Representations and Linear Actions


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1 Representations and Linear Actions Definition 0.1. Let G be an Sgroup. A representation of G is a morphism of Sgroups φ G GL(n, S) for some n. We say φ is faithful if it is a monomorphism (in the category of Sschemes). We must take the categorical definition of a monomorphism here because Sschemes commonly do not have enough points for the settheoretic definition to make sense. A morphism of φ X Y of Sschemes is called monic (equivalently, a monomorphism) if for all T S the induced map of sets φ(t ) X(T ) Y (T ) is injective. Hence, φ is faithful if and only if G(T ) is a subgroup of GL n (T ) for all T S. Example 1. Consider the case when S is affine, say, S = Spec k. If φ is a monomorphism of algebraic groups G H, then a result in the theory of algebraic groups implies that φ is a closed embedding. However, if the dimension of S is greater than 0, this need not be true for trivial reasons. For instance, if S = Spec Z, G = Spec Z {±1}, and H = G/{(2) { 1}}, where here we ve arbitrarily picked the point (2) { 1} G to delete. Certainly, H is a subgroup of G and H G is an immersion, but this is clearly not a closed immersion. Linear actions on affine space Suppose E is a locally free sheaf on S. The associated (geometric) vector bundle is V(E) = Spec(Sym E ). Then Aut S (E) acts on V(E) in a canonical way: for any T valued point t T S any g G(T ) defines an automorphism of O T modules t E g t F, which in turn induces an automorphism Spec(Sym t E) = V(E) S T by forming the pullback. We then have an action morphism Aut S (E) V(E) V(E). If E is free, then V(E) A n S, after choosing a basis for E and E. Hence, Aut(E) = GL(n, S), so we get an action of GL(n, S) on A n S. Denote this action by σ GL(n, S) An S An S ; it is dual to the homomorphism O S [T 1,..., T n ] O S [x ij, D 1 ][T 1,..., T n ] T i n j=1 x ij T j Let s simplify our discussion by assuming that S = Spec O S and G/S are affine. Suppose we have a representation φ G GL(n, S). Then GL(n, S) acts on A n S as defined above, so composing with φ defines an action of G on A n φ 1 G A n S GL(n, S) A n S σ G σ A n S We then have the maps O GL(n,S) O G σ # G O S[T 1,..., T n ] O G [T 1,..., T n ] x ij f ij T i n j=1 f ij T j 1
2 In this way, we can define what it means for a group to act linearly on affine space. Definition 0.2. We say G acts linearly on A n S if there are global sections f ij O G such that σ # G (T i) = n j=1 f ij T j for each i = 1,..., n. Example 2. The additive group G a (S) is defined as Spec S[T ] and the Hopf algebra structure is defined by µ # S[T ] S[T ] S[T ], e # S[T ] S, i # S[T ] S[T ] T (T 1) (1 T ) T 0 T T Note that this is not a reductive group. The group G a (S) acts nonlinearly on A S n as follows: The map O S [T 1,..., T n ] O S [T ][T 1,..., T n ] by fixing an i and sending T i to T i + T while acting like the identity on T j for j i defines and action of G a on A n. Since the image of T i is not a linear combination of the T j s, this action is not linear. On the level of points the action is given by the affine (nonlinear) transformation (t, t 1,..., t n ) (t 1,..., t i 1, t i + t, t i+1,..., t n ). However we know that G a (S) is a linear group. We will show that it is a closed subgroup of GL(2). The map G a (S) GL(2, S), is given by the corresponding map S[x 11, x 12, x 21, x 22, D 1 ] φ# S[T ] x 11 1, x 12 T, x 21 0, x 22 1 It is easy to check that this map is compatible with comultiplication, and G a (S) acts linearly on A 2 S map O S [T 1, T 2 ] O S [T ][T 1, T 2 ] defined by by the 2 T 1 φ # (x ij )T j = φ # (x 11 )T 1 + φ # (x 12 )T 2 = T 1 + T T 2, T 2 T 2 j=1 On the level of points this map corresponds to identifying G a with the subgroup of matrices of the form ( 1 a 0 1 ) and then the action is given by the usual matrix multiplication ( 1 a 0 1 ) ( x y ). This example provides motivation for the following proposition. Proposition 0.3. Giving a representation φ G GL(n, S) is equivalent to giving a linear action of G on A n S. Proof. A morphism by definition gives a linear action, so suppose G acts linearly on A n S, that is, for each i = 1,..., n, there are f ij O G such that T i maps to n j=1 f ij T j. This then defines a map of coordinate rings O S [x ij ] O G by mapping x ij to f ij. We then only need to check that D maps to a unit in O G, which is easy to do. 2
