Linearly reductive algebraic groups and Hilbert s theorem

Transcription

1 Linearly reductive algebraic groups and Hilbert s theorem Matei Ionita, Nilay Kumar May 15, 2014 Throughout this note, we fix an algebraically closed field k of characteristic zero. All algebraic varieties are taken to be affine and irreducible. 1 Basic Definitions We begin with the definition of an algebraic group: Definition 1.1. Let G be an (affine) variety equipped with a morphism µ : G G G such that the set of points of G with the operation given by µ is a group, and the operation of inversion i : G G is a morphism. In this case we say that (G, µ, i) is an algebraic group. We take morphisms between two algebraic groups to be maps that are simultaneously morphisms of varieties and group homomorphisms. In particular, we define a linear algebraic group to be a closed subgroup of some GL(n). The basic object of study is, of course, a representation: Definition 1.2. Let V be a k-vector space. We say that ρ : G GL(V ) is an (algebraic) representation of G if ρ is a morphism of algebraic groups. By abuse of notation we will denote the representation by V and ρ(g)v by g v. A subrepresentation of V is a vector subspace W V such that ρ(g)w = W for all g G, i.e. W is G-invariant as a subspace of V. We say that V is reducible if it contains a non-trivial subrepresentation and irreducible otherwise. Finally, we denote by V G = {v V g v = v for all g G} the invariant subspace of V under G. We take morphisms between two representations of G to be G-equivariant linear maps, i.e. linear maps commuting with ρ. It will be convenient for our purposes to consider instead the category dual to the category of G-representations above. Definition 1.3. Let G be an affine algebraic group over k, and A be the coordinate ring of G. An algebraic corepresentation of the group G is a vector space V over k together with a linear map µ V : V V k A, which satisfies the following two conditions: 1. Let ɛ : A k be the coidentity, i.e. the map that evaluates f A on the identity of G. The composition µ V 1 ɛ V V k A V is the identity in V. 1

2 2. Let µ A : A A k A be the comultiplication. Then the following diagram commutes: V µ V V k A µ V µv 1 A V k A 1 V µ A V k A k A Remark. There are several remarks to be made here. First let us make explicit the duality between corepresentations and representations. We regard the vector space in Definition 1.3 as the dual V of the vector space V in Definition 1.2, which is allowed because V = V. Then a corepresentation is a linear map µ V : V V k A. This can be extended to a ring homomorphism on the polynomial algebra of V as µ V : Sym V Sym V k A. Note that Sym V and A are finitely generated k-algebras, which are the affine coordinate rings of V and G respectively. Then µ V induces a morphism ρ of affine varieties: ρ : G V V. The duals of properties (1) and (2), stated below, make this morphism into a group action: 1. Let e G be the identity. Then the following composition is the identity in V : V (e,v) G V ρ V i.e. ρ(e, v) = v. 2. Let m : G G G be the group operation. Then the following diagram commutes: G G V (e,ρ) G V In other words, ρ(g 1 g 2, v) = ρ(g 1, ρ(g 2, v)). (m,1 V ) ρ G V V ρ The fact that µ V is linear shows that ρ is linear on the V factor. Moreover, using (1) and (2) we see that ρ(g 1, ρ(g, v)) = v. These facts allow us to think of ρ as a homomorphism from G to GL(V ), so we obtain a representation of G on V. This construction is clearly invertible, i.e. starting from a representation ρ and dualizing will give a ring homomorphism ρ : Sym V Sym V k A, where ρ is the pullback by ρ. Since ρ is linear on the V factor, ρ is linear, so in particular it takes V to V k A. The map µ V in the definition of a corepresentation is precisely the restriction of ρ to V. Definition 1.4. Let µ : V V A be a corepresentation of a group G. We say that a vector x V is G-invariant if µ(x) = x 1 and we denote the G-invariant subspace of all such vectors by V G. Moreover, a subspace U V is called a subrepresentation if µ(u) U A and V is called irreducible if it has no non-trivial subrepresentations. The following property demonstrates the utility of corepresentations. Proposition 1.5. Every corepresentation V of G is locally finite-dimensional, i.e. every x V is contained in a finite-dimensional corepresentation of G. Proof. See [2], Proposition

