Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Capelli's identity wikipedia , lookup
Matrix multiplication wikipedia , lookup
Orthogonal matrix wikipedia , lookup
Matrix calculus wikipedia , lookup
Exterior algebra wikipedia , lookup
Perron–Frobenius theorem wikipedia , lookup
Four-vector wikipedia , lookup
ALGEBRAIC GROUPS 43 3.5. Separable morphisms. Recall that a morphism φ : X → Y of irreducible varieties is called dominant if its image is dense in Y . In the case of affine varieties, that is equivalent to φ∗ : k[Y ] → k[X] being injective. Hence, even for arbitrary irreducible X, Y , the map φ∗ : k(Y ) → k(X) induced by a dominant morphism is injective. So we can view k(X) as an extension field of k(Y ). The morphism φ is called separable if k(X) is a separable extension of k(Y ). Go back to the case that X, Y are affine. The composite of φ∗ and dX : k[X] → ΩX is a derivation k[Y ] → ΩX . So by the universal property of differentials, we get induced a k[Y ]-module map φ̄∗ : ΩY → ΩX such that dX ◦ φ∗ = φ̄∗ ◦ dY . Let x ∈ X, y = φ(x). The k[X]-module kx viewed as a k[Y ]-module via φ∗ is ky . We used this before to see that φ∗ induced the linear map dφx : Tx (X) = Derk (k[X], kx ) → Ty (Y ) = Derk (k[Y ], ky ), D $→ D ◦ φ∗ . Equivalently, we can view dφx as the map dφx : Homk[X] (ΩX , kx ) → Homk[Y ] (ΩY , ky ), θ $→ θ ◦ φ̄∗ . Applying adjointness of tensor and hom, we can even view dφx as a linear map Homk (ΩX (x), k) → Homk (ΩY (y), k). Theorem 3.15. Let φ : X → Y be a morphism of irreducible varieties. (i) Assume that x ∈ X and y = φ(x) ∈ Y are simple points and that dφx is surjective. Then, φ is a dominant separable morphism. (ii) Assume φ is a dominant separable morphism. Then the points x ∈ X with the property of (i) form a non-empty open subset of X. Proof. We may assume X and Y are affine and ΩX , ΩY are free k[X]- resp. k[Y ]-modules of rank d = dim X resp. e = dim Y . In particular, X and Y are smooth. The map φ̄∗ : ΩY → ΩX of k[Y ]-modules induces a homomorphism of free k[X]-modules ψ : k[X] ⊗k[Y ] ΩY → ΩX . We can represent ψ as a d × e matrix A with entries in k[X], fixing bases for ΩX , ΩY . Suppose that dφx is surjective. Then, A(x), which represents the dual map (dφx )∗ : ΩY (y) → Ωx (x), is injective hence a matrix of rank e. Hence the rank of A is at least e, hence equal to e since rank cannot be more than the number of columns. This shows that ψ is injective. Hence φ̄∗ is injective too. Since ΩX and ΩY are free modules, this implies that φ∗ : k[Y ] → k[X] must be injective, so φ is dominant. Moreover, injectivity of ψ implies injectivity of k(X) ⊗k[Y ] ΩY → k(X) ⊗k(X) ΩX . 44 ALGEBRAIC GROUPS This is the map α in the exact sequence β α k(X) ⊗k(Y ) Ωk(Y )/k −→ Ωk(X)/k −→ Ωk(X)/k(Y ) −→ 0. Hence k(X) is a separable extension of k(Y ) by the differential criterion for separability. The proof of (ii) is similar to the proof of (ii) in 3.12. ! Corollary 3.16. Let G be a connected algebraic group. (i) If X is a variety on which G acts transitively, then X is irreducible and smooth. In particular G is smooth. (ii) Let φ : X → Y be a G-equivariant morphism between two varieties on which G acts transitively. Then φ is separable if and only if dφx is surjective for some x ∈ X which is if and only if dφx is surjective for all x ∈ X. (iii) Let φ : G → H be a surjective homomorphism of algebraic groups. Then, φ is separable if and only if dφe is surjective. Proof. (i) Take x ∈ X. The orbit map g $→ gx is onto, so X is irreducible as G is. Since at least one point of x is simple, and the action is transitive, we see that all points of x are simple, so X is smooth. (ii) This follows at once from the theorem. (iii) Apply (ii) to X = G, Y = G" . ! 