Representation theory Lecture Notes: Chapter 3

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 6 November 2013

Restriction and Induction

Let 𝒞 and 𝒟 be categories and let f*:𝒞𝒟 and f*:𝒟𝒞 be functors. Then f* is a right adjoint to f* and f* is a left adjoint to f* if, for each D𝒟 and C𝒞 there is a natural vector space isomorphism Hom𝒞(f*D,C) ΦHom𝒟 (D,f*C). For C𝒞 and D𝒟 define τC=Φ-1 (idf*C) Hom𝒞(f*f*C,C) and φD=Φ (idf*D) Hom𝒟(D,f*f*D). In general τC and ϕD are neither injective nor surjective, see the examples below in ??? and ???.

The functors HomB(M,·) and MZ·

Let B and Z be algebras and let M be a left B-module and a right Z-module. If N is a left Z-module then HomB(M,N) is a left Z-module with Z-action given by (zϕ)(m)=ϕ (mz),forϕ HomA(M,N), mM,zZ. If P is a left Z-module then MZP is the left B-module which as a -module is given by generators mp, mM, pP, and relations (m1+m2)p = m1p+m2p, form1,m2M, pP, m(p1+p2) = mp1+mp2, formM,p1, p2P, rmp = mrp=r(mp), forr,mM, pP, and which has B-action given by b(mp)=bmp, formM,pP, andbB. The covariant functors HomB(M,·): {leftB-modules} {leftZ-modules} MZ·: {leftZ-modules} {leftB-modules} are adjoint since the map HomZ(P,HomB(M,N)) Φ HomB(MZP,N) ψ: P HomB(M,N) p ψp: M N m ψp(m) Φ(ψ): MZP N mp ϕp(m) Φ-1(ϕ): P HomB(M,N) p ϕp: M N m ϕ(mp) ϕ: MZP N mp ϕ(mp) is a -module isomorphism.

The functor HomB(M,-) is very different from the functor HomB(-,M). There is always a canonical isomorphism HomB(B,M) M ϕ ϕ(1), butthedual module toM, HomB(M,B), can, in general, be much larger than M (take B= and M an infinite dimensional vector space over ).

The functor HomB(M,-) is left exact and the functor MZ- is right exact, i.e. if 0PPP is exact then 0HomB(M,P) HomB(M,P) HomB(M,P) is exact, and if NNN0 is exact then MZNM ZNMN 0is exact. A left A-module M is projective if the functor HomA(M,·) is exact and a right Z-module M is flat if the functor MZ· is exact.

Define Ext and Tor here.

Example 1. Let ι:ZB be a injective algebra homomorphism. Then M=B is a left B-module and right Z-module. The adjointness of the two functors ResZB=HomB (B,-)=ι* andIndZB=BZ -=ι*, is HomB (IndZB(P),N) HomZ(P,ResZBN).

Example 2. Let ι:ZB be a injective algebra homomorphism. Then M=B is a left Z-module and right B-module. The adjointness of the two functors ResZB=BB- =ι!and coIndZB=HomZ (B,-)=ι!, is HomZ (ResZB(N),P) HomB (N,coIndZB(P)).

Example 3. Let π:ZB be a surjective algebra homomorphism. Then M=B is a left B-module and right Z-module. The adjointness of the two functors InfBZ=HomB (B,-)=π* andDefZB= BZ-=π*, is HomB(DefBZ(P),N) HomZ(P,InfBZ(N)). If K=kerπ then DefZB(P)=BZP P/KP, since the map ϕ:PBZP given by ϕ(p)=1p has kernel KP.

Example 4. Let π:ZB be a surjective algebra homomorphism. Then M=B is a left Z-module and right B-module. The adjointness of the two functors InfBZ=BZ- =π!and coDefBZ=HomZ (B,-)=π!, is HomZ(InfBZ(N),P) HomB(N,coDefBZ(P)). If K=kerπ then HomZ(B,P) HomZ(Z/K,P) { fHomZ(Z,P) |f(k)=0 forkK } , and so HomZ(B,P) {pP|Kp=0}.

Example 5. Every algebra homomorphism ϕ:ZBis a composition ϕ:Zπim(ϕ) ιB of a surjective and an injective algebra homomorphism.

Example 6. Let B be an algebra and let eB be an idempotent. Let M=BeandZ=e BeEndB(M). Then the Z-module obtained by applying the functor HomB(M,-) to a B-module N is HomB(M,N)= HomB(Be,N) =eN.

(What does BeZP=BeeBeN look like?)

Computations for finite groups

Let G be a finite group and let H be a subgroup. If P is an H-module then IndHG(P) coIndHG(P) asG-modules. To construct the isomorphism fix a basis {pj} of P and a set {gj} of coset representatives of the cosets in G/H so that IndHG(P)=G HPhas basis {gipj}. An element ψcoIndHG(P)=HomH(G,P) is given by the values ψij given by ψ(gi-1)= jψijpj ,sinceψ(hgi-1) =ψ(gi-1)for all hH. The elements ψGHP are given by the values ψij determined by ψ=i,jψij (gipj)= igi (jψijpj) =igiψ (gi-1)= 1|H|gG gψ(g-1). Hence the isomorphism must be given by GHP = IndHG(P) coIndHG(P) = HomH(G,P) 1|H|gGgψ(g-1) ψ

Consider ι:HG. Then, for an H-module V, ι*V=IndHG(V) =GHVand ι*V=ResHG(V) =HomG(G,V)V, where the isomorphism HomG(G,V)V is given by ϕϕ(1). The isomorphisms HomH(ResHGV,ResHGV)= HomH(ι*V,ι*V)Ψ HomG(ι*ι*V,V)= HomG(GHV,V) and HomG(IndHGV,IndHGV)= HomG(ι*V,ι*V)Φ HomG(V,ι*ι*V)= HomH(V,GHV) are given explicitly by Ψ(b): GHV V gv gb(v), forb HomH(V,V)= HomH(ResHGV,ResHGV), and Φ(b): V HomG(G,GHV)GHV v b(1v) forbHomG (ι*V,ι*V). Thus Ψ(id): GHV V gv gv and Φ(id): V GHV v 1v.

Consider ι:HG. Then ι!V=coIndHG(V) IndHGand ι!V=ResHGV= GGVV, where the isomorphism coIndHGV= HomH(G,V) GHV= IndHGV is given by (???) and the isomorphism VGGV=ResHGV is given by v1v. The isomorphisms HomH(ResHGV,ResHG)= HomH(ι!V,ι!V)Ψ HomG(V,ι!ι!V)= HomG(V,coIndHGV) and HomG(coIndHGV,coIndHGV)= HomG(ι!V,ι!V)Φ HomH(ι!ι!V,V)= HomH(coIndHGV,V) are given explicitly by Ψ(b): V HomH(G,V)GHV v 1|H|gGgb(g-1v) ,forbHomH (ResHGV,ResHGV) =HomH(V,V), and Φ(b): GHV 1VV gv b(gv)|1V, forbHomG (GHV,GHV). Thus Ψ(id): V GHV v 1|H|gGgg-1v, and Φ(id): GHV V gv { v, ifgH, 0, ifgH.

Now define e: ι*ι*V Ψ(id) V Φ(id) ι!ι!V gv gv 1|H|G-1gv and define ε(b): V Φ(b) ι!ι!V=ι*ι*V Ψ(id) V v 1|H|gGgb(g-1v) 1|H|gGgbg-1v, for each bHomH(ι!V,ι!V)=HomH(ResHGV,ResHGV). Then eEndG(IndHGResHGV) andε:EndH (ResHGV) EndG(V,V).

Similarly, define e: ι!ι!V Ψ(id) V Φ(id) ι*ι*V gv { v, ifgH, 0, ifgH, { 1v, ifgH, 0, ifgH, and define ε(b): V Φ(b) ι*ι*V=ι!ι!V Ψ(id) V v b(1v) b(1v)|1V, for each bHomG(ι*V,ι*V)=HomG(IndHGV,IndHGV). Then eEndH(ResHGIndHGV) andε:EndG (IndHGV) EndH(V,V).

Thus, hopefully the general picture is that we can consider a sequence of injective algebra homomorphisms EndB(V)ι! EndZ(ι!V,ι!V)ι! EndB(ι!ι!V,ι!ι!V) and define e:ι*ι*Ψ(id) VΦ(id) ι!ι!Vand ε(b):VΦ(b) ι!ι!V=ι*ι* VΦ(id)V, for bEndZ(ι!V,ι!V) so that eEndB(ι!ι!V,ι!ι!V) and ε:EndZ(ι!V,ι!V) EndB(V). We then want to show

(a) e2=e,
(b) ea=ae, for aEndB(V),
(c) ebe=ε(b)e=eε(b), for bEndZ(ι!V,ι!V).
Note that (a) can be proved by applying (c) if one knows that ε(1)=1, i.e. ε(id)=id. Then e2=e·1·e=ε (1)e=1·e=e. Also (b) can be proved by applying (c) if one knows that ε(a)=a for aEndB(V). Then ae=ε(a)e=eε (a)=ea. Certainly we don’t expect all this to hold in a completely general situation. A better question is to ask what assumptions/setup we need that makes it hold.

Generators and relations for the partition algebra

Let PPk(n) be the subalgebra of the partition algebra generated by the planar diagrams. There is a bijection between diagrams in PPk(n) and Temperly-Lieb diagrams on 2k vertices such that pie2i-1 andbie2i. In general, the Temperley-Lieb diagram can be obtained by placing crosses on each side of each vertex of the partition algebra diagram and connecting the crosses according to the boundary of the blocks in the partition diagram. PICTURE PICTURE PICTURE Then the relations ei2=xei and eiei±1ei=ei in TL2k(x) correspond to the relations pi2=npi, bi2=bi, and pibipi=pi, bipibi=bi, bipi+1bi=bi, pi+1bipi+1= pi+1, respectively. Hence these relations should provide a full set of generators and relations for the planar partition algebra.

