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 $f_{*}:𝒞\to 𝒟$ and $f^{*}:𝒟\to 𝒞$ be functors. Then $f_{*}$ is a right adjoint to $f^{*}$ and $f^{*}$ is a left adjoint to $f_{*}$ if, for each $D\in 𝒟$ and $C\in 𝒞$ there is a natural vector space isomorphism $${\text{Hom}}_{𝒞}(f^{*}D,C)\stackrel{\Phi }{\to }{\text{Hom}}_{𝒟}(D,f_{*}C)\text{.}$$ For $C\in 𝒞$ and $D\in 𝒟$ define $${\tau }_C={\Phi }^{-1}\left({\text{id}}_{f_{*}C}\right)\in {\text{Hom}}_{𝒞}(f^{*}{f}_{*}C,C)\text{and}{\phi }_D=\Phi \left({\text{id}}_{f^{*}D}\right)\in {\text{Hom}}_{𝒟}(D,f_{*}{f}^{*}D)\text{.}$$ In general ${\tau }_C$ and ${\varphi }_D$ are neither injective nor surjective, see the examples below in ??? and ???.

The functors ${\text{Hom}}_B(M,·)$ and $M{\otimes }_Z·$

Let $B$ and $Z$ be algebras and let $M$ be a left $B\text{-module}$ and a right $Z\text{-module.}$ If $N$ is a left $Z\text{-module}$ then ${\text{Hom}}_B(M,N)$ is a left $Z\text{-module}$ with $Z\text{-action}$ given by $$\left(z\varphi \right)\left(m\right)=\varphi \left(mz\right),\text{for}\hspace{0.17em}\varphi \in {\text{Hom}}_A(M,N),\hspace{0.17em}m\in M,z\in Z\text{.}$$ If $P$ is a left $Z\text{-module}$ then $M{\otimes }_{Z}P$ is the left $B\text{-module}$ which as a $\mathbb{Z}\text{-module}$ is given by generators $m\otimes p,$ $m\in M,$ $p\in P,$ and relations $$\begin{array}{cccc}(m_1+m_2)\otimes p& =& m_1\otimes p+m_2\otimes p,& \text{for}\hspace{0.17em}{m}_1,m_2\in M,p\in P,\\ m\otimes (p_1+p_2)& =& m\otimes p_1+m\otimes p_2,& \text{for}\hspace{0.17em}m\in M,p_1,p_2\in P,\\ rm\otimes p& =& m\otimes rp=r(m\otimes p),& \text{for}\hspace{0.17em}r\in \mathbb{Z},\hspace{0.17em}m\in M,p\in P,\end{array}$$ and which has $B\text{-action}$ given by $$b(m\otimes p)=bm\otimes p,\text{for}\hspace{0.17em}m\in M,p\in P,\hspace{0.17em}\text{and}\hspace{0.17em}b\in B\text{.}$$ The covariant functors $$\begin{array}{cccc}{\text{Hom}}_B(M,·):& \left\{\text{left}\hspace{0.17em}B\text{-modules}\right\}& ⟶& \left\{\text{left}\hspace{0.17em}Z\text{-modules}\right\}\\ M{\otimes }_Z·:& \left\{\text{left}\hspace{0.17em}Z\text{-modules}\right\}& ⟶& \left\{\text{left}\hspace{0.17em}B\text{-modules}\right\}\end{array}$$ are adjoint since the map $$\begin{array}{ccc}{\text{Hom}}_Z(P,{\text{Hom}}_B(M,N))& \stackrel{\Phi }{⟶}& {\text{Hom}}_B(M{\otimes }_{Z}P,N)\\ \begin{array}{cccc}\psi :& P& \to & {\text{Hom}}_B(M,N)\\ & p& \mapsto & \begin{array}{cccc}{\psi }_p:& M& \to & N\\ & m& \mapsto & {\psi }_p\left(m\right)\end{array}\end{array}& ⟼& \begin{array}{cccc}\Phi \left(\psi \right):& M{\otimes }_{Z}P& \to & N\\ & m\otimes p& \mapsto & {\varphi }_p\left(m\right)\end{array}\\ \begin{array}{cccc}{\Phi }^{-1}\left(\varphi \right):& P& \to & {\text{Hom}}_B(M,N)\\ & p& \mapsto & \begin{array}{cccc}{\varphi }_p:& M& \to & N\\ & m& \mapsto & \varphi (m\otimes p)\end{array}\end{array}& ⟼& \begin{array}{cccc}\varphi :& M{\otimes }_{Z}P& \to & N\\ & m\otimes p& \mapsto & \varphi (m\otimes p)\end{array}\end{array}$$ is a $\mathbb{Z}\text{-module}$ isomorphism.

The functor ${\text{Hom}}_B(M,-)$ is very different from the functor ${\text{Hom}}_B(-,M)\text{.}$ There is always a canonical isomorphism $$\begin{array}{ccc}{\text{Hom}}_B(B,M)& ⟶& M\\ \varphi & ⟼& \varphi \left(1\right),\end{array}\text{but}\text{the}\hspace{0.17em}\text{dual module}\hspace{0.17em}\text{to}\hspace{0.17em}M,{\text{Hom}}_B(M,B),$$ can, in general, be much larger than $M$ (take $B=ℂ$ and $M$ an infinite dimensional vector space over $ℂ\text{).}$

The functor ${\text{Hom}}_B(M,-)$ is left exact and the functor $M{\otimes }_Z-$ is right exact, i.e. if $0\to P\prime \to P\to P″$ is exact then $$0\to {\text{Hom}}_B(M,P\prime )\to {\text{Hom}}_B(M,P)\to {\text{Hom}}_B(M,P″)\hspace{0.17em}\text{is exact,}$$ and if $N\prime \to N\to N″\to 0$ is exact then $$M{\otimes }_{Z}N\prime \to M{\otimes }_{Z}N\to M\otimes N″\to 0\hspace{0.17em}\text{is exact.}$$ A left $A\text{-module}$ $M$ is projective if the functor ${\text{Hom}}_A(M,·)$ is exact and a right $Z\text{-module}$ $M$ is flat if the functor $M{\otimes }_Z·$ is exact.

Define Ext and Tor here.

Example 1. Let $\iota :Z\to B$ be a injective algebra homomorphism. Then $M=B$ is a left $B\text{-module}$ and right $Z\text{-module.}$ The adjointness of the two functors $${\text{Res}}_Z^B={\text{Hom}}_B(B,-)={\iota }_{*}\text{and}{\text{Ind}}_Z^B=B{\otimes }_Z-={\iota }^{*},$$ is $${\text{Hom}}_B({\text{Ind}}_Z^B\left(P\right),N)\cong {\text{Hom}}_Z(P,{\text{Res}}_Z^{B}N)\text{.}$$

Example 2. Let $\iota :Z\to B$ be a injective algebra homomorphism. Then $M=B$ is a left $Z\text{-module}$ and right $B\text{-module.}$ The adjointness of the two functors $${\text{Res}}_Z^B=B{\otimes }_B-={\iota }_{!}\text{and}{\text{coInd}}_Z^B={\text{Hom}}_Z(B,-)={\iota }^{!},$$ is $${\text{Hom}}_Z({\text{Res}}_Z^B\left(N\right),P)\cong {\text{Hom}}_B(N,{\text{coInd}}_Z^B\left(P\right))\text{.}$$

Example 3. Let $\pi :Z\to B$ be a surjective algebra homomorphism. Then $M=B$ is a left $B\text{-module}$ and right $Z\text{-module.}$ The adjointness of the two functors $${\text{Inf}}_B^Z={\text{Hom}}_B(B,-)={\pi }_{*}\text{and}{\text{Def}}_Z^B=B{\otimes }_Z-={\pi }^{*},$$ is $${\text{Hom}}_B({\text{Def}}_B^Z\left(P\right),N)\cong {\text{Hom}}_Z(P,{\text{Inf}}_B^Z\left(N\right))\text{.}$$ If $K=\text{ker}\hspace{0.17em}\pi$ then $${\text{Def}}_Z^B\left(P\right)=B{\otimes }_{Z}P\stackrel{\sim }{⟶}P/KP,$$ since the map $\varphi :P\to B{\otimes }_{Z}P$ given by $\varphi \left(p\right)=1\otimes p$ has kernel $KP\text{.}$

