Grothendieck groups of invariant rings: linear actions of finite groups

Kenneth A. Brown and Martin Lorenz

May 9, 1995

Abstract

Let G be a finite subgroup of GL_{n}(k), where k is a field of characteristic not dividing |G|. Then G acts linearly on the vector space ∑_{i=1}^{n} kX_{i} of homogeneous polynomials of degree 1 of the polynomial ring k[X_{1}, …, X_{n}], and hence by algebra automorphisms on k[X_{1}, …, X_{n}]. The Grothendieck group G_{0}(R) of the fixed ring R of this action is known to be finitely generated of torsion free rank 1. We give an explicit formula for G_{0}(R) and use it to calculate this group for various classes of examples.

Introduction

Let G be a finite group and let k be a field with |G|^{−1} ∈ k. Let V be a finite dimensional vector space over k, with basis {X_{1}, …, X_{n}}, and suppose that G acts as automorphisms of V. This action extends to give an action of G as a group of k–algebra automorphisms of the symmetric algebra S = S(V) = k[X_{1}, …, X_{n}]. Denote the fixed ring S^{G} under this action by R. The purpose of this article is to describe the Grothendieck group G_{0}(R) of finitely generated R–modules.

We shall present G_{0}(R) as a factor group of the Grothendieck ring G_{0}(kG) of finitely generated modules over the group algebra kG. (Recall that the ring structure of G_{0}(kG) is afforded by ⊗_{k}.) To do so we need the following notation. For each subgroup H of G let V (H) be the subspace of V spanned by {v − v^{h} : v ∈ V, h ∈ H}. This is a kN_{G}(H)–module, and hence so is ∧^{i}V (H), where ∧^{i} denotes the i–th exterior power. Thus we may define

α_{H} := ∑_{i≥0}(−1)^{i}[∧^{i}V (H)],

an element of G_{0}(kN_{G}(H)). Further, define a subgroup of G_{0}(kN_{G}(H)),

G^{∗}(H) := ⟨[X] : X an irreducible kN_{G}(H)–module with X^{H} = 0⟩,

where X^{H} denotes the space of H–invariants in X. For a subgroup N of G let Ind_{kN}^{kG} denote the group homomorphism from G_{0}(kN) to G_{0}(kG) afforded by the induction functor (.) ⊗_{kN} kG. Define a subgroup U of G_{0}(kG) by

U := ∑_{1≠H⊆G} Ind_{kN_{G}(H)}^{kG}(α_{H}G^{∗}(H)).

Our main result can now be stated as Theorem G_{0}(R) ≅ G_{0}(kG)/U.

The strategy of the proof is as follows. Let T = S ∗ G, the skew group ring corresponding to the given action, let t = ∑_{g∈G} g ∈ T and I = T tT. The fixed point functor (.)^{G} yields an exact

sequence ([Br-L], Corollary 1.3(i))

G_{0}(T/I) ⟶^{Infl_{T /I}^{T}} G_{0}(T) ⟶^{(.)^{G}} G_{0}(R) → 0  (1)

where Infl_{T/I}^{T} is the inflation map (viewing T/I–modules as T –modules). By Quillen’s induction theorem ([Q], Theorem 7), there is an isomorphism

Ind_{kG}^{T} : G_{0}(kG) →^{≅} G_{0}(T).

Thus G_{0}(R) is isomorphic to G_{0}(kG) modulo the image of the map (Ind_{kG}^{T})^{−1} ∘ Infl_{T/I}^{T}. The issue is to describe the image.

The layout of the paper is as follows. Some preliminary results and proofs are given in §1. These show, first, that the subgroup U of G_{0}(kG) sought for the main theorem is generated by subgroups K_{H}, for 1 ≠ H ≤ G, (Prop. 1.2); and then that, by means of an inductive argument, we may concentrate on those subgoups H of G which are normal (Prop. 1.6). The contribution of the normal subgroups of G is analysed in §2, and the proof of the theorem is completed in §3. (We also point out in 3.4 that it is not in general necessary to consider all the non-identity subgroups H of G in calculating K.)

Partial results on the structure of G_{0}(R) have been obtained previously by various authors ([A-R2], [H-S], [H-M-W] and [M] for example). Most of this work has used the formal power series ring Ŝ = k[[X_{1}, …, X_{n}]] rather than the polynomial algebra S(V), although, with minor amendments, methods and results can be switched as desired between these two settings. We explain briefly in 4.5 that our results and proofs deal with this case also, and indicate in 4.2 and 4.7(1) how some special cases obtained by the above authors can be deduced from the main theorem. In particular, we point out that the main result of [H-M-W] is the case of the above theorem for G abelian. Also included in §4 are various worked examples ( 4.4, 4.6, 4.7(2) and 4.9), and a discussion (4.4) of the question: When is G_{0}(R) ≅ ℤ? As we show by examples, the classical sufficient condition stemming from the Sheppard-Todd-Chevalley theorem is not necessary.

In 4.8 we observe that the Sheppard-Todd-Chevalley theorem permits one first to factor out the subgroup of pseudoreflections from G before applying the Main Theorem; in particular this leads to improved bounds for the exponent of the torsion subgroup of G_{0}(R).

Notations and Conventions

We shall retain throughout the paper the notation defined in the introduction. Note in particular that exponential notation (s ↦ s^{g}) is used to denote the action of g ∈ G on s ∈ S (and on v ∈ V). Thus the skew group ring T = S ∗ G is the free left S-module with the elements of G as a basis and with multiplication given by the rule sg = gs^{g}(s ∈ S, g ∈ G). For a subset of X of G, write t_{X} = ∑_{x∈X} x ∈ T; when X = G we usually omit the suffix. The ideal of a ring A generated by a collection of subsets B, … and elements b, … is denoted by (B, …, b, …); the corresponding notation for subgroups of a group will use the brackets ⟨…⟩.

Acknowledgements

The second author’s research was supported in part by a grant from the NSF. He would also like to thank Temple University for granting him a research leave during the Fall Semester 1992, when much of the work for this article was done. Thanks are also due to the Universities of Delhi and Glasgow for visiting appointments during this period. It is a pleasure to thank Professor V.N. Dixit for arranging the former; the latter visit was supported by grants from the Edinburgh Mathematical Society and the London Mathematical Society.

1 Some Reductions

1.1 For each subgroup H of G, define an ideal of S,

I(H) := (s − s^{h} : s ∈ S, h ∈ H) = (v − v^{h} : v ∈ V, h ∈ H).

The following easily checked facts will be used freely:

• H ⊆ D ⇒ I(H) ⊆ I(D) ;

• I(H) + I(D) = I(⟨H, D⟩) ;

• I(H)^{g} = I(H^{g}), for all g ∈ G .

Recall that I := T tT = StS. In view of the exact sequence (1) of the Introduction, this ideal and the ramification locus I ∩ S are fundamental to our analysis. They are closely approximated by

Lemma ([Br-L], Lemma 2.2) ∏_{1≠H⊆G} I(H) ⊆ I ∩ S ⊆ ⋂_{1≠H⊆G} I(H).

We also define

I^{0}(H) := ⋂_{g∈G} I(H)^{g} and S^{0}(H) := S/I^{0}(H) ;
J^{0}(H) := I^{0}(H)T  and T^{0}(H) := T/J^{0}(H) ≅ S^{0}(H) ∗ G;
J(H) := J^{0}(H) + I  and T (H) := T/J(H) ≅ T^{0}(H)/(t) .

The above lemma implies that ∏_{1≠H⊆G} J(H) ⊆ I. Hence

G_{0}(T/I) = ∑_{1≠H⊆G} Infl_{T (H)}^{T} (G_{0}(T (H))) .

1.2 Note that G_{0}(T) is a (left) G_{0}(kG)–module. Thus, if X_{kG} and W_{T} are finitely generated modules then X ⊗_{k} W is a finitely generated right T –module under the action (x ⊗ w)sg =

xg ⊗ wsg for x ∈ X, w ∈ W, s ∈ S and g ∈ G. Exactness of ⊗_{k} ensures that this defines a

module action on G_{0}(T). Note also that kG is a T –kG–bimodule, with T acting via the ring epimorphism T → T/V T ≅ kG, and T is a kG–free

kG–T –bimodule via the embedding of rings kG ↪ T. At the level of Grothendieck groups these maps yield inverse isomorphisms:

Lemma (i) (Quillen [Q], [H], Theorem 1.2)

Ind_{kG}^{T} : G_{0}(kG) → G_{0}(T) : [M] ↦ [M ⊗_{kG} T]

is an isomorphism of G_{0}(kG)–modules.