3 We illustrate the concepts just introduced for two simple examples: the multiplicative group G m and the additive group G a. Definition 1.6. Given a vector space V and an integer m Z, it s easy to see that the map V V k[t, t 1 ] v v t m is a corepresentation of G m. It is called the corepresentation of weight m, and we denote it by V (m). Proposition 1.7. Every corepresentation V of G m is a direct sum V = m Z V (m) Proof. See [2], Proposition 4.7. Proposition 1.8. Every corepresentation V of G a is given by µ(v) = n=0 f n (v) sn n! for some endomorphism f End(V ) which is locally nilpotent (that is, for every v, there exists some n such that f n (v) = 0). Proof. See [2], Proposition Linear Reductivity We now restrict ourselves to linearly reductive algebraic groups, whose representations and corepresentations, as we shall see, decompose quite simply. Definition 2.1. An algebraic group G is said to be linearly reductive if for every epimorphism φ : V W of G-corepresentations the induced map on G-invariants φ G : V G W G is surjective. There are a few equivalent definitions that are useful. Proposition 2.2. The following are equivalent: 1. G is linearly reductive. 2. For every epimorphism φ : V W of finite-dimensional representations the induced map on G-invariants φ G : V G W G is surjective. 3. If V is any finite-dimensional representation and v V is G-invariant modulo a proper subrepresentation U V, then the coset v + U contains a nontrivial G-invariant vector. Proof. See [2], Proposition As a first example, we note the following: Proposition 2.3. Every finite group G is linearly reductive. 3

4 Proof. Let V be a finite-dimensional corepresentation and v V be a vector invariant modulo a subrepresentation U V. Defining v = 1 g v, G we find that v is G-invariant and v v = 1 G g G (g v v), which is contained in U. This satisfies condition (3) above, and hence G is linearly reductive. g G Example 2.4. The additive group G a is not linearly reductive. Indeed, consider the two-dimensional representation given by ( ) 1 t t. 1 Then the restriction to the x-axis k[x, y] k[x, y]/(y) = k[x] is a surjective homomorphism of G a -representations. But k[x, y] Ga = k[y] and k[x] Ga = k[x] and hence the corresponding map on invariants is not surjective. Theorem 2.5. G is linearly reductive if and only if every finite-dimensional corepresentation V of G is completely reducible, i.e. V splits as a direct sum of irreducible corepresentations. Proof. It suffices to show that any finite-dimensional corepresentation V of G with a subrepresentation W decomposes as a direct sum of corepresentations V = W W. Consider first Hom k (V, W ) and Hom k (W, W ), as spaces of linear maps, as G-corepresentations. Linear reductivity of G implies that the surjective restriction map Hom k (V, W ) Hom k (W, W ) induces a surjective map Hom k (V, W ) G Hom k (W, W ) G. Note now that Hom k (V, W ) G is precisely those morphisms that are G-equivariant, i.e. morphisms of G-corepresentations. Consider now the identity morphism Id W Hom k (W, W ) G, which lifts via surjectivity to a φ Hom k (V, W ). Denoting W = ker φ, we obtain a short exact sequence of G-corepresentations 0 W φ V W 0. Treating the identity Id W as the inclusion ι : W V, we find that φ ι = Id W : W W, and hence we obtain a splitting V = W W. Consider a surjective map φ : V W of G-corepresentations. By hypothesis, the above sequence splits. W and W are each stable under G, therefore V G = W G W G, and hence the induced map V G W G is surjective. This characterization of linearly reductive groups shows their importance; one can hope to classify the irreducible (co)representations and express arbitrary (co)representations as direct sums of these. Moreover, the linearly reductive groups are the ones Hilbert s theorem holds for. For the rest of this section, we focus on discussing examples of linearly reductive groups, so that, when we prove Hilbert s theorem, we already know what kind of groups it applies to. As a first example, Proposition 1.8 shows that G m is linearly reductive. Furthermore, it is a basic fact in the representation theory of Lie groups that compact Lie groups are linearly reductive (where compactness is taken in the sense of the Euclidean topology). The proof of this is analytic, and we do not give it here. The idea is to generalize the proof of Proposition 2.3, by replacing the sum over 4