Example 4. Let $\pi :Z\to B$ be a surjective algebra homomorphism. Then $M=B$ is a left $Z\text{-module}$ and right $B\text{-module.}$ The adjointness of the two functors $${\text{Inf}}_B^Z=B{\otimes }_Z-={\pi }_{!}\text{and}{\text{coDef}}_B^Z={\text{Hom}}_Z(B,-)={\pi }^{!},$$ is $${\text{Hom}}_Z({\text{Inf}}_B^Z\left(N\right),P)\cong {\text{Hom}}_B(N,{\text{coDef}}_B^Z\left(P\right))\text{.}$$ If $K=\text{ker}\hspace{0.17em}\pi$ then $${\text{Hom}}_Z(B,P)\stackrel{\sim }{⟶}{\text{Hom}}_Z(Z/K,P)\stackrel{\sim }{⟶}\{f\in {\text{Hom}}_Z(Z,P)\hspace{0.17em}|\hspace{0.17em}f\left(k\right)=0\hspace{0.17em}\text{for}\hspace{0.17em}k\in K\},$$ and so $${\text{Hom}}_Z(B,P)\stackrel{\sim }{⟶}\{p\in P\hspace{0.17em}|\hspace{0.17em}Kp=0\}\text{.}$$

Example 5. Every algebra homomorphism $$\varphi :Z\to B\text{is a composition}\varphi :Z\stackrel{\pi }{⟶}\text{im}\left(\varphi \right)\stackrel{\iota }{⟶}B$$ of a surjective and an injective algebra homomorphism.

Example 6. Let $B$ be an algebra and let $e\in B$ be an idempotent. Let $$M=Be\text{and}Z=eBe\cong {\text{End}}_B\left(M\right)\text{.}$$ Then the $Z\text{-module}$ obtained by applying the functor ${\text{Hom}}_B(M,-)$ to a $B\text{-module}$ $N$ is $${\text{Hom}}_B(M,N)={\text{Hom}}_B(Be,N)=eN\text{.}$$

(What does $Be{\otimes }_{Z}P=Be{\otimes }_{eBe}N$ look like?)

Computations for finite groups

Let $G$ be a finite group and let $H$ be a subgroup. If $P$ is an $H\text{-module}$ then $${\text{Ind}}_H^G\left(P\right)\cong {\text{coInd}}_H^G\left(P\right)\text{as}\hspace{0.17em}G\text{-modules.}$$ To construct the isomorphism fix a basis $\left\{p_j\right\}$ of $P$ and a set $\left\{g_j\right\}$ of coset representatives of the cosets in $G/H$ so that $${\text{Ind}}_H^G\left(P\right)=ℂG{\otimes }_{ℂH}P\text{has basis}\{g_i\otimes p_j\}\text{.}$$ An element $\psi \in {\text{coInd}}_H^G\left(P\right)={\text{Hom}}_{ℂH}(ℂG,P)$ is given by the values ${\psi }_{ij}\in ℂ$ given by $$\psi \left(g_i^{-1}\right)=\sum _j{\psi }_{ij}{p}_j,\text{since}\hspace{0.17em}\psi \left(h{g}_i^{-1}\right)=\psi \left(g_i^{-1}\right)\hspace{0.17em}\text{for all}\hspace{0.17em}h\in H\text{.}$$ The elements $\psi \in ℂG{\otimes }_{ℂH}P$ are given by the values ${\psi }_{ij}\in ℂ$ determined by $$\psi =\sum _{i,j}{\psi }_{ij}(g_i\otimes p_j)=\sum _i{g}_i\otimes \left({\sum }_j{\psi }_{ij}{p}_j\right)=\sum _i{g}_i\otimes \psi \left(g_i^{-1}\right)=\frac{1}{\left|H\right|}\sum _{g\in G}g\otimes \psi \left(g^{-1}\right)\text{.}$$ Hence the isomorphism must be given by $$\begin{array}{ccccccc}ℂG{\otimes }_{ℂH}P& =& {\text{Ind}}_H^G\left(P\right)& \stackrel{\sim }{⟶}& {\text{coInd}}_H^G\left(P\right)& =& {\text{Hom}}_{ℂH}(ℂG,P)\\ & & \frac{1}{\left|H\right|}\sum _{g\in G}g\otimes \psi \left(g^{-1}\right)& ⟼& \psi \end{array}$$

Consider $\iota :ℂH\hookrightarrow ℂG\text{.}$ Then, for an $H\text{-module}$ $V,$ $${\iota }^{*}V={\text{Ind}}_H^G\left(V\right)=ℂG{\otimes }_{ℂH}V\text{and}{\iota }_{*}V={\text{Res}}_H^G\left(V\right)={\text{Hom}}_G(ℂG,V)\cong V,$$ where the isomorphism ${\text{Hom}}_G(ℂG,V)\stackrel{\sim }{⟶}V$ is given by $\varphi \mapsto \varphi \left(1\right)\text{.}$ The isomorphisms $${\text{Hom}}_H({\text{Res}}_H^{G}V,{\text{Res}}_H^{G}V)={\text{Hom}}_H({\iota }_{*}V,{\iota }_{*}V)\stackrel{\Psi }{⟶}{\text{Hom}}_G({\iota }^{*}{\iota }_{*}V,V)={\text{Hom}}_G(ℂG{\otimes }_{ℂH}V,V)$$ and $${\text{Hom}}_G({\text{Ind}}_H^{G}V,{\text{Ind}}_H^{G}V)={\text{Hom}}_G({\iota }^{*}V,{\iota }^{*}V)\stackrel{\Phi }{⟶}{\text{Hom}}_G(V,{\iota }_{*}{\iota }^{*}V)={\text{Hom}}_H(V,ℂG{\otimes }_{ℂH}V)$$ are given explicitly by $$\begin{array}{cccc}\Psi \left(b\right):& ℂG{\otimes }_{ℂH}V& \to & V\\ & g\otimes v& \mapsto & gb\left(v\right),\end{array}\text{for}\hspace{0.17em}b\in {\text{Hom}}_H(V,V)={\text{Hom}}_H({\text{Res}}_H^{G}V,{\text{Res}}_H^{G}V),$$ and $$\begin{array}{cccc}\Phi \left(b\right):& V& \to & {\text{Hom}}_G(ℂG,ℂG{\otimes }_{ℂH}V)\cong ℂG{\otimes }_{ℂH}V\\ & v& \mapsto & b(1\otimes v)\end{array}\text{for}\hspace{0.17em}b\in {\text{Hom}}_G({\iota }^{*}V,{\iota }^{*}V)\text{.}$$ Thus $$\begin{array}{cccc}\Psi \left(\text{id}\right):& ℂG{\otimes }_{ℂH}V& \to & V\\ & g\otimes v& \mapsto & gv\end{array}\text{and}\begin{array}{cccc}\Phi \left(\text{id}\right):& V& \to & ℂG{\otimes }_{ℂH}V\\ & v& \mapsto & 1\otimes v\text{.}\end{array}$$

Consider $\iota :ℂH\hookrightarrow ℂG\text{.}$ Then $${\iota }^{!}V={\text{coInd}}_H^G\left(V\right)\cong {\text{Ind}}_H^G\text{and}{\iota }_{!}V={\text{Res}}_H^{G}V=ℂG{\otimes }_{ℂG}V\cong V,$$ where the isomorphism ${\text{coInd}}_H^{G}V={\text{Hom}}_H(ℂG,V)\stackrel{\sim }{⟶}ℂG{\otimes }_{ℂH}V={\text{Ind}}_H^{G}V$ is given by (???) and the isomorphism $V\stackrel{\sim }{⟶}ℂG{\otimes }_{ℂG}V={\text{Res}}_H^{G}V$ is given by $v\mapsto 1\otimes v\text{.}$ The isomorphisms $${\text{Hom}}_H({\text{Res}}_H^{G}V,{\text{Res}}_H^G)={\text{Hom}}_H({\iota }_{!}V,{\iota }_{!}V)\stackrel{\Psi }{⟶}{\text{Hom}}_G(V,{\iota }^{!}{\iota }_{!}V)={\text{Hom}}_G(V,{\text{coInd}}_H^{G}V)$$ and $${\text{Hom}}_G({\text{coInd}}_H^{G}V,{\text{coInd}}_H^{G}V)={\text{Hom}}_G({\iota }^{!}V,{\iota }^{!}V)\stackrel{\Phi }{⟶}{\text{Hom}}_H({\iota }_{!}{\iota }^{!}V,V)={\text{Hom}}_H({\text{coInd}}_H^{G}V,V)$$ are given explicitly by $$\begin{array}{cccc}\Psi \left(b\right):& V& \to & {\text{Hom}}_H(ℂG,V)\cong ℂG{\otimes }_{ℂH}V\\ & v& \mapsto & \frac{1}{\left|H\right|}\sum _{g\in G}g\otimes b\left(g^{-1}v\right)\end{array},\text{for}\hspace{0.17em}b\in {\text{Hom}}_H({\text{Res}}_H^{G}V,{\text{Res}}_H^{G}V)={\text{Hom}}_H(V,V),$$ and $$\begin{array}{cccc}\Phi \left(b\right):& ℂG{\otimes }_{ℂH}V& \to & 1\otimes V\cong V\\ & g\otimes v& \mapsto & b(g\otimes v){|}_{1\otimes V},\end{array}\text{for}\hspace{0.17em}b\in {\text{Hom}}_G(ℂG{\otimes }_{ℂH}V,ℂG{\otimes }_{ℂH}V)\text{.}$$ Thus $$\begin{array}{cccc}\Psi \left(\text{id}\right):& V& \to & ℂG{\otimes }_{ℂH}V\\ & v& \mapsto & \frac{1}{\left|H\right|}\sum _{g\in G}g\otimes g^{-1}v,\end{array}\text{and}\begin{array}{cccc}\Phi \left(\text{id}\right):& ℂG{\otimes }_{ℂH}V& \to & V\\ & g\otimes v& \mapsto & \{\begin{array}{cc}v,& \text{if}\hspace{0.17em}g\in H,\\ 0,& \text{if}\hspace{0.17em}g\notin H\text{.}\end{array}\end{array}$$

