Transformation Groups, Vol. ?, N. ?, ??, pp. 1–?? \begin{matrix}c \\ ◯ \end{matrix}Birkhäuser Boston (??)

A CHARACTERIZATION OF LINEARLY REDUCTIVE GROUPS BY THEIR INVARIANTS

GREGOR KEMPER

IWR Universität Heidelberg Im Neuenheimer Feld 368 69 120 Heidelberg Germany Gregor.Kemper@iwr.uniheidelberg.de

Abstract. The theorem of Hochster and Roberts says that for every module V of a linearly reductive group G over a field K the invariant ring K[V]^{G} is Cohen-Macaulay. We prove the following converse: if G is a reductive group and K[V]^{G} is Cohen-Macaulay for every module V, then G is linearly reductive.

Introduction

Linearly reductive groups play a prominent role in invariant theory. Hilbert’s original proof of the finiteness theorem [3] works for linearly reductive groups. Much later, Hochster and Roberts [5] proved that if G is a linearly reductive group and V a G-module, then the invariant ring K[V]^{G} is Cohen-Macaulay, i.e., a free module over the subalgebra generated by a homogeneous system of parameters. The goal of this note is to prove a converse. Namely, we will show that if G is a reductive group and K[V]^{G} is Cohen-Macaulay for every G-module V, then G must be linearly reductive. This becomes false if the hypothesis that G be reductive is dropped (see Remark 8).

Our result ties in nicely with theorems of Nagata and Popov. In fact, Nagata extended Hilbert’s finiteness theorem to the case where G is geometrically reductive (see Nagata [10]). He also found an example of an algebraic group whose invariant ring is not finitely generated. This was extended by Popov [12], who obtained the following converse of the finiteness theorem: if K[X]^{G} is finitely generated for every affine scheme X on which G acts, then G is (geometrically) reductive. Together with these results, we obtain a characterization of linearly reductive groups by the structure of their invariant rings, which is stated as Corollary 9 at the end of the paper.

I would like to thank Hanspeter Kraft, Vladimir Popov, and Jean-Pierre Serre for very valuable comments on an earlier version of this note.

Received . Accepted .

Let us fix some notation. G will be a linear algebraic group (i.e., a reduced affine group scheme of finite type) over a field K. By an affine G-scheme X we mean an affine scheme X of finite type over K with an action ρ: G × X → X satisfying the usual diagrams (see Mumford et al. [9, Definition 0.3]). If X is an affine n-space and the action is given by a morphism G → GL_{n}(K), then X is called a G-module and will usually be denoted by V. Recall that G is called linearly reductive if all Gmodules are completely reducible. Given an affine G-scheme X, let K[G] and K[X] be the coordinate rings (i.e., the rings of global sections) of G and X, respectively. Then we have a map

ρ^{∗}: K[X] → K[G] ⊗_{K} K[X]

of K-algebras. An f ∈ K[X] is called an invariant if ρ^{∗}(f) = 1⊗f, and the invariant ring K[X]^{G} is the subalgebra of K[X] consisting of all invariants (see Mumford et al. [9, Definition 1.3]). Let K̄ be an algebraic closure of K. Since K[X]^{G} is the kernel of a K-linear map, we have

K̄[Spec K̄ × X]^{Spec K̄×G} = K̄ ⊗ K[X]^{G}.  (1)

In many situations this allows a reduction to the case that K is algebraically closed. We have a natural group structure on the set G(K) = Mor_{K}(Spec(K), G) of K-valued points of G. G(K) acts on K[X] by

σ(f) = (σ^{∗} ⊗ id)(ρ^{∗}(f)) for σ ∈ G(K), f ∈ K[X],

where σ^{∗}: K[G] → K is the map corresponding to σ. It is easily seen that if K is algebraically closed, then the invariant ring consists of exactly those elements of K[X] which are fixed by the action of G(K): K[X]^{G} = K[X]^{G(K)}. In the case that K is algebraically closed, we will sometimes write G for G(K) by abuse of notation.

1. First cohomology of algebraic groups

Let G be a linear algebraic group over a field K, which we assume to be algebraically closed in this section, and let V be a G-module. A morphism g: G → V gives rise to a map G(K) → V (K), σ ↦ g_{σ}, between the sets of K-valued points. We call g a 1-cocycle if g_{στ} = σ(g_{τ}) + g_{σ} for σ, τ ∈ G(K) (note that V (K) is a vector space over K with a G(K)-action). We write Z^{1}(G, V) for the additive group of all 1-cocycles. For v ∈ V (K), a 1-cocycle is given by the map G(K) → V (K), σ ↦ σ(v)−v. Denote the group of all 1cocycles of this type by B^{1}(G, V), and the factor group Z^{1}(G, V)/B^{1}(G, V) by H^{1}(G, V). After these preliminaries we can simplify our notation and write G for G(K), V for V (K), and so on.