3 We claim that A n S has a group structure isomorphic to G a(s) n. The addition map of the group structure will be dual to O S [T 1,..., T n ] + O S [T 1,..., T n ] O S [T 1,..., T n ] T i (1 T i ) + (T i 1) while the multiplication map A 1 S An S A n S will be dual to O S [T 1,..., T n ] O S [T ][T 1,..., T n ] T i T T i This then restricts to a map G m (S) A n S A1 S, which is the action by scaling. If k is any field, then we can identify k with A 1 S (Spec k) and obtain a kvector space structure on An S (Spec k) = (A1 S (Spec k))n. The fact that G a acts linearly on A n S is equivalent to saying that the following diagram for addition commutes 1 G + G A n S An S A G n S σ A n S (σ, ) µ A n S An S and likewise for scalar multiplication. At the field level, we then get a linear action of G(k) on A n S (k) = kn, where A n S (k) has the canonical k = A1 S (k) vector space structure. To summarize what we have so far: giving a morphism G GL(n, S) is equivalent to giving a linear action A n S, which is equivalent to a linear action of G(k) on the vector space A n S (k) for Spec k S. Definition 0.4. A character of a group G/S is a map G G m (S) = GL(1, S). For instance, the determinant map det GL(n, S) G m (S) defines a character because on the ring level it s the map O S [T, T 1 ] O S [x ij, D 1 ] defined by T D. Example 3. Consider µ p = ω, where ω = e 2πi/p for some prime p (here, we are working over C). Then µ p acts on A 1 by χ a (v) = ω a v, with a {0,..., p 1}. This action can be extended to an action µ p (Z) = Spec Z[T ] T p 1 G m on A 1 Z[T ] Z = Spec Z[x] by considering the map Z[x] T p 1 [x], defined by sending x to T a x, and looking at the corresponding map on Specs. Now let s base change to F p and we obtain a map F p [x] Fp[T,x] by x T a x. The base change of µ T p 1 p is the nonreduced scheme G = Spec Fp[T ] T p 1 which has a single F p valued point. Then G acts on A 1 F p, and for a field k of characteristic p, G(k) = {e} acts trivially on A 1 (k). 3
4 Now let Z = Spec Fp[U] U p 1. We will look at the action of G(Z) on A1 (Z). First observe that we may identify the group G(Z) with the finite group Z/pZ. Explicitly given b Z/pZ we have a point of G(Z) corresponding to the homomorphism g F p[t ] defined by T U b. Similarly the set A 1 (Z) can be identified with the set of ordered ptuples (a p 1,..., a 0 ) in F p with (a p 1,..., a 0 ) corresponding to the Zvalued point associated to the map f fp[x] F p[u] U p 1, x a p 1U p a 0. T p 1 Fp[U] U p 1 The image of the point of A 1 (Z) corresponding to f # under muplitplication by the point of G(Z) corresponding to g is the point of A 1 (Z) corresponding to the composite morphism F p [x] σ # F p[t ][x] T p 1 g # f # F p[u] U p 1 x T a x (U b ) a (a p 1 U p a 0 ) = b p 1 U p b 0 for some b i F p, which are a cyclic permutation of the a i. Thus we see that G(Z) acts by cyclic permutation on A 1 (Z). When G acts on X there s no general nonsense way of defining the Gfixed locus X G. However, we can categorically define the inertia group of an action, which is {(g, x) g x = x} G X. Note that since (id, x) is in the inertia for all x X, the invariants are contained in the inertia. Definition 0.5. Given an Sgroup G acting on a scheme X by the morphism σ G X X, the inertia group I G (X) is defined by the Cartesian diagram I G (X) X G S X σ 1 X X S X That is, as a set, I G (X) = {((g, x), x ) (g x, x) = (x, x )} = {(g, x) g x = x} G X. Note that I G (X) is a subgroup of G X, and if X/S is separated, then is a closed embedding, making the pullback I G (X) G S X a closed embedding as well (hence, a monomorphism). We want to define a multiplication I G X I G X I G X. That is, for all T X we want a group structure on I G X(T ) = Hom(T, I G X). Giving such a map is equivalent to giving maps T x X and T (g,y) G such that (σ 1 X ) (g, y) = x (by universality) and such that the maps are compatible with the action and, making y = x T X and σ (g, x) = x. So a point is a pair g G(T ), x X(T ) such that g x = x. We then have I G X X I G X(T ) = {(g, x), (g, x ) x = x} = (I G X X I G X)(T ). Multiplication can then be defined on the Xgroup G X as (g, x) (g, x) = (gg, x). 4