Now define $$\begin{array}{cccccc}e:& {\iota }^{*}{\iota }_{*}V& \stackrel{\Psi \left(\text{id}\right)}{⟶}& V& \stackrel{\Phi \left(\text{id}\right)}{⟶}& {\iota }^{!}{\iota }_{!}V\\ & g\otimes v& \mapsto & gv& \mapsto & \frac{1}{\left|H\right|}\sum _{\ell \in G}\ell \otimes {\ell }^{-1}gv\end{array}$$ and define $$\begin{array}{cccccc}\epsilon \left(b\right):& V& \stackrel{\Phi \left(b\right)}{⟶}& {\iota }^{!}{\iota }_{!}V={\iota }^{*}{\iota }_{*}V& \stackrel{\Psi \left(\text{id}\right)}{⟶}& V\\ & v& \mapsto & \frac{1}{\left|H\right|}\sum _{g\in G}g\otimes b\left(g^{-1}v\right)& \mapsto & \frac{1}{\left|H\right|}\sum _{g\in G}gb{g}^{-1}v,\end{array}$$ for each $b\in {\text{Hom}}_H({\iota }_{!}V,{\iota }_{!}V)={\text{Hom}}_H({\text{Res}}_H^{G}V,{\text{Res}}_H^{G}V)\text{.}$ Then $$e\in {\text{End}}_G\left({\text{Ind}}_H^G{\text{Res}}_H^{G}V\right)\text{and}\epsilon :{\text{End}}_H\left({\text{Res}}_H^{G}V\right)⟶{\text{End}}_G(V,V)\text{.}$$

Similarly, define $$\begin{array}{cccccc}e:& {\iota }_{!}{\iota }^{!}V& \stackrel{\Psi \left(\text{id}\right)}{⟶}& V& \stackrel{\Phi \left(\text{id}\right)}{⟶}& {\iota }_{*}{\iota }^{*}V\\ & g\otimes v& \mapsto & \{\begin{array}{cc}v,& \text{if}\hspace{0.17em}g\in H,\\ 0,& \text{if}\hspace{0.17em}g\notin H,\end{array}& \mapsto & \{\begin{array}{cc}1\otimes v,& \text{if}\hspace{0.17em}g\in H,\\ 0,& \text{if}\hspace{0.17em}g\notin H,\end{array}\end{array}$$ and define $$\begin{array}{cccccc}\epsilon \left(b\right):& V& \stackrel{\Phi \left(b\right)}{⟶}& {\iota }_{*}{\iota }^{*}V={\iota }_{!}{\iota }^{!}V& \stackrel{\Psi \left(\text{id}\right)}{⟶}& V\\ & v& \mapsto & b(1\otimes v)& \mapsto & b(1\otimes v){|}_{1\otimes V},\end{array}$$ for each $b\in {\text{Hom}}_G({\iota }^{*}V,{\iota }^{*}V)={\text{Hom}}_G({\text{Ind}}_H^{G}V,{\text{Ind}}_H^{G}V)\text{.}$ Then $$e\in {\text{End}}_H\left({\text{Res}}_H^G{\text{Ind}}_H^{G}V\right)\text{and}\epsilon :{\text{End}}_G\left({\text{Ind}}_H^{G}V\right)⟶{\text{End}}_H(V,V)\text{.}$$

Thus, hopefully the general picture is that we can consider a sequence of injective algebra homomorphisms $${\text{End}}_B\left(V\right)\stackrel{{\iota }_{!}}{⟶}{\text{End}}_Z({\iota }_{!}V,{\iota }_{!}V)\stackrel{{\iota }^{!}}{⟶}{\text{End}}_B({\iota }^{!}{\iota }_{!}V,{\iota }^{!}{\iota }_{!}V)$$ and define $$e:{\iota }^{*}{\iota }_{*}\stackrel{\Psi \left(\text{id}\right)}{⟶}V\stackrel{\Phi \left(\text{id}\right)}{⟶}{\iota }^{!}{\iota }_{!}V\text{and}\epsilon \left(b\right):V\stackrel{\Phi \left(b\right)}{⟶}{\iota }^{!}{\iota }_{!}V={\iota }^{*}{\iota }_{*}V\stackrel{\Phi \left(\text{id}\right)}{⟶}V,$$ for $b\in {\text{End}}_Z({\iota }_{!}V,{\iota }_{!}V)$ so that $$e\in {\text{End}}_B({\iota }^{!}{\iota }_{!}V,{\iota }^{!}{\iota }_{!}V)\text{and}\epsilon :{\text{End}}_Z({\iota }_{!}V,{\iota }_{!}V)⟶{\text{End}}_B\left(V\right)\text{.}$$ We then want to show

(a)	$e^2=e,$
(b)	$ea=ae,$ for $a\in {\text{End}}_B\left(V\right),$
(c)	$ebe=\epsilon \left(b\right)e=e\epsilon \left(b\right),$ for $b\in {\text{End}}_Z({\iota }_{!}V,{\iota }_{!}V)\text{.}$

Note that (a) can be proved by applying (c) if one knows that $\epsilon \left(1\right)=1,$ i.e. $\epsilon \left(\text{id}\right)=\text{id.}$ Then $$e^2=e·1·e=\epsilon \left(1\right)e=1·e=e\text{.}$$ Also (b) can be proved by applying (c) if one knows that $\epsilon \left(a\right)=a$ for $a\in {\text{End}}_B\left(V\right)\text{.}$ Then $$ae=\epsilon \left(a\right)e=e\epsilon \left(a\right)=ea\text{.}$$ 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 $P{P}_k\left(n\right)$ be the subalgebra of the partition algebra generated by the planar diagrams. There is a bijection between diagrams in $P{P}_k\left(n\right)$ and Temperly-Lieb diagrams on $2k$ vertices such that $$p_i⟷e_{2i-1}\text{and}{b}_i⟷e_{2i}\text{.}$$ 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. $$\text{PICTURE}⟶\text{PICTURE}⟶\text{PICTURE}$$ Then the relations $e_i^2=x{e}_i$ and $e_i{e}_{i\pm 1}{e}_i=e_i$ in ${TL}_{2k}\left(x\right)$ correspond to the relations $$p_i^2=n{p}_i,b_i^2=b_i,$$ and $$p_i{b}_i{p}_i=p_i,b_i{p}_i{b}_i=b_i,b_i{p}_{i+1}{b}_i=b_i,p_{i+1}{b}_i{p}_{i+1}=p_{i+1},$$ respectively. Hence these relations should provide a full set of generators and relations for the planar partition algebra.