Example 1. A B-module M is projective if and only if there is a B-module M such that MM is a free B-module.

Simple modules

Assume that

(a) the action of B on M generates EndZ(M),
(b) the action of Z on M generates EndB(M).
Then there are surjective maps BβEndZ(M)and ZζEndB(M) and so every EndZ(M)-module is a B-module and every EndB(M)-module is a Z-module.

(a) HomB(M,P) is always an EndB(M)=Z module.
(b) MZN is always an EndZ(M)=B module.
(c) If P is a simple B-module and HomB(M,P)0 then P is a simple B-module.
(d) If N is a simple Z-module then MZN0.

Proof.

(a) If zAnnZ(M) then (zϕ)(m)=ϕ(zm)=0, for all mM. So AnnZ(M) acts by 0 on HomB(M,P). So HomB(M,P) is an EndB(M)=Z module.

(b) If bAnnB(M) then b(mn)=bmn=0, for all mM, nN. So AnnB(M) acts by 0 on MZN. So MZN is an EndZ(M)=B module.

(c) Suppose P is a simple B-module and HomB(M,P)0. Let ϕHomB(M,P), ϕ0. Then MϕP0 is an exact sequence since P is simple. So HomB(M,M) HomB(M,P) 0 z zϕ is an exact sequence. So Zϕ=N where Z=HomB(M,M)=EndB(M). So HomB(M,P) is a simple Z-module.

(c) Assume bAnnB(M) and ϕHomB(M,P)0, Since P is simple ϕ is surjective and so every element pP can be written in the form p=ϕ(m) for some mM. So bp=bϕ(m)=ϕ(bm)=0 for all pP. So AnnB(M) acts by 0 on P. So P is an EndZ(M)=B module.

(d) Let N be a simple Z module. Let nN be a nonzero element of N. Then every element of N can be written in the form zn for some zZ. So, for every zZ such that zn0, mzn=mzn0, since z is a nonzero element of EndB(M). So MZN0.

Thus we have shown that

(a) If P is a simple B-module such that HomB(M,P)0 then P is a simple B-module,
(b) If N is a simple Z-module then MZN0 and MZN is a B-module.

(a) If M is projective and P is a simple B-module such that HomB(M,P)0 then HomB(M,P) is simple.
(b) If N is a simple Z-module and MZN has a unique maximal proper submodule then there is a simple B-module P such that HomB(M,P)N.

Proof.

Step 1. HomB(M,MZN)N.

We shall show that the canonical homomorphism (???) is an isomorphism. Since MZN0, some ϕn0 and so αN is injective (since N is simple).

Let ϕHomB(M,MZN). Fix a nonzero element n1N. Then every element nN is of the form n=zn1,for some zZ. For each mM write ϕ(m) = mn = mzn1 = mzn1 = mn1. Then the map ϕ: M M m m is an element of EndA(M). So ϕ(m)=mz, for some zZ (and all mM). So ϕ: M MZN m mzn1=mzn1 and thus ϕ=ϕzn1. So αN is surjective. So αN is an isomorphism.

Step 2. Let P be a proper submodule of MZN. Then 0PMZN and so 0HomB(M,P) HomB(M,MZN) N. So HomB(M,P) is a submodule of N. If HomB(M,P)=N then every element of HomB(M,MZN) has image in the proper submodule PMZN. This is a contradiction to the fact that all the generators mn of MZN, i.e in the image of some map ϕn: M MZN m mn in HomB(M,MZN). So HomB(M,P) is a proper submodule. Since N is simple it follows that HomB(M,P)=0.

Step 3. Let P be the unique maximal proper submodule of P. Then HomB(M,P)=0 and P/P is simple. The canonical map 0PP P/P gives rise to a map 0HomB(M,P) HomB(M,P) HomB(M,P/P) and both the left hand side and the right hand side are simple (since M is projective). So HomB(M,P/P) HomB(M,P)N. So P=P/P is simple and HomB(M,P)N.

If M is finitely generated and projective then MZN has a unique maximal proper submodule.

Proof.

We have shows that if P is a proper submodule of MZN then HomB(M,P)=0.

Let P1,P2 be two proper submodules of MZN. Then the exact sequence 0P1P2 P1P2P1 +P20 yields an exact sequence 0HomB(M,P1P2) HomB(M,P1P2) HomB(M,P1+P2)=0. Since HomB(M,P1P2)= HomB(M,P1) HomB(M,P2) =0, HomB(M,P1P2) =0and HomB(M,P1+P2)=0. By Bourbaki Algébre Ch. II, §6, Ex. 4, HomB(M,limPi)= limHomB(M,Pi)=0 as Pi ranges over all proper submodules of MZN, ordered by inclusion. Thus, if P=sum of proper submodules ofMZN then HomB(M,P)=0. Since HomB(M,MZN)=N, P must be a proper submodule of MZN. So P is the unique maximal proper submodule of M.

M is projective as a B-module if and only if there is a B-module M such that MM=BI, i.e. MM is a free B-module.

Proof.

: Let I be a set of generators of M and let ϕ:BIM be the canonical map. Let K=kerϕ. The exact sequence 0KBI ϕM0 (0.5) gives an exact sequence 0HomB(M,K) HomB(M,BI) ϕ* HomB(M,M)0. Since ϕ* is surjective there is a map ψ:MBI such that ϕ*(ψ)=IdM. So ϕψ=IdM. So the first exact sequence splits, 0KBI ϕψ M0. So BI=imψKMK.

: If MMBI and 0PPP 0 is an exact sequence of B-modules then 0HomB(BI,P) HomB(BI,P) HomB(BI,P) 0 is the same as 0iIHomB (B,P) iIHomB (B,P) iIHomB (B,P)0 which is the same as 0iIP iIP iIP 0 which is exact since the first sequence is. So 0HomB(MM,P) HomB(MM,P) HomB(MM,P)0 which is the same as HomB(M,P) HomB(M,P) HomB(M,P) 0 0 HomB(M,P) HomB(M,P) HomB(M,P) is exact. This forces that the sequences 0 HomB(M,P) HomB(M,P) HomB(M,P)0 and 0 HomB(M,P) HomB(M,P) HomB(M,P)0 are both exact. So both M and M are projective.

(a) If M is a projective B-module and P is a simple B-module such that HomB(M,P)0 then HomB(M,P) is a simple Z module.
(b) If N is a simple Z-module and M is a finitely generated projective B-module then there is a simple B-module P such that HomB(M,P)N.

Proof.

Let P1=MZN. It is possible that MZN is not simple (too big). Let τ be the B-module generated by all the images of the maps in HomB(M,P). Then τP and HomB(M,P)= HomB(M,τ). Let t be the sum of all the submodules ττ such that HomB(M,τ)=0. Since M is finitely generated and projective HomB(M,t)=0. Now tτ and HomB(M,t/τ)= HomB(M,τ)= HomB(M,P1). Since M is finitely generated and projective the canonical map N HomB(M,MZN) n ϕn: M MZN m mn is a bijection. So HomB(M,P1)N.

Let P=τ/t. We know HomB(M,P)N. Let PP be a proper submodule of P. Then the injection 0PP gives us an injection 0HomB(M,P) HomB(M,P)N. Since N is simple HomB(M,P)=0 of HomB(M,P)=HomB(M,P). If HomB(M,P)=HomB(M,P) then every map in HomB(M,P) has its image in P. This is impossible since P is a proper submodule of P. So HomB(M,P)=0. This means P is 0 in P (by construction of P as τ/t). So P is simple.

Example 1. Let B be an algebra and let eB be an idempotent. If M=BethenEndB (M)=eBe, where eBe acts on M=Be by right multplication.

(a) M is finitely generated and projective B-module.
(b) If P is a B-module then HomB(M,P)eP.
(c) If P is a simple B-module then eP is a simple eBe-module and every simple eBe-module can be obtained this way.

Proof.

(a) The element e is a generator of Be=M as a B-module. So M is finitely generated. Since (1-e)2=1-e, e(1-e)=(1-e)e =0,and1=e+ (1-e), it follows that B=Be+B(1-e) and BeB(1-e)=0. So B=BeB(1-e) and so, by ???, Be is projective.

(b) Consider the map eP Φ HomB(M,P) x ϕx: M P be bex If ϕHomB(Be,P) then ϕ(be)=ϕ (bee)=beϕ(e) =be(eϕ(e)), and so ϕ=ϕx where x=eϕ(e)eP. So Φ is surjective. If ep1,ep2eP and ϕep1=ϕep2 then ϕep1(e) = eep1 = =p1 = ϕep2(e) = eep2 = =p2, and so Φ is injective. If ebeeBe then ((ebe)ϕ)(m) = ϕen (mebe) = meben=ϕeben (m), for all ϕenHomB(M,P). So Φ is a eBe-module homomorphism.

(c) follows immediately from (a) and (b) and Theorem ????.

Example 2. Let G be a group and let C(G) be an algebra of functions which is closed under convolution with respect to Haar measure μ on G. Let B be a measurable subgroup of G and assume that μ is normalized so that μ(B)=1. Then

(a) χB is an idempotent in C(G),
(b) C(G/B)=C(G)*χB, and
(c) C(B\G/B)=χB*C(G)*χB, is the Hecke algebra of the pair (G,B). From Proposition ???, if P is a simple C(G)-module then χB*P is either 0 or a simple C(B\G/B)-module. Every simple C(B\G/B)-module is obtained in this way.

Example 3. Let R be an algebra and let G be a finite group acting on R by automorphisms. The skew group ring is RG= { gGrgg |rgR } with multiplication given by the relation gr=g(r)g, forgG,rR. Clifford theory is a mechanism for constructing the simple RG modules from those of R and subgroups HG. Note that R is an RG module since R acts on R by left multiplication and G acts on R by automorphisms. Let e=1|G| gGg. Then e is an idempotent in RG. Let RG= { rR|g(r) =r, for allgG } .

