VERY SIMPLE REPRESENTATIONS: VARIATIONS ON A THEME OF CLIFFORD

YURI G. ZARHIN

Abstract. We discuss a certain class of absolutely irreducible group representations that behave nicely under the restriction to normal subgroups and subalgebras.

1. Introduction

We start this paper with the following natural definition. Definition 1.1. Let V be a vector space over a field k, let G be a group and ρ : G → Aut_{k}(V) a linear representation of G in V. Suppose R ⊂ End_{k}(V) is an k-subalgebra containing the identity operator Id. We say that R is normal if

ρ(s)Rρ(s)^{−1} ⊂ R ∀s ∈ G.

Examples 1.2. Clearly, End_{k}(V) is normal. The algebra k · Id of scalars is also normal. If H is a normal subgroup of G then the image of the group algebra k[H] in End_{k}(V) is normal.

The following assertion is a straightforward generalization of well-known Clifford’s theorem [1]; [2, §49]; [5, §8.1]; [3, Ch. 6]. Lemma 1.3 (Lemma 7.4 of [7]). Let G be a group, k a field, V a non-zero k-vector space of finite dimension n and

ρ : G → Aut_{k}(V)

an irreducible representation. Let R ⊂ End_{F}(V) be a normal subalgebra. Then:

(i) The faithful R-module V is semisimple.

(ii) Either the R-module V is isotypic or there exists a subgroup G′ ⊂ G of index r dividing n and a G′-module V′ of finite k-dimension n/r such that r > 1 and the G-module V is induced from V′.

The following notion was introduced by the author in [7] (see also [8]); it proved to be useful for the construction of abelian varieties with small endomorphism rings [10, 9]. Definition 1.4. Let V be a non-zero finite-dimensional vector space over a field k, let G be a group and ρ : G → Aut_{k}(V) a linear representation of G in V. We say that the G-module V is very simple if it enjoys the following property:

If R ⊂ End_{k}(V) is a normal subalgebra then either R = k · Id or R = End_{k}(V). Here is (obviously) an equivalent definition: if R ⊂ End_{k}(V) is a normal subal-

gebra then either dim_{k}(R) = 1 or dim_{k}(R) = (dim_{k}(V))^{2}.

Partially supported by the NSF.

In this paper we prove that very simple representations over an algebraically closed field are exactly those absolutely irreducible representations that are not induced from a representation of a subgroup and do not split non-trivially into a tensor product of projective representations. We also give a certain criterion that works for any ground field with trivial Brauer group.

The paper is organized as follows. In §2 we list basic properties of very simple representations. In §3 we discuss certain natural constructions of representations that are not very simple. The Section 4 contains the statement of main results and their proof.

Acknowledgements. This paper germinated during author’s short stay in Manchester in August of 2002. The author would like to thank UMIST Department of Mathematics for its hospitality. My special thanks go to Professor A. V. Borovik for helpful encouraging discussions.

2. Very simple representations

Remarks 2.1. (i) Clearly, if dim_{k}(V) = 1 then the G-module V is very sim-

ple. In other words, every one-dimensional representation is very simple. (ii) Clearly, the G-module V is very simple if and only if the corresponding

ρ(G)-module V is very simple. (iii) Clearly, if V is very simple then the corresponding algebra homomorphism

k[G] → End_{k}(V)

is surjective. Here k[G] stands for the group algebra of G. In particular, a very simple module is absolutely simple.

(iv) If G′ is a subgroup of G and the G′-module V is very simple then the

G-module V is also very simple.

(v) Suppose W is a one-dimensional k-vector space and

κ : G → k^{∗} = Aut_{k}(W)

is a one-dimensional representation of G. Then the G-module V ⊗_{k} W is very simple if and only if the G-module V is very simple. Indeed, there are the canonical k-algebra isomorphisms

End_{k}(V) = End_{k}(V) ⊗_{k} k = End_{k}(V) ⊗_{k} End_{k}(W) ≅ End_{k}(V ⊗_{k} W),

which are isomorphisms of the corresponding G-modules. (vi) Let G′ be a normal subgroup of G. If V is a very simple G-module then

either ρ(G′) ⊂ Aut_{k}(V) consists of scalars (i.e., lies in k · Id) or the G′- module V is absolutely simple. Indeed, let R′ ⊂ End_{k}(V) be the image of the natural homomorphism k[G′] → End_{k}(V). Clearly, R′ is normal. Hence either R′ consists of scalars and therefore ρ(G′) ⊂ R′ consists of scalars or R′ = End_{k}(V) and therefore the G′-module V is absolutely simple.