(ii) ([Br-L2], Lemma 2.2) The inverse isomorphism to Ind_{kG}^{T} is

π : G_{0}(T) → G_{0}(kG) : [W] ↦ ∑_{i≥0}(−1)^{i}[Tor_{i}^{T} (W, kG)].

(iii) ([Br-L2], Lemma 2.3) The endomorphism

π ∘ Infl_{T/V T}^{T} : G_{0}(kG) → G_{0}(kG)

is given by multiplication by α_{V} := ∑_{i≥0}(−1)^{i}[∧^{i}V] ∈ G_{0}(kG).

For a subgroup H of G the isomorphism from G_{0}(S ∗ H) to G_{0}(kH) given by (ii) of the above lemma will be denoted by π_{H}.

We now write

I := Infl_{T/I}^{T}(G_{0}(T/I)) ⊆ G_{0}(T)  and K := π(I) ⊆ G_{0}(kG);
I_{H} := Infl_{T (H)}^{T}(G_{0}(T (H))) ⊆ G_{0}(T) and K_{H} := π(I_{H}) ⊆ G_{0}(kG).

With this notation, the exact sequence (1) of the introduction coupled with the isomorphism π of Lemma 1.2 (ii) yields (i) of the following result. The first equality in (ii) has already been noted in 1.1 and the second equality follows by applying π.

Proposition (i) G_{0}(R) ≅ G_{0}(T)/I ≅ G_{0}(kG)/K.

(ii) I = ∑_{1≠H⊆G} I_{H} and K = ∑_{1≠H⊆G} K_{H}.

In view of the above proposition, our goal will be to describe I_{H} and K_{H} for a fixed subgroup H ⊆ G.

1.3 For a fixed subgroup H ⊆ G, put

S^{∗}(H) := ∏_{g∈T_{H}}S/I(H^{g}),

where T_{H} is a set of right coset representatives of N_{G}(H) in G with 1 ∈ T_{H}.

Lemma The canomical embedding φ : S^{0}(H) ↪ S^{∗}(H) makes S^{∗}(H) into a finitely generated S^{0}(H)–module. The factor coker φ is annihilated by Y^{h−1}, where h = [G : N_{G}(H)] and

Y = ⋂_{1≠x∈T_{H}}I^{0}(⟨H, H^{x}⟩).

Furthermore, there is a canonical action of G on S^{∗}(H) which makes φ a G-equivariant map.

Proof The idempotents corresponding to the components S/I(H^{g}) of S^{∗}(H) generate S^{∗}(H) as S^{0}(H)–module. For x ∈ G and σ = (s_{g} + I(H^{g}))_{g∈T_{H}} ∈ S^{∗}(H) put

σ^{x} = (s_{g}^{x} + I (H^{gx}))_{g∈T_{H}} ∈ S^{∗}(H) .

This defines an action of G on S^{∗}(H) so that φ is G-equivariant.

Note that

Y = ⋂_{g∈G} ( ⋂_{1≠x∈T_{H}} I(⟨H, H^{x}⟩))^{g} = ⋂ ; \begin{matrix}x, y ∈ T_{H} \\ x ≠ y \end{matrix} (I(H^{x}) + I(H^{y})) .

Thus the assertion about coker φ is a direct consequence of the Chinese Remainder Theorem below.

Chinese Remainder Theorem. Suppose P_{1}, …, P_{h} are ideals of the ring A. Put Q =

⋂_{i≠j}(P_{i} + P_{j}). Then the cokernel of the canonical homomorphism φ : A → B := ∏_{i=1}^{h}A/P_{i} is

annihilated by Q^{h−1}.

Proof Letting e_{j} ∈ B denote the idempotent corresponding to the component A/P_{j} of B, it suffices to show that e_{j}Q^{h−1} ∈ φA holds for all j. By symmetry, we may concentrate on j = 1. Let q_{1}, q_{2}, …, q_{h−1} ∈ Q be given and write

q_{l} = p_{1,l} + p_{l} (l = 1, …, h − 1),

where p_{1,l} ∈ P_{1}, p_{l} ∈ P_{l+1} . Then

q_{1}q_{2} ⋯ q_{h−1} = a + p ,

with a ∈ P_{1} and p = p_{1}p_{2} ⋯ p_{h−1} ∈ ⋂_{i=2}^{h} P_{i} . Therefore,

e_{1}· q_{1}q_{2} ⋯ q_{h−1} = (a + p + P_{1}, 0, …, 0)
= (p + P_{1}, p + P_{2}, …, p + P_{h})
= φ(p) .

Thus e_{1}Q^{h−1} ∈ φ(A), as required. □

1.4 Using the notation of 1.3 we can define, for H ⊆ G,

T^{∗}(H) = S^{∗}(H) ∗ G ,

the skew group ring of G over S^{∗}(H). The embedding φ : S^{0}(H) ↪ S^{∗}(H) extends to an embedding

φ : T^{0}(H) = S^{0}(H) ∗ G ↪ T^{∗}(H)

which makes T^{∗}(H) a finitely generated module (left or right) over T^{0}(H). Furthermore, since the ideal Y of 1.3 is G–invariant, Y T is a twosided ideal of T and Lemma 1.3 implies that

(coker φ)· (Y T)^{h−1} = (0) .

Lemma (i) There is a ring homomorphism

ψ : T (H) = T^{0}(H)/(t) ⟶ T^{∗}(H)/(t)

which makes T^{∗}(H)/(t) a finitely generated module (left or right) over T (H). Furthermore,

(coker ψ)· (Y T)^{h−1} = (0)

and

(Y T)^{h−1}· (ker ψ)· (Y T)^{h−1} = (0) .

(ii) Let

ρ : G_{0}(T^{∗}(H)/(t)) ⟶ G_{0}(T (H))

denote restriction along ψ. Then the image of the map Infl_{T (H)}^{T} ∘ρ in G_{0}(T) coincides with the image of the map

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N) ⟶^{Ind_{S∗N}^{T}} G_{0}(T) ,

where N = N_{G}(H) and t_{N} = ∑_{g∈N} g ∈ S ∗ N ⊆ T.

Proof (i) The map ψ is obtained from φ by factoring out the ideals generated by t = ∑_{g∈G} g. Thus the claim about coker ψ is clear from the corresponding fact for φ . As to ker ψ, note that

ker ψ = S^{∗}tS^{∗} ∩ T^{0}/S^{0}tS^{0},

where we have written S^{∗} for S^{∗}(H), T^{0} for T^{0}(H), etc. Since

Y^{h−1}S^{∗}tS^{∗}Y^{h−1} ⊆ S^{0}tS^{0},

the statement about ker ψ follows.

(ii) To prove the assertions about Grothendieck groups, we first establish the following

Claim: T^{∗}(H) ≅ M_{h}((S/I(H)) ∗ N), and under this isomorphism t = ∑_{g∈G} g gets mapped to the matrix with t_{N} in every component.

Proof of claim Let e_{g} ∈ S^{∗} denote the idempotent corresponding to the component S/I(H^{g}) of S^{∗} and note that

e_{g} = e_{1}^{g} = g^{−1}e_{1}g (g ∈ T_{H}) .

Therefore,

e_{1}T^{∗}e_{1} = e_{1}(S^{∗} ∗ G)e_{1} = e_{1}S^{∗}e_{1} ∗ N ≅ (S/I(H)) ∗ N .

Also, as right T^{∗}–modules,

T^{∗} = ⨁_{g∈T_{H}} e_{g}T^{∗} = ⨁_{g∈T_{H}} g^{−1}e_{1}T^{∗} ≅ (e_{1}T^{∗})^{h} ,

and so, as rings,

T^{∗} ≅ End_{T^{∗}}(T^{∗}) ≅ M_{h}(End_{T ∗}(e_{1}T^{∗})) ≅ M_{h}(e_{1}T^{∗}e_{1}) .