(a) The map RG Φ e(RG)e r re is a ring isomorphism.
(b) The ring R is an (RG,RG) bimodule, where RG acts on the left and RG acts on R by right multiplication. The map R (RG)e r re is an isomorphism of (RG,RG) bimodules, where RG is identified with e(RG)e via (a).

Proof.

(a) Let rRG. Then ere=1|G| gGgre= 1|G| gGg(r) ge=1|G| gGre= re. (0.9) So Φ(r)=re is an element of e(RG)e and thus Φ is well defined.

Let e(gGrgg)e be a general element of e(RG)e. Then egGrgge= 1|G|gG hGh(rg) e=1|G|gG Φ(xg), where xg=hGh(rg) is an element of RG since k(xg)=hG kh(rg)=G (rg)=xg, for all kG. So Φ is surjective.

Let r,sRG. Then Φ(r)Φ(s) =rese=rse,and so Φ(r)Φ (s)=Φ(rs). So Φ is a ring homomorphism.

(b) Let aRG and bRG and rR. then, by ???, Φ(arb)= arbe=arebe=a (re)(ebe)= aΦ(r)be. So Φ is an (RG,RG) bimodule homomorphism. Let gGrgg be a general element of RG. Then (gGrgg) e=gGrgge =gGrge= gGΨ(rg). So Φ is surjective. The injectivity of Φ is proved as in (a).

Proposition ??? implies that all simple RG modules can be constructed in the following way: Let P be a simple RG module. Then HomRG(M,P) =HomRG ((RG)e,P) =eP, is either 0 or a simple e(RG)e=RG module.

Example 4. Recall that G(r,1,n)=(/r)nSn and G(r,p,n) is a normal subgroup of index p in G(r,1,n), G(r,1,n) = { tλw|λ= (λ1,,λn) (/r)n, wSn } , G(r,p,n) = { tλwG(r,1,n) |λ1+λn =0modp } , where the multiplication is determined by wtλ=twλw, λ(/r)n ,wSn. Let ζ=e2πi/p and define an action of /p on the group algebra G(r,1,n) by the powers of the map ρ: G(r,1,n) G(r,1,n) tλw ζ|λ|tλw, where|λ|= λ1++λn. The map ρ is a vector space isomorphism since it is a diagonal map with nonzero diagonal entries with respect to the basis {tλw|λ(/r)n,wSn} of G(r,1,n). Since ρ(tλwtμv)= ζ|λ|+|μ| tλwtμv=ρ (tλw)ρ(tμv), for all λ,μ(/r)n and w,vSn, the map ρ is an automorphism of G(r,1,n). Since ρp=Id the action of ρ on G(r,1,n) defines an action of /p on G(r,1,n). G(r,p,n)= (G(r,1,n))/p is the subalgebra of fixed points for this action since ρ(tλw)= ζ|λ|tλw =tλwif and only if tλwG(r,p,n). It follows that the irreducible representations of G(r,p,n) can be obtained in the following way: Let R=G(r,1,n), G=/p,RG= G(r,1,n)/p ,e=1p k=0p-1 ζk. Let P be a simple RG module (we know these by Clifford theory). Then either eP=0oreP is a simpleG(r,p,n) -module and all simple G(r,p,n) modules are obtained this way.

Induction and restriction

Let A be a subalgebra of an algebra B. Let M be a B-module. The A-module ResAB(M) is M viewed simply as and A-module. Let N be an A-module. The vector space BAN is the vector space BN with the additional relations ban=ban, for allaA,bB, nN. The B-module IndAB(N) is the vector space BAN with B-action given by b(b1n)=bb1 n,for allb,b1B, nN.

(Frobenius reciprocity) ResAB and IndAB are adjoint functors, i.e. HomB(IndAB(N),P) HomA(N,ResAB(P)), for all A-modules N and all B-modules P.

Proof.

This result is the special case M=B of the more general result HomA(N,HomB(M,P)) HomB(MAN,P), (2.2) where A and B are algebras, N is a left A-module, P is a left B-module and M is a (B,A)-bimodule. The left A-module HomB(M,P) has A-action given by (aϕ)(m)=ϕ(am), foraA,ϕHomB (M,P), andmM. The map HomB(B,P) ResAB(P) ϕ ϕ(1) is an A-module isomorphism since (aϕ)(1)=ϕ(1·a)=ϕ(a)=a(ϕ(1)), for aA, ϕHomB(B,P), Thus the statement of the theorem is a special case of the isomorphism in (???).

Example 0. Let ρ:AB be a homomorphism of algebras. Then ρ induces a functor ρ*:B-modA-mod. The functor ResAB is a special case of this functor. Induction IndAB is the left adjoint of ResAB and coinduction coIndAB is the right adjoint.

Example 1. Let G be a group and let H be a subgroup. Let G and H be the group algebras of G and H respectively and let N be a H-module. Let

(a) BN={nj} be a basis of N, and
(b) RG/H={ri} a set of coset representatives for G/H.
Then {rinj|riRG/H,njBn} is a basis ofIndHG (N)=GHN. Informally, this is because riHN= riHN= riN for all riRG/H.

Example 2. Let A be a subalgebra of an algebra B. Let aA. The left ideal Aa is an A-module and IndAB(Aa)Ba, asB-modules. This can be justified informally by BAAa=BAAa =BaA1=Ba. More formally one must show that the map BAa Ba bp bp is defined and has well defined inverse, so that it is an isomorphism. These checks are straightforward.

Example 3. Suppose that A is a subalgebra of B and that both A and B are semisimple. Let Aλ, λAˆ, be the simple A-modules, and
Bμ, μBˆ, be the simple B-modules.
Define nonnegative integers cλμ and dλμ by ResAB(Bμ)= (Aλ)cλμ andIndAB(Aλ) =(Bμ)dλμ. Then cλμ=dλμ, for allλAˆ andμBˆ. since cλμ=dim (HomA(ResAB(Bμ),Aλ)) =dim(HomB(Bμ,IndAB(Aλ))) =dλμ, by Schur’s lemma and Frobenius reciprocity.

Example 4. Let A be a subalgebra of B and let M be a B-module which is semisimple both as a B-module and as an A-module, MμBˆM (Bμ)mμ andM λAˆM (Aλ)nλ, where Bμ, μBˆM, are the simple B-modules that appear in M, and
Aλ, λAˆM, are the simple A-modules that appear in M. MμBˆM (Bμ)mμ. Let ZA=EndA(M) and ZB=EndB(M). Then ABand ZAZB and M μBˆM BμZBμ, asBZB modules, and M λAˆM AλZAλ, asAZA modules, where
ZAλ, λAˆM, be the simple ZA-modules, and
ZBμ, μBˆM, the simple ZB modules. Since ZA and ZB are both semisimple algebras there are positive integers cλμ such that ResZBZA (ZAλ)= μBˆM (ZBμ)cλμ. (2.3)

Each Bμ, μBˆM, is semisimple as an A-module and ResAB(Bμ) λAˆM (Aλ)cλμ, where the positive integers cλμ are as in (???).

Proof.

Since μBˆM ResAB(Bμ) ZBμ ResAZBBZB(M) ResAZBAZA(M) λAˆM AλResZBZA (ZAλ) λAˆM μBˆM Aλ (ZBμ)cλμ λAˆM μBˆM (AλZBμ)cλμ λAˆM μBˆM (Aλ)cλμ ZBμ it follows that ResAB(Bμ) HomZB (ZBμ,M) λAˆM (Aλ)cλμ as A-modules

Then HomZB (ResZBZA(ZAλ),ZBμ) HomA(ResAB(Bμ),Aλ). Thus, if cλμ and dλμ are the nonnegative integers such that ResAB(Bμ)= (Aλ)cλμ and IndZAZB(ZBμ) =(ZAλ)dλμ then cλμ=dλμ for allλAˆM andμBˆM.

Example 5. Let A be a semisimple subalgebra of a semisimple algebra B. Let trA be the trace of the regular representation of A. Let 𝒜 be a basis of A and let {a*} be the dual basis to 𝒜 with respect to the form on A defined by a1,a2A =trA(a1a2), fora1,a2A. Let trB be the trace of the regular representation of B. Let be a basis of B and let {b*} be the dual basis to with respect to the form on B defined by b1,b2A =trB(b1b2), forb1,b2B. If cB define [c]=b bcb*. Let N be an A-module. Then the character of IndAB(N) is given by χN(b)= a𝒜χN (a)b,[a*]B, where χN is the character of N.

Proof.

First note that, by Schur’s lemma, dim(HomA(Aλ,Aμ))=δλμ if Aλ and Aμ are simple A-modules. By the orthgonality relation for characters, δλμ=a𝒜χλ(a)χμ(a*), where χλ and χμ are the characters of Aλ and Aμ respectively. It follows that dim(HomA(M,P))= a𝒜χM(a) χN(a*), for any two module M and P with corresponding characters χM and χP.

There are two things to prove:

(a) χN, as defined in the statement, is a character,
(b) bχN(n)χP(b*)= a𝒜χN(a)χP(a*) for any B-module P.

(a) χN(b1b2) = a𝒜 χN(a) b1b2,[a*]B = a𝒜 χN(a) trB ( b1b2 b ba*b* ) = a𝒜 χN(a) trB (b1[a*]b2) = a𝒜 χN(a) trB ([a*]b2b1) = a𝒜 χN(a) [a*],b2b1B = χN(b2b1).

(b) a𝒜 χN(a) χP(a*) = χP ( a𝒜 χN(a) a* ) = χP ( a𝒜 χN(a) [a*] ) = χP ( a𝒜 χN(a) b [a*],b b* ) = b ( a𝒜 χN(a) [a*],b ) χP(b*) = b χN(b) χP(b*) since, for any character χ of B and any element cB, χ([c])= χ(bbcb*)= χ(bcb*b)= χ(c).