(vii) Suppose F is a discrete valuation field with valuation ring O_{F}, maximal ideal m_{F} and residue field k = O_{F} /m_{F}. Suppose V_{F} is a finite-dimensional F -vector space and

ρ_{F} : G → Aut_{F} (V_{F})

is a F -linear representation of G. Suppose T is a G-stable O_{F} -lattice in V_{F} and the corresponding k[G]-module T/m_{F} T is isomorphic to V. Assume that the G-module V is very simple. Then the G-module V_{F} is also very simple. In other words, a lifting of a very simple module is also very simple. (See [10], Remark 5.2(v).)

Example 2.2. Suppose V is a finite-dimensional vector space over a finite field k of characteristic ℓ and G is a perfect subgroup of Aut(V) enjoying the following properties:

(i) If Z is the center of G then the quotient Γ := G/Z is a simple non-abelian group.

(ii) Every nontrivial projective representation of Γ in characteristic ℓ has di-

mension ≥ dim_{k}(V). Then the G-module V is very simple. See Cor. 5.4 in [10].

3. Counterexamples

Throughout this section, k is a field, V a non-zero finite-dimensional k-vector space and ρ : G → Aut_{k}(V) is a linear representation of G in V. Example 3.1. (i) Assume that there exist k[G]-modules V_{1} and V_{2} such that

dim_{k}(V_{1}) > 1, dim_{k}(V_{2}) > 1 and the G-module V is isomorphic to V_{1} ⊗_{k} V_{2}. Then V is not very simple. Indeed, the subalgebra

R = End_{k}(V_{1}) ⊗ Id_{V_{2}} ⊂ End_{k}(V_{1}) ⊗_{k} End_{k}(V_{2}) = End_{k}(V_{1} ⊗_{k} V_{2}) = End_{k}(V)

is normal but coincides neither with k · Id nor with End_{k}(V). (Here Id_{V_{2}} stands for the identity operator in V_{2}.) Clearly, the centralizer of R in End_{k}(V) coincides with Id_{V_{1}} ⊗ End_{k}(V_{2}) and is also normal. (Here Id_{V_{1}} stands for the identity operator in V_{1}.)

(ii) Let X ↠ G be a surjective group homomorphism. Assume that there exist k[X]-modules V_{1} and V_{2} such that dim_{k}(V_{1}) > 1, dim_{k}(V_{2}) > 1 and V, viewed as X-module, is isomorphic to V_{1} ⊗_{k} V_{2}. Then the X-module V is not very simple. Since X and G have the same image in Aut_{k}(V), the G-module V is also not very simple.

Definition 3.2. Let V be a vector space over a field k, let G be a group and ρ : G → Aut_{k}(V) a linear representation of G in V. Let

1 → C ↪ X ↠^{π} G → 1

be a central extension of G, i.e., C is a central subgroup of G which coincides with the kernel of surjective homomorphism π : X → G. Suppose that the representation

X ↠^{π} G ⟶^{ρ} Aut_{k}(V)

of X is isomorphic to a tensor product ρ_{1} ⊗ρ_{2} : X → Aut_{k}(V_{1} ⊗_{k} V_{2}) of two k-linear representations

ρ_{1} : X → Aut_{k}(V_{1}), ρ_{2} : X → Aut_{k}(V_{2})

with

dim_{k}(V_{1}) > 1, dim_{k}(V_{2}) > 1.

Then we say that the G-module V splits and call the triple (X ↠^{π} G; ρ_{1}, ρ_{2}) a splitting of the G-module V.

We say that V splits projectively if both images ρ_{1}(C) ⊂ Aut_{k}(V_{1}) and ρ_{2}(C) ⊂ Aut_{k}(V_{2}) consist of scalars. In other words, one may view both ρ_{1} and ρ_{2} as projective representations of G. In this case we call (X ↠^{π} G; ρ_{1}, ρ_{2}) a projective splitting of the G-module V.

We call a splitting (X ↠^{π} G; ρ_{1}, ρ_{2}) absolutely simple if both ρ_{1} and ρ_{2} are absolutely irreducible representations of X.

Remarks 3.3. We keep the notations of Definition 3.2.

(i) Clearly,

End_{X}(V_{1}) ⊗_{k} End_{X}(V_{2}) ⊂ End_{X}(V_{1} ⊗_{k} V_{2}) = End_{X}(V) = End_{G}(V).