Example 1. A $B\text{-module}$ $M$ is projective if and only if there is a $B\text{-module}$ $M\prime$ such that $M\oplus M\prime$ is a free $B\text{-module.}$

Simple modules

Assume that

(a)	the action of $B$ on $M$ generates ${\text{End}}_Z\left(M\right),$
(b)	the action of $Z$ on $M$ generates ${\text{End}}_B\left(M\right)\text{.}$

Then there are surjective maps $$B\stackrel{\beta }{⟶}{\text{End}}_Z\left(M\right)\text{and}Z\stackrel{\zeta }{⟶}{\text{End}}_B\left(M\right)$$ and so every ${\text{End}}_Z\left(M\right)\text{-module}$ is a $B\text{-module}$ and every ${\text{End}}_B\left(M\right)\text{-module}$ is a $Z\text{-module.}$

(a)	${\text{Hom}}_B(M,P)$ is always an ${\text{End}}_B\left(M\right)=\bar{Z}$ module.
(b)	$M{\otimes }_{Z}N$ is always an ${\text{End}}_Z\left(M\right)=\bar{B}$ module.
(c)	If $P$ is a simple $B\text{-module}$ and ${\text{Hom}}_B(M,P)\ne 0$ then $P$ is a simple $\bar{B}\text{-module.}$
(d)	If $N$ is a simple $\bar{Z}\text{-module}$ then $M{\otimes }_{Z}N\ne 0\text{.}$

		Proof.
	(a) If $z\in {\text{Ann}}_Z\left(M\right)$ then $\left(z\varphi \right)\left(m\right)=\varphi \left(zm\right)=0,$ for all $m\in M\text{.}$ So ${\text{Ann}}_Z\left(M\right)$ acts by $0$ on ${\text{Hom}}_B(M,P)\text{.}$ So ${\text{Hom}}_B(M,P)$ is an ${\text{End}}_B\left(M\right)=\bar{Z}$ module. (b) If $b\in {\text{Ann}}_B\left(M\right)$ then $b(m\otimes n)=bm\otimes n=0,$ for all $m\in M,$ $n\in N\text{.}$ So ${\text{Ann}}_B\left(M\right)$ acts by $0$ on $M{\otimes }_{Z}N\text{.}$ So $M{\otimes }_{Z}N$ is an ${\text{End}}_Z\left(M\right)=\bar{B}$ module. (c) Suppose $P$ is a simple $B\text{-module}$ and ${\text{Hom}}_B(M,P)\ne 0\text{.}$ Let $\varphi \in {\text{Hom}}_B(M,P),$ $\varphi \ne 0\text{.}$ Then $$M\stackrel{\varphi }{⟶}P⟶0$$ is an exact sequence since $P$ is simple. So $$\begin{array}{ccccc}{\text{Hom}}_B(M,M)& ⟶& {\text{Hom}}_B(M,P)& ⟶& 0\\ z& ⟼& z\varphi \end{array}$$ is an exact sequence. So $\bar{Z}\varphi =N$ where $\bar{Z}={\text{Hom}}_B(M,M)={\text{End}}_B\left(M\right)\text{.}$ So ${\text{Hom}}_B(M,P)$ is a simple $Z\text{-module.}$ (c) Assume $b\in {\text{Ann}}_B\left(M\right)$ and $\varphi \in {\text{Hom}}_B(M,P)\ne 0,$ Since $P$ is simple $\varphi$ is surjective and so every element $p\in P$ can be written in the form $p=\varphi \left(m\right)$ for some $m\in M\text{.}$ So $bp=b\varphi \left(m\right)=\varphi \left(bm\right)=0$ for all $p\in P\text{.}$ So ${\text{Ann}}_B\left(M\right)$ acts by $0$ on $P\text{.}$ So $P$ is an ${\text{End}}_Z\left(M\right)=\bar{B}$ module. (d) Let $N$ be a simple $\bar{Z}$ module. Let $n\in N$ be a nonzero element of $N\text{.}$ Then every element of $N$ can be written in the form $zn$ for some $z\in \bar{Z}\text{.}$ So, for every $z\in \bar{Z}$ such that $zn\ne 0,$ $$m\otimes zn=mz\otimes n\ne 0,$$ since $z$ is a nonzero element of ${\text{End}}_B\left(M\right)\text{.}$ So $M{\otimes }_{Z}N\ne 0\text{.}$ $\square$

Thus we have shown that

(a)	If $P$ is a simple $B\text{-module}$ such that ${\text{Hom}}_B(M,P)\ne 0$ then $P$ is a simple $\bar{B}\text{-module,}$
(b)	If $N$ is a simple $\bar{Z}\text{-module}$ then $M{\otimes }_{Z}N\ne 0$ and $M{\otimes }_{Z}N$ is a $\bar{B}\text{-module.}$

(a)	If $M$ is projective and $P$ is a simple $B\text{-module}$ such that ${\text{Hom}}_B(M,P)\ne 0$ then ${\text{Hom}}_B(M,P)$ is simple.
(b)	If $N$ is a simple $\bar{Z}\text{-module}$ and $M{\otimes }_{Z}N$ has a unique maximal proper submodule then there is a simple $B\text{-module}$ $P$ such that ${\text{Hom}}_B(M,P)\cong N\text{.}$

		Proof.
	Step 1. ${\text{Hom}}_B(M,M{\otimes }_{Z}N)\cong N\text{.}$ We shall show that the canonical homomorphism (???) is an isomorphism. Since $M{\otimes }_{Z}N\ne 0,$ some ${\varphi }_n\ne 0$ and so ${\alpha }_N$ is injective (since $N$ is simple). Let $\varphi \in {\text{Hom}}_B(M,M{\otimes }_{Z}N)\text{.}$ Fix a nonzero element $n_1\in N\text{.}$ Then every element $n\in N$ is of the form $$n=z{n}_1,\text{for some}\hspace{0.17em}z\in \bar{Z}\text{.}$$ For each $m\in M$ write $$\varphi \left(m\right)=m\prime \otimes n=m\prime \otimes z{n}_1=m\prime z\otimes n_1=m″\otimes n_1\text{.}$$ Then the map $$\begin{array}{cccc}\stackrel{\sim }{\varphi }:& M& ⟶& M\\ & m& ⟼& m″\end{array}$$ is an element of ${\text{End}}_A\left(M\right)\text{.}$ So $\stackrel{\sim }{\varphi }\left(m\right)=mz,$ for some $z\in \bar{Z}$ (and all $m\in M\text{).}$ So $$\begin{array}{cccc}\varphi :& M& ⟶& M{\otimes }_{Z}N\\ & m& ⟼& mz\otimes n_1=m\otimes z{n}_1\end{array}$$ and thus $\varphi ={\varphi }_{z{n}_1}\text{.}$ So ${\alpha }_N$ is surjective. So ${\alpha }_N$ is an isomorphism. Step 2. Let $P\prime$ be a proper submodule of $M{\otimes }_{Z}N\text{.}$ Then $$0⟶P\prime \hookrightarrow M{\otimes }_{Z}N$$ and so $$0⟶{\text{Hom}}_B(M,P\prime )⟶{\text{Hom}}_B(M,M{\otimes }_{Z}N)\cong N\text{.}$$ So ${\text{Hom}}_B(M,P\prime )$ is a submodule of $N\text{.}$ If ${\text{Hom}}_B(M,P\prime )=N$ then every element of ${\text{Hom}}_B(M,M{\otimes }_{Z}N)$ has image in the proper submodule $P\prime \subseteq M{\otimes }_{Z}N\text{.}$ This is a contradiction to the fact that all the generators $m\otimes n$ of $M{\otimes }_{Z}N,$ i.e in the image of some map $$\begin{array}{cccc}{\varphi }_n:& M& ⟶& M{\otimes }_{Z}N\\ & m& ⟼& m\otimes n\end{array}$$ in ${\text{Hom}}_B(M,M{\otimes }_{Z}N)\text{.}$ So ${\text{Hom}}_B(M,P\prime )$ is a proper submodule. Since $N$ is simple it follows that ${\text{Hom}}_B(M,P\prime )=0\text{.}$ Step 3. Let $\stackrel{\sim }{P}$ be the unique maximal proper submodule of $P\prime \text{.}$ Then ${\text{Hom}}_B(M,\stackrel{\sim }{P})=0$ and $P\prime /\stackrel{\sim }{P}$ is simple. The canonical map $$0⟶\stackrel{\sim }{P}⟶P\prime ⟶P\prime /\stackrel{\sim }{P}$$ gives rise to a map $$0⟶{\text{Hom}}_B(M,\stackrel{\sim }{P})⟶{\text{Hom}}_B(M,P\prime )⟶{\text{Hom}}_B(M,P\prime /\stackrel{\sim }{P})$$ and both the left hand side and the right hand side are simple (since $M$ is projective). So $${\text{Hom}}_B(M,P\prime /\stackrel{\sim }{P})\cong {\text{Hom}}_B(M,P\prime )\cong N\text{.}$$ So $P=P\prime /\stackrel{\sim }{P}$ is simple and ${\text{Hom}}_B(M,P)\cong N\text{.}$ $\square$

If $M$ is finitely generated and projective then $M{\otimes }_{Z}N$ has a unique maximal proper submodule.