5 Example 4. There is a natural conjugation action of G on I G X. On points, it s given by h (g, x) = (hgh 1, hx). Clearly, this is an action of G on G X and it is invariant for the subscheme I G X. This action will play an important role later in the discussion of quotient stacks. As an example, let S 3 act on A 2 C, giving the two dimensional irreducible representation of S 3. Now S 3 acts on A 3 by permutation of coordinates, so A 2 = {(x 1, x 2, x 3 ) x 1 + x 2 + x 3 = 0} is invariant under this action. Then I S3 A 2 S 3 A 2, the latter which can be viewed as six disjoint linear subspaces. That is, we view each element of S 3 as a plane and determine the fixed locus of the plane under the action. For instance, the fixed locus of the identity is the entire plane while that of (12) must satisfy (12)(x 1, x 2, x 1 x 2 ) = (x 2, x 1, x 1 x 2 ), so the fixed locus is the line x 1 = x 2. Similarly, we determine that the fixed locus of (13) is the line 2x 1 + x 2 = 0, that of (23) is x 1 + 2x 2 = 0, and that of the three cycles are just points. Notice that the fixed loci of the transpositions are in one orbit, the ones from the three cycles are in another, while the entire plane coming from the identity is in another. This is because, in general, if G is finite, then I G X = ψ I ψ, where the disjoint union ranges over all conjugacy classes of G and I ψ = {(g, x) g x = x, g ψ}. This gives a Ginvariant decomposition of the inertia group. Example 5. The group GL(2, C) has the twodimensional defining representation on V = C 2 given by the standard action ( a 11 a 12 a 21 a 22 ) ( x 1 x 2 ) = ( a 11x 1 + a 12 x 2 a 21 x 1 + a 22 x 2 ) The stabilizer of the origin is all of GL(2, C) and all other fibers are conjugate to the stabilizer of ( 1 ) since the 0 subgroup H = {( 1 b 0 λ ) λ G m, b G a } fixes this vector (and is an extension of G a by G m ). Thus, there are only two Gorbits of the action since the stabilizer of gx is always conjugate to the stabilizer of x. Now let s see what happens when we delete the origin, that is, let GL(2, C) act on P(V ) = (A 2 /{0})/C = P 1. This action is welldefined because it commutes with scalar multiplication on V and because V is a representation of G. Now all orbits are conjugate, so P 1 is a homogeneous space for G, meaning there is only one Gorbit. Since the stabilizer ( 1 ) only has to fix this vector up to scalar multiplication, the stabilizer is the Borel subgroup 0 B = {( λ 1 b ) λ 0 λ 1, λ 2 0} 2 The inertia is then the group scheme whose fibers are all possible Borel subgroups since all Borel subgroups are conjugate to each other. This shows that I GL2 P 1 P 1 is smooth. Similar calculations can be done for the action of GL n on P n 1. Now suppose GL 2 acts on 2 n matrices (that is, X = A 2n ) by left multiplication. Let U X be the open set where the matrix has maximal rank. Then the inertia group of any point in U is trivial, that is, the action is 5
6 free on this set. To find the stabilizers, it suffices to let U 0 = {0}, U 1 be the locus where the rank is equal to 1 (and hence dimension n+1), and U 2 = X/U 1. The stabilizer of U 0 is GL 2 and over any point in U 1 the stabilizer is the group H defined above. We note that GL m acts freely on the open set U m A m+n corresponding to the matrices of maximal rank and the quotient U m /GL m is the Grassmannian Gr(m, n). This will play an important role later. Before stating the next result, we need the fact that if a group G acts on X and f Y X is a Gequivariant map, then the stabilizer of f(x) is a subgroup of the stabilizer of x for any x X. Hence, if G acts freely on X, then it also acts freely on Y for an Gequivariant map Y X. Proposition 0.6. Let G be an Sgroup, X, Y Sschemes, and φ Y X a Gequivariant map. Then base change is a canonical monomorphism of Y groups I G Y Y X I G X, which is a closed embedding if φ is separated. Proof. Recall that φ being Gequivariant implies that the following diagram commutes. G Y σ Y 1 Y Y 1 G φ φ φ G X σ X 1 X X Hence, by universality, we have the commutative diagram θ G Y (G X) X X (Y Y ) σ Y 1 p 2 Y Y where, on points, θ(g, y) = ((g, φ(y)), (gy, y)). From the universality of fiber products (G X) X X (Y Y ) G S (Y X Y ). By Yoneda s lemma, the isomorphism of schemes is induced by the canonical isomorphism of sets given by ((g, x), (y 1, y 2 )) (g, g 1 y 1, y 2 ) with inverse (g, y 1, y 2 ) ((g, φ(y 1 )), (gy 1, y 2 )). Under this isomorphism, the map θ G S Y G S (Y X Y ) is the same as the map 1 G φ. By taking fiber products, we have the commutative diagram 6
7 I G Y Y X I G X Y φ G Y 1 G φ G S (Y X Y ) Y Y X φ φ G X X X where the bottom square (and hence the square above it) is Cartesian. The morphism I G Y Y I G X is then given by the pullback and hence must be monic since it s the base change of 1 G φ. This is a closed embedding if φ is, which is the case when φ is separated. 7
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