This implies that if End_{G}(V) = k then End_{X}(V_{1}) = k and End_{X}(V_{2}) = k. (ii) Suppose W is a proper X-invariant subspace in V_{1} (resp. in V_{2}). Then

W ⊗_{k} V_{2} (resp. V_{1} ⊗_{k} W) is a proper X-invariant subspace in V_{1} ⊗_{k} V_{2} = V and therefore the corresponding X-module V is not simple. This implies that the G-module V is also not simple.

(iii) Suppose that the G-module V is absolutely simple and splits. It follows from (i) and (ii) that both ρ_{1} and ρ_{2} are also absolutely simple. In other words, every splitting of an absolutely simple module is absolutely simple.

Now the centrality of C combined with the absolute irreducibility of ρ_{1} and ρ_{2} implies, thanks to Schur’s Lemma, that both images ρ_{1}(C) ⊂ Aut_{k}(V_{1}) and ρ_{2}(C) ⊂ Aut_{k}(V_{2}) consist of scalars. In other words, every splitting of an absolutely simple G-module is projective.

(iv) Suppose G is a finite perfect group. We write γ : G̃ ↠ G for the universal central extension of G [6, Ch. 2, §9] also known as the representation group or the primitive central extension of G. It is known [6, Ch. 2, Th. 9.18] that G̃ is also a finite perfect group and for each central extension X ↠^{π} G there exists a surjective homomorphism ϕ : G̃ ↠ [X, X] to the derived subgroup [X, X] of X such that the composition

G̃ ↠^{ϕ} [X, X] ⊂ X ↠^{π} G

coincides with γ : G̃ ↠ G. This implies that while checking whether the Gmodule V admits a projective absolutely simple splitting, one may always restrict oneself to the case of X = G̃ and π = γ, and deal exclusively with absolutely irreducible linear representations of G̃ over k.

(v) We refer the reader to [4] for a study of projective representations of arbitrary finite groups over not necessarily algebraically closed fields.

Example 3.4. Assume that there exists a subgroup G′ in G of finite index m > 1 and a G′-module V′ such that the k[G]-module V is induced from the k[G′]-module V′. Then V is not very simple. Indeed, one may view W as a G′-submodule of V such that V coincides with the direct sum ⊕_{σ∈G/G′}σW and G permutes all σW ’s. We write Pr_{σ} : V ↠ σW ⊂ V for the corresponding projection maps. Then R = ⊕_{σ∈G/G′}k · Pr_{σ} is the algebra of all operators sending each σW into itself and acting on each σW as scalars. Clearly, R is normal but coincides neither with k · Id nor with End_{k}(V).

Remark 3.5. Suppose k′/k is a finite algebraic extension of fields. We write Aut(k′/k) for the group of k-linear automorphisms of the field k′. It is well-known that Aut(k′/k) is finite and its order divides the degree [k′ : k]; the equality holds if and only if k′/k is Galois.

Suppose there exists a homomorphism χ : G → Aut(k′/k) enjoying the following property:

There exists a structure of k′-vector space on V such that

ρ(s)(av) = (χ(s)(a))v ∀s ∈ G, a ∈ k′, v ∈ V.

We claim that if the G-module V is absolutely simple then k′/k is Galois and χ is surjective. Indeed, let us consider the image χ(G) ⊂ Aut(k′/k). We write k_{0} for the subfield of χ(G)-invariants in k′. We have

k ⊂ k_{0} ⊂ k′;

the degree [k′ : k_{0}] coincides with the order of χ(G) and therefore divides the order of Aut(k′/k). Clearly, G acts on V by k_{0}-linear automorphisms, i.e., k_{0} commutes with the action of G on V. Now the absolute irreducibility of V implies that k_{0} = k. This implies that [k_{0} : k] = 1. Since [k′ : k] = [k′ : k_{0}][k_{0} : k], we conclude that [k′ : k] = [k′ : k_{0}] coincides with the order of χ(G) and therefore divides the order of Aut(k′/k). This implies that [k′ : k] coincides with the order of Aut(k′/k) and therefore k′/k is Galois. Since [k′ : k] coincides with the order of χ(G), the order of the group Aut(k′/k) must coincide with the order of its subgroup χ(G) and therefore χ(G) = Aut(k′/k), i.e., χ is surjective. Definition 3.6. We say that the G-module V admits a twisted multiplication if there exist a nontrivial Galois extension k′ of k and a surjective homomorphism χ : G → Gal(k′/k) enjoying the following property:

There exists a structure of k′-vector space on V such that

ρ(s)(av) = (χ(s)(a))v ∀s ∈ G, a ∈ k′, v ∈ V.