Let

0 ⟶ V ⟶ Ṽ ⟶^{π} K ⟶ 0

be an exact sequence of G-modules, where K has trivial G-action. Choose v_{0} ∈ π^{−1}(1), then g_{σ} = (σ − 1)v_{0} ∈ V defines a 1-cocycle, and the sequence splits if and only if (g_{σ}) ∈ B^{1}(G, V). Conversely, every (g_{σ}) ∈ Z^{1}(G, V) defines an exact sequence as above: set Ṽ = V ⊕ K as a vector space and define the G-action on Ṽ by σ(v, α) = (σ(v) + α · g_{σ}, α) for σ ∈ G, v ∈ V, α ∈ K.

In Springer [15] we find five conditions on G which are equivalent to linear reductivity. The following proposition adds a sixth equivalent condition in terms of the first cohomology. As Hanspeter Kraft explained to me, this seems to be almost folklore, see for example Kraft and Kutzschenbauch [8], where it appears hidden in the proof of Proposition 1.

Proposition 1. With the above notation, G is linearly reductive if and only if H^{1}(G, V) = 0 for every G-module V.

Proof. If G is linearly reductive, then every exact sequence 0 → V → Ṽ → K → 0 of G-modules splits, hence by the above H^{1}(G, V) = 0.

Conversely, suppose that H^{1}(G, V) = 0 for every G-module V. Then every exact sequence as above splits, and it follows that for any epimorphism V → W of G-modules the induced map V^{G} → W^{G} is also surjective. Indeed, for 0 ≠ w ∈ W^{G} the G-module ⟨w⟩ spanned by w is isomorphic to K, thus the epimorphism V_{w} → ⟨w⟩ (with V_{w} the preimage of ⟨w⟩) splits, which provides a preimage of w in V^{G}. Now consider any exact sequence

0 → U → V → W → 0  (2)

of G-modules. This gives rise to an exact sequence

0 → Hom_{K}(W, U) → Hom_{K}(V, U) → Hom_{K}(U, U) → 0.

By the above, the induced map Hom_{G}(V, U) → Hom_{G}(U, U) is surjective. In particular, there exists a preimage of id_{U} which provides the desired splitting of (2). Thus every exact sequence of G-modules splits, so G is linearly reductive.

I thank Hanspeter Kraft for providing a substantial simplification of my original proof of Proposition 1. In order to use Proposition 6, we have to produce invariants which annihilate a given element from H^{1}(G, V). This is the purpose of the following proposition. For g ∈ H^{1}(G, V) and w ∈ W^{G} with V and W G-modules, we write w ⊗ g for the element in H^{1}(G, W ⊗ V) given by the cocycle (w ⊗ g)_{σ} := w ⊗ g_{σ}.

Proposition 2. Let G be a linear algebraic group over an algebraically closed field K and V a G-module. Then for any g ∈ H^{1}(G, V) there exists a G-module W and a nonzero w ∈ W^{G} such that w ⊗ g = 0 as an element of H^{1}(G, W ⊗ V).

More precisely, a G-module W contains 0 ≠ w ∈ W^{G} with w ⊗ g = 0 if and only if there exists a morphism W^{∗} → Ṽ of G-modules into the extension Ṽ given by g such that the composition W^{∗} → Ṽ → K is surjective.

Proof. Let W be a G-module and w ∈ W^{G}. By definition w ⊗ g is the image of g under the homomorphism H^{1}(G, V) → H^{1}(G, W ⊗V) induced by V → W ⊗V, v ↦ w ⊗v. By considering the construction of the cup product by projective resolutions (see Benson [1, p. 52]), we see that w ⊗ g = φ ∪ g, where φ ∈ H^{0}(G, W) = Hom_{KG}(K, W) is given by φ(1) = w. On the other hand, g ∪ φ is equal to the Yoneda composite of g ⊗ id_{W} and φ (Benson [1, Proposition 3.2.1]). More precisely, if

0 ⟶ V ⟶ Ṽ ⟶^{π} K ⟶ 0

is an extension given by g, then g ∪ φ defines an extension which is given by the upper row of

