b) : µ(a, b) ab it easil follows that (Spec C) Z 4. 8/6/5 ( ) 3. Let Spec Q[], we can still define a multiplication, identit, inverse morphism the same wa as in eample. But the group scheme structure on will be a ( 4 ) bit different ( since the ) polnomial ( 4 ) does not split over Q. Observe that Q Spec Q[] Q[] Spec Q[s,t] Q[s,t]. The ring is a reduced Artin ring, ( 4 ) ( 4 ) (s 4,t 4 ) (s 4,t 4 ) Q[s,t] hence is the product of fields. The ring has ten maimal ideals. The are (s 4,t 4 ) (s, t ), (s +, t ), (s, t + ), (s +, t + ), (s, t + ), (s +, t + ), (s +, t ), (s +, t + ), (s +, t s) (t +, t s), (s +, t + s) (t +, t + s). Note that (s +, t s) does indeed coincide with the ideal (t +, t s), the ke observation is that t + s + + (t + s)(t s). Therefore, (s 4, t 4 ) (s, t ) (s +, t ) (s, t + ) (s +, t + ) (s, t + ) (s +, t + ) (s +, t ) (s +, t + ) (s +, t s) (s +, t + s). To underst the multiplication map µ : Q we onl need to underst what each (µ # ) Q[s,t] (m) is for each of the maimal ideals m of. For eample, (s 4,t 4 ) (µ# ) ((s, t )) ( ). To verif this observe µ # ( ) st s + s(t ). Hence the maimal ideal ( ) (µ # ) ((s, t )) equalit of the ideals must follow. Similarl, µ((s, s t)) (+) since µ # (+) st+ s ++s(t s) (s +, s(t s)). Affine Sgroups An affine Sgroup is simpl an Sgroup satisfing the etra condition that S is an affine morphism. Notice that to give an affine Sgroup is equivalent to giving an affine 3
4 morphism S morphisms of sheaves of O S modules µ # : O O OS O e # : O O S i # : O O so that if we take the diagrams in the definition of a Sgroup, replace the schemes with their sheaves of regular functions, reverse the arrows we get a commutative diagrams. Caution: Not all Sgroups are affine, but a Chevalle s Theorem on Algebraic roups sas that ever algebraic group over a field is constructed or built up from an affine algebraic group an Abelian variet. A classical eample of an Abelian variet is a nonsingular cubic in P whose multiplication is given b the group law. We will mostl be interested in affine algebraic groups. A ke eample of an affine algebraic group is L(n, Z) : Spec ( Z[{ ij } n i,j, D ] ) where D det( ij ). iven an arbitrar scheme S we define L(n, S) b base change, i.e., L(n, S) : L(n, Z) Z S. The multiplication map L(n, Z) L(n, Z) L(n, Z) is given b the map on rings which sends ij n k ( ik )( kj ), which mimics matri multiplication. The identit map e : Z L(n, Z) is given b sending i j δ ij, which is inspired b the identit matri. Lastl, the inverse map i : L(n, Z) L(n, Z) is given b the map on rings which sends i j ( ) j+i M ji D where M ji is the determinant of the jiminor of the n n matri ( ij ). This map comes from the formula that gives the inverse of a nonsingular matri. Notation: We use L(n, S) L(n, Z) S, which is a scheme, we use L n (S) to denote the Svalued points of L(n, S), i.e., L n (S) : L(n, S)(S) Hom Z (S, L(n, S)). Take a look at the set L n (Z). To give an element of L n (Z) is equivalent to giving a homomorphism of rings Z[{ ij } n i,j, D ] Z. This is equivalent to assigning each of the n variables ij to an integer in such a wa that D is mapped to unit, else D has nowhere to go. Therefore to give an element of L n (Z) is equivalent to giving an integer valued n n matri A (a ij ) so that det(a) ±. Similarl, if R is a commutative ring, then to give an element of L n (R) : L n (Spec(R)) is equivalent to giving an n n matri with entries in R so that the determinant of that matri is a unit in R. Notation: We use m to denote the affine algebraic group L(, Z) Spec(Z[, ]). We sa that an Sgroup is linear if S is isomorphic to a closed subgroup of L(n, S) for some n. 8/8/5 We can define a group functor L(n, Z) : Schemes roups b sending T L n (T ). Let F be a rank n free Zmodule. Let T be a scheme denote b t the morphism T t Spec(Z). Then we can define another group functor from the categor of ) schemes to the categor of groups, denoted Aut Z (F ), b sending a scheme T to Aut Z (t F where t F is the ) pullback of the free Spec(Z)module F. If T Spec(R) for some ring R, then Aut Z (t F Aut R (F Z R) since t F is the coherent module F Z R. 4