(Here Gal(k′/k) stands for the Galois group of k′/k.) In other words, G acts on V by k′-semi-linear automorphisms. Example 3.7. Let us assume that V admits a twisted multiplication. Then the degree [k′ : k] divides dim_{k}(V) and therefore dim_{k}(V) > 1. Then the G-module V is not very simple. Indeed, k′ is obviously normal but does coincide neither with k · Id nor with End_{k}(V), since k′ ≠ k and End_{k}(V) is noncommutative. Remark 3.8. Let us assume that either k is algebraically closed or G is perfect and k is finite. Then V never admits a twisted multiplication, because either every algebraic extension k′/k is trivial or G is perfect and every Galois group Gal(k′/k) is abelian. In the latter case every homomorphism from perfect G to abelian Gal(k′/k) must be trivial.

4. Main Theorem

Theorem 4.1. Suppose the Brauer group of a field k is trivial (e.g., k is either finite or algebraically closed). Suppose V is a non-zero finite-dimensional k-vector space and

ρ : G → Aut_{k}(V)

is a linear representation of a group G over k. Then the G-module V is very simple if and only if all the following conditions hold:

(i) The G-module V is absolutely simple; (ii) The G-module V does not admit a projective absolutely simple splitting; (iii) The G-module V is not induced from a representation of a subgroup G′ ≠ G

of finite index in G; (iv) The G-module V does not admit a twisted multiplication.

Corollary 4.2. Let us assume that either k is algebraically closed or G is perfect and k is finite. Suppose V is a non-zero finite-dimensional k-vector space and

ρ : G → Aut_{k}(V)

is a linear representation of a group G over k. Then the G-module V is very simple if and only if all the following conditions hold:

(i) The G-module V is absolutely simple; (ii) The G-module V does not admit a projective absolutely simple splitting; (iii) The G-module V is not induced by a representation of a subgroup G′ ≠ G

of finite index in G.

Proof of Corollary 4.2. Indeed, the Brauer group of k is trivial. Now the proof follows readily from Theorem 4.1 combined with Remark 3.8.

□

Proof of Theorem 4.1. It follows from results of §3 that every very simple representation enjoys all the properties (i)-(iv). Now suppose that an absolutely irreducible representation

ρ : G → Aut_{k}(V)

enjoys the properties (ii)-(iv). It follows from Remark 3.3(iii) that the G-module V does not split.

Let R ⊂ End_{k}(V) be a normal subalgebra. Since the G-module V is not induced, it follows from Lemma 1.3 that the faithful R-submodule V is isotypic. This means that there exist a simple R-module W, a positive integer d and an isomorphism

ψ : V ≅ W^{d}

of R-modules. Let us put

V_{1} = W, V_{2} = k^{d}.

The isomorphism ψ gives rise to the isomorphism of k-vector spaces

V = W^{d} = W ⊗_{k} k^{d} = V_{1} ⊗_{k} V_{2}.

We have

d · dim_{k}(W) = dim_{k}(V)

Clearly, End_{R}(V) is isomorphic to the matrix algebra Mat_{d}(End_{R}(W)) of size d over End_{R}(W).

Let us put

k′ = End_{R}(W).

Since W is simple, k′ is a finite-dimensional division algebra over k. Since the Brauer group of k is trivial, k′ must be a field. Clearly, k′ is a finite algebraic extension of k.

We have

End_{k}(V) ⊃ End_{R}(V) = Mat_{d}(k′) ⊃ k′.

In particular,

k ⊂ k′ ⊂ End_{k}(V).

Clearly, End_{R}(V) ⊂ End_{k}(V) is stable under the adjoint action of G. This induces a homomorphism

α : G → Aut_{k}(End_{R}(V)) = Aut_{k}(Mat_{d}(k′)), α(s)(u) = ρ(s)uρ(s)^{−1} ∀s ∈ G, u ∈ End_{R}(V).

Since k′ is the center of Mat_{d}(k′), it is stable under the conjugate action of G. Thus we get a homomorphism χ : G → Aut(k′/k) such that

χ(s)(a) = α(s)(a) = ρ(s)aρ(s)^{−1} ∀s ∈ G, a ∈ k′.

I claim that the absolute irreducibility of V implies that k′/k is Galois and χ is surjective. Indeed, the inclusion k′ ⊂ End_{k}(V) provides V with a natural structure of k′-vector space and it is clear that

ρ(s)(av) = (χ(s)(a))v ∀s ∈ G, a ∈ k′, v ∈ V.