Example 6. Suppose that G is a finite group and H is a subgroup. Then G is a basis of G and trG(g)= { |G|, ifg=1, 0, otherwise, is the trace of the regular representation of G. The basis {g-1/|G|}gG is the dual basis to G with respect to ,G. If kG then [k]=1|G| gGgkg-1, and, for k1, k2G, [k1],k2G= k1,[k2]G= { 1, ifk1is conjugate to k2-1, 0, otherwise, Let N be an H-module. Then, by (???), the character χN of IndHG(N) is given by χN = hH χN(h) [h-1/|H|],gG = 1|H| hH χN(h) h-1,[g]G = 1|H| hHh𝒞g χN(h), where 𝒞g is the conjugacy class of g in G.

Example 7. Let H be a subgroup of G. If P is a G-module then the subgroup H acts on P and so P is an H-module. This H-module is denote ResHG(P). This provides a functor ResHG: {G-modules} {H-modules} and we define IndHG: {H-modules} {G-modules} to be the left adjoint functor to ResHG, i.e. HomH(M,ResHG(P)) HomG(IndHG(M),P) (*) as vector spaces.

If M is an H-module then IndHG(M)= { f:GM| f(hg)=hf(g), hH,gG } with G-action given by (gf)(x)=f (xg),for gG,fIndHG(M) ,xG. If fIndHG(M) then the value of f at any element of the coset Hg is determined by the value of f at g. Thus, we sometimes view fIndHG(M) as a function f:H\GM. For a proof of Frobenius reciprocity (i.e. (*)) in this setting see [Bump, Prop. 4.5.1].

Example 8. Let A be a subalgebra of B and let p be an idempotent in A. Then Ap is an A-module and IndAB(Ap)= BAApBp.

Let p1 and p2 be idempotents in A. Then Ψ: p1Bp2 HomB(Bp1,Bp2) x ψx where ψx: Bp1 Bp2 bp1 bp1x .

Proof.

(a) If xp1Bp2 then x=p1b1p2 for some b1B. Then ψx(bp1)=bp1p1b1p2Bp2 and so ψx is well defined.

(b) Since ψx(bp1)= bp1b1p2= bψx(p), for ψxHomB(Bp1,Bp2), Ψ is well defined.

(c) If ψx=0 then ψx(p1) =p1x= p1p1b1p2 =p1b1p2= 0. So x=0. So Ψ is injective.

(d) If ψHomB(Bp1,Bp2) then ψ(p1)= ψ(p1p1)= p1ψ(p1)= p1ψ(p1)p2 p1Bp2, since ψ(p1)Bp2. Let x=p1ψ(p1)p2. Then ψ(bp1)= bψ(p1)= bp1p1ψ(p1)p2 =bp1x=ψx(bp1). So ψ=ψx. So Ψ is surjective.

Example 9. If X is a measure space with a positive measure μ then L2(μ)= { f:X| X |f(x)|2 dμ(x)< } is a Hilbert space with inner product f1,f2= Xf1(x) f2(x) dμ(x). More generally, if V is a Hilbert space with inner product ,V then L2(μ)= { f:XV| X|f(x)|V2 dμ(x)< } is a Hilbert space with inner product f1,f2= X f1(x), f2(x) V dμ(x). A unitary representation of a Lie group G on a Hilbert space V is an action of G on V such that

(a) gv1,gv2= v1,v2, for all v1,v2V,
(b) The map G×VV (g,v)gv giving the action of G on V is continuous.
Thus, if H is a Lie subgroup of G and V is a unitary representation of H then we can define IndHG= { f:GV| f(hg)=hf(g), for allhH,and G |f(g)|V2 dμ(g)< } with G-action given by (gf)(x)= f(xg), forxG,f IndHG(V), gG. Then the inner product on IndHG(V) given by f1,f2 = G f1(g),f2(g)V dμ(g) satisfies gf1,gf2 = G gf1(x),gf2(x)V dμ(x) = G f1(xg),f2(xg)V dμ(x) = G f1(y),f2(y)V dμ(yg-1) = G f1(y),f2(y)V dμ(y) = f1,f2 if the measure μ on G is right invariant, i.e. if μ is a Haar measure on G.

Example 10. (Some “leftovers”) If G is a finite group then (G) G f gGf(g)g is an isomorphism of G-modules, but, if G is infinite then (G) is much larger than G.

Let G be a finite group and let H be a subgroup of G. Let ϕ:H* be a one-dimensional representation of H. Then pϕ=1|H| hHϕ (h-1)h is an element of G such that hpϕand pϕ2=pϕ. Thus Hpϕ=pϕϕ as H-modules.

Let G be a finite group and let B be a subgroup of G. Let 1B be the trivial representation of B and let p be the corresponding idempotent in B as defined in the previous paragraph. Then 1BG=IndBG(1B) (G)pand EndG(1BG)= p(G)p where p(G)p acts on (G)p by multiplication on the right.

Hecke algebras.

General Hecke algebras.

Let B be an algebra. The subspace pBpB is closed under multiplication and is an algebra with identity p. The Hecke algebra of the pair (B,p) is the algebra pBp.

Let B be an algebra and let p be an idempotent in B. Then EndB(Bp)=pBp, where pBp acts on Bp on the right. More precisely, EndB(Bp)= {ϕh|hpBp}, whereϕh is given byϕh= bph,for allbpBp.

Proof.

Let ϕEndB(Bp) and let bpBp be such that ϕ(p)=bp. For all bpBp, ϕ(bp)= ϕ(bpp)= bpϕ(p)= bpbp= (bp)(pbp), and so ϕ=ϕh with h=pbp. The multiplication of the ϕhEndB(Bp) corresponds to the action of pBp on Bp on the right since (ϕh1ϕh2)(bp)= ((bph2)h1)= (bp)(h2h1), for all h1,h2pBp and bpBp.

Recall that if A is a subalgebra of an algebra B and p is an idempotent in A then the left ideal Ap is an A-module and IndAB(Ap)Bp. (3.2)

Haar measure on G

Let G be a group. The vector space of functions f:G is a G-module with G action (gf)(h)=f (g-1h), for allg,hG,f . Fix a G-submodule C(G) of the space of all functions on G. A Haar measure on G is a linear functional μ:C(G) such that

(a) (continuity) μ is continuous with respect to the topology on C(G) given by f= sup{|f(g)||gG},
(b) (postivity) If f is such that f(g)0 for all gG then μ(f)0.
(c) (left invariance) For all gG and gC(G), μ(gf)=μ(f).
Use the notation μ(f)=Gf(g) dμ(g). The left invariance of μ means that Gf(g)dμ(g)= μ(f)=μ(hf)= Gf(h-1g)dμ (g)=Gf(k)dμ (hk). In general it is not true that for a Haar measure μ, Gf(g)dμ(g)= Gf(g)dμ(gh) or Gf(g-1)dμ(g)= Gf(g)dμ(g), hold for all fC(G). A group G is unimodular if one of the (equivalent) conditions in ??? hold.

The convolution of f1,f2C(G) is the function f1*f2 on G given by f1*f2(h)= Gf1(hg) f2(g-1)dμ (g),for allhG. (3.3)

Assume that C(G) is closed under convolution and that Fubini’s theorem holds.

(a) C(G) is an associative algebra.
(b) If G is unimodular then the map t:C(G) given by t(f)=f(1) is a trace on C(G).

Proof.

(a) Let f1,f2,f3C(G). Then ((f1*f2)*f3)(h) = G(f1*f2) (hg1)f3(g1-1) dμ(g1) = G G f1(hg1g2) f2(g2-1) f3(g1-1) dμ(g2)dμ(g1) and (f1*(f2*f3))(h) = Gf1(hk)f2 *f3(k-1)dμ (k) = Gf1(hk) Gf2 (k-1g1) f3(g1-1) dμ(g1)dμ(k) = GGf1(hg1g2) Gf2(g2-1) f3(g1-1)dμ (g1)dμ(g1g2) = GGf1 (hg1g2)f2 g2-1f3 (g1-1)dμ (g1)dμ(g2), where the last equality is a consequence of the left invariance of μ. Thus, by Fubini’s theorem (f1*f2)*f3= f1*(f2*f3).

(b) Since t(f1*f2) = (f1*f2)(1) = Gf1(p) f2(p-1) dμ(p), t(f2*f1) = (f2*f1)(1) = Gf2(p) f1(p-1) dμ(p), and G is unimodular, t(f1*f2)= t(f2*f1).

Remark. The algebra C(G) has an identity element only when G is finite.

The proof of the following proposition was contributed by Sarah Witherspoon.

Let H1 and H2 be subgroups of G and let W1 and W2 be representations of H1 and H2, respectively. Define = { f:GHom(W1,W2) |f(h1gh2) =W1(h1)f(g) W2(h2),h1 H1,h2H2 } . Then HomG ( IndH1G(W1), IndH2G(W2) ) .

Proof.

HomG ( IndH1G(W1), IndH2G(W2) ) HomH1 ( W1,ResH1G IndH2G(W2) ) . Then define HomH1(W1,IndH2G(W2)) ψ: W1λ IndH2G(W2) w1 ψw1 T:GHom(W1,W2) by ψw1(g)= T(g)w1. Then (h1ψw1)(g) =ψw1(gh1)=T (gh1)w1,and ψh1w1(g) =T(g)h1w1. So ϕHomH1(W1,IndH2G(W2)) if and only if T is right invariant under H1. Also ψw1(h2g)= T(h2g)w1 andh2ψw1 (g)=h2T(g) w1, so ψw1IndH2G(W2) if and only if T is left H2 invariant.

The Hecke algebra H(G,B,χ)

Fix a group G and a subgroup B and suppose that μ:C(G) is a Haar measure on G. Let χ:B* be a character of B. Define C(G/B) = { fC(G)| f(gb)=f(g) χ(b)for allb B,gG } , C(B\G/B) = { fC(G)| f(b1gb2)= χ(b1)f(g) χ(b2)for all b1,b2B, gG } . The following proposition puts these objects into the context of Section ??? It shows that C(B\G/B) is an algebra under convolution and that C(G/B) is a right C(B\G/B)-module. The Hecke algebra of the pair (G,B) is the algebra C(B\G/B).