5 elements of g with an integral over the compact group G. The technical difficulty lies in defining an invariant measure on G, known as the Haar measure. We are interested in proving that certain noncompact linear algebraic groups are linearly reductive. An important family of groups for which this holds are the complex special linear groups SL(n, C). Our proof is based on Weyl s unitary trick, which is a way of using the representation theory of compact subgroups, known to be linearly reductive, to get information about noncompact groups, such as SL(n, C). We begin by proving the following lemma about linear algebraic groups. For the remainder of the section, we assume that k = C. Lemma 2.6. Let G be a linear algebraic group, and K a subgroup of G that is Zariski-dense in G and is compact in the Euclidean topology. Then G is linearly reductive. Proof. Let V W be a surjective morphism of G-corepresentations. Clearly V and W are also corepresentations of K. But K is compact, and we assumed that all compact groups are linearly reductive. This gives a surjection V K W K. We claim that, in fact, V K = V G and W K = W G. Note that this proves that V G W G is surjective, which is enough to conclude that G is linearly reductive. Clearly V G V K as K is a subgroup of G. To see that V K V G, we fix v V K and, using the equivalence of representations and corepresentations, we have a morphism of affine varieties φ : G {v} V φ is continuous, therefore φ 1 v is Zariski closed in G {v}. But, since v V K, K {v} φ 1 v, so K {v} φ 1 v. Since K is Zariski dense in G, this means G {v} φ 1 v, so v V G. This lemma gives an easy way to prove that a linear algebraic group G is linearly reductive: we simply need to exhibit a compact, Zariski dense subgroup K. For example, an easy proof that G m = GL(1) is linearly reductive uses the compact subgroup U(1) of GL(1). Since GL(1) is 1- dimensional, any closed subset is either finite or the whole of GL(1). U(1) is infinite, therefore U(1) = GL(1). In the next proposition, we apply Lemma 2.6 to show that SL(n) is linearly reductive by taking K = SU(n). The proof of density of SU(n) in SL(n) follows [1]. The following elementary lemma will be useful. Lemma 2.7. Let V be an R-vector space and let f : V R C C be a holomorphic function such that f(v R R) = 0. Then f is identically zero. Proof. Suppose dim V = 1. Then it is a basic complex-analytic fact that if f : C C holomorphic vanishes on the real line, f must be identically zero. If dim V = n, we have f holomorphic satisfying f(x 1,..., x n ) for all x i R. Viewing f as a function of the first variable by fixing x 2,..., x n, we find using the reasoning of the n = 1 case that f(z 1, x 2,..., x n ) = 0 for all z 1 C. Repeating this argument, we find that f(z 1,..., z n ) = 0 for all z i C. Proposition 2.8. SL(n, C) is linearly reductive. Proof. Using Lemma 2.6, it suffices to show that SU(n) is Zariski-dense in SL(n). Suppose not then there exists an f k[sl(n)] not identically zero but satisfying f(su(n)) = 0. Consider now the Lie algebra su(n), which is a real subspace of sl(n) with the property that su(n) R C = sl(n). Note now that the exponential map exp : sl(n) SL(n) is holomorphic and maps su(n) sl(n) to SU(n). Thus the composition f exp : sl(n) C is holomorphic and satisfies f(exp su(n)) = 0. By the previous lemma, f exp must be identically zero on sl(n), whence f must be identically zero, contradicting our original assumption. 5