An explicit isomorphism is given as follows:

T^{∗} → M_{h}(e_{1}T^{∗}e_{1}) : a ↦ (e_{1}xay^{−1}e_{1})_{x,y∈T_{H}} .

Under this isomorphism t maps to the matrix whose (x, y)–position equals ∑_{g∈G} e_{1}xgy^{−1}e_{1}. Since e_{1}xgy^{−1}e_{1} = e_{1}xgy^{−1} if xgy^{−1} ∈ N and e_{1}xgy^{−1}e_{1} = 0 otherwise, the above sum equals e_{1}t_{N} or, identifying e_{1}T^{∗}e_{1} with (S/I(H)) ∗ N, just t_{N}. This proves the claim.

The claim implies in particular that

T^{∗}/(t) ≅ M_{h}(S ∗ N/(I(H), t_{N})) .

Consider the following commutative diagram of rings:

[figure: \begin{matrix}T & T (H) & T^{∗}/(t) ≅ M_{h}(S ∗ N/(I(H), t_{N})) \\ T & T_{0} & T^{∗} ≅ M_{h}((S/I(H)) ∗ N) \end{matrix}; id_{T}; ψ; can.]

This yields a commutative diagram

[figure: \begin{matrix}G_{0}(T) & G_{0}(T (H)) & G_{0}(T^{∗}/(t)) ≅ G_{0}(S ∗ N/(I(H), t_{N})) \\ G_{0}(T) & G_{0}(T^{0}) & G_{0}(T^{∗}) ≅ G_{0}((S/I(H)) ∗ N) \end{matrix}; id; ^{Infl{T \atop T (H)}}; ρ; Infl; _{Infl{T \atop T^{0}}}; Res]

Thus it is enough to show that the image of

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}((S/I(H)) ∗ N) ≅ G_{0}(T^{∗}) ⟶^{Res} G_{0}(T^{0}) ⟶^{Infl} G_{0}(T)

coincides with the image of

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N) ⟶^{Ind_{S∗N}^{T}} G_{0}(T) .

This will follow if we can show that the diagram

[figure: \begin{matrix}G_{0}((S/I(H)) ∗ N) & G_{0}(T^{∗}) & G_{0}(T^{0}) \\ G_{0}(S ∗ N) & & G_{0}(T) \end{matrix}; Infl; ≅; Res; Infl; _{Ind{T \atop S∗N}}]

commutes. So let W be a finitely generated (S/I(H)) ∗ N–module. Proceeding clockwise, [W] successively gets mapped as follows:

[W] ⟼^{≅} [W ⊗_{S/I(H)∗N} e_{1}T^{∗}] ⟼^{Infl ∘ Res} [W ⊗_{S/I(H)∗N} e_{1}T_{T}^{∗}] ∈ G_{0}(T) .

Now, as right T –modules,

e_{1}T^{∗} = (S/I(H))T^{∗} ≅ T/I(H)T ≅ Ind_{S∗N}^{T}((S/I(H)) ∗ N) ,

and so

W ⊗_{S/I(H)∗N} e_{1}T_{T}^{∗} ≅ Ind_{S∗N}^{T} ∘ Infl_{S/I(H))∗N}^{S∗N}(W)

as required. This completes the proof of the lemma. □

1.5 Lemma Let ψ : A → B be a homomorphism of right Noetherian rings so that B is finitely generated as right A–module. Suppose further that, for some ideal X of A and some n ≥ 1,

X^{n}· ker ψ · X^{n} = (0) and coker ψ · X^{n} = (0) .

Then

G_{0}(A) = ρ(G_{0}(B)) + Infl G_{0}(A/X) ,

where ρ denotes restriction along ψ.

Proof Since inflation yields isomorphisms G_{0}(A/X) ≅ G_{0}(A/X^{n}) for every n ≥ 1, we may assume that n = 1. Let M be a finitely generated right A–module. Then [M] − [MX] = [M/MX] ∈ Infl G_{0}(A/X), and so it suffices to show that

[MX] ∈ ρ(G_{0}(B)) + Infl G_{0}(A/X) .

Applying MX ⊗_{A}( .) to the A–A–exact sequence

0 → ψA → B → coker ψ → 0

gives the following exact sequence of finitely generated right A–modules:

Tor_{1}^{A}(MX, coker ψ) → MX ⊗_{A} ψA → MX ⊗_{A} B → MX ⊗_{A} coker ψ → 0 .

Here, the first and the last term are A/X–modules, and MX ⊗_{A} B is the restriction of a B–

module. Thus

[MX ⊗_{A} ψA] ∈ ρ(G_{0}(B)) + Infl G_{0}(A/X) .

Finally, MX ⊗_{A} ψA ≅ MX/MX· ker ψ and MX· ker ψ is a finitely generated right A/X–

module. Thus

[MX] = [MX ⊗_{A} ψA] + [MX· ker ψ] ∈ ρ(G_{0}(B)) + Infl G_{0}(A/X) ,

and the proof is complete. □

1.6 Proposition (i) Let X_{H} denote the image of the map

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N) ⟶^{Ind} G_{0}(T) ,

where N = N_{G}(H). Then

I_{H} = X_{H} + ∑_{1≠x∈T_{H}} I_{⟨H,H^{x}⟩} .

(ii) Let Y_{H} denote the image of the map

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N)⟶_{≅}^{π_{N}}G_{0}(kN) ⟶^{Ind} G_{0}(kG) ,

where N = N_{G}(H). Then

K_{H} = Y_{H} + ∑_{1≠x∈T_{H}} K_{⟨H,H^{x}⟩} .

Proof (i) By Lemmas 1.4 and 1.5,

G_{0}(T (H)) = ρ(G_{0}(T^{∗}(H)/(t))) + Infl G_{0}(T (H)/Z),

where we have put Z = (Y T + J(H))/J(H). Since Z ⊆ J(⟨H, H^{x}⟩)/J(H) and ∏_{1≠x∈T_{H}} J(⟨H, H^{x}⟩)/J(H) ⊆ Z, by definition of Y, we have

G_{0}(T (H)/Z) = ∑_{1≠x∈T_{H}} Infl G_{0}(T/J(⟨H, H^{x}⟩))
= ∑_{1≠x∈T_{H}} Infl G_{0}(T (⟨H, H^{x}⟩)) .

Therefore,

Infl_{T (H)/Z}^{T} G_{0}(T (H)/Z) = ∑_{1≠x∈T_{H}} Infl G_{0}(T (⟨H, H^{x}⟩))
= ∑_{1≠x∈T_{H}} I_{⟨H,H^{x}⟩} .

Since (Infl_{T (H)}^{T} ∘ρ)(G_{0}(T^{∗}(H)/(t))) = X_{H}, by Lemma 1.4, the proof of (i) is complete. (ii) is immediate from (i), using the commutative diagram

[figure: \begin{matrix}G_{0}(S ∗ N) & G_{0}(kN) \\ G_{0}(T) & G_{0}(kG) \end{matrix}; _{Ind{T \atop S∗N}}; π_{N}; _{Ind{kG \atop kN}}; π] .

To check commutativity, it suffices to test a projective S ∗ N–module X, since S ∗ N has finite global dimension ([McC-R], Theorem 7.5.6(iii)). But then, proceeding clockwise, [X] gets mapped as follows:

[X] ↦ [X ⊗_{S∗N} kN] ⟼ [X ⊗_{S∗N} kN ⊗_{kN} kG] = [X ⊗_{S∗N} kG] ,

and the counterclockwise route gives

[X] ↦ [X ⊗_{S∗N} T] ⟼ [X ⊗_{S∗N} T ⊗_{T} kG] = [X ⊗_{S∗N} kG] ,

as needed. □

1.7 Note that the algebra S ∗ N/(I(H), t_{N}) which features in the terms X_{H} and Y_{H} in Proposition 1.6 can be thought of as T (H), but with G replaced by N = N_{G}(H). Thus the image of the map

G_{0}(S ∗ N/(I(H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N)

corresponds to I_{H}, again with N = N_{G}(H) instead of G. In order to describe X_{H} and Y_{H} in more detail, we can therefore assume that H is normal in G. This case will be dealt with in the next section.