It follows from Remark 3.5 that k′/k is Galois and χ : G → Aut(k′/k) = Gal(k′/k) is surjective. Since V does not admit a twisted multiplication, k′ = k.

This implies that End_{R}(V) = Mat_{d}(k) and one may rewrite α as

α : G → Aut_{k}(Mat_{d}(k)) = Aut(End_{k}(V_{2})) = Aut_{k}(V_{2})/k^{∗} = PGL(V_{2}).

It follows from the Jacobson density theorem that R = End_{k}(W) = End_{k}(V_{1}). The adjoint action of G on R gives rise to a homomorphism

β : G → Aut_{k}(End_{k}(W)) = Aut_{k}(End_{k}(V_{1})) = Aut_{k}(V_{1})/k^{∗} = PGL(V_{1}).

Notice that

R = End_{k}(V_{1}) = End_{k}(V_{1}) ⊗ Id_{V_{2}} ⊂ End_{k}(V_{1}) ⊗_{k} End_{k}(V_{2}) = End_{k}(V).

Clearly, there exists a central extension π : X ↠ G such that one may lift projective representations α and β to linear representations

ρ′_{2} : X → Aut_{k}(V_{2}), ρ_{1} : X → Aut_{k}(V_{1}).

respectively. For instance, one may take as X the subgroup of G × Aut_{k}(V_{1}) × Aut_{k}(V_{2}) which consists of all triples (g, u_{1}, u_{2}) such that α(g) coincides with the image of u_{2} in Aut_{k}(V_{2})/k^{∗} and β(g) coincides with the image of u_{1} in Aut_{k}(V_{1})/k^{∗}. In this case the homomorphisms π, ρ_{1}, ρ′_{2} are just the corresponding projection maps

(g, u_{1}, u_{2}) ↦ g; (g, u_{1}, u_{2}) ↦ u_{1}; (g, u_{1}, u_{2}) ↦ u_{2}.

Now I am going to check that the tensor product ρ_{1} ⊗ ρ′_{2} coincides with the composition

ρπ : X ↠ G → Aut_{k}(V)

up to a twist by a linear character of X.

In order to do that, notice that if x ∈ X and g = π(x) ∈ G then the conjugation by ρ(g) in End_{k}(V) = End_{k}(V_{1} ⊗_{k} V_{2}) leaves stable R = End_{k}(V_{1}) ⊗ Id_{V_{2}} and coincides on R with the conjugation by ρ_{1}(x)⊗Id_{V_{2}} (by the definition of β and ρ_{1}). Since the centralizer of End_{k}(V_{1}) ⊗ Id_{V_{2}} in

End_{k}(V) = End_{k}(V_{1}) ⊗_{k} End_{k}(V_{2})

coincides with Id_{V_{1}} ⊗ End_{k}(V_{2}), there exists u ∈ Aut_{k}(V_{2}) such that

ρ(g) = ρ_{1}(x) ⊗ u.

Since the conjugation by ρ(g) leaves stable the centralizer of R, i.e. Id_{V_{1}} ⊗End_{k}(V_{2}) and coincides on it with the conjugation by Id_{V_{1}} ⊗ ρ′_{2}(x) (by the definition of α and ρ′_{2}), there exists a non-zero constant λ = λ(x) ∈ k^{∗} such that u = λρ′_{2}(x). This implies that for each x ∈ X there exists a non-zero constant λ = λ(x) such that

ρπ(x) = ρ(g) = ρ_{1}(x) ⊗ u = λ · ρ_{1}(x) ⊗ ρ′_{2}(x).

Since both

ρπ : X → Aut_{k}(V), ρ_{1} ⊗ ρ′_{2} : X → Aut_{k}(V),

are group homomorphisms, one may easily check that the map

X → k^{∗}, x ↦ λ = λ(x)

is a group homomorphism (linear character). Let us define ρ_{2} as the twist

ρ_{2} : X → Aut_{k}(V), ρ_{2}(x) = λ(x) · ρ′_{2}(x) ∀x ∈ X.

Clearly, ρ_{2} is a linear representation of X and

ρπ = ρ_{1} ⊗ ρ_{2}.

Since the G-module V does not split, either dim_{k}(V_{1}) = 1 or dim_{k}(V_{2}) = 1. If dim_{k}(V_{1}) = 1 then R = End_{k}(W) = End_{k}(V_{1}) = k consists of scalars. If dim_{k}(V_{2}) = 1 then d = dim_{k}(V_{2}) = 1, i.e., V = W and R = End_{k}(W) = End_{k}(V). □

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Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA

  E-mail address: zarhin@math.psu.edu