		Proof.
	We have shows that if $P\prime$ is a proper submodule of $M{\otimes }_{Z}N$ then ${\text{Hom}}_B(M,P\prime )=0\text{.}$ Let $P_1,P_2$ be two proper submodules of $M{\otimes }_{Z}N\text{.}$ Then the exact sequence $$0⟶P_1\cap P_2⟶P_1\oplus P_2⟶P_1+P_2⟶0$$ yields an exact sequence $$0⟶{\text{Hom}}_B(M,P_1\cap P_2)⟶{\text{Hom}}_B(M,P_1\oplus P_2)⟶{\text{Hom}}_B(M,P_1+P_2)=0\text{.}$$ Since ${\text{Hom}}_B(M,P_1\oplus P_2)={\text{Hom}}_B(M,P_1)\oplus {\text{Hom}}_B(M,P_2)=0,$ $${\text{Hom}}_B(M,P_1\cap P_2)=0\text{and}{\text{Hom}}_B(M,P_1+P_2)=0\text{.}$$ By Bourbaki Algébre Ch. II, §6, Ex. 4, $${\text{Hom}}_B(M,\stackrel{⟶}{lim}{P}_i)=\stackrel{⟶}{lim}{\text{Hom}}_B(M,P_i)=0$$ as $P_i$ ranges over all proper submodules of $M{\otimes }_{Z}N,$ ordered by inclusion. Thus, if $$\stackrel{\sim }{P}=\text{sum of proper submodules of}\hspace{0.17em}M{\otimes }_{Z}N$$ then ${\text{Hom}}_B(M,\stackrel{\sim }{P})=0\text{.}$ Since ${\text{Hom}}_B(M,M{\otimes }_{Z}N)=N,$ $\stackrel{\sim }{P}$ must be a proper submodule of $M{\otimes }_{Z}N\text{.}$ So $\stackrel{\sim }{P}$ is the unique maximal proper submodule of $M\text{.}$ $\square$

$M$ is projective as a $B\text{-module}$ if and only if there is a $B\text{-module}$ $M\prime$ such that $M\oplus M\prime =B^{\oplus I},$ i.e. $M\oplus M\prime$ is a free $B\text{-module.}$

		Proof.
	$⟹\text{:}$ Let $I$ be a set of generators of $M$ and let $\varphi :B^{\oplus I}\to M$ be the canonical map. Let $K=\text{ker}\hspace{0.17em}\varphi \text{.}$ The exact sequence $$\begin{array}{cc}0⟶K⟶B^{\oplus I}\stackrel{\varphi }{⟶}M⟶0& \text{(0.5)}\end{array}$$ gives an exact sequence $$0⟶{\text{Hom}}_B(M,K)⟶{\text{Hom}}_B(M,B^{\oplus I})\stackrel{{\varphi }^{*}}{⟶}{\text{Hom}}_B(M,M)⟶0\text{.}$$ Since ${\varphi }^{*}$ is surjective there is a map $\psi :M\to B^{\oplus I}$ such that ${\varphi }^{*}\left(\psi \right)={\text{Id}}_M\text{.}$ So $\varphi \circ \psi ={\text{Id}}_M\text{.}$ So the first exact sequence splits, $$0⟶K⟶B^{\oplus I}\underset{\stackrel{\varphi }{⟶}}{\overset{\stackrel{\psi }{⟵}}{}}M⟶0\text{.}$$ So $B^{\oplus I}=\text{im}\hspace{0.17em}\psi \oplus K\cong M\oplus K\text{.}$ $⟸\text{:}$ If $M\oplus M\prime \cong B^{\oplus I}$ and $$0⟶P\prime ⟶P⟶P″⟶0$$ is an exact sequence of $B\text{-modules}$ then $$0⟶{\text{Hom}}_B(B^{\oplus I},P\prime )⟶{\text{Hom}}_B(B^{\oplus I},P)⟶{\text{Hom}}_B(B^{\oplus I},P″)⟶0$$ is the same as $$0⟶\coprod _{i\in I}{\text{Hom}}_B(B,P\prime )⟶\coprod _{i\in I}{\text{Hom}}_B(B,P)⟶\coprod _{i\in I}{\text{Hom}}_B(B,P″)⟶0$$ which is the same as $$0⟶\coprod _{i\in I}P\prime ⟶\coprod _{i\in I}P⟶\coprod _{i\in I}P″⟶0$$ which is exact since the first sequence is. So $$0⟶{\text{Hom}}_B(M\oplus M\prime ,P\prime )⟶{\text{Hom}}_B(M\oplus M\prime ,P)⟶{\text{Hom}}_B(M\oplus M\prime ,P″)⟶0$$ which is the same as $$\begin{array}{ccccccc}& & {\text{Hom}}_B(M,P\prime )& & {\text{Hom}}_B(M,P)& & {\text{Hom}}_B(M,P″)\\ 0& ⟶& \oplus & ⟶& \oplus & ⟶& \oplus & ⟶& 0\\ & & {\text{Hom}}_B(M\prime ,P\prime )& & {\text{Hom}}_B(M\prime ,P)& & {\text{Hom}}_B(M\prime ,P″)\end{array}$$ is exact. This forces that the sequences $$0⟶{\text{Hom}}_B(M,P\prime )⟶{\text{Hom}}_B(M,P)⟶{\text{Hom}}_B(M,P″)⟶0$$ and $$0⟶{\text{Hom}}_B(M\prime ,P\prime )⟶{\text{Hom}}_B(M\prime ,P)⟶{\text{Hom}}_B(M\prime ,P″)⟶0$$ are both exact. So both $M$ and $M\prime$ are projective. $\square$

(a)	If $M$ is a projective $B\text{-module}$ and $P$ is a simple $B\text{-module}$ such that ${\text{Hom}}_B(M,P)\ne 0$ then ${\text{Hom}}_B(M,P)$ is a simple $Z$ module.
(b)	If $N$ is a simple $\bar{Z}\text{-module}$ and $M$ is a finitely generated projective $B\text{-module}$ then there is a simple $B\text{-module}$ $P$ such that ${\text{Hom}}_B(M,P)\cong N\text{.}$

		Proof.
	Let $P_1=M{\otimes }_{Z}N\text{.}$ It is possible that $M{\otimes }_{Z}N$ is not simple (too big). Let $\tau$ be the $B\text{-module}$ generated by all the images of the maps in ${\text{Hom}}_B(M,P)\text{.}$ Then $\tau \subseteq P$ and $${\text{Hom}}_B(M,P)={\text{Hom}}_B(M,\tau )\text{.}$$ Let $t$ be the sum of all the submodules $\tau \prime \subseteq \tau$ such that ${\text{Hom}}_B(M,\tau \prime )=0\text{.}$ Since $M$ is finitely generated and projective $${\text{Hom}}_B(M,t)=0\text{.}$$ Now $t\subseteq \tau$ and $${\text{Hom}}_B(M,t/\tau )={\text{Hom}}_B(M,\tau )={\text{Hom}}_B(M,P_1)\text{.}$$ Since $M$ is finitely generated and projective the canonical map $$\begin{array}{ccc}N& ⟶& {\text{Hom}}_B(M,M{\otimes }_{Z}N)\\ n& ⟼& \begin{array}{cccc}{\varphi }_n:& M& \to & M{\otimes }_{Z}N\\ & m& \mapsto & m\otimes n\end{array}\end{array}$$ is a bijection. So ${\text{Hom}}_B(M,P_1)\cong N\text{.}$ Let $P=\tau /t\text{.}$ We know ${\text{Hom}}_B(M,P)\cong N\text{.}$ Let $P\prime \subseteq P$ be a proper submodule of $P\text{.}$ Then the injection $$0⟶P\prime ⟶P$$ gives us an injection $$0⟶{\text{Hom}}_B(M,P\prime )⟶{\text{Hom}}_B(M,P)\cong N\text{.}$$ Since $N$ is simple ${\text{Hom}}_B(M,P\prime )=0$ of ${\text{Hom}}_B(M,P\prime )={\text{Hom}}_B(M,P)\text{.}$ If ${\text{Hom}}_B(M,P\prime )={\text{Hom}}_B(M,P)$ then every map in ${\text{Hom}}_B(M,P)$ has its image in $P\prime \text{.}$ This is impossible since $P\prime$ is a proper submodule of $P\text{.}$ So ${\text{Hom}}_B(M,P\prime )=0\text{.}$ This means $P\prime$ is $0$ in $P$ (by construction of $P$ as $\tau /t\text{).}$ So $P$ is simple. $\square$

Example 1. Let $B$ be an algebra and let $e\in B$ be an idempotent. If $$M=Be\text{then}{\text{End}}_B\left(M\right)=eBe,$$ where $eBe$ acts on $M=Be$ by right multplication.

