Clifford theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 28 February 2013

Clifford theory

Let R be an algebra over and let G be a finite group acting by automorphisms on R. The skew group ring is

RG= { gGrgg rgR }

with multiplication given by the distributive law and the relation

gr=g(r)g, forgG andrR.

Let N be a (finite dimensional) left R-module. For each gG define an R-module gN, which has the same underlying vector space N but such that

gNhasR -action given byrn= g-1(r)n, (A.1)

for rR,nN. If W is an R-submodule of N then gW is an R-submodule of gN and so gN is simple if and only if N is simple. Thus there is an action of G on the set of simple R-modules.

Let Rλ be a simple R-module. The inertia group of Rλ is

H= { hG RλhRλ } . (A.2)

If hH then Schur’s lemma implies that the isomorphism RλhRλ is unique up to constant multiples (since both Rλ and hRλ are simple). For each hH fix an isomorphism ϕh:Rλ h-1Rλ. Then, as operators on Rλ,

ϕhr=h(r) ϕh,and ϕgϕh=α (g,h)ϕgh , (A.3)

where α(g,h)* are determined by the choice of the isomorphisms ϕh. The resulting function α:H×H* is called a factor set [CRe1981, 8.32].

Let (H)α-1 be the algebra with basis {chhH} and multiplication given by

cgch=α (g,h)-1 cgh,for g,hH. (A.4)

Let Hμ be a simple (H)α-1-module. The putting

rh(mn)=r ϕhmchn, forrR,h H,mRλ,n Hμ, (A.5)

defines an action of RH on RλHμ.

Theorem A.6 (Clifford theory) Let Rλ be a simple R-module and let H be the inertia group of Rλ. Let Hμ be a simple (H)α-1-module where α:H×H* is the factor set determined by a choice of isomorphisms ϕh:Rλ hRλ. Define an action of RH on RλHμ as in (A.5) and define

RGλ,μ= IndRHRG (RλHμ)= (RG) RH (RλHμ).


  1. RGλ,μ is a simple RG-module.
  2. Every simple RG-module is obtained by this construction.
  3. If RGλ,μ RGν,γ then Rλ and Rν are in the same G-orbit of simple R-modules and HμHγ as (H)α-1-modules.


The proof of this theorem is as in [Mac1980] except that the consideration of the factor set α:H×HC* is necessary to correct an error there. We thank P. Deligne for pointing this out to us. A sketch of the proof is as follows.

Let M be a simple RG-module and let Rλ be a simple R-submodule of M. Then gRλgRλ as R-modules and M=gG gRλ since the right hand side is an RG-submodule of M. Then

M=giG/H giN= Ind RH RG (N),whereN= hHhRλ,

and the first sum is over a set {gi} of coset representatives of the cosets G/H.

The R-module N is semisimple and by [Bou1958]

NRλHμ, (A.7)

where Hμ=HomR(Rλ,N). It can be checked that the vector space Hμ has a (H)α-1-action given by

(chψ)(m)=α (h,h-1)hψ (ϕh-1(m)) ,forhH,ψ HomR(Rλ,N),

where ch is as in (A.4). Then, with RH-action on RλHμ given by (A.5), the isomorphism in (A.7) is an isomorphism of RH-modules (see [CRe1981, Thm. (11.17) (ii)]).

If P is an (H)α-1-submodule of Hμ then RλP is an RH-submodule of RλHμ and Ind RH RG (RλP) is an RG-submodule of M. Thus Hμ must be a simple (H)α-1-module.

This argument shows that every simple RG-module is of the form RGλ,μ. The uniqueness follows as in [Mac1980, App.].

Remark A.8. A different choice ψh:RλhRλ of the isomorphisms in (A.3) may yield a factor set β:H×H* which is different from the factor set α. However, the algebras (H)β-1 and (H)α-1 are always isomorphic (a diagonal change of basis suffices).

Lemma A.9. Define RG= { rRg (r)=rfor all gG } and let

e=(1/G) gGgRG.
  1. The map θ: RG e(RG)e s se is a ring isomorphism.
  2. Left multiplication by elements of R and the action of G by automorphisms make R into a left RG-module. Right multiplication makes R a right RG-module. The rings RG and e(RG)e act on (R×G)e by left and right multiplication, respectively. The map ψ: R (RG)e r re is an isomorphism of (RG,RG)-bimodules.


(a) If rRG then

ere=1G gGg(r) ge=1G gGre=re.

Thus the map θ is well defined and if r,sRG then rese=rse, so θ is a homomorphism. If re=se then r=s since RG is a free R-module with basis G. Thus θ is injective. If gGrgg is a general element of RG then

e(gGrgg) e=g,hGh (rg)hge= (g,hG)h (rg)e,

and, for each gG, hGh(rg)RG. So θ is surjective.