Define a function p:G by p(g)= { χ(g)μ(B), ifgB, 0, otherwise.

(a) p is an idempotent in C(G).
(b) C(G/B)=C(G)p,
(c) C(B\G/B)=pC(G)p,

Proof.

(a) If hG, (p*p)(h)= Gp(hk)p (k-1)dμ (k)= { 0 ifhB, (χ(h)μ(B)2) μ(B), ifhB =p(h), since μ(B)=1.

(b) Let fC(G) and let hG, bB. Then, by the left invariance of μ, (f*p)(hb) = Gf(hbk)p (k-1)dμ (k)=Gf (hg)p(g-1b) dμ(b-1g) = χ(b)Gf (hg)p(g-1) dμ(g)= (f*p)(h) χ(b). So f*pC(G/B) and thus C(G)*pC(G/B).

Let fC(G/B) and hG. Then (f*p)(h)= Gf(hk)p (k-1)dμ(k) =Gf(h)χ(k) χ(k-1)dμ (k)=f(h)μ (B)=f(h). So f=f*pC(G)*p. Thus C(G/B)C(G)*p.

The proof of (c) is similar to the proof of (b).

Assume that

(a) Each function in C(G/B) is supported on only a finite number of cosets of B,
(b) Each function in C(B\G/B) is supported on only a finite number of cosets BwB in B\G/B,
(c) For each BwBB\G/B the number of B cosets in BwB is finite.
Fix a set R of coset representatives of G/B and a subset WR of coset representatives of B\G/B. For each xR and each wW define functions pxC(G/B) and TwC(B\G/B) by px(g)= { χ(b)μ(B), ifg=xbxB, 0, otherwise, andTw(g)= { χ(b1)χ(b2) μ(B) , ifg=b1wb2 BwB, 0, otherwise. Then C(G/B)has basis {px|xR}, andC(B\G/B) has basis {Tw|wW}. The following proposition completely determines the structure of the Hecke algebra C(B\G/B) and its action on C(G/B) in terms of the combinatorics of cosets.

(a) If xR and wW then px*Tw= yBG/ByBxBwB μ(B)Tw (x-1y)py.
(b) If v,wW then Tu*Tv=wW cu,vwTw, wherecu,vw =μ(B)2 xBBuBwBv-1B Tu(x)Tv (x-1w).

Proof.

(a) Let yR. Then (px*Tw) (y) = Gpx(g)Tw (g-1y)dμ(g) =Bpx(xb) Tw(b-1x-1y) dμ(xb) = { px(x)Tw (x-1y) μ(B), ifx-1y BwB, 0, otherwise, = { μ(B)Tw (x-1y) py(y), ifyBxBwB, 0, otherwise,

(b) Let wW. Then (Tu*Tv)(w) = GTu(x) Tv(x-1w) dμ(x)=μ(B) xBwBv-1BBuB Tu(x)Tv(x-1w) = μ(B)2 ( xBwBv-1BBuB Tu(x)Tv (x-1w) ) Tw(w).

By ???, the map t:C(G)C(G) given by t(h)=h(1), hC(G), is a trace on C(G) and, by restriction, t is a trace on C(B\G/B). Let , be the bilinear form on C(B\G/B) given by h1,h2= t(h1h2), h1,h2C(B\G/B).

Let W be a set of coset representatives of the cosets in B\G/B and define ind(w)=μ (BwB)= (number of cosets ofBin BwB), for each wW. Then

(a) {Tw-1ind(w)}wW is the dual basis to {Tw}wW with respect to ,.
(b) If |G/B| is finite then t=1|G/B|tr, where tr is the trace of the action of C(B\G/B) on C(G/B).

Proof.

(a) If u,vW then t(TuTv)= wWcu,vw t(Tw)= cu,v1= { ind(u), ifBuB=Bv-1B, 0, otherwise.

(b) If wW, tr(Tw) = xBG/B (px*Tw) |px= xBG/B yBxBwB Tw(x-1y) py|px = { xBG/B1, ifw=1, 0, otherwise, = { |G/B|, ifw=1, 0, otherwise.

Examples.

1. Let G be a finite group. Then the delta functions δg, gG, given by δg(h)= { 1, ifg=h, 0, otherwise, for allhG, form a basis of the space C(G) of all functions on G. If μ is a Haar measure on C(G) then μ(f)=μ ( gGf(g) δg ) =gGf(g) μ(gδ1)=μ (δ1)gG f(g), and so there is, up to multiplication by constants, a unique Haar measure on the space of all functions on G. Choosing μ(δ1)=1/|G| normalizes μ so that μ(G)=1. The vector space C(G) is closed under convolution with respect to μ and the associative algebra is isomorphic to the group algebra G of G via the map C(G) G f μ(δ1)gGf(g)g.

2. Let G = GL2()+= { (abcd) |a,b,c,d, ad-bc>0 } , B = SL2()= { (abcd) |a,b,c,d, ad-bc=1 } . Then the matrices in W= { (d100d2) |d1,d2, d1/d2>0 } are a set of coset representatives of the cosets in B\G/B. The Hecke algebra of the pair (G,B) is commutative and acts on the space of modular forms of weight k on the upper half plane.

3. Let 𝔽q be a finite field with q elements and let G = GLn(𝔽q)= {AMn(𝔽q)|det(A)0}, W = { upper triangular matrices inGLn (𝔽q) } . Then the set W=Sn= {permutation matrices inGLn(𝔽q)} is a set of coset representatives of the cosets in B\G/B and the Hecke algebra of the pair (G,B) is a deformation of the group algebra of the symmetric group. If si=(i,i+1) is the transposition switching i and i+1 in the symmetric group and Ti=Tsi=χBsiB denotes the corresponding basis element of the Hecke algebra then TiTj = TjTi,if |I-j|>1, TiTi+1Ti = Ti+1TiTi+1, 1in-1, Ti2 = (q-1)Ti+q, 1in. This example is a special case of Example 4.

4. Let G be a Chevalley group over the finite field 𝔽q with q elements and let B be a Borel subgroup of G. Then the cosets in B\G/B are indexed by the elements of the Weyl group W and the Hecke algebra of the pair (G,B) is a deformation of the group algebra of W. These algebras were introduced by Iwahori in [Iw].

5. Let G be a p-adic group and let B be an Iwahori subgroup of G. Then the cosets in B\G/B are indexed by the elements of the affine Weyl group W and the Hecke algebra of the pair (G,B) is a deformation of the group algebra of W. These algebras were introduced by Iwahori and Matsumoto in [IM].

Clifford Theory

Let R be an algebra over , K a finite group, and fix β:K×KR and automorphisms ck:RR, kK, such that the semidirect product algebra R(K)β is associative, where R(K)β= { kKrk tk|rk R } , with multiplication determined by the multiplication in R and the relations tk1tk2=β (k1,k2) tk1k2, tkr=ck(r) tk,andt1 =1, for k,k1,k2K and rR.

Let N be an R module. Define an new R-module kN to have the same underlying vector space N but with R-action given by ck(r)n=rn, for allrR,nN. If PN is an R-submodule of N then kP is an R-submodule of kN, and so, N is a simple R-module if and only if kN is a simple R-module. In this way K acts on the (isomorphism classes of) simple R-modules.

The inertia group of a simple R-module Rλ is H= { hK| hRλ Rλ } . For each hH fix an R-module isomorphism ϕh:hRλRλ. The condition that ϕh is an R-module homomorphism is the same as saying that, as linear transformations on the vector space Rλ, ch(r)ϕh= ϕhr,for allr R,hH. (Proof: (ch(r)ϕhm)= ϕh(ch(r)m)= ϕhrm.) Thus β(h1,h2)-1 ϕh1ϕh2r= β(h1,h2)-1 ch1(ch2(r)) ϕh1ϕh2= ch1h2(r) β(h1,h2)-1 ϕh1ϕh2. By Schur’s lemma ϕh is unique up to constant multiples and so β(h1,h2)-1 ϕh1φh2=α (h1,h2)ϕh1h2, for someα(h1,h2) . Let (H)1/α be the algebra over with basis {bh|hH} and multiplication given by bh1bh2=α (h1,h2)-1 bh1h2, h1,h2H. (4.1) If Hμ is a (H)1/α-module then RλHμ is an R(H)β module with action given by rth(mn)=r ϕhmbhn, for all rR, hH, mRλ and nHμ. This is an R(H)β action since thr(mn) = ϕhrmbhn= ch(r)ϕh mbhn, th1th2 (mn) = ϕh1ϕh2 mbh1bh2 n=β(h1,h2) α(h1,h2)α (h1,h2)-1 ϕh1h2m bh1h2n.

(Clifford Theory)

(a) Let M be a finite dimensional simple R(K)β-module. Then M Ind R(H)β R(K)β (RλHμ), where Rλ is a simple R submodule of M,
H={hK|hRλRλasR-modules},
α:H×H is determined by choices of R module isomorphisms ϕh:thRλRλ, and
Hμ is the simple (H)1/α-module given by Hμ=HomR(Rλ,M) with
(bhψ)(m)= α(h,h-1)-1 thψ ( ϕh-1β (h,h-1)-1 m ) ,hH,ψ Hμ,mRλ, (4.3) where {bh|hH} is the basis of (H)1/α in (???).

Proof.

Let M be a simple R(K)β module. Let Rλ be a simple R-submodule of M. Then tkRλ is an R-submodule of M isomorphic to kRλ. The sum kKtkRλ is an R(K)β submodule of M, and, since M is simple, M=kKtk Rλ=kiK/H tkiN,whereN= hHthRλ and the second sum is over a set of coset representatives of the cosets in K/H. So M Ind R(H)β R(K)β (N), as R(K)β-modules. We shall define a (H)1/α action on Hμ=HomR(Rλ,N)=HomR(Rλ,M) so that Θ: RλHomR(Rλ,N) N mψ ψ(m) is an R(K)β module isomorphism. The condition that Θ is an R(H)β-module homomorphism is that th(Θ(mψ)) =Θ(th(mψ)) ,for allmM, ψHμ,hH, and so thψ(m)=Θ (ϕhmbhψ) =(bhψ)(ϕhm) ,for allmM,hH ,ψHμ. Thus the appropriate formula for the action of (H)1/α on Hμ must be (bhψ)(m)= thψ(ϕh-1m) =α(h,h-1)-1 thψ(ϕh-1β(h,h-1)-1m). Given these formulas, it is straightforward to check that Hμ is a well defined (H)1/α module and Θ is an R(H)β module isomorphism.