2 Normal Subgroups

2.1 Throughout this section, we assume that H is a normal subgroup of G. Thus, using the notation of 1.1, I(H) = I^{0}(H) is a G–stable ideal of S, and

T^{0}(H) = T/I(H)T ≅ (S/I(H)) ∗ G .

We define

I_{1}(H) := I(H)T + t_{H}T ,

where t_{H} = ∑_{h∈H} h ∈ T and

I_{2}(H) := I(H)T + ∑_{h∈H}(h − 1)T + T t′_{H}T ,

where t′_{H} ∈ T is the sum over a fixed transversal for H in G. The choice of this transversal is immaterial because of the term ∑_{h∈H}(h − 1)T. Note that t_{H} is central modulo I(H)T and that ∑_{h∈H}(h − 1)T ≡ ∑_{h∈H} T (h − 1) modulo I(H)T. Thus I_{1}(H) and I_{2}(H) are twosided ideals of T. Moreover, t_{H}(h − 1) = 0 holds for all h ∈ H, and t = t_{H}t′_{H}. Therefore, with J(H) = I(H)T + I as in 1.1, we have

I_{1}(H) ∩ I_{2}(H) ⊇ J(H)

and

I_{1}(H)I_{2}(H) ⊆ J(H) .

Recall that T (H) = T/J(H) = T^{0}(H)/(t). Putting

T_{1}(H) = T/I_{1}(H) ≅ T^{0}(H)/(t_{H})

and

T_{2}(H) = T/I_{2}(H) ≅ (S/I(H)) ∗ (G/H)/(t_{G/H})

we obtain that

G_{0}(T (H)) = Infl G_{0}(T_{1}(H)) + Infl G_{0}(T_{2}(H)) .

Hence we have the following

Lemma Let K_{H} be as in 1.2. Then K_{H} = Im φ_{1} + Im φ_{2}, where

φ_{ℓ} : G_{0}(T_{ℓ}(H)) ⟶^{Infl} G_{0}(T) ⟶^{π} G_{0}(kG) (ℓ = 1, 2). □

2.2 Put

V (H) := k–subspace of V generated by the elements v − v^{h} (v ∈ V, h ∈ H).

This is a G–invariant subspace of V with I(H) = V (H)T. Moreover, V = V (H) ⊕ V^{H}, since kG is semisimple, and

S/I(H) ≅ S(V/V (H)) ≅ S(V^{H}) ,

the symmetric algebra of V/V (H) ≅ V^{H}. Define

α_{H} := ∑_{i≥0}(−1)^{i}[∧^{i}V (H)] ∈ G_{0}(kG) ,

and

G^{∗}(H) := image of Infl : G_{0}(kG/(t_{H})) → G_{0}(kG)

= subgroup of G_{0}(kG) generated by the irreducible kG–modules X with X^{H} = (0).

Indeed, the above modules X provide a ℤ–basis for G^{∗}(H).

Lemma Im φ_{1} = α_{H}· G^{∗}(H) .

Proof Put e_{H} = |H|^{−1}t_{H} ∈ kG ⊆ T^{0}(H), a central idempotent of T^{0}(H) with T_{1}(H) ≅ T^{0}(H)/(e_{H}) ≅ (1 − e_{H})T^{0}(H). The Quillen induction isomorphism

Ind_{kG}^{T^{0}(H)} : G_{0}(kG) ⟶^{≅} G_{0}(T^{0}(H))

(see Lemma 1.2(i)) maps the direct factor G_{0}(kG/(t_{H})) ≅ G_{0}((1 − e_{H})kG) = G^{∗}(H) of G_{0}(kG) to the direct factor G_{0}((1 − e_{H})T^{0}(H)) ≅ G_{0}(T_{1}(H)) of G_{0}(T^{0}(H)). Therefore, Im φ_{1} is identical with the image of the map

λ : G_{0}(kG/(t_{H}))⟶_{≅}^{Ind}G_{0}(T_{1}(H)) ⟶^{φ_{1}} G_{0}(kG) .

To describe the latter, put

T̃ = S(V (H)) ∗ G ⊆ T

and consider the following commutative diagram of rings:

[figure: \begin{matrix}T̃ & kG/(t_{H}) = T̃/(V (H), t_{H}) \\ T & T_{1}(H) = T/(V (H), t_{H}) \end{matrix}; can.; can.] .

This yields a commutative diagram

[figure: \begin{matrix}G_{0}(T̃) & G_{0}(kG/(t_{H})) \\ G_{0}(T) & G_{0}(T_{1}(H)) \end{matrix}; Ind; Ind; Infl; Infl] .

To check commutativity, note that

T_{1}(H) ≅ (T̃/(V (H), t_{H})) ⊗ ; _{T} T ≅ kG/(t_{H}) ⊗ ; _{T} T

as kG/(t_{H})–T –bimodules. Thus, for any [W] ∈ G_{0}(kG/(t_{H})),

IndT ; _{T} ∘ Infl_{T} ; kG/(t_{H})([W]) = [W ⊗ ; _{T} T] = [W ⊗_{kG/(t_{H})} kG/(t_{H}) ⊗ ; _{T} T]
= Infl_{T_{1}(H)}^{T}([W ⊗_{kG/(t_{H})} T_{1}(H)])
= Infl_{T_{1}(H)}^{T} ∘ Ind_{kG/(t_{H})}^{T_{1}(H)}([W]) ,

as required.

Therefore, we obtain

λ = π ∘ IndT ; _{T} ∘ Infl_{T} ; kG/(t_{H})
= ∑_{i≥0}(−1)^{i} Tor_{i}^{T} (IndT ; _{T} ∘ Infl_{T} ; kG/(t_{(H)}( .), kG)

= ∑_{i≥0}(−1)^{i} Tor_{T} ; i(Infl_{T} ; kG/(t_{H})( .), kG)

= π̃ ∘ Infl_{T} ; kG/(t_{H}) ,

where the third equality follows from Shapiro’s Lemma and where π̃ : G_{0}(T̃) → G_{0}(kG) denotes the usual isomorphism. By Lemma 1.2(iii), with V replaced by V (H) and α_{V} by α_{V (H)} = α_{H}, we know that

π̃ ∘ Infl_{T} ; kG = multiplication by α_{H} .

Therefore,

λ = π̃ ∘ Infl_{T} ; kG ∘ Infl_{kG/(t_{H})}^{kG}
= (mult. by α_{H}) ∘ Infl_{kG/(t_{H})}^{kG} ,

which proves the lemma. □

2.3 For φ_{2}, note that

T_{2}(H) ≅ S(V/V (H)) ∗ (G/H) /(t_{G/H})

is analogous with T/I, but with V replaced by V/V (H) and G by G/H. Thus, putting

T_{G/H} := S(V/V (H)) ∗ (G/H) ≅ T /(I(H)T + ∑_{h∈H}(h − 1)T )

and letting π_{G/H} : G_{0}(T_{G/H}) → G_{0}(k(G/H)) denote the usual isomorphism, we will be able to assume in §3 that, for 1 ≠ H,

K(G/H) := Im {ψ_{G/H} : G_{0}(T_{2}(H)) ⟶^{Infl} G_{0}(T_{G/H}) ⟶^{π_{G/H}} G_{0}(k(G/H))}

is known by induction (on |G|).

Lemma Im φ_{2} = α_{H}· Infl_{k(G/H)}^{kG}(K(G/H)) .

Proof We will show that

φ_{2} = (multiplication by α_{H}) ∘ Infl_{k(G/H)}^{kG} ∘ ψ_{G/H} .

Since φ_{2} = π ∘ Infl_{T_{G/H}}^{T} ∘ Infl_{T_{2}(H)}^{T_{G/H}}, it suffices to show that

π ∘ Infl_{T_{G/H}}^{T} = (mult. by α_{H}) ∘ Infl_{k(G/H)}^{kG} ∘ π_{G/H} .

Using the fact that π_{G/H} is inverse to Ind_{k(G/H)}^{T_{G/H}} (Lemma 1.2), this amounts to proving that

π ∘ Infl_{T_{G/H}}^{T} ∘ Ind_{k(G/H)}^{T_{G/H}} = (mult. by α_{H}) ∘ Infl_{k(G/H)}^{kG} .

So let X be a finite-dimensional k(G/H)–module and put

W = Ind_{k(G/H)}^{T_{G/H}}(X) = X ⊗_{k(G/H)} T_{G/H} ≅ X ⊗_{k} S(V/V (H)) .