3.6. Lie algebras. Notation: if G is an algebraic group, let g denote the tangent space Te (G) to G at the identity. At the moment, g is just a vector space of dimension dim G (we will see in a while that it has additional structure as a Lie algebra). For example, the tangent spaces to GLn , SLn , Sp2n , SOn , . . . at the identity will be denoted gln , sln , sp2n , son , . . . . At the moment these are all just vector spaces! Example 3.17. (1) First, GLn . Then: gln = Derk (k[GLn ], ke ) is the n2 dimensional vector space on basis {ei,j | 1 ≤ i, j ≤ n} where ei,j is the point derivation f $→ ∂f (e). ∂Ti,j Here, Ti,j is the ij-coordinate function, and k[GLn ] is the localization of the polynomial ring k[Ti,j ] at determinant. I will always identify the vector space gln with the vector space of n × n matrices over k, so that ei,j is identified with the ij matrix unit. (2) SLn . Since SLn = V (det −1) inside of GLn , sln is a canonically embedded subspace of gln . Indeed, it will be all matrices ! X= ai,j ei,j i,j ALGEBRAIC GROUPS 45 such that X(det −1) = 0. But (calculate!) ! ! ! ( ai,j ei,j )( sgn(w)T1,w1 . . . Tn,wn − 1) = ai,i . i,j w∈Sn i So sln is all matrices in gln of trace zero. (3) Sp2n . Recall that Sp2n = {x ∈ GLn | xt Jx = J} where " # 0 In J= −In 0 as a closed subgroup of GL2n . Let T be the 2n × 2n matrix with ij entry Ti,j , the ij-coordinate function on GL2n . Then, Sp2n = V (T t KT −J) (4n2 polynomial equations written as one matrix equation!). Then sp2n is the X ∈ gl2n such that X(T t JT − J) = 0. You calculate: this is exactly the condition that X t J + JX = 0. Now you can calculate dim Sp2n by linear algebra... (4) Similarly, son (in characteristic )= 2) is all X ∈ gln satisfying X T + X = 0. (5) Also recall that Tn is all upper triangular invertible matrices. The tangent space tn will be all upper triangular matrices embedded into gln . Similarly you can compute the tangent spaces of Dn , Un , . . . . Okay, now let’s introduce on g the structure of a Lie algebra. It is convenient to go via an intermediate: L(G). This is defined to be the vector space L(G) = {D ∈ Derk (k[G], k[G]) | D(λx f ) = λx D(f ) for all f ∈ k[G], x ∈ G. (Recall: (λx f )(g) = f (x−1 g)). We call L(G) the left invariant derivations because they are the derivations commuting with the left regular action of G on k[G]. Now, Derk (k[G], k[G]) is a Lie algebra with operation being the commutator: [D, D" ] = D ◦ D" − D" ◦ D. Obviously, if D and D" are left invariant derivations, so is [D, D" ]. Therefore L(G) is a Lie subalgebra of Derk (k[G], k[G]). Lemma 3.18. Let G be a connected algebraic group, g = Te (G). The maps L(G) → g, D $→ eve ◦ D and g → L(G), X $→ X̃, where X̃ : k[G] → k[G] is the derivation with (X̃f )(g) = X(λg−1 f ) for all g ∈ G, are mutually inverse isomorphisms. 46 ALGEBRAIC GROUPS Proof. We’d better first make sure the maps make sense, i.e. that eve ◦ D is a point derivation and that X̃ is a left invariant derivation. Only the last thing is tricky: (X̃(λg f ))(h) = X(λh−1 λg f ) = X(λh−1 g f ) = (X̃f )(g −1 h) = (λg (X̃f ))(h). Now let’s compute eve ◦ X̃: (eve ◦ X̃)(f ) = (X̃f )(e) = X(f ) for all f ∈ k[G], hence eve ◦ X̃ = X. ! Finally, let’s compute ev e ◦ D for a left invariant derivation D: ! (ev e ◦ D(f ))(g) = (eve ◦D)(λg −1 f ) = (D(λg −1 f ))(e) = (λg −1 (Df ))(e) = (Df )(g). ! Hence ev e ◦ D = D. ! Because of the lemma, we get on g an induced structure as a Lie