(a)	$M$ is finitely generated and projective $B\text{-module.}$
(b)	If $P$ is a $B\text{-module}$ then ${\text{Hom}}_B(M,P)\cong eP\text{.}$
(c)	If $P$ is a simple $B\text{-module}$ then $eP$ is a simple $eBe\text{-module}$ and every simple $eBe\text{-module}$ can be obtained this way.

		Proof.
	(a) The element $e$ is a generator of $Be=M$ as a $B\text{-module.}$ So $M$ is finitely generated. Since $${(1-e)}^2=1-e,e(1-e)=(1-e)e=0,\text{and}1=e+(1-e),$$ it follows that $B=Be+B(1-e)$ and $Be\cap B(1-e)=0\text{.}$ So $B=Be\oplus B(1-e)$ and so, by ???, $Be$ is projective. (b) Consider the map $$\begin{array}{ccc}eP& \stackrel{\Phi }{⟶}& {\text{Hom}}_B(M,P)\\ x& ⟼& \begin{array}{cccc}{\varphi }_x:& M& \to & P\\ & be& \mapsto & bex\end{array}\end{array}$$ If $\varphi \in {\text{Hom}}_B(Be,P)$ then $$\varphi \left(be\right)=\varphi \left(bee\right)=be\varphi \left(e\right)=be\left(e\varphi \left(e\right)\right),$$ and so $\varphi ={\varphi }_x$ where $x=e\varphi \left(e\right)\in eP\text{.}$ So $\Phi$ is surjective. If $e{p}_1,e{p}_2\in eP$ and ${\varphi }_{e{p}_1}={\varphi }_{e{p}_2}$ then $$\begin{array}{cccccc}& {\varphi }_{e{p}_1}\left(e\right)& =& ee{p}_1& =& =p_1\\ =& {\varphi }_{e{p}_2}\left(e\right)& =& ee{p}_2& =& =p_2,\end{array}$$ and so $\Phi$ is injective. If $ebe\in eBe$ then $$\begin{array}{ccc}\left(\left(ebe\right)\varphi \right)\left(m\right)& =& {\varphi }_{en}\left(mebe\right)\\ & =& meben={\varphi }_{eben}\left(m\right),\end{array}$$ for all ${\varphi }_{en}\in {\text{Hom}}_B(M,P)\text{.}$ So $\Phi$ is a $eBe\text{-module}$ homomorphism. (c) follows immediately from (a) and (b) and Theorem ????. $\square$

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

(a)	${\chi }_B$ is an idempotent in $C\left(G\right),$
(b)	$C(G/B)=C\left(G\right)*{\chi }_B,$ and
(c)	$C(B\backslash G/B)={\chi }_B*C\left(G\right)*{\chi }_B,$ is the Hecke algebra of the pair $(G,B)\text{.}$ From Proposition ???, if $P$ is a simple $C\left(G\right)\text{-module}$ then ${\chi }_B*P$ is either $0$ or a simple $C(B\backslash G/B)\text{-module.}$ Every simple $C(B\backslash G/B)\text{-module}$ 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 $$R⋉G=\left\{\sum _{g\in G}{r}_{g}g\hspace{0.17em}\right|\hspace{0.17em}{r}_g\in R\}$$ with multiplication given by the relation $$gr=g\left(r\right)g,\text{for}\hspace{0.17em}g\in G,r\in R\text{.}$$ Clifford theory is a mechanism for constructing the simple $R⋉G$ modules from those of $R$ and subgroups $H\subseteq G\text{.}$ Note that $R$ is an $R⋉G$ module since $R$ acts on $R$ by left multiplication and $G$ acts on $R$ by automorphisms. Let $$e=\frac{1}{\left|G\right|}\sum _{g\in G}g\text{.}$$ Then $e$ is an idempotent in $R⋉G\text{.}$ Let $$R^G=\{r\in R\hspace{0.17em}|\hspace{0.17em}g\left(r\right)=r\text{, for all}\hspace{0.17em}g\in G\}\text{.}$$

(a)	The map $$\begin{array}{ccc}{R}^G& \stackrel{\Phi }{⟶}& e(R⋉G)e\\ r& ⟼& re\end{array}$$ is a ring isomorphism.
(b)	The ring $R$ is an $(R⋉G,R^G)$ bimodule, where $R⋉G$ acts on the left and $R^G$ acts on $R$ by right multiplication. The map $$\begin{array}{ccc}R& ⟶& (R⋉G)e\\ r& ⟼& re\end{array}$$ is an isomorphism of $(R⋉G,R^G)$ bimodules, where $R^G$ is identified with $e(R⋉G)e$ via (a).

		Proof.
	(a) Let $r\in R^G\text{.}$ Then $$\begin{array}{cc}ere=\frac{1}{\left|G\right|}\sum _{g\in G}gre=\frac{1}{\left|G\right|}\sum _{g\in G}g\left(r\right)ge=\frac{1}{\left|G\right|}\sum _{g\in G}re=re\text{.}& \text{(0.9)}\end{array}$$ So $\Phi \left(r\right)=re$ is an element of $e(R⋉G)e$ and thus $\Phi$ is well defined. Let $e\left({\sum }_{g\in G}{r}_{g}g\right)e$ be a general element of $e(R⋉G)e\text{.}$ Then $$e\sum _{g\in G}{r}_{g}ge=\frac{1}{\left|G\right|}\sum _{g\in G}\sum _{h\in G}h\left(r_g\right)e=\frac{1}{\left|G\right|}\sum _{g\in G}\Phi \left(x_g\right),$$ where $x_g={\sum }_{h\in G}h\left(r_g\right)$ is an element of $R^G$ since $$k\left(x_g\right)=\sum _{h\in G}kh\left(r_g\right)=\sum _{\ell \in G}\ell \left(r_g\right)=x_g,$$ for all $k\in G\text{.}$ So $\Phi$ is surjective. Let $r,s\in R^G\text{.}$ Then $$\Phi \left(r\right)\Phi \left(s\right)=rese=rse,\text{and so}\Phi \left(r\right)\Phi \left(s\right)=\Phi \left(rs\right)\text{.}$$ So $\Phi$ is a ring homomorphism. (b) Let $a\in R⋉G$ and $b\in R^G$ and $r\in R\text{.}$ then, by ???, $$\Phi \left(arb\right)=arbe=arebe=a\left(re\right)\left(ebe\right)=a\Phi \left(r\right)be\text{.}$$ So $\Phi$ is an $(R⋉G,R^G)$ bimodule homomorphism. Let ${\sum }_{g\in G}{r}_{g}g$ be a general element of $R⋉G\text{.}$ Then $$\left(\sum _{g\in G}{r}_{g}g\right)e=\sum _{g\in G}{r}_{g}ge=\sum _{g\in G}{r}_{g}e=\sum _{g\in G}\Psi \left(r_g\right).$$ So $\Phi$ is surjective. The injectivity of $\Phi$ is proved as in (a). $\square$

Proposition ??? implies that all simple $R^G$ modules can be constructed in the following way: Let $P$ be a simple $R⋉G$ module. Then $${\text{Hom}}_{R⋉G}(M,P)={\text{Hom}}_{R⋉G}((R⋉G)e,P)=eP,$$ is either $0$ or a simple $e(R⋉G)e=R^G$ module.