Here, the last term has the T_{G/H}–module structure given by the diagonal action of G/H and the regular action of S(V/V (H)) on itself, so S(V/V (H)) ≅ k ⊗_{k(G/H)} T_{G/H} as T_{G/H}–modules.

Thus, Infl_{T_{G/H}}^{T} S(V/V (H)) ≅ k ⊗ ; _{T} T, where T̃ = S(V (H)) ∗ G is as in the proof of Lemma 2.2.

Using the G_{0}(kG)–module structure on G_{0}(T) as explained in 1.2, we obtain

Infl_{T_{G/H}}^{T} [W] = Infl_{k(G/H)}^{kG}([X])· [k ⊗ ; _{T} T] .

Using G_{0}(kG)–linearity of π (Lemma 1.2), one computes

π(Infl_{T_{G/H}}^{T} [W]) = Infl_{kG/H}^{kG}([X])· π([k ⊗ ; _{T} T])
= Infl_{k(G/H)}^{kG}([X])· ∑_{i≥0}(−1)^{i}[Tor_{i}^{T} (k ⊗ ; _{T} T, kG)]

= Infl_{k(G/H)}^{kG}([X])· ∑_{i≥0}(−1)^{i}[Tor_{T} ; i(k, kG)]
= Infl_{k(G/H)}^{kG}([X])· π̃([k])
= Infl_{k(G/H)}^{kG}([X])· α_{H} ,

where the third equality uses Shapiro’s Lemma and the last follows from Lemma 1.2(iii). This proves the desired equality. □

2.4 Combining Lemmas 2.1, 2.2 and 2.3 we obtain the following

Proposition If H ◁ G then K_{H} = α_{H}· G^{∗}(H) + α_{H}· Infl_{k(G/H)}^{kG}(K(G/H)) .

3 Proof of the Main Result

3.1 Let H be a subgroup of G and put

N = N_{G}(H) .

Let V (H) be defined as in 2.2.. Then V (H) is a module over kN. Also, the remaining definitions of 2.2 now become

α_{H} = ∑_{i≥0}(−1)^{i}[∧^{i}V (H)] ∈ G_{0}(kN)

and

G^{∗}(H) = image of Infl : G_{0}(kN/(t_{H})) → G_{0}(kN)

= subgroup of G_{0}(kN) generated by the irreducible kN–modules X with X^{H} = (0).

In addition, we put

R_{H} := Ind_{kN}^{kG}(α_{H}G^{∗}(H)) ⊆ G_{0}(kG).

Lemma Let Y_{H} be as in Proposition 1.6(ii). Then

Y_{H} = R_{H} + Ind_{kN}^{kG} (α_{H} Infl_{k(N/H)}^{kN}(K(N/H))) ,

where K(N/H) is as in 2.3. In particular,

R_{H} ⊆ K_{H} ⊆ K .

Proof By definition of Y_{H} and R_{H}, the expression for Y_{H} will follow if we can show that the map

G_{0}(S ∗ N/(V (H), t_{N})) ⟶^{Infl} G_{0}(S ∗ N) ⟶^{π_{N}} G_{0}(kN)

has image

α_{H}G^{∗}(H) + α_{H} Infl_{k(N/H)}^{kN}(K(N/H)) .

But this is the content of Proposition 2.4, with N in the role of G. (See also 1.7.) □

3.2 Lemma Suppose that H ⊆ E ⊆ Ê = N ∩ N_{G}(E).

(i) Let ᾱ_{E/H} ∈ G_{0}(k(Ê/H)) be defined in the same way as α_{E}, but for E/H acting on V/V (H). Then

(Res_{k_{E}}^{kN_{G}(H)} α_{H})· (Infl_{k(_{E}/H)}^{k_{E}} ᾱ_{E/H}) = Res_{k_{E}}^{kN_{G}(E)} α_{E} .

(ii) Let G^{∗}(E/H) denote the subgroup of G_{0}(k(Ê/H)) generated by all irreducible k(Ê/H)–modules X with X^{E/H} = (0). Then

Ind_{k_{E}}^{kN_{G}(E)} ∘ Infl_{k(_{E}/H)}^{k_{E}}(G^{∗}(E/H)) ⊆ G^{∗}(E) .

Proof (i) Note that (V/V (H))(E/H) = V (E)/V (H) and V (E) ≅ V (H) ⊕ (V (E)/V (H)) as kÊ–modules, since kÊ is semisimple. Thus (i) is a consequence of the following general fact which holds for any finite group D. For any finite-dimensional kD–module U, define

α_{U} = ∑_{i≥0}(−1)^{i}[∧^{i}U] ∈ G_{0}(kD) .

Claim. Let U and W be finite-dimensional kD–modules. Then α_{U⊕W} = α_{U}· α_{W}.

Proof of claim For all i ≥ 0, ∧^{i}(U ⊕ W) ≅ ⊕_{a+b=i} ∧^{a}U ⊗_{k} ∧^{b}W as kD–modules, and so

[∧^{i}(U ⊕ W)] = ∑_{a+b=i}[∧^{a}U][∧^{b}W]

holds in G_{0}(kD). The claim follows from this, and hence (i) is proved.

(ii) Clearly, Infl_{k(_{E}/H)}^{k_{E}}(G^{∗}(E/H)) is contained in the subgroup of G_{0}(kÊ) that is generated

by all finite-dimensional kÊ–modules X with X^{E} = (0). Letting E denote a set of right coset representatives of Ê in N_{G}(E) we have, for any such X,

(Ind_{k_{E}}^{kN_{G}(E)} X)|_{E} ≅ ⊕_{g∈E} X^{g}|_{E} ,

where X^{g} denotes the g–conjugate of X. Taking E–fixed points we obtain

(Ind_{k_{E}}^{kN_{G}(E)} X)^{E} ≅ ⊕_{g∈E}(X^{g})^{E} ≅ (X^{E})^{|E|} = (0) .

This proves (ii). □

3.3 Since G_{0}(R) ≅ G_{0}(kG)/K by 1.1, the following proves the main result of this article.

Theorem K = ∑_{1≠H⊆G} R_{H} .

Proof In view of Lemma 3.1, we need only show that K ⊆ ∑_{H} R_{H}. Furthermore, since K = ∑_{1≠H⊆G} K_{H}, by Proposition 1.2, it suffices to establish the following assertion.

(∗) For each 1 ≠ H ≤ G, K_{H} ⊆ ∑_{H⊆E⊆G} R_{E}.

We argue by induction, first on |G| and then on [G : H]. The assertion being vacuous for |G| = 1, assume that G ≠ ⟨1⟩ and that (∗) is true for all nonidentity subgroups of groups of smaller order than G. In particular, the theorem holds for all these groups. Furthermore, if H = G, then Proposition 1.6(ii) and Lemma 3.1 yield

K_{G} = Y_{G} = R_{G} ,

since K(⟨1⟩) = (0). Thus (∗) holds for H = G, and hence we may assume that G ≠ H and that (∗) holds for all subgroups of G having smaller index than H, in particular for the subgroups of the form ⟨H, H^{x}⟩ with x ∈ G\N (where N = N_{G}(H), as usual). Thus, from Proposition 1.6(ii) and Lemma 3.1, we obtain that

K_{H} ⊆ R_{H} + Ind_{N}^{G} (α_{H} Infl_{k(N/H)}^{kN}(K(N/H))) + ∑_{x∈G\N} ∑_{<H,H^{x}>⊆E⊆G} R_{E} .

Therefore, it suffices to show that

Z_{H} := Ind_{kN}^{kG} (α_{H} Infl_{k(N/H)}^{kN}(K(N/H))) ⊆ ∑_{H⊆E⊆G} R_{E} .

Since N/H has smaller order than G, we know by induction that

K(N/H) = ∑_{H⊆E⊆N} R_{E/H}.