algebra, induced by the Lie algebra structure on L(G): [X, Y ] := eve ◦ [X̃, Ỹ ]. Example 3.19. Let’s compute the Lie algebra structure on gln . We need to work out ẽi,j . Well, for g ∈ G, ∂ ! ( Tp,r (g)Tr,q )(e) = δq,j Tp,i (g). (ẽi,j Tp,q )(g) = ei,j (λg−1 Tp,q ) = ∂Ti,j r Therefore ẽi,j Tp,q = δj,q Tp,i . Now you can compute the commutator [ẽi,j , ẽk,l ]: it acts on each Tp,q in the same way as δj,k ẽi,l − δi,l ẽk,j . Hence we’ve worked out [ei,j , ek,l ] = δj,k ei,l − δi,l ek,j . The right hand side is just the commutator ei,j ek,l − ek,l ei,j of the matrix units. So in general: [X, Y ] = XY − Y X for X, Y ∈ gln – the usual commutator as matrices! Lemma 3.20. Let H be a closed subgroup of G and set I = I(H). Then, h = {X ∈ g | XI = 0} = {X ∈ g | X̃I ⊆ I}. In particular, h is a Lie subalgebra of g. Proof. That h = {X ∈ g | XI = 0} is just the definition of tangent space of a closed subvariety... If XI = 0, consider X̃f for f ∈ I. Evaluating at h ∈ H, (X̃f )(h) = X(λh−1 f ) which is zero as λh−1 f is still in I (as H is a subgroup!). So X̃f vanishes on H, i.e. is containing in I. This shows X̃I ⊆ I. The opposite, namely that if X̃I ⊆ I then XI = {0} is easy. ALGEBRAIC GROUPS 47 Finally, observe that if D, D" ∈ L(G) satisfy DI ⊆ I, D" I ⊆ I, then the commutator [D, D" ] also does. This implies that h is a subalgebra of g. ! Now we get the Lie algebra structure in all the above examples: sln , son , . . . , since they are all just Lie subalgebras of gln , where the Lie bracket is just the commutator as matrices. The next lemma I’m going to leave to you to supply the proof (hint: its easier to rephase things in terms of L(G) and L(H))... Lemma 3.21. Let φ : G → H be a morphism of connected algebraic groups. Then dφe : g → h is a Lie algebra homomorphism. 3.7. Some differential calculations. I now want to compute differentials of various natural morphisms. In all cases, since an arbitrary algebraic group can be embedded as a closed subgroup of some GLn , it is enough to make the computation in the case of GLn , when we can be very explicit. Example 3.22. (1) Let µ : G × G → G be multiplication. Then, dµ(e,e) : g ⊕ g → g is addition. Proof: Suppose G = GLn . Recall g⊕g ∼ = Derk (k[G]⊗k[G], k(e,e) ), the isomorphism mapping (X, Y ) to the map (f ⊗ g $→ X(f )g(e) + f (e)Y (g). By definition, ! (dµ(e,e) (ei,j , ek,l ))(Tr,s ) = (ei,j , ek,l )( Tr,t ⊗ Tt,s ) = δi,r δj,s + δk,s δr,k t = (ei,j + ek,l )(Tr,s ). Hence dµ(e,e) (ei,j , ek,l ) = ei,j + ek,l , i.e. dµ(e,e) is addition. (2) Let i : G → G be the inverse map. Then, die : g → g is the map X $→ −X. Proof: Consider the composite G → G × G → G, g $→ (g, i(g)) $→ gi(g) = e. The composite is a constant function, so its differential is zero. But the differential of a composite is the composite of the differentials, so appying (1), 0 = d ide +die . The differential of the identity map is the identity map, so we are done. (3) Fix x ∈ G. Let Int x : G → G be the automorphism of algebraic groups g $→ xgx−1 . The differential d(Int x)e : g → g is a Lie algebra automorphism. It is usually denoted Ad x : g → g. For example, suppose G = GLn . Then, for a matrix X ∈ gln , (Ad x)(X) = xXx−1 , i.e Ad x is just the Lie algebra automorphism given by conjugation by x. Hence, for H and closed subgroup of G and x ∈ H, Ad x : h → h is just conjugation by x too – it leaves the subspace h of gln invariant. Proof: Let us compute (Int x)∗ Ti,j : ((Int x)∗ Ti,j )(g) = Ti,j (xgx−1 ) = (µ∗ (µ∗ Ti,j ))(x, g, x−1 ) ! = Ti,k (x)Tk,l (g)Tl,j (x−1 ). k,l 48 ALGEBRAIC GROUPS Hence: (Int x)∗ Ti,j = ! xi,k Tk,l (x−1 )l,j . k,l The ij-entry of (Ad x)(X) is ! (Ad x)(X)(Ti,j ) = xi,k X(Tk,l )(x−1 )i,j k,l which is the ij-entry of xXx−1 . (4) Here is a consequence of (3). For each x ∈ G, Ad x : g → g is an invertible linear map (even a Lie algebra automorphism) So you can think of Ad as a group homomorphism from G to GL(g) (or even to Aut(g) which is a closed subgroup of GL(g)). I claim that Ad : G → GL(g) is a morphism of algebraic groups. Proof: Embed G as a closed subgroup of some GLn . Then by (3), Ad x is conjugation by x, which is clearly a morphism of varieties – since it is given by matrix multiplication and inversion which are polynomial operations. (5) The image of Ad : G → GL(g) is a closed connected subgroup of Aut(g), denoted Ad G. It is interesting to consider the kernel of Ad, a closed normal subgroup of G. Obviously, every element of Z(G) belongs to ker Ad, i.e. Z(G) ⊆ ker Ad. I warn you that equality need not hold here. However it usually does, for example it always does in characteristic 0. For example, if G = GLn , ker Ad = Z(G) (check directly: ker Ad is the invertible matrices which commute with all other matrices), the scalar matrices. In this case Ad G is the group known as P GLn . As an abstract group, P GLn ∼ = GLn /{scalars}. Similarly, if G = SLn , ker Ad is the scalar matrices of determinant one, i.e. the matrices ω 0 0 0 ... 0 0 0 ω where ω runs over all nth roots of unity. The group Ad SLn is known as P SLn . As an abstract group P SLn ∼ = SLn /{scalars}. Warning. In characteristic p|n, Z(SLn ) is trivial. So P SLn ∼ = SLn as an abstract group, the isomorphism being the map g $→ Ad g. However, SLn is not isomorphic to P SLn as an algebraic group: the problem is that Ad is a bijective morphism that is NOT an isomorphism of varieties, i.e. its inverse is not a morphism of varieties. (6) Because of (4), we have a morphism Ad : G → GL(g). So we can consider its differential again, d Ad : g → gl(g). ALGEBRAIC GROUPS 49 The map d Ad is usually denoted ad. I claim that ad X ∈ gl(g) is the map Y $→ [X, Y ] (in particular, ad X is a derivation of g, so ad is a Lie algebra homomorphism g → Der(g)). In other words, The differential of Ad is ad. The proof is so nasty I’m not going to type it in. . . Exercise 3.23. (8) Here is an example to show that ker Ad may be larger than Z(G) if chark = p > 0. Let G be the two dimensional closed subgroup of GL3 consisting of all matrices of the form a 0 0 0 ap b 0 0 1 where a )= 0 and b are arbitrary elements of k. Describe the Lie algebra g explicitly as a subspace of gl3 . Now compute Z(G) and ker Ad and show that they are not equal. (9) Consider the morphism Ad : SL2 → P SL2 . Both SL2 and P SL2 are three dimensional algebraic groups, so their Lie algebras are three dimensional too. Consider the differential ad : sl2 → psl2 . Using Example (8) above, compute the kernel of ad, an ideal in the Lie algebra sl2 . Deduce by 3.16 that the morphism Ad is separable if and only if chark )= 2. (So in characteristic 2 – when Ad is a bijective morphism – it is inseparable so it cannot be an isomorphism). (10) Let chark = p > 0. Take X ∈ g = Derk (k[G], ke ). Recall g is isomorphic to L(G), the left invariant derivations of k[G], via the map X $→ X̃. Show that (X̃)p is a left invariant derivation of k[G]. Hence, there is a unique element X [p] ∈ g with * [p] = (X) + p. X This gives an extra operation on g in characteristic p, the map X $→ X [p] , which makes g into what is known as a restricted Lie algebra. Describe this map explicitly in the one dimensional cases G = Ga and G = Gm .