[figure: 0; V ⊗ W; X; K; 0; id; φ; 0; V ⊗ W; Ṽ ⊗ W; π⊗id_{W}; W; 0,]

where X is the pullback of Ṽ ⊗ W and K over W. The universal property of the pullback implies that the extension splits if and only if there exists a homomorphism K → Ṽ ⊗ W of G-modules such that the diagram

[figure: K; φ; Ṽ ⊗ W; π⊗id_{W}; W]

commutes. Using the natural isomorphism Ṽ ⊗ W ≅ Hom_{K}(W^{∗}, Ṽ) and tracing 1 ∈ K in the above diagram yields the equivalent condition that there exists a homomorphism W^{∗} → Ṽ such that

[figure: W^{∗}; φ^{∗}; Ṽ; π; K]

commutes. This yields the condition on W from the second statement. The condition is satisfied by W = Ṽ^{∗}.

Remark.

(a) A proof of the first statement of Proposition 2 can also be given by explicit computations involving cocycles.

(b) With the appropriate definition of H^{1}(G, V), Propositions 1 and 2 also hold without the hypothesis that K be algebraically closed.

2. Geometry of invariants of reductive groups

Recall that a linear algebraic group G is called reductive if its unipotent radical (the maximal connected unipotent normal subgroup) is trivial. By the theorem of Haboush [2], G is reductive if and only if G is geometrically reductive, i.e., for any G-module V and any one-dimensional submodule W ≤ V on which G acts trivially, there exists a homogeneous invariant f ∈ K[V]^{G} of positive degree such that the restriction f|_{W} is nonzero. In particular, any linearly reductive group is reductive. If the characteristic of K is zero, then the converse also holds (see Springer [15, V. Satz 1.1]). In positive characteristic, the notions of linear and geometric reductivity do differ: for example, any finite group is reductive, and it is linearly reductive if and only if the characteristic of K does not divide its order.

The following geometric property of reductive groups is well known.

Lemma 3. Let G be a reductive group and X an affine G-scheme. Then for any finitely generated ideal I ⊆ K[X]^{G} we have

\sqrt{IK[X]} ∩ K[X]^{G} = \sqrt{I},

where IK[X] denotes the ideal in K[X] generated by I.

Proof. For f ∈ \sqrt{I} we have f^{n} ∈ I for some n ≥ 1, hence f^{n} ∈ IK[X]. It follows that f ∈ \sqrt{IK[X]}∩K[X]^{G}, thus we have the inclusion \sqrt{I} ⊆ \sqrt{IK[X]}∩ K[X]^{G}. The reverse inclusion follows by Newstead [11, Lemma 3.4.2].

I thank Vladimir Popov for providing the reference to Newstead’s book. Note that in the above situation we have IK[X]∩K[X]^{G} = I if G is linearly reductive.

A sequence a_{1}, …, a_{m} of elements of a commutative ring R is said to be a partial system of parameters if (a_{1}, …, a_{m}) ≠ R and the ideal (a_{1}, …, a_{k}) has height k for k = 1, …, m.

Lemma 4. Suppose that G is a reductive group and X is an affine Gscheme. If a_{1}, …, a_{k} ∈ K[X]^{G} form a partial system of parameters in K[X] then they also form a partial system of parameters in K[X]^{G}.

Proof. Clearly the ideal (a_{1}, …, a_{m})K[X]^{G} in K[X]^{G} generated by the a_{i} is not equal to K[X]^{G}, since 1 is not contained in this ideal. So we have to prove that a_{k} lies outside every minimal associated prime ideal of (a_{1}, …, a_{k−1})K[X]^{G}, for k = 1, …, m, and this is also necessary (see, for example, Kemper [6, Lemma 1.5]).

Take any k ∈ {1, …, m} and let P_{1}, …, P_{r} ⊆ K[X] be the minimal associated prime ideals of (a_{1}, …, a_{k−1})K[X]. Then

\sqrt{(a_{1}, …, a_{k−1})K[X]} = ⋂_{i=1}^{r} P_{i},

and Lemma 3 yields

\sqrt{(a_{1}, …, a_{k−1})K[X]^{G}} = \sqrt{(a_{1}, …, a_{k−1})K[X]} ∩ K[X]^{G} =

= ⋂_{i=1}^{r}(P_{i} ∩ K[X]^{G}).

Observe that the P_{i}∩K[X]^{G} are prime ideals in K[X]^{G}, hence every minimal associated prime ideal of (a_{1}, …, a_{k−1})K[X]^{G} lies in {P_{i} ∩ K[X]^{G} | i = 1, …, r}. But by assumption a_{k} lies outside every P_{i} and in particular outside P_{i} ∩ K[X]^{G}. This completes the proof.