The proof of (b) is straightforward.

Let (H)α be the algebra with basis {bhhH} and multiplication given by

bgbh=α (g,h)bgh, forg,hH,

and let (H)α-1 be as in (A.4). Let M be a (H)α-module. The dual of M is the (H)α-1-module given by the vector space M*=Hom(M,) with action

(chψ)(m)=α (h,h-1)-1 ψ(bh-1m), forhH,ψM* .

This is a (H)α-1 action since, for all g,hH, ψM*,

(cgchψ)(m) = α(h,h-1)-1 α(g,g-1)-1 ψ(bh-1bg-1m) = α(h,h-1)-1 α(g,g-1)-1 α(h-1,g-1) ψ(b(gh)-1m) = α(h,h-1)-1 α(g,g-1)-1 α(h-1,g-1) α(gh,h-1g-1) (cghψ)(m) = α(g,h)-1 (cghψ)(m),

where the last equality follows from the associativity of the product bgbhbh-1bg-1 in (H)α. If ρ:(H)αEnd(M) is the representation corresponding to M then the representation ρ*:(H)α-1End(M*) corresponding to M* is

ρ*(ch)= α(h,h-1)-1 ρ(bh-1)t= (ρ(bh)-1)t. (A.11)

If M is a (H)α-module and N is a (H)α-1-module then MN is an H-module with action defined by

h(mn)=bhm chn,for hH,mMand nN. (A.12)

The following lemma is a version of Schur’s lemma which will be used in the proof of Theorem A.13.

Lemma A.12. Suppose that M and N are simple (H)α-modules and let N* be the (H)α-1-module which is the dual of N. Let eH=(1/H) hHh. Then

dim (eH(MN*)) = { 1, ifMN, 0, otherwise.


Identify MN* with Hom(N,M). Then, by (A.10) and (A.11), the action of H on Hom(N,M) is given by,

hA=ρ(bh)Aρ (bh)-1, forhHandA Hom(N,M),

where ρ:(H)αEnd(M) is the representation corresponding to M. If AHom(N,M) and gH then

eHA=geHA= g(eHA)=ρ (bg) (eHA)ρ (bg)-1,

and so ρ(bg)(eHA) =(eHA)ρ(bg) for all gH. Then, by Schur's lemma, eHA=0 if MN and eHA is a constant if M=N.

Theorem A.13. Let Rλ be a simple R-module and let H be the inertia group of Rλ. The ring RG acts on Rλ (by restriction) and (H)α acts on Rλ (by the R-module isomorphisms ϕh:RλhRλ of (A.3)) and these two actions commute. Thus there is a decomposition

RλνH^α Rλ,ν (Hν)*,

where H^α is an index set for the simple (H)α-modules, (Hν)* is the dual of the simple (H)α-1-module Hν, and Rλ,ν is an RG-module.

  1. If Rλ,μ0 then it is a simple RG-module.
  2. Every simple RG-module is isomorphic to some Rλ,μ.
  3. The nonzero Rλ,μ are pairwise nonisomorphic.


The setup of Lemma A.9(b) puts us in the situation of [Gre1980, §6.2]. If e is the idempotent used in Lemma A.9 then the functor

RG-modules RG-modules M eM

is an exact functor such that if M is a simple RG-module then eM is either 0 or a simple RG-module. Furthermore, every simple RG-module arises as eM for some simple RG-module M.

Let RGλ,μ be a simple RG-module as given by Theorem A.6. From the definition of RGλ,μ we obtain

eEGλ,μ = e(RG) RH (RλHμ) = e (RλHμ) = eeH (RλHμ) = eeH (RλHμ),

where eH=(1/H)hHh. Using the decomposition in the statement of the Theorem, we conclude that, as RG-modules,

eRGλ,μ = eeH ( νH^ Rλ,ν (Hν)* Hμ ) = eeH ( νH^ Rλ,ν eH ((Hν)*Hμ) ) Rλ,μ.

The last isomorphism is a consequence of Lemma A.12. The statement of the Theorem now follows from the results of J.A. Green quoted above.

Remark A.14. It follows from Theorem A.13 that Rλ is semisimple as an RG-module and the action of (H)α on Rλ generates EndRG(Rλ).

Notes and References

This is an excerpt of the paper entitled Affine Hecke Algebras, Cyclotomic Hecke algebras and Clifford Theory authored by Arun Ram and Jacqui Ramagge. It was dedicated to Professor C.S. Seshadri on the occasion of his 70th birthday.

Research partially supported by the Natioanl Science Foundation (DMS-0097977) and the National Security Agency (MDA904-01-1-0032).

page history