For each such E, put Ê = N ∩ N_{G}(E). Then

R_{E/H} = Ind_{k(_{E}/H)}^{k(N/H)} (ᾱ_{E/H}G^{∗}(E/H)) ,

where ᾱ_{E/H} and G^{∗}(E/H) are as in Lemma 3.2. Therefore, since Infl_{k(N/H)}^{kN} ∘ Ind_{k(_{E}/H)}^{k(N/H)} =

Ind_{k_{E}}^{kN} ∘ Infl_{k(_{E}/H)}^{k_{E}}, we have

Z_{H} = ∑_{H⊆E⊆N} Ind_{kN}^{kG} (α_{H}· (Infl_{k(N/H)}^{kN} ∘ Ind_{k(_{E}/H)}^{k(N/H)})(ᾱ_{E/H}G^{∗}(E/H)))

= ∑_{H⊆E⊆N} Ind_{kN}^{kG} (α_{H}· Ind_{k_{E}}^{kN}(Infl_{k(_{E}/H)}^{k_{E}} ᾱ_{E/H}· Infl_{k_{E}/H}^{k_{E}} G^{∗}(E/H)))

= ∑_{H⊆E⊆N} Ind_{k_{E}}^{kG} (α_{H}|_{E}· Infl_{k(_{E}/H)}^{k_{E}} ᾱ_{E/H}· Infl_{k(_{E}/H)}^{k_{E}} G^{∗}(E/H)) ,

where the last equality follows from Mackey’s tensor product theorem ([C-R], Corollary 10.20). By Lemma 3.2(i),

α_{H}|_{E}· Infl_{k(_{E}/H)}^{k_{E}} ᾱ_{E/H} = α_{E}|_{E}

and so, using Mackey’s tensor product theorem again, the last sum above can be rewritten as follows:

∑_{H⊆E⊆N} Ind_{k_{E}}^{kG} (α_{E}|_{E}· Infl_{k(_{E}/H)}^{k_{E}} G^{∗}(E/H)) =

∑_{H⊆E⊆N} Ind_{kN_{G}(E)}^{kG} (α_{E}· Ind_{k_{E}}^{kN_{G}(E)} Infl_{k(_{E}/H)}^{k_{E}} G^{∗}(E/H))

, and by Lemma 3.2(ii) this last expression is contained in

∑_{H⊆E⊆N} Ind_{kN_{G}(E)}^{kG} (α_{E}G^{∗}(E)) = ∑_{H⊆E⊆N} R_{E} .

This shows that Z_{H} ⊆ ∑_{H⊆E⊆N} R_{E}, and the proof of the theorem is complete. □

3.4 Not all subgroups H ⊆ G are actually required in 3.3. First, R_{⟨1⟩} = (0) because G^{∗}(⟨1⟩) = (0). Next, R_{H^{g}} = R_{H} holds for all H ⊆ G and g ∈ G. Finally, for a given H ⊆ G, put

H̃ = {g ∈ G : V^{1−g} ⊆ V (H)} .

Clearly, H̃ is the unique largest subgroup of G with V (H̃) = V (H). It follows easily that N_{G}(H̃) = Stab_{G} V (H) and so, in particular, N_{G}(H) ⊆ N_{G}(H̃). Using Lemma 3.2(i) with E = H̃ (and so Ê = N_{G}(H)) we obtain

α_{H} = Res_{kN_{G}(H)}^{kN_{G}(_{H})} α_{H} ,

and so Mackey’s tensor product theorem ([C-R], Corollary 10.20) yields

R_{H} = Ind_{N_{G}(H)}^{kG} (α_{H}G^{∗}(H)) = Ind_{kN_{G}(_{H})}^{kG} (α_{H} Ind_{kN_{G}(H)}^{kN_{G}(_{H})} G^{∗}(H)) .

As shown in the proof of Lemma 3.2(ii) (here with E = H̃, Ê = N_{G}(H)), we have

Ind_{kN_{G}(H)}^{kN_{G}(_{H})} G^{∗}(H) ⊆ G^{∗}(H̃) .

Therefore

R_{H} ⊆ R_{H} ,

and consequently

K = ∑_{H} R_{H} ,

where H runs over a full set of non-conjugate non-identity subgroups of G.

4 Remarks, Examples, and Questions

4.1 The elements α_{H}. Recall that α_{H} = ∑(−1)^{−1}[∧^{i}V (H)] ∈ G_{0}(kN_{G}(H)). More generally, let G be any finite group, k any field and W an aribitrary finite dimensional kG– module. Define, as in the proof of Lemma 3.2,

α_{W} := ∑_{i≥0}(−1)^{i}[∧^{i}W] ∈ G_{0}(kG) .

We determine the Brauer character χ_{α_{W}} of α_{W}. Fix a p–modular system (K, A, k) and a finitely generated A–free AG–module W̃ with W ≅ W̃ ⊗_{A} k ([C-R], Theorem 22.1). Put W_{0} =

W̃ ⊗_{A} K and, for each g ∈ G, let g_{W_{0}} denote the K–endomorphism of W_{0} given the action of

g. Then

χ_{α_{W}} (g) = ∑(−1)^{i} Trace(∧^{i}g_{W_{0}}) = det(1 − g_{W_{0}}) .  (1)

The second of the above equalities follows from the fact that the characteristic polynomial of

any ϕ ∈ End_{K}(W_{0}) is given by ∑_{i=0}^{m}(−1)^{i} Trace(∧^{i}ϕ)X^{m−i}, where m = dim_{K} W_{0} = dim_{k} W.

Note in particular that it follows from (1) that if W is kG–isomorphic to W_{1} ⊕ … ⊕ W_{t} then

α_{W} = ∏_{i=1}^{t} α_{W_{i}} ;  (2)

this was already established by other means in the proof of Lemma 3.2.

4.2 The subgroup R_{G}. By definition, R_{G} = α_{G}G^{∗}(G), where G^{∗}(G) is the subgroup of G_{0}(kG) generated by all the non-trivial irreducible kG–modules. Thus [kG] = 1 − β where β ∈ G^{∗}(G) and 1 denotes the class of the trivial module. Since α_{G}· [kG] = Ind_{⟨1⟩}^{kG}(α_{G}|_{⟨1⟩}) = 0, it follows that α_{G} = α_{G}β ∈ R_{G}. Hence

R_{G} = α_{G}G_{0}(kG) ;

in particular R_{G} is an ideal of G_{0}(kG) contained in K. As the following result demonstrates, this inclusion is usually strict. (This is in fact a restatement of [Br-L2], Theorem 2.4(ii). See also [M].)

Proposition The equality K = α_{G}G_{0}(kG) holds if and only if G acts fixed-point-freely on V (G).

Proof Suppose G acts fixed-point-freely on V (G). Then, for each non-identity subgroup H of G, V (H) = V (G). Thus, in the notation of 3.4, H̃ = G for each such H; and hence K = R_{G}.

Suppose now that G does not act fixed-point-freely on V (G), so that V (G)^{H} ≠ {0} for some non-identity subgroup H of G. Since, as kH–modules, V (G) = V (G)^{H} ⊕ V (H), 4.1(2) implies that

Res_{H}^{G} α_{G} = α_{V (G)^{H}} · α_{V (H)} = 0· α_{V (H)} = 0 .

In other words, Res_{H}^{G} : G_{0}(kG) → G_{0}(kH) factors through G_{0}(kG)/(α_{G}). However, Res_{H}^{G}[k] = [k_{H}] and Res_{H}^{G}[kG] = [G : H][kH] are ℤ–independent, so Im(Res_{H}^{G}) has torsion-free rank at least 2. Hence, G_{0}(kG)/(α_{G}) has rank at least 2, whereas G_{0}(R) ≅ G_{0}(kG)/K has rank 1 by [Br-L2], Theorem 2.5(i). Therefore (α_{G}) must be properly contained in K. □

4.3 Deleting trivial constituents. Since V = V^{G} ⊕ V (G), V (H) = V (G)(H) for every subgroup H of G. Thus K (and so also G_{0}(R)) are unchanged on replacing V by V (G). An alternative way to observe this is as follows: Write R_{1} = S(V (G))^{G}, so that R = R_{1} ⊗_{k} S(V^{G}), a

polynomial ring in dim_{k} V^{G} variables over the coefficient ring R_{1}. By Grothendieck’s theorem ([B], Theorem XII.4.1), G_{0}(R) ≅ G_{0}(R_{1}).