Remark 5. Lemma 4 becomes false if the assumption that G be reductive is dropped. In fact, let G = G_{a}(ℂ) be the additive group over ℂ and consider the linear action on variables x_{i}, y_{i} (i = 1, 2, 3) given by x_{i} ↦ x_{i}, y_{i} ↦ y_{i} + tx_{i} for t ∈ G = ℂ. Then an easy computation shows that the invariant ring ℂ[V]^{G} is generated by the six invariants

x_{i} (i = 1, 2, 3) and u_{ij} = x_{i}y_{j} − x_{j}y_{i} (1 ≤ i < j ≤ 3),

with the only relation

x_{1}u_{23} − x_{2}u_{13} + x_{3}u_{12} = 0.

In particular, ℂ[V]^{G} has dimension 5. Clearly x_{1}, x_{2}, x_{3} form a partial system of parameters in ℂ[V]. But ℂ[V]^{G}/(x_{1}, x_{2}, x_{3}) is isomorphic to the polynomial ring in the indeterminates u_{ij}, hence ht(x_{1}, x_{2}, x_{3}) = 5 − 3 = 2, which means that x_{1}, x_{2}, x_{3} do not form a partial system of parameters in ℂ[V]^{G}.

We also remark that the converse of Lemma 4 is false. As an example, consider the action of the multiplicative group G_{m}(ℂ) with weights 1, −1, −1, i.e., the representation given by

t ↦ (\begin{matrix}t & 0 & 0 \\ 0 & t^{−1} & 0 \\ 0 & 0 & t^{−1} \end{matrix}) for t ∈ ℂ \ \{0\}.

The invariant ring is generated by x_{1}x_{2} and x_{1}x_{3}, hence these two invariants form a partial system of parameters in the invariant ring, but not in the polynomial ring.

3. A characterization of linearly reductive groups

If G is a linear algebraic group and X an affine G-scheme, we write H^{1}(G, K[X]) for the usual first cohomology group H^{1}(G(K), K[X]) of the group G(K) with values in K[X]. The following proposition appeared in a more general form in Kemper [6, Theorem 1.4]. Since the proof in our situation is very much simpler, it is included here for the convenience of the reader.

Proposition 6. Let G be a reductive algebraic group over an algebraically closed field K, and let V be a G-module such that K[V]^{G} is Cohen-Macaulay. Suppose that a_{1}, a_{2}, a_{3} ∈ K[V]^{G} form a partial system of parameters in K[V]. Then the map

H^{1}(G, K[V]) → H^{1}(G, K[V]^{3})

given by the multiplication with (a_{1}, a_{2}, a_{3}) is injective.

Proof. By Lemma 4, a_{1}, a_{2}, a_{3} is a partial system of parameters in K[V]^{G}, so by assumption it must be a K[V]^{G}-regular sequence. Now take a cocycle (g_{σ}) ∈ Z^{1}(G, K[V]) whose class in H^{1}(G, K[V]) is mapped to zero under the multiplication with (a_{1}, a_{2}, a_{3}). This means that there are b_{i} ∈ K[V] such that

(σ − 1)b_{i} = a_{i} · g_{σ} for σ ∈ G(K), i = 1, 2, 3.

Set u_{ij} = a_{i}b_{j} − a_{j}b_{i} for 1 ≤ i < j ≤ 3, then u_{ij} is fixed by all σ ∈ G(K), hence u_{ij} ∈ K[V]^{G}. We have

u_{23}a_{1} − u_{13}a_{2} + u_{12}a_{3} = |\begin{matrix}a_{1} & a_{2} & a_{3} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}| = 0.

The regularity of a_{1}, a_{2}, a_{3} implies that u_{12} is contained in the ideal generated by a_{1} and a_{2} in K[V]^{G}, so

a_{1}b_{2} − a_{2}b_{1} = f_{1}a_{1} + f_{2}a_{2}

with f_{1}, f_{2} ∈ K[V]^{G}. Since a_{1} and a_{2} are coprime polynomials, it follows that a_{1} divides f_{2} + b_{1}, so f_{2} + b_{1} = a_{1} · h with h ∈ K[V]. Now

a_{1} · (σ − 1)h = (σ − 1)(a_{1}h) = (σ − 1)(f_{2} + b_{1}) = (σ − 1)b_{1} = a_{1}g_{σ}

for σ ∈ G(K), hence g_{σ} = (σ − 1)h. So the class of (g_{σ}) in H^{1}(G, K[V]) is indeed 0, and we have shown the required injectivity.