6 Remark. The proof of Proposition 2.8 did not use any particular features of SL(n), apart from the fact that its Lie algebra sl(n) has a compact real form su(n). Therefore the same proof generalizes to other linear algebraic groups G, as long as we can find a compact real form for Lie(G). In particular, for G = GL(n), we have u(n) R C = gl(n), so we can take K = U(n), which proves that GL(n) is linearly reductive. Similarly, for the symplectic group SL(2n, C), let U(n, H) be the quaternionic unitary group, which is compact; we have u(n, H) R C = sp(2n, C), which proves that Sp(2n, C) is linearly reductive. 3 Hilbert s Theorem In this section we obtain an impressive payoff for the somewhat nontransparent definitions of corepresentations and linear reductivity. Under this formalism, the proof of Hilbert s theorem on the finite generation of rings of invariants is almost trivial. Before we study the theorem itself, we give some motivation for our interest in this result. Suppose we have an action of a linear algebraic group G on an affine variety X. We look at the quotient X/G, i.e. the set of orbits in X under G, and ask whether it is also an affine variety. In most cases it will not: if there exists an orbit which is not Zariski closed in X, then the corresponding point in X/G is not closed, so we need the formalism of schemes in order to talk about the quotient. However, in the case that X/G is an affine variety, the easiest way to describe it is through its coordinate ring. An obvious candidate for this is (A[X]) G, the ring of invariants under G. We would like, then, for this ring to be finitely generated over k, since the equivalence of categories between finitely generated k-algebras and affine varieties will give an affine variety structure on X/G. That being said, we turn to Hilbert s theorem. We follow the proof in [2], Theorem Theorem 3.1 (Hilbert). Let G be a linearly reductive algebraic group, which acts on a polynomial ring S, preserving the grading. Then the ring of invariant polynomials S G is finitely generated. Proof. The invariant ring inherits a grading from S: S G = e 0 S G S e It s useful to consider S G +, the span of the invariants of positive degree, and the ideal J that S G + generates in S. Since S is Noetherian, J is generated by finitely many polynomials f 1,..., f N S G +. This means that the following S-module homomorphism is surjective. φ : S S J N (h 1,..., h N ) h i f i We claim that f 1,..., f N generate S G. To see this, we pick a homogenous invariant h S G. We use induction on deg h to show that h k[f 1,..., f N ]. If deg h = 0, then this is clear. For deg h > 0, h S G +, so in paticular h J G. The map φ above is a surjective morphism of G corepresentations, so since G is linearly reductive the induced map i=1 S G S G J G 6

7 is also surjective. Therefore there exist h 1,..., h N SG such that h = N h if i i=1 The f i have positive degree, therefore deg h i < deg h for all i. By the inductive hypothesis, this means h i k[f 1,..., f N ], so h k[f 1,..., f N ]. We are now interested in generalizing Hilbert s theorem by replacing the polynomial ring S with an arbitrary finitely generated k-algebra R. The following lemma will be useful in the proof. Lemma 3.2. Let G be a linear algebraic group that acts on a finitely generated k-algebra R. Then there exists a set of generators r 1,..., r N for R such that the linear span of r 1,..., r N is a G- invariant vector subspace of R. Proof. Let f 1,..., f M be any set of generators. By Proposition 1.6, each f i is contained in a finite dimensional corepresentation V i R. Consider V = i V i; it is G-invariant because each of its factors is, and it is finite dimensional. Therefore we can take r 1,..., r N to be any basis of V. Apart from being useful in the following proof, this lemma has a useful geometric interpretation. Let X be the affine algebraic variety corresponding to R. The lemma gives a surjective morphism of corepresentations Sym(V ) R. This corresponds to an injective, G-equivariant morphism of affine varieties X V = A N In other words, every affine variety X can be equivariantly embedded in an affine space on which G acts linearly. Theorem 3.3. Let G be a linearly reductive algebraic group, which acts on a finitely generated k-algebra R. Then the invariant ring R G is finitely generated. Proof. As in the remark above, the choice of generators r 1,..., r N homomorphism: gives a surjective k-algebra S = k[x 1,..., x N ] φ R x i r i This map induces a corepresentation of G on S, by composition with the corepresentation µ of G on R: S φ R µ R A S A Note that R A S A is generally not a homomorphism, since R is a quotient of S, and not a subalgebra. However, the definition of a corepresentation only requires a linear map between vector spaces, which is clearly satisfied here. By Hilbert s theorem, S G is finitely generated; let f 1,..., f k be generators. Now, since G is linearly reductive, the map S G R G induced by φ is surjective, so R G is generated by φ(f 1 ),..., φ(f k ). 7

8 References [1] Darij Grinberg, Symmetric subspace of linear operators (answer), MathOverflow, URL: http: //mathoverflow.net/a/43474 (visited on ). [2] Shigeru Mukai, An introduction to invariants and moduli, Cambridge University Press,