4.4 The torsion subgroup of G_{0}(R). Let the dimension homomorphism dim : G_{0}(kG) → ℤ : [W] ↦ dim_{k} W have kernel Ω = Ω_{G}. Thus Ω is an ideal of G_{0}(kG), with ℤ–module generators {dim_{k}(W)1 − [W] : W irreducible}, and

G_{0}(kG) = ℤ· ⟨1⟩ ⊕ Ω .

Since dim α_{H} = ∑_{i}(−1)^{i} dim_{k} ∧^{i}V (H) = ∑(−1)^{i} ( \begin{matrix}dim_{k} V (H) \\ i \end{matrix} ) = 0, we have α_{H}G^{∗}(H) ⊆

Ω_{N_{G}(H)}. Furthermore, Ind_{N_{G}(H)}^{G} maps Ω_{N_{G}(H)} to Ω_{G}, so by Theorem 3.3

K ⊆ Ω.

Thus,

G_{0}(R) ≅ G_{0}(kG)/K ≅ ℤ · 1 ⊕ Ω/K.

As explained in the Introduction, (or see also [Br–L2], 2.4), the first of these isomorphisms, from right to left, is given by μ := ( .)^{G}∘Ind_{kG}^{T}. Thus [1] is mapped by it to [(k ⊗_{kG} T)^{G}] = [S^{G}] = [R].

Since G_{0}(R) = ℤ · [R] ⊕ F where F is finite by [Br-L2], Theorem 2.5(i), μ identifies F with Ω/K; so by the same result again, (and with n = dim_{k} V),

Ω/K ≅ G_{0}(R)torsion is a finite group of exponent dividing |G|^{n}.  (1)

Recall that an element g of G is a pseudoreflection (on V) if the endomorphism 1 − g of V has rank 1 (or, equivalently, if V (⟨g⟩) has dimension 1). Let H be the (normal) subgroup of G generated by the pseudoreflections on V. By [S], we have

Cl(R) ≅ H^{1}(G/H, U(S^{H})) ≅ Hom(G/H, k^{∗}),  (2)

where Cl(R) is the class group of R and U(X) is the group of units of the ring X. (In fact, Singh’s result is stated and proved in the context of a local ring S, but it is easy to check that his proof can be adapted to the present setting.) Now by [Br-L], Cl(R) is a factor of Ω/K, and indeed is the top factor given by a filtration of Ω/K whose successive factors are quotients of the higher class groups of R – for details, see [Br–L]. At most (dim_{k} V) − 1 of these class groups can be non-trivial when k is algebraically closed; in particular, when dim_{k} V = 2 and k is algebraically closed,

Ω/K ≅ Cl(R).  (3)

This fact was already observed in [H-S], where it was used to list the Grothendieck groups for the canonical action of the finite subgroups of SL(2, ℂ) on ℂ^{(2)}.

Singh’s result quoted above determines in particular when R is a UFD. In this context, it is natural to ask the

Question: When is Ω/K trivial?

By the Sheppard–Todd–Chevalley theorem [Sp], this certainly happens when G is generated by pseudoreflections. Indeed when G is abelian and k is a splitting field for G it is easy to see that Ω/K is trivial if and only if G is generated by pseudoreflections. However the following example (a simplified treatment of [Br-L2], Theorem 6.3) shows that this fails in the absence of the hypothesis on the field.

Example A: Let p be a prime, k a field of characteristic not p which does not contain a root of unity of order p, G a group of order p and V = kG, the regular representation. Thus

V ≅ k[X]/(X^{p} − 1) ≅ k[X]/(X − 1) ⊕ W = k ⊕ W,

where W = k[X]/(1 + X + … + X^{p−1}) is irreducible. Thus

G_{0}(kG) = ℤ[k] ⊕ ℤ[W] ≅ ℤ^{(2)}.

Now V (G) = W and G acts fixed-point-freely on this module. Hence, by Proposition 4.2,

G_{0}(R) ≅ G_{0}(kG)/(α) with α = ∑(−1)^{i}[∧^{i}W].

We claim that

α = (p − 1)[k] − [W].

Denote the right hand side here by β, so that β = p[k] − [kG]. Fix a p–modular system (K, A, k); the Brauer character χ of β is thus given by

χ_{β}(1) = 0, χ_{β}(g) = p, 1 ≠ g ∈ G.

To compute the Brauer character χ_{α} of α, put W_{0} = K[X]/(1 + X + … + X^{p−1}) and let g_{W_{0}} denote the K–endomorphism of W_{0} given by the action of g ∈ G. Clearly χ_{α}(1) = 0, and for 1 ≠ g ∈ G 4.1(1) shows that

χ_{α}(g) = ∑(−1)^{i}Trace(∧^{i}g_{W_{0}}) = det(1 − g_{W_{0}}) = (1 − ω)(1 − ω)^{2} … (1 − ω^{p−1}),

where ω is a primitive pth root of unity in some algebraic closure K. Since (1−ω)(1−ω^{2}) … (1− ω^{p−1}) = p, we get χ_{β} = χ_{α} and so α = β as claimed. Therefore,

G_{0}(R) ≅ (ℤ[k] ⊕ ℤ[W])/((p − 1)[k] − [W]) ≅ ℤ.

For non-abelian groups we do not know the answer to the above question, even over algebraically closed fields. As the following example illustrates, the triviality of Ω/K does not force G to be generated by pseudoreflections.

Example B: Let G be the binary icosahedral group. Then it is well known that G has an irreducible fixed point free complex representation V of dimension 2; see, for example, [W, page 181]. Since dim_{ℂ} V = 2, (2) and (3) show that Ω/K is Hom(G/H, ℂ^{∗}), where H is the subgroup of G generated by the pseudoreflections on V. But H = 1 since the action is fixed point free, and so, since G is perfect, Ω/K is trivial.

4.5 Power series rings. This case has predominated in much recent work on Grothendieck groups of invariant rings; see, for example, [A-R2], [H-S], [H-M-W] and [M]. Here, the notation is as in the Introduction, but one considers the invariants R̂ of the action of G on the power series ring Ŝ = k[[X_{1}, …, X_{n}]]. All the results proved in this paper are valid also in this setting, with essentially the same proofs. Note that in this case the isomorphism of G_{0}(kG) with G_{0}(Ŝ ∗ G) via the induction map is afforded by [B], Proposition IX. 1.3.(1), since Ŝ ∗ G is a complete semilocal ring with radical V Ŝ ∗ G.

4.6 Abelian groups with a splitting field. Let G be an Abelian group and assume that k contains a root of unity of order equal to the exponent of G. Then

V ≅ ⨁_{i=1}^{n} k_{χ_{i}}

for suitable χ_{i} ∈ G^{∗} = Hom(G, k^{∗}). Here, k_{χ_{i}} = k with cg = cχ_{i}(g) for c ∈ k, g ∈ G. We may assume that G acts faithfully, or – equivalently – that G^{∗} = ⟨χ_{i} : 1 ≤ i ≤ n⟩. Now, G^{∗} ≅ G and there is a ring isomorphism

G_{0}(kG) ⟶^{≅} ℤG^{∗} : [k_{χ}] ↦ χ,

where ℤG^{∗} is the integral group ring. Note that the ideal Ω of 4.4 corresponds to the augmentation ideal of ℤG^{∗} under this isomorphism.

For each subgroup H of G, write

G^{∗}(H) = {χ ∈ G^{∗} : χ|_{H} ≠ 1|_{H}},

so that G^{∗}(H) = ℤG^{∗}(H) ⊆ ℤG^{∗}. Then

α_{H} = ∏_{{i=1 \atop χ_{i}∈G^{∗}(H)}}^{n}(1 − χ_{i})

as follows from 4.1, since V (H) ≅ ⨁_{i}{k_{χ_{i}} : k_{χ_{i}} ∈ G^{∗}(H)}. For a given subgroup H of G, the

subgroup H̃ of 3.4 is given by

H̃ = ⋂_{i=1}^{n}{ker χ_{i} : χ_{i} ∉ G^{∗}(H)}

Thus, in this notation, 3.3 becomes

K = ∑_{1≠H⊆G} α_{H}ℤG^{∗}(H̃).  (1)

4.7 Further remarks on the Abelian case. (1) The result 4.6(1) for G abelian was the main theorem of [H-M-W]. (As mentioned in the introduction, one can without difficulty pass from polynomial rings to power series rings, and vice versa.)