The R(H)β module N is simple, since if P is an R(H)β submodule of N then IndR(H)βR(K)β(P) is and R(K)β submodule of M. Since M is simple P must be equal to N. Since N is simple the map Φ is surjective. If Φ:RλN is a nonzero R module homomorphism then it must be injective, since Rλ is simple. So Φ(m)0 for all nonzero mRλ. So the map Φ is injective. Thus RλHomR (Rλ,N)N. It follows that HomR(Rλ,N) is a simple (H)1/α modules since any submodule P would yield a submodule Φ(RλP) of N.

Remark. Note that the map kRλ tkRλ n tkn is an isomorphism.

(a) The simple R(K)β modules RK(λ,μ) are indexed by pairs (λ,μ) with λRˆ/K, μ(H)a/αˆ, where Rˆ/K is a set of representatives of the K orbits on Rˆ, and
(H)1/αˆ is an index set for the simple (H)1/α modules.
(b) dim(RK(λ/μ))= dim(Rλ)dim(Hμ)Card(K/H).
(c) The irreducible R(K)β module RK(λ,μ) is given by RK(λ,μ) Ind R(H)β R(K)β (RλHμ).

Proof.

Let Rλ be a simple R module. If the inertia group of Rλ is H then the inertia group of kRλ is kHk-1H. If hH and ϕh:hRλ Rλis an isomorphism, then ψghg-1: ghRλ gRλ m ϕh(m) is anRmodule isomorphism, since rψghg-1 (m) = rϕh(m)= g-1(r) ϕh(m) = ϕh(g-1(r)m) =ϕh(h-1g-1(r)m) = ψghg-1(rm). So, in fact we may choose ϕghg-1=ϕh. Then the factor set for gHg-1 is α:H×H and clearly (gH)1/α (H)1/α and RλHμ gRλgHμ rm rm is an R-module isomorphism.

A different choice ϕh:h RλRλ may yield a different factor set α:H×H, ϕh1 ϕh2= α(h1,h2) ϕh1h2. By Schur’s lemma ϕh=γ (h)ϕh, for some constantγ(h) *. So ϕh1 ϕh2= γ(h1) γ(h2) ϕh1 ϕh2= γ(h1) γ(h2) α(h1,h2) ϕh1h2 implies α(h1,h2)= γ(h1)γ(h2) γ(h1h2) α(h1,h2). Then the algebra (H)1/α given by (H)1/α= span{dh|hH}, withdh1dh2 =α(h1,h2)-1 dh1h2, is isomorphic to (H)1/α via the isomorphism (H)1/α Φ (H)1/α dh chγ(h)-1. Just to check: Φ(dh1dh2)= ch1γ(h1)-1 ch2γ(h2)-1= γ(h1-1) γ(h2-1) ch1h2 α(h1,h2)-1= α(h1,h2)-1 Φ(dh1h2).

Example 1. Let G be a group and let N be a normal subgroup of G. Let K be the quotient group N\G and choose a set {tk|kK} of the representatives of the cosets in N\G. If the map β:K×KN and automorphisms ck:NN are tk1tk2= β(k1,k2) tk1k2and ck(n)= tkntk-1,then GN(K)β.

Example 2. G(r,1,n)=(/r)nSn.

Recall that Sn acts on (/r)n by permuting the factors. The simple /r representations are indexed by λ=0,1,2,,r-1 and are given by xλ: /r e2πik/r e2πiλk/r The simple (/r)n modules are indexed by n-tuples λ=(λ1,,λn) with λi{0,1,2,,r-1}. The group Sn acts on the simple (/r)n-modules and on the indexes λ=(λ1,,λn) by permuting the factors. Each orbit under this action has a unique representative of the form λ= ( 0,0,,0 m0times , 1,1,,1 m1times , r-1,r-1,,r-1 mr-1times ) . The inertia group of this representation is Sm= Sm0× Sm1×× Smr-1 and (/r)n module isomorphisms ϕh:hRλRλ can be fixed to be ϕh=IdRλ for all hH. So the factor set is trivial in this case.

The simple Sm0×Sm1××Smr-1 modules are indexed by r-tuples of partitions μ= (μ(0),,μ(r-1)) withμ(i)mi. So the simple G(r,1,n)-modules are indexed by pairs (λ,μ),with λas in ??? andμas in ???. Given μ=(μ(0),,μ(r-1)) the information in λ is redundant and so we may index the simple G(r,1,n) modules by r-tuples of partitions μ=(μ(0),,μ(r-1)) with n boxes total.

The same type of analysis can be carried through for the affine symmetric group Sn=nSn. The simple -modules are given by xλ: 1 λ λ*. Thus the representations of Sn are indexed by *-tuples of partitions μ=(μ(λ))λ* with n boxes total. Alternatively the simple Sn modules are indexed by functions from * to the set 𝒫 of partitions μ: * 𝒫 λ μ(λ) ,such thatn= λ* |μ(λ)|.

The same analysis works for GnSn=GSn where G is any finite group. The general statement is that the simple GSn modules are indexed by functions μ: Gˆ 𝒫 λ μ(λ) such thatn= λGˆ |μ(λ)|, where Gˆ is an index set for the simple G-modules.

Monomial groups

A composition series of G is a sequence {1}=G0G1Gn=G with Gi-1 normal in Gi.

(a) A group is solvable if there exists composition series of G with Gi/Gi+1 abelian.
(b) A group is supersolvable if there exists a composition series with Gi normal in G and Gi/Gi-1 cyclic.
(c) A group is nilpotent if there exists a composition series of G with Gi/Gi-1Z(G/Gi-1). nilpotent supersolvablesolvable.
(d) A group is monomial if every irreducible representation of G can be obtained by induction from a one dimensional representation of some subgroup.

Supersolvable groups are monomial.

Proof.

Assume that G is supersolvable and let M be a simple G-module.

Case 1. Suppose that K=kerM{1}. Then M is a G/K-module. M= Ind K(G/K)β K(G/K)β (1KM)= 1KM. Since |G/K|<|G|, by induction, M=IndH/KG/K(1ξ), for some one dimensional representation 1ξ of a subgroup H/K.

Case 2. If K=kerM={1} then M is a faithful representation of G. Since G is supersolvable, G/Z(G) is supersolvable and has a composition series in which the first nontrivial term (G/Z(G))1 is a cyclic subgroup of G/Z(G). The inverse image of (G/Z(G))1 in G is a normal abelian subgroup A of G which is not contained in the center of G. Then M= Ind A(I/A)β A(G/A)β (1ξM)= IndIG (1ξM), where I is the inertia group of M and 1ξM is a simple I module. Since A is not contained in Z(G) and M is a faithful representation of G the group A does not act on M by scalars, but A does act on 1ξM by scalars. So 1ξMM and therefore IG. Thus, by induction, there is a one dimensional representation 1η of a subgroup H such that 1ξM= IndHI(1η). SoM=IndIG (IndHI(1η)) =IndHG(1η).

Nilpotent groups are monomial.

Proof.

Let M be an irreducible representation of G. Let us assume that the theorem is proved for

(a) groups of lower order (in the finite group case),
(b) groups of lower dimension (in the Lie group case).
We need to show that M is obtained by induction from a one dimensional representation of a subgroup.

Let ρ:GGL(M) be the hommorphism determined by M. Let K=kerρ. Then G/K acts on M and M is an irreducible representation of G/K. So, if

(a) G/K has lower order than G (in the finite case),
(b) G/K has lower dimension than G (Lie case),
then MIndH/KG/K (1ξ), as aG/Kmodule. But then MIndHG (1ξ), since IndHG(1ξ) and IndH/KG/K(1ξ) can both be viewed as functions on G/H(G/K)(H/K). So the theorem is proved for K{1}.

Now assume that K={1}, i.e. ρ is injective. The last nontrivial term 𝒞k(G) of the lower central series of G is Z(G) and 𝒞k-1 is a subgroup of G containing Z(G). Let x𝒞k-1(G) be such that xZ(G), xZ(𝒞k-1(G)/Z(G)). Then let A = { xkz| k0, zZ(G) } ,(finite case) A = { etXz| t,zZ(G) } ,(Lie case) so that
A is cyclic (finite case),
A is one dimensional (Lie case).
Then A is abelian since xk1z1 xk2z2= zk1+k2 z1z2= xk2z2 xk1z1, forki 0,zi Z(G). Let L be an irreducible A-submodule of M. Then dimL=1 (since A is abelian). If gG then gL is a representation of A that looks just like L except a(g)=g (g-1ag) ,forL. Since L is a simple A module so is gL. Now gGgL is aG-submodule ofM. Since M is simple M=gGgL. Let A= { gG| gLLasA modules } . Then L= aA aL is an A-module. Then M=gGgL= giG/A giL, where gi runs over a set of coset representatives of G/A. This is a decomposition of M as an A-module such that

(a) every irreducible A submodule of giL is isomorphic,
(b) The irreducible A submodules of giL and gjL are not isomorphic.
Since A acts on L by scalars, A acts on L by scalars. So LM, since ρ is injective and AZ(G), AZ(G). So AG and M=IndAG(L). Since A is smaller than G (since G is not abelian) LIndHA (1ξ). So M=IndAG (IndHA(1ξ)) IndHG(1ξ). The theorem now follows from the fact that all representations of an abelian group are 1 dimensional.

{supersolvable groups} {monomial groups} {solvable groups}.

Exercise. Give examples to show that these inclusions are strict.