5 Eercise: A choice of ordered basis for the free Zmodule F defines an isomorphism of functors Aut Z (F ) L(n, Z). Yoneda s Lemma Representable Functors Let C be the categor of Sschemes. For an X C we can define a contravariant functor X from the categor C to sets b X(R) Hom S (T, X). So X X defines a functor from C to the categor of contravariant functors C Sets, which we denote Func C. Yoneda s Lemma sas that this functor is full faithful. So if X, X objects in C then there is a bijection between Hom S (X, X ) Hom FuncC (X, X). Consequentl, if X X are schemes we can construct an isomorphism of functors X X, there is an isomorphism of Sschemes X X. We sa that a functor F : Schemes Sets is representable if F X for some scheme X. Observe that the above eercise sas that if F is a free Zmodule of rank n then Aut Z (F ) is representable b the scheme Spec(Z[{X ij } n i,j, D ]). Fi a scheme S a vector bundle E of rank n on S, i.e. a locall free sheaf of rank n on S. Now consider the group functor Aut S (E) which sends T Aut T (t E). Proposition.. The functor Aut S (E) is representable b an affine Sgroup Aut S (E). Sketch of proof. The result follows b the above eercise if E is trivial. Otherwise we find an open cover {U i } i I which trivializes E with trivializations ϕ i : E Ui L(n, U i ). Then Aut Ui (E Ui ) L(n, U i ) b choice of basis, i.e., b the trivializations ϕ i. Then the isomorphism ϕ j ϕ i : E Ui U j E Ui U j induce automorphisms of Aut Ui U j (E Ui U j ) which patch together to define a scheme Aut S (E). One can now show Aut S (E) is naturall represented b Aut S (E). Proposition.. If S Spec(A) is affine, E P where P is a projective Amodule of rank n, then Aut S (E) is a closed subgroup of L(N, S) for some N, i.e., Aut S (E) is linear. Ke Idea. Since P is projective we can write P K F for some free Amodule F. Then Aut S (E) is a subgroup of Aut S (F ) L(N, S) for some N. In fact ϕ Aut S (F ) is in Aut S (E) if onl if ϕ fies P acts identicall on K. The following eample is an eplicit calculation of Aut S (E) as a closed subgroup of L(, S) where E is a coherent line bundle over the affine scheme S Spec(C[, ]/( ( 3 ))). C[,] ( ( 3 )) Eample: Consider A look at the projective Amodule (, ). Note that A is regular, in fact A is the coordinate ring of an elliptic curve in P { } A. The ideal (, ) is projective since it is locall principal, i.e., locall free of rank. We can also check that (, ) is not globall generated b single element. But I is the quotient of A via the map π which sends ( ) ( ). Since (, ) is projective, there must be a map (, ) A which splits π. To find this map, we first find K A such that A /K (, ), i.e. we find first find the kernel of π. 5
6 ( ) ( ) We claim kernel of the map π : A (, ) is generated b ( ). It is clear that these two elements( lie in the kernel of π. To see these two elements generate the kernel of π we suppose that r r + r ( ( 3 )), i.e. there is a s C[, ] r ) such that r + r s + s( 3 ). This is equivalent to (r s) + (r s + s) in C[, ]. As, is a regular sequence in ( C[, ) ] there ( is ) a t ( C[, ] such ) that r s t r s r + s t. It follows that r t + s ( ), proving that ker π ) ( ) is generated b. ( ( ) ( To ) give a map ( (, ) ) A is equivalent to giving a map A A so that the two ( elements ) ( ) are mapped to. Consider f : A A which maps ( ) ( ) ( ( ) ( ) ( ) ( ) ). Then ( 3 ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) + ( ( 3 )). Thus ( ) ( there is an Alinear map f : (, ) A sending ). (( We now )) check that f splits π, i.e. π f is the identit map (( on )) (, ). Well, π(f()) π ( ) + () π(f()) π ( ) + () 3, hence f( splits) π. We ( now have) that A ker π (, ) where ker π is generated b the two elements ( ) of A (, ) is generated b the two elements ( ) ( ) of A. Denote b S Spec(C[, ]/( ( 3 ))) I (, ). We can now compute Aut S (Ĩ) as a closed subgroup of AutS(Ã ) Spec (A[,,,, (D) ]) where D. B the above Proposition this is equivalent to giving an automorphism of A which fies (, ) acts identicall on the kernel of π. To give an automorphism of ( A which)( acts identicall ) ( ) on the ( kernel of π)( is represented ) b the two matri equations ( ) ( ) in A[,,,, (D) ]. To give a map (, ) (, ) is equivalent ( ) to giving a map A A such that all elements of ker π are mapped to elements in ker (( π. So to give )( an automorphism )) of (( A which)( fies (, )) ) is represented b the equations π π. Eping this information, we see that the scheme Aut S (Ĩ) is cut out b the following equations in L(n, S); 6
7 () () (3) (4) (5) (6) + ( ) ( ) + ( ) ( ( ) + ) + ( ( ) + ) ( + ) + ( + ). 7
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