(2) Groups of prime order ([Br–L2],§3) Let G have order p, prime, and assume that k contains a primitive p-th root of unity. By 4.3, we may assume that V ≅ ⊕_{i=1}^{n} k_{χ_{i}}, with each

irreducible character χ_{i} non-trivial. Thus we are in the situation of both 4.2 and 4.6, and so

K = α_{G}ℤG^{∗} = ∏_{i=1}^{n} ℤG^{∗}(1 − χ_{i}) = Ω^{n}.

Therefore, by 3.3,

G_{0}(R) ≅ ℤ ⊕ (Ω/Ω^{n}).

Let n − 1 = (p − 1)q + u, where q and u are non–negative integers with 0 ≤ u < p − 1. Then it was shown by Passi [P] that

Ω/Ω^{n} ≅ (ℤ/p^{q+1}ℤ)^{u} ⊕ (ℤ/p^{q}ℤ)^{p−1−u}.

4.8 Factoring out pseudoreflections. Let G, k, V and S be as in the introduction, and let H be the subgroup of G generated by the pseudoreflections on V. The theorem of Sheppard, Todd and Chevalley guarantees that S^{H} is also a polynomial ring (in n variables since H is finite), and of course R = S^{G} = (S^{H})^{G/H}. The following easy and presumably well-known lemma, (whose proof is left to the reader), shows that the action of G/H on S^{H} is linear.

Lemma Let a group B act on a commutative affine graded k-algebra W = ∑_{i≥0} ⊕W_{i}, with

W_{0} = k, such that B respects the grading and the action is completely reducible. Then the action of B on W is linearisable.

It follows that in calculating G_{0}(R) we may apply the Main Theorem 3.3 to G/H acting on S^{H}. In particular, we can improve 4.4(1):

Ω/K ≅ G_{0}(R)torsion is a finite group of exponent dividing |G/H|^{n}.

4.9 Dihedral groups of order 2p (p an odd prime). Let

G = ⟨x, y : x^{p} = 1 = y^{2}, xy = yx^{−1}⟩,

and assume that char k ∤ 2p and that k contains a root of unity ω of order p. Let L = k_{ϕ} be the non-trivial one-dimensional kG–module (so ϕ(y) = −1 and ϕ(x) = 1). For i = 0, …, p − 1 define ψ_{i} ∈ ⟨x⟩^{∗} by ψ_{i}(x) = ω^{i} and set

M_{i} := Ind_{k⟨x⟩}^{kG} k_{ψ_{i}},

so M_{i} is irreducible for i ≠ 0, M_{0} ≅ k ⊕ L and M_{i} ≅ M_{p−i}. In G_{0}(kG) write 1 for [k] and μ_{i} for [M_{i}], 0 ≤ i ≤ q := (p − 1)/2, so that

G_{0}(kG) = ℤ1 ⊕ ⨁_{i=0}^{q} ℤμ_{i} .

Reading subscripts modulo p, multiplication in G_{0}(kG) is given by

μ_{i}μ_{j} = μ_{i+j} + μ_{i−j} (0 ≤ i, j ≤ q).

Let

[V] = r1 + s[L] + ∑_{i=1}^{q} t_{i}μ_{i}

for r, s, t_{i} ≥ 0, where we shall assume that some t_{i} (say t_{1}) is non-zero. (Otherwise, G/⟨x⟩ acts on V). Up to conjugacy the non-identity subgroups of G are ⟨x⟩, ⟨y⟩ and G; we determine R_{H} for each of these.

(i) \underline{H = ⟨x⟩}: Here N_{G}(⟨x⟩) = G, V (⟨x⟩) ≅ ⊕_{i=1}^{q} M_{i}^{(t_{i})}, and (in the notation of 4.1),

α_{M_{i}} = 1 − μ_{i} + det M_{i} = μ_{0} − μ_{i} .

Thus, writing α_{i} := μ_{0} − μ_{i} (i = 1, …, q),

α_{⟨x⟩} = ∏_{i=1}^{q} α_{i}^{t_{i}} ;

since G^{∗}(⟨x⟩) = ⟨μ_{i} : i = 1, …, q⟩, we have

R_{⟨x⟩} = ⟨α_{⟨x⟩}μ_{j} : j = 1, …, q⟩.

(ii) \underline{H = ⟨y⟩}: Here N_{G}(⟨y⟩) = ⟨y⟩ and V (⟨y⟩) ≅ k_{χ}^{(s+∑ t_{i})} where χ = ϕ|_{⟨y⟩}. Thus G_{0}(k⟨y⟩) = ℤ1 ⊕ ℤχ and

α_{⟨y⟩} = (1 − χ)^{s+∑ t_{i}} = 2^{s−1+∑ t_{i}}(1 − χ).

Since G^{∗}(⟨y⟩) = ⟨χ⟩ and α_{⟨y⟩}χ = −α_{⟨y⟩}, we obtain

R_{⟨y⟩} = ⟨2^{s−1+∑ t_{i}} Ind_{k⟨y⟩}^{kG}(1 − χ)⟩
= ⟨2^{s−1+∑ t_{i}}(2 − μ_{0})⟩ ,

where the second equality follow from

Ind_{k⟨y⟩}^{G}(1 − χ) = (1 − [L]) Ind_{k⟨y⟩}^{kG} 1_{⟨y⟩} = (2 − μ_{0})(1 + ∑_{i=1}^{q} μ_{i}) = 2 − μ_{0} .

(iii) \underline{H = G}: Since V (G) ≅ L^{(s)} ⊕ ⊕_{i=1}^{q} M_{i}^{(t_{i})}, we have

α_{G} = (1 − [L])^{s}α_{⟨x⟩} = \{ \begin{matrix}α_{⟨x⟩} & if s = 0 \\ 0 & if s ≠ 0 \end{matrix} ,

since [L]μ_{i} = (μ_{0} − 1)μ_{i} = μ_{i} for i = 0, …, q, and hence [L]α_{⟨x⟩} = α_{⟨x⟩}. Furthermore, G^{∗}(G) = ⟨[L], μ_{j} : j = 1, …, q⟩ and so α_{⟨x⟩}G^{∗}(G) = ⟨α_{⟨x⟩}μ_{j} : j = 1, …, q⟩ = R_{⟨x⟩}. Thus

R_{G} = α_{G}G^{∗}(G) = \{ \begin{matrix}0 & if s ≠ 0 \\ R_{⟨x⟩} & if s = 0 \end{matrix} .

Combining the contributions (i), (ii) and (iii) gives

K = ⟨2^{s−1+∑ t_{i}}(2 − μ_{0}), α_{⟨x⟩}μ_{j} : j = 1, …, q⟩ .

Using the ℤ–basis {1, 2 − μ_{0}, α_{1}, …, α_{q}} for G_{0}(kG) and noting that ⟨α_{1}, …, α_{q}⟩ is an ideal of G_{0}(kG) which therefore contains all but the first of the above relations, we find

G_{0}(R) ≅ G_{0}(kG)/K ≅ ℤ ⊕ (ℤ /2^{s−1+∑ t_{i}}ℤ) ⊕ A ,

where

A ≅ ( ⊕_{i=1}^{q} ℤα_{i}) /⟨α_{⟨x⟩}μ_{j} : j = 1, …, q⟩ .

We claim that

|A| = p^{∑ t_{i}−1} .

To see this, put α′_{⟨x⟩} = α_{1}^{t_{1}−1} ∏_{i=2}^{q} α_{i}^{t_{i}} and β_{j} = α_{1}μ_{j} = α_{j+1}+α_{j−1}−2α_{j}. Then α_{⟨x⟩}μ_{j} = α′_{⟨x⟩}β_{j}

for j = 1, …, q. One checks that {β_{1}, …, β_{q}} forms a ℤ–basis for ⟨α_{1}, …, α_{q}⟩ and that the endomorphsim of this group given by multiplication by a fixed α_{i} has determinant p. The claim follows.

In particular, when G = S_{3} (so p = 3, q = 1), then

A ≅ ℤ /3^{∑ t_{i}−1}ℤ .

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Department of Mathematics  Department of Mathematics
University of Glasgow  Temple University
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e-mail: kab@maths.gla.ac.uk  e-mail: lorenz@euclid.math.temple.edu