A Lie group is exponential if the map 𝔤expG is a diffeomorphism. {nilpotent Lie groups} {exponential Lie groups} {solvable Lie groups}, and the inclusions are strict.

A nilpotent Lie group is exponential.

Proof.

First note that if X𝔤 then eX=1+X+X22!+X33!+ is a finite sum, since Xn=0 for sufficiently large n. Further X=log(1+(eX-1)) =(eX-1)- (eX-1)22+ (eX-1)33- is also finite, and so the map exp is invertible if 𝔤𝔫𝔤𝔩n for some n.

Note that the proof of this theorem uses the fact that any nilpotent Lie algebra can be imbedded in the nilpotent Lie algebra 𝔫n of strictly upper triangular matrices for some n. This fact is proved in [CorGrn, Thm. 1.1.11]. See also [Bou, I, §7] where this fact is called Ado’s theorem. This statement is analogous to the statements that (a) for a finite group G,GSn for some n, and (b) for an algebraic group G,GGLn, for some n. The main idea in the proofs of all these statements is to get G to “act on itself”.

Every irreducible representation of an abelian group is one dimensional.

Proof.

Let M be an irreducible G-module. Let ρ:GEnd(M) be the corresponding homomorphism. If gG then ρ(g)EndG(M) and so, by Schur’s lemma, ρ(g)=αIdM for some α. So every element gG acts on M by scalars. Thus, if mM then mM is a submodule of M. Since M is irreducible, mM, and therefore M is one dimensional.

Kirillov’s classification

Let G be a simply connected nilpotent Lie group. The coadjoint representation of G is the action of G on 𝔤*=Hom(𝔤,) given by (gϕ)(x)=ϕ (Adg-1(x)) ,forgG,x 𝔤andϕ𝔤*. A coadjoint orbit is a set Gϕ, ϕ𝔤*. Then {coadjoint orbits} {irreducible unitary representations ofG}

Let Ω be a coadjoint orbit (Ω𝔤*). Let ϕΩ so that Ω=Gϕ. A Lie subalgebra 𝔥𝔤 is subordinate to ϕ if ϕ|[𝔥,𝔥]=0, or, equivalently,ϕ ([x,y])=0 for allx,y𝔥. A subalgebra 𝔥 is subordinate to 𝔤 if and only if H eX e2πiϕ(X) defines a representation of H=exp(𝔥).

Choose a maximal dimensional subalgebra 𝔥𝔤 which is subordinate to ϕ. (Choosing a maximal subalgebra 𝔥 with respect to inclusion turns out to be the same thing.) Let H=exp(𝔥) and define a one dimensional representation of H by 1ϕ: H * ex e2πiϕ(x) for x𝔥. Then VΩ= IndHG(1ϕ) is the irreducible representation of G associated to Ω. This says that every irreducible unitary representation of G is obtained by inducing from a one dimensional representation of some subgroup H. We will want to show that

(a) V is an irreducible representation of G,
(b) V does not depend on the choice of ϕ and 𝔥 (only on the orbit Ω).
and answer the questions
(c) Why is it sufficient to define Wϕ by its values on eX for X𝔥?
(d) What is IndHG(W)?

Let 𝔥𝔥 be Lie subalgebras of 𝔤 and let ϕ𝔤* be such that 𝔥 and 𝔥 are both subordinate to ϕ. Then, since HH there are fewer functions in IndHG(Wϕ) than in IndHG(Wϕ). The G-actions on the two spaces are defined in the same way so IndHG(Wϕ) is a submodule of IndHG(Wϕ). So VΩ is not irreducible unless 𝔥 is maximal.

Suppose Ω is a coadjoint orbit, ϕΩ and 𝔥 is a maximal dimensional subalgebra of 𝔤 subordinate to ϕ. Suppose ϕ is another element of Ω. Then there is a gG such that ϕ=gϕ- Adg*ϕ.Let 𝔥=g𝔥= Adg𝔥. Then, for gXg𝔥, ϕ(gX) = (gϕ)(gX)= ϕ(g-1gX)= ϕ(X),and since g[X,Y] = Adg([X,Y])= Rg[X,Y] Rg-1 = Rg(XY-YX)Rg-1 = (RgXRg-1) (RgYRg-1)- (RgYRg-1) (RgXRg-1) = [gX,gY], it follows that 𝔥 is subordinate to ϕ if and only if 𝔥 is subordinate to ϕ. If H=exp(𝔥) and H=exp(𝔥) then H=gHg-1, since Adg is the differential of Intg. So we get one dimensional representations Wϕ: H h e2πiϕ(X) and Wϕ: gHg-1 ghg-1 e2πiϕ(gX) = e2πiϕ(X) if h=eX. Define a map IndHG(Wϕ) Φ IndHG(Wϕ) f f byf(t)= f(g-1t)˜. Then f(ghg-1t)= f(hg-1t)˜= hf(g-1t)˜= hf(t) and xf˜(t)= xf(g-1t)˜= f(g-1tx)˜= f(tx)= (xf)(t). So Φ is a well defined G-module isomorphism.

In general, how do we construct maximal subordinate subalgebras?

(Vergne’s lemma) Let V be a vector space and let , be a skew symmetirc bilinear form on V. f=(0V1V2Vn=V) ,dimVi=i. Let W(f,,)= i=0nVi Vi, Then

(a) W(f,,) is a maximal isotropic subspace of V,
(b) If x,y=ϕ,[x,y] for some ϕ𝔤* then 𝔥 is a subalgebra of 𝔤.

Let 𝔤 be a nilpotent Lie algebra such that dim(Z(𝔤))=1. Then 𝔤=xyzW, where z=Z(𝔤), [x,y]=z, and [y,w]=0 for wW.

Proof.

Let y𝔤 such that the image of y in 𝔤/Z(𝔤) is a nonzero element of Z(𝔤/Z(𝔤)). Then ady: 𝔤 Z(𝔤) t [y,t], has im(ady)=Z(𝔤), and ker(ady)=Zg(y). Thus dim(ker(ady))=dim(𝔤)-1 and kerady is an ideal of 𝔤 since ady([x,t])= [y,[x,t]]=- [t,[y,x]]- [x,[t,y]]=0, fortkerady andx𝔤. So Z𝔤(y)=y zW,where [y,w]=0for wW, is an ideal of 𝔤. Let x𝔤 such that [x,y]0 and then normalize x so that [x,y]=z.

Note that, in the previous lemma, x+y+z is a subalgebra isomorphic to a Heisenberg algebra: 𝔤= { (0xz0y0) |x,y,z } . and let x=(010000), y=(000010), z=(001000). Then 𝔤=x+y+z, [x,y]=z and z=Z(𝔤). So, if G is a nilpotent Lie group with one dimensional center the G0=exp(Z𝔤(y)), whereZ𝔤(y)= yzW, and G0 is normal since etxehe-tx=et[x,h]+, where t[x,h]+Z𝔤(y) since only involves brackets of x and h and Z𝔤(y) is an ideal. So G= { g0etx |g0 G0,t } . Since Z𝔤(y) is an ideal of 𝔤, G0=exp(Z𝔤(y)) is a normal subgroup ofG, and, in fact, every element of G can be written uniquely in the form g0etx, g0G0,t. Since G0 is normal in G, the group G acts on G0 be automorphisms. Let V0 be an irreducible representation of G0. Define gV0 to be the G0 module with the same vector space gV0=V and with G0 action g0·v= (g-1g0g) v,forg0 G0andvV0. Then gV0 is a new G0 representation and gV0 is irreducible if and only if V0 is (since, if PV0 is a submodule of V0 then gPgV0 is a submodule of gV0). So G acts on the irreducible representations of G0.

Why consider Lie algebras with one dimensional center? Suppose we are trying to prove that nilpotent Lie groups are monomial. Let M be an irreducible representation. Then Z(G) acts on M by scalars. Let z1,,zk be a basis of Z(𝔤). Let ϕ:Z(𝔤) be the map such that e2πiϕ(z)= ρ(ez),forz Z(𝔤),and let Z0= {zZ(𝔤)|ϕ(z)=0}. So zZ0 if i=1kciϕ(zi)=0. So dim(z0)dimZ-1 and exp(Z0) acts trivially on M. Let ρ:GGL(M) has a nontrivial kernel unless dimZ(𝔤)=1. In the case when dimZ(𝔤)=1 we can use Kirillov’s lemma to obtain the normal subgroup G0=exp(Z𝔤(y)).

The Heisenberg group

The Heisenberg group G= { (1xz1y1) |x,y,z } . has the Lie algebra 𝔤= { (0xz0y0) |x,y,z } , where we exhibit 𝔤 as a Lie subalgebra of Mn() since H is a subgroup of GLn(). Setting X=(010000), Y=(000010), Z=(001000). 𝔥=span-{X,Y,Z} with bracket determined by [X,Y]=Z, [X,Z]=0, [Y,Z]=0.

Since [𝔤,[𝔤,𝔤]]=0, the lower central series of 𝔤 is 𝔤[𝔤,𝔤]0, where[𝔤,𝔤]= { (00z000) |z } , is the center of 𝔤. The exponential map is given by 𝔤 G (0xz0y0) (1xz+xy/21y1) , since exp(0xz0y0) =1+(0xz0y0) +12=(1xz+xy/21y1). Let X=(0xz0y0) 𝔤,g=eX= (1xz+xy/21y1) G,andY= (0ac0b0) 𝔤. Then Ad𝔤Y=gYg-1= (0ac+bx0b0) (1-x-z+xy/21-y1)= (0a-ay+bx+c0b0) describes the adjoint representation of G. (Alternatively one can use the formula (AdeX)(Y)=(eadX)(Y) to do this calculation.)