Example 4. Recall that $G(r,1,n)={(\mathbb{Z}/r\mathbb{Z})}^n⋉S_n$ and $G(r,p,n)$ is a normal subgroup of index $p$ in $G(r,1,n),$ $$\begin{array}{ccc}G(r,1,n)& =& \left\{t_{\lambda }w\hspace{0.17em}\right|\hspace{0.17em}\lambda =({\lambda }_1,\dots ,{\lambda }_n)\in {(\mathbb{Z}/r\mathbb{Z})}^n,w\in S_n\},\\ G(r,p,n)& =& \{t_{\lambda }w\in G(r,1,n)\hspace{0.17em}|\hspace{0.17em}{\lambda }_1+\cdots {\lambda }_n=0\hspace{0.17em}\text{mod}\hspace{0.17em}p\},\end{array}$$ where the multiplication is determined by $$w{t}_{\lambda }=t_{w\lambda }w,\lambda \in {(\mathbb{Z}/r\mathbb{Z})}^n,w\in S_n\text{.}$$ Let $\zeta =e^{2\pi i/p}$ and define an action of $\mathbb{Z}/p\mathbb{Z}$ on the group algebra $ℂG(r,1,n)$ by the powers of the map $$\begin{array}{cccc}\rho :& ℂG(r,1,n)& ⟶& ℂG(r,1,n)\\ & t_{\lambda }w& ⟼& {\zeta }^{\left|\lambda \right|}{t}_{\lambda }w,\end{array}\text{where}\left|\lambda \right|={\lambda }_1+\cdots +{\lambda }_n\text{.}$$ The map $\rho$ is a vector space isomorphism since it is a diagonal map with nonzero diagonal entries with respect to the basis $\left\{t_{\lambda }w\hspace{0.17em}\right|\hspace{0.17em}\lambda \in {(\mathbb{Z}/r\mathbb{Z})}^n,w\in S_n\}$ of $ℂG(r,1,n)\text{.}$ Since $$\rho \left(t_{\lambda }w{t}_{\mu }v\right)={\zeta }^{\left|\lambda \right|+\left|\mu \right|}{t}_{\lambda }w{t}_{\mu }v=\rho \left(t_{\lambda }w\right)\rho \left(t_{\mu }v\right),$$ for all $\lambda ,\mu \in {(\mathbb{Z}/r\mathbb{Z})}^n$ and $w,v\in S_n,$ the map $\rho$ is an automorphism of $ℂG(r,1,n)\text{.}$ Since ${\rho }^p=\text{Id}$ the action of $\rho$ on $ℂG(r,1,n)$ defines an action of $\mathbb{Z}/p\mathbb{Z}$ on $G(r,1,n)\text{.}$ $$ℂG(r,p,n)={\left(ℂG(r,1,n)\right)}^{\mathbb{Z}/p\mathbb{Z}}$$ is the subalgebra of fixed points for this action since $$\rho \left(t_{\lambda }w\right)={\zeta }^{\left|\lambda \right|}{t}_{\lambda }w=t_{\lambda }w\text{if and only if}{t}_{\lambda }w\in G(r,p,n)\text{.}$$ It follows that the irreducible representations of $G(r,p,n)$ can be obtained in the following way: Let $$R=ℂG(r,1,n),G=\mathbb{Z}/p\mathbb{Z},R⋉G=ℂG(r,1,n)⋉\mathbb{Z}/p\mathbb{Z},e=\frac{1}{p}\sum _{k=0}^{p-1}{\zeta }^k\text{.}$$ Let $P$ be a simple $R⋉G$ module (we know these by Clifford theory). Then either $$eP=0\text{or}eP\hspace{0.17em}\text{is a simple}\hspace{0.17em}G(r,p,n)\text{-module}$$ and all simple $G(r,p,n)$ modules are obtained this way.

Induction and restriction

Let $A$ be a subalgebra of an algebra $B\text{.}$ Let $M$ be a $B\text{-module.}$ The $A\text{-module}$ ${\text{Res}}_A^B\left(M\right)$ is $M$ viewed simply as and $A\text{-module.}$ Let $N$ be an $A\text{-module.}$ The vector space $B{\otimes }_{A}N$ is the vector space $B\otimes N$ with the additional relations $$ba\otimes n=b\otimes an,\text{for all}\hspace{0.17em}a\in A,b\in B,n\in N\text{.}$$ The $B\text{-module}$ ${\text{Ind}}_A^B\left(N\right)$ is the vector space $B{\otimes }_{A}N$ with $B\text{-action}$ given by $$b(b_1\otimes n)=b{b}_1\otimes n,\text{for all}\hspace{0.17em}b,b_1\in B,n\in N\text{.}$$

(Frobenius reciprocity) ${\text{Res}}_A^B$ and ${\text{Ind}}_A^B$ are adjoint functors, i.e. $${\text{Hom}}_B({\text{Ind}}_A^B\left(N\right),P)\cong {\text{Hom}}_A(N,{\text{Res}}_A^B\left(P\right)),$$ for all $A\text{-modules}$ $N$ and all $B\text{-modules}$ $P\text{.}$

		Proof.
	This result is the special case $M=B$ of the more general result $$\begin{array}{cc}{\text{Hom}}_A(N,{\text{Hom}}_B(M,P))\cong {\text{Hom}}_B(M{\otimes }_{A}N,P),& \text{(2.2)}\end{array}$$ where $A$ and $B$ are algebras, $N$ is a left $A\text{-module,}$ $P$ is a left $B\text{-module}$ and $M$ is a $(B,A)\text{-bimodule.}$ The left $A\text{-module}$ ${\text{Hom}}_B(M,P)$ has $A\text{-action}$ given by $$\left(a\varphi \right)\left(m\right)=\varphi \left(am\right),\text{for}\hspace{0.17em}a\in A,\varphi \in {\text{Hom}}_B(M,P)\text{, and}\hspace{0.17em}m\in M\text{.}$$ The map $$\begin{array}{ccc}{\text{Hom}}_B(B,P)& ⟶& {\text{Res}}_A^B\left(P\right)\\ \varphi & ⟼& \varphi \left(1\right)\end{array}$$ is an $A\text{-module}$ isomorphism since $\left(a\varphi \right)\left(1\right)=\varphi (1·a)=\varphi \left(a\right)=a\left(\varphi \left(1\right)\right),$ for $a\in A,$ $\varphi \in {\text{Hom}}_B(B,P),$ Thus the statement of the theorem is a special case of the isomorphism in (???). $\square$

Example 0. Let $\rho :A\to B$ be a homomorphism of algebras. Then $\rho$ induces a functor ${\rho }^{*}:B\text{-mod}\to A\text{-mod.}$ The functor ${\text{Res}}_A^B$ is a special case of this functor. Induction ${\text{Ind}}_A^B$ is the left adjoint of ${\text{Res}}_A^B$ and coinduction ${\text{coInd}}_A^B$ is the right adjoint.

Example 1. Let $G$ be a group and let $H$ be a subgroup. Let $ℂG$ and $ℂH$ be the group algebras of $G$ and $H$ respectively and let $N$ be a $H\text{-module.}$ Let

(a)	$B_N=\left\{n_j\right\}$ be a basis of $N,$ and
(b)	$R_{G/H}=\left\{r_i\right\}$ a set of coset representatives for $G/H\text{.}$

Then $$\{r_i\otimes n_j\hspace{0.17em}|\hspace{0.17em}{r}_i\in R_{G/H},n_j\in B_n\}\text{is a basis of}{\text{Ind}}_H^G\left(N\right)=ℂG{\otimes }_{ℂH}N\text{.}$$ Informally, this is because $r_{i}H\otimes N=r_i\otimes HN=r_i\otimes N$ for all $r_i\in R_{G/H}\text{.}$

Example 2. Let $A$ be a subalgebra of an algebra $B\text{.}$ Let $a\in A\text{.}$ The left ideal $Aa$ is an $A\text{-module}$ and $${\text{Ind}}_A^B\left(Aa\right)\cong Ba,\text{as}\hspace{0.17em}B\text{-modules.}$$ This can be justified informally by $$B{\otimes }_{A}Aa=BA{\otimes }_{A}a=Ba{\otimes }_{A}1=Ba\text{.}$$ More formally one must show that the map $$\begin{array}{ccc}B\otimes Aa& ⟶& Ba\\ b\otimes p& ⟼& bp\end{array}$$ 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 $A^{\lambda },$ $\lambda \in \hat{A},$ be the simple $A\text{-modules,}$ and
$B^{\mu },$ $\mu \in \hat{B},$ be the simple $B\text{-modules.}$ Define nonnegative integers $c_{\lambda \mu }$ and $d_{\lambda \mu }$ by $${\text{Res}}_A^B\left(B^{\mu }\right)={\left(A^{\lambda }\right)}^{\oplus c_{\lambda \mu }}\text{and}{\text{Ind}}_A^B\left(A^{\lambda }\right)={\left(B^{\mu }\right)}^{\oplus d_{\lambda \mu }}\text{.}$$ Then $$c_{\lambda \mu }=d_{\lambda \mu },\text{for all}\hspace{0.17em}\lambda \in \hat{A}\hspace{0.17em}\text{and}\hspace{0.17em}\mu \in \hat{B}\text{.}$$ since $$c_{\lambda \mu }=\text{dim}\left({\text{Hom}}_A({\text{Res}}_A^B\left(B^{\mu }\right),A^{\lambda })\right)=\text{dim}\left({\text{Hom}}_B(B^{\mu },{\text{Ind}}_A^B\left(A^{\lambda }\right))\right)=d_{\lambda \mu },$$ by Schur’s lemma and Frobenius reciprocity.