We are now ready to prove the main result, where we will use both cohomology with values in a G-module and cohomology with values in the polynomial ring.

Theorem 7. Let G be a reductive algebraic group over a field K such that the invariant ring K[V]^{G} is Cohen-Macaulay for any G-module V. Then G is linearly reductive.

Proof. Suppose that G is not linearly reductive. We will construct a Gmodule V such that K[V]^{G} is not Cohen-Macaulay. First observe that also Spec K̄ ×G is not linearly reductive, since the existence of a complement of a G-submodule is equivalent to the solvability of a system of linear equations over K. Moreover, the Cohen-Macaulay properties of K[V]^{G} and K̄⊗K[V]^{G} are equivalent. Hence we can assume without loss that K is algebraically closed.

By Proposition 1 there exists a G-module U such that H^{1}(G, U) ≠ 0. Choose a nonzero element g ∈ H^{1}(G, U). By Proposition 2 there exists a G-module W and a nonzero vector w ∈ W^{G} such that w ⊗ g = 0 in H^{1}(G, W ⊗ U). Choose V = U^{∗} ⊕ W^{∗} ⊕ W^{∗} ⊕ W^{∗}, so

R := K[V] = S(U ⊕ W ⊕ W ⊕ W)

(the symmetric algebra of U ⊕ W ⊕ W ⊕ W), and let a_{i} ∈ R be the copy of w in the i-th summand W in the direct sum (i = 1, 2, 3). U is a direct summand of R, hence we have an embedding of H^{1}(G, U) into H^{1}(G, R). We also have three embeddings of H^{1}(G, W ⊗ U) into H^{1}(G, R), given by the three copies of W^{∗} in V. The image of w ⊗ g under these embeddings is a_{i} · g (i = 1, 2, 3). But w ⊗ g = 0, so the image of g in H^{1}(G, R) is mapped to zero by the map

H^{1}(G, R) → H^{1}(G, R)^{3} ≅ H^{1}(G, R^{3})

induced by the multiplication with a_{1}, a_{2}, a_{3}. Hence this map is not injective, and Proposition 6 yields that R^{G} is not Cohen-Macaulay.

Remark 8. Theorem 7 becomes false if the hypothesis that G be reductive is dropped. As above, consider the additive group G = G_{a}(ℂ), and let V be any G-module. The action of G_{a}(ℂ) on V can be extended to an action of SL_{2}(ℂ) (see, for example, Kraft [7, III.3.9]), and we have an isomorphism

ℂ[V]^{G_{a}(ℂ)} ≅ ℂ[V ⊕ ℂ^{2}]^{SL_{2}(ℂ)},

where SL_{2}(ℂ) acts naturally on ℂ^{2}. This isomorphism was proved 1861 by Roberts (see Springer [14, p. 69, Remark 4] or Schur [13, Satz 1.14]). Now the Theorem of Hochster and Roberts [5] and the above isomorphism imply that ℂ[V]^{G} is Cohen-Macaulay, although G is not (linearly) reductive.

I thank David Wehlau and Vladimir Popov for drawing my attention to Robert’s isomorphism.

Combining Theorem 7 with theorems of Nagata, Hochster and Roberts, and Popov yields the following characterization of linearly reductive groups by their invariant rings.

Corollary 9. Let G be a linear algebraic group over a field K. Then G is linearly reductive if and only if both of the following two conditions are satisfied:

(a) For every affine G-scheme X the invariant ring K[X]^{G} is finitely generated, and

(b) for every G-module V the invariant ring K[V]^{G} is Cohen-Macaulay.

Proof. Suppose that G is linearly reductive. Then (a) holds by the theorem of Hilbert and Nagata (see Nagata [10]), and (b) follows by Hochster and Roberts [5].

Conversely, if (a) and (b) are satisfied, then also K̄ ⊗ K[X]^{G} is finitely generated, hence Spec K̄ × G is reductive by Popov [12], and the reductivity of G follows. Therefore Theorem 7 applies and it follows that G is linearly reductive.

For the special case of finite groups we obtain the following converse of the theorem of Hochster and Eagon [4].

Corollary 10. Let G be a finite group and K a field. Then the following two conditions are equivalent:

(a) K[V]^{G} is Cohen-Macaulay for every finitely generated KG-module V.

(b) The group order |G| is not a multiple of the characteristic of K.

This result was also obtained by Kemper [6].

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