To compute the coadjoint representation we need a good way of looking at 𝔤*=Hom(𝔤,). Use the trace on Mn() Tr: Mn() X Tr(X) to define a symmetric bilinear form ,:Mn()×Mn() by X,Y= Tr(XY),for X,Y𝔤. The form , provides a vector space isomorphism Mn() Mn()* X X,· where X,·: Mn() Y X,Y ,forXMn (). Use this isomorphism to identify Mn() and Mn()*. Our favourite basis of Mn() is {Eij|1i,jn}, where Eij denotes the matrix with a 1 in the (i,j)th entry and 0 everywhere else. Then {Eij*=Eji|1i,jn} is the dual basis with respect to ,. Since 𝔤=span-{E12,E13,E23}, 𝔤*=span- {E12*,E13*,E23*}= { (000α00γβ0) |α,β,γ } . Let g=eX= (1xz+xy/21y1) G,ϕ= (000α00γβ0) 𝔤*,Y= (0ac0b0) 𝔤. Then (Adg*)(Y) = ϕ(Adg-1Y) =ϕ(g-1Yg)= ϕ,g-1Yg = (000α00γβ0), (0a-ay+bx+c0b0) = αa+βb+γay- γbx+γc = (0α+γy0γβ-γx0), (0ac0b0) . Thus Adg* (α,β,γ)= (α+γy,β-γx,γ), where we use the more concise notation (α,β,γ) for the matrix (000α00γβ0)𝔤*.

To compute the coadjoint orbits note that

(a) If γ0 then we can choose x,y so that Adg*(α,β,γ)=(0,0,γ).
So (0,0,γ) is in the orbit of (α,β,γ).
(b) If γ=0 then Adg*(α,β,γ)= (0,0,γ), for all gG.
So (α,β,0) is in an orbit all by itself.
Thus there are two kinds of coadjoint orbits Ωγ= { (α,β,γ) |α,β } ,γ\{0}, andΩα,β ={(α,β,0)}, α,β, with dimΩγ=2 and dimΩα,β=0.

Note that dim𝔤=3 and { (00c0b0) |b,c } is a two dimensional subalgerba of 𝔤 such that [𝔥,𝔥]=0. So 𝔥 is a subalgebra of 𝔤 which is subordinate to any ϕ𝔤*, ϕ([x,y])= ϕ(0)=0, for allx,y𝔥. So, if ϕ𝔤* then either (a) there is a 3 dimensional subalgebra subordinate to ϕ (which must be all of 𝔤), or (b) 𝔥 is a maximal subordinate subalgebra to ϕ.

(a) If ϕ=(α,β,0)= (000α000β0) Ωα,β then 𝔤 is subordinate to ϕ and G * eX e2πiϕ,X is the representation VΩα,β associated to the orbit Ωα,β. More precisely, VΩα,β=span-{v} and (1xz+xy/21y1) v=e2πi(αx+βy)v,forx,y,z, since ϕ,X= (000α000β0) , (0xzα0y0) =αx+βy.

(b) If ϕ=(α,β,γ)= (000α00γβ0) Ωγ,γ0, then 𝔤 is not subordinate to ϕ and so 𝔥 is a maximal subordinate subalgebra for ϕ. The one dimensional representation of H=exp(𝔥) is given by Wϕ: H * (10z1y1) e2πi(yβ+γz) since exp((00z0y0)) =(10z1y1) and (00z0y0) , (0α0γβ0) =yβ+γz. The representation of G corresponding to the orbit Ωγ is VΩγ= IndHG(Wϕ), and we can view elements of IndHG(Wϕ) as functions on H\G. Since H= { (10z1y1) |y,z } andG= { (1xz1y1) |y,z } , the elements (1t0101) ,t, are coset representatives of the cosets in H\G. So we may view elements of VΩγ as functions f: = Wϕ t f(t) . If X=(0xz0y0) then (eXf)(t)=f (teX)=f ( (1t0101) (1xz+xy/21y1) ) =f ( (10a21a11) (1s0101) ) , where a1=y, s=t+x, and a2=z+(xy)/2+ty. So (eXf)(t) = f ( (1xz+xy/2+ty1y1) (1t+x0101) ) = (1xz+xy/2+ty1y1)f ((1t+x0101)) = e2πi(yβ+γz+γ(xy)/2+γty) f(t+x), and this formula describes the action of G on VΩγ.

Summary: For each orbit Ωα,β, α,βRR, we get a one dimensional representation VΩα,β=v with action eXv= e2πi(αx+βy) v,ifX= (0xz0y0) 𝔤, and for each orbit Ωy, γ0, we get an action of G on functions WΩγ= {f:} given by(eXf)(t) =e2πi(yβ+γz+γ(xy)/2+γty) f(t+x).

Let us show that the representations Vγ are irreducible by computing HomG(VΩg,VΩg)= HomG(IndHG(Wϕ),IndHG(Wϕ)) { T:GHom(Wϕ,Wϕ) |T(h1gh2) =T(g) } . Then T(1xz1y1) = T ( (10z1y1) (1x0101) ) =e2πi(βy+γz) T(x) = T ( (1x0101) (10z-xy1y1) ) =e2πi(βy+γz-γxy) T(x). So T(x)=0 unless e2πi(-γxy)=1 for all y. So T(x)=0 unless x=0. So T is determined by its value at the identity matrix. So dimHomG(VΩg,VΩg)=1. So VΩg is irreducible.

The following is an attempt (correct??) to make this same argument work in general.

Let ϕ𝔤* and let 𝔥 be a maximal subalgebra of G subordinate to ϕ. Then (if Vϕ is unitary) Vϕ=IndHG(Wϕ) is irreducible.

Proof.

Let x be a coset representative for a coset in H\G. Let t:GHom(Wϕ,Wϕ) be such that T(h1xh2)= h1T(x)h2, for allh1,h2H. Let η𝔥. Then e2πiϕ(η) T(x)=T(eηx) =T(xx-1eηx) =T(xeAdx-1η) =T(x)e2πiϕ(Adx-1η). So T(x)=0 unless ϕ(Adx-1η)=ϕ(η) for all η𝔥. So T(x)=0 unless ϕ(ead(-X)η)=ϕ(η). So T(x)=0 unless ϕ((ead(-X)-1)η)=0. Now ad(-X)=log (1+(ead(-X)-1)) =(ead(-X)-1)- (ead(-X)-1)22 +. So ϕ(ad(-X)η)=0 for all η𝔥. So ϕ(-[X,η])=0 for all η𝔥.

Let 𝔥=X𝔥. We know X𝔥 since x is a coset representative of H\G (not 1). Then 𝔥 is a subalgebra of 𝔤 subordinate to ϕ. This is a contradiction to the maximality of 𝔥. So T(x)=0 unless x=1. So T is determined by its value at e0=1. So dimHomG(VΩ,VΩ)=1.

Remark 1. Let be the subgroup of G given by = { (10n101) |n } . The group G/ is often called the Heisenberg group. This group has a subgroup S1= { (10t101) |t } {z||z|=1} /, which is contained in Z(G/)=[G/,G/], and this can be used to show that G/ does not have a faithful finite dimensional representation. The Lie group G is the simply connected cover of G/. There is an exact sequence 1S1G/ R21and if ξ,η= ξ1η2-ξ2η1, then G/= { uexp(ξ)|u S1,ξ2 } ,withuexp(ξ) ·vexp(η)=uv eiξ,η exp(ξ+η).

Let P,Q be such that PQ-QP=-i. Then eiaPeibQ=- eiabeibQ eiaP and G/ is the group generated by eiaP and eibQ. Let Ta, Mb and Uc be the operators on L2() given by (Taf)(x)= f(x-a),Mb f=e2πiλxf, Ucf=e2πiλc f,for fixedλ*. Then this is a unitary representation of G/ for each λ and G/{TaMbUc|a,b,c}.

Remark 2. If P,Q are such that PQ-QP=-i and if a=12(P+iQ) anda*=12 (P-iQ),then [a,a*]=1. Then, on L2(), a*acts by -i2 (ddx+x) anda*Ω =0forΩ= e-12x2. So Ω is a lowest weight vector and {anΩ}n0 is a basis of L2(). 𝔰𝔩2() ( Lie algebra generated by { (i/2)P2, (i/2)(PQ+QP) ,(i/2)Q2 } ) 𝔤.

Remark 3. If P=(010000), Q=(000010), Z=(001000), then [P,Q]=Z and these act on functions VΩγ= {f:}, via the differential of the representation VΩγ of G. Then (Zf)(t)= 2πiγf(t). In particular, by considering the action of P,Q on VΩγ when γ=-1/2π, we have operators P,Q on functions f: which satisfy [P,Q]=PQ-QP=-i. This solves a basic problem in quantum mechanics, see [Dirac] and [Heis].

Remark 4. Define an action of SL2()= { (abcc) |ad-bc=1 } on 𝔤 by (abcc) P=aP+cQ, (abcc) Q=bP+dQ, (abcc) Z=Z. Then [ (abcc) P, (abcc) Q ] =[aP+cQ,bP+dQ]= ad[P,Q]+cb [Q,P]=(ad-bc) [P,Q]=[P,Q], and so SL2() acts on 𝔤 by automorphisms. Let mSL2() and let ϕ𝔤*. Let 𝔥 be subordinate to ϕ. Say ϕ=(0,0,γ). Then m𝔥 is subordinate to ϕ (since m𝔥 is, after all, abelian). Then let us make the isomorphism IndHG(Wϕ) IndHG(Wϕ) f f where H=exp(m𝔥)....

Notes and References

[Bou] N. Bourbaki, Lie groups and Lie algebras, Hermann, Paris, ???.

[Bump] D. Bump, Automorphic forms and representations, Cambridge University Press, ????.

[CorGrn] L. Corwin and F.P. Greenleaf, Representations of Nilpotent Lie groups and their applications, Cambridge University Press, ????.

[Dirac] P.A.M. Dirac, ?????????, ??????????, ????.

[Heis] W. Heisenberg, The Physical Principles of the Quantum Theory, Dover, 1930.

[CG] N. Chriss and V. Ginzburg, Representation Theory and Complex Geometry, Birkhäuser, 1997.

Notes and references

This is a typed exert of Representation theory Lecture notes: Chapter 3 by Arun Ram. Research supported in part by National Science Foundation grant DMS-9622985.

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