Example 4. Let $A$ be a subalgebra of $B$ and let $M$ be a $B\text{-module}$ which is semisimple both as a $B\text{-module}$ and as an $A\text{-module,}$ $$M\cong \underset{\mu \in {\hat{B}}_M}{⨁}{\left(B^{\mu }\right)}^{\oplus m_{\mu }}\text{and}M\cong \underset{\lambda \in {\hat{A}}_M}{⨁}{\left(A^{\lambda }\right)}^{\oplus n_{\lambda }},$$ where $B^{\mu },$ $\mu \in {\hat{B}}_M,$ are the simple $B\text{-modules}$ that appear in $M,$ and
$A^{\lambda },$ $\lambda \in {\hat{A}}_M,$ are the simple $A\text{-modules}$ that appear in $M\text{.}$ $$M\cong \underset{\mu \in {\hat{B}}_M}{⨁}{\left(B\mu \right)}^{\oplus m_{\mu }}\text{.}$$ Let $Z_A={\text{End}}_A\left(M\right)$ and $Z_B={\text{End}}_B\left(M\right)\text{.}$ Then $$A\subseteq B\text{and}{Z}_A\supseteq Z_B$$ and $$\begin{array}{ccc}M& \cong & \underset{\mu \in {\hat{B}}_M}{⨁}{B}^{\mu }\otimes Z_B^{\mu },\text{as}\hspace{0.17em}B\otimes Z_B\hspace{0.17em}\text{modules, and}\\ M& \cong & \underset{\lambda \in {\hat{A}}_M}{⨁}{A}^{\lambda }\otimes Z_A^{\lambda },\text{as}\hspace{0.17em}A\otimes Z_A\hspace{0.17em}\text{modules, where}\end{array}$$ $Z_A^{\lambda },$ $\lambda \in {\hat{A}}_M,$ be the simple $Z_A\text{-modules,}$ and
$Z_B^{\mu },$ $\mu \in {\hat{B}}_M,$ the simple $Z_B$ modules. Since $Z_A$ and $Z_B$ are both semisimple algebras there are positive integers $c_{\lambda \mu }$ such that $$\begin{array}{cc}{\text{Res}}_{Z_B}^{Z_A}\left(Z_A^{\lambda }\right)=\underset{\mu \in {\hat{B}}_M}{⨁}{\left(Z_B^{\mu }\right)}^{\oplus c_{\lambda \mu }}\text{.}& \text{(2.3)}\end{array}$$

Each $B^{\mu },$ $\mu \in {\hat{B}}_{},$ is semisimple as an $A\text{-module}$ and $${\text{Res}}_A^B\left(B^{\mu }\right)\cong \underset{\lambda \in {\hat{A}}_{}}{⨁}{\left(A^{\lambda }\right)}^{\oplus c_{\lambda \mu }},$$ where the positive integers $c_{\lambda \mu }$ are as in (???).

		Proof.
	Since $$\begin{array}{ccc}\underset{\mu \in {\hat{B}}_{}}{⨁}{\text{Res}}_A^B\left(B^{\mu }\right)\otimes Z_B^{\mu }& \cong & {\text{Res}}_{A\otimes Z_B}^{B\otimes Z_B}\left(M\right)\cong {\text{Res}}_{A\otimes Z_B}^{A\otimes Z_A}\left(M\right)\\ & \cong & \underset{\lambda \in {\hat{A}}_{}}{⨁}{A}^{\lambda }\otimes {\text{Res}}_{Z_B}^{Z_A}\left(Z_A^{\lambda }\right)\\ & \cong & \underset{\lambda \in {\hat{A}}_{}}{⨁}\underset{\mu \in {\hat{B}}_{}}{⨁}{A}^{\lambda }\otimes {\left(Z_B^{\mu }\right)}^{\oplus c_{\lambda \mu }}\\ & \cong & \underset{\lambda \in {\hat{A}}_M}{⨁}\underset{\mu \in {\hat{B}}_{}}{⨁}{(A^{\lambda }\otimes Z_B^{\mu })}^{\oplus c_{\lambda \mu }}\\ & \cong & \underset{\lambda \in {\hat{A}}_{}}{⨁}\underset{\mu \in {\hat{B}}_{}}{⨁}{\left(A^{\lambda }\right)}^{\oplus c_{\lambda \mu }}\otimes Z_B^{\mu }\end{array}$$ it follows that $${\text{Res}}_A^B\left(B^{\mu }\right)\cong {\text{Hom}}_{Z_B}(Z_B^{\mu },M)\cong \underset{\lambda \in {\hat{A}}_{}}{⨁}{\left(A^{\lambda }\right)}^{\oplus c_{\lambda \mu }}$$ as $A\text{-modules}$ $\square$

Then $${\text{Hom}}_{Z_B}({\text{Res}}_{Z_B}^{Z_A}\left(Z_A^{\lambda }\right),Z_B^{\mu })\cong {\text{Hom}}_A({\text{Res}}_A^B\left(B^{\mu }\right),A^{\lambda })\text{.}$$ Thus, if $c_{\lambda \mu }$ and $d_{\lambda \mu }$ are the nonnegative integers such that $${\text{Res}}_A^B\left(B^{\mu }\right)={\left(A^{\lambda }\right)}^{\oplus c_{\lambda \mu }}\text{and}{\text{Ind}}_{Z_A}^{Z_B}\left(Z_B^{\mu }\right)={\left(Z_A^{\lambda }\right)}^{\oplus d_{\lambda \mu }}$$ then $$c_{\lambda \mu }=d_{\lambda \mu }\text{for all}\hspace{0.17em}\lambda \in {\hat{A}}_{}\hspace{0.17em}\text{and}\hspace{0.17em}\mu \in {\hat{B}}_{}\text{.}$$

Example 5. Let $A$ be a semisimple subalgebra of a semisimple algebra $B\text{.}$ Let ${\text{tr}}_A$ be the trace of the regular representation of $A\text{.}$ Let $𝒜$ be a basis of $A$ and let $\left\{a^{*}\right\}$ be the dual basis to $𝒜$ with respect to the form on $A$ defined by $${⟨a_1,a_2⟩}_A={\text{tr}}_A\left(a_1{a}_2\right),\text{for}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$ Let ${\text{tr}}_B$ be the trace of the regular representation of $B\text{.}$ Let $ℬ$ be a basis of $B$ and let $\left\{b^{*}\right\}$ be the dual basis to $ℬ$ with respect to the form on $B$ defined by $${⟨b_1,b_2⟩}_A={\text{tr}}_B\left(b_1{b}_2\right),\text{for}\hspace{0.17em}{b}_1,b_2\in B\text{.}$$ If $c\in B$ define $$\left[c\right]=\sum _{b\in ℬ}bc{b}^{*}\text{.}$$ Let $N$ be an $A\text{-module.}$ Then the character of ${\text{Ind}}_A^B\left(N\right)$ is given by $${\chi }_{N\uparrow }\left(b\right)=\sum _{a\in 𝒜}{\chi }_N\left(a\right){⟨b,\left[a^{*}\right]⟩}_B,$$ where ${\chi }_N$ is the character of $N\text{.}$

		Proof.
	First note that, by Schur’s lemma, $\text{dim}\left({\text{Hom}}_A(A^{\lambda },A^{\mu })\right)={\delta }_{\lambda \mu }$ if $A^{\lambda }$ and $A^{\mu }$ are simple $A\text{-modules.}$ By the orthgonality relation for characters, ${\delta }_{\lambda \mu }={\sum }_{a\in 𝒜}{\chi }^{\lambda }\left(a\right){\chi }^{\mu }\left(a^{*}\right),$ where ${\chi }^{\lambda }$ and ${\chi }^{\mu }$ are the characters of $A^{\lambda }$ and $A^{\mu }$ respectively. It follows that $$\text{dim}\left({\text{Hom}}_A(M,P)\right)=\sum _{a\in 𝒜}{\chi }_M\left(a\right){\chi }_N\left(a^{*}\right),$$ for any two module $M$ and $P$ with corresponding characters ${\chi }_M$ and ${\chi }_P\text{.}$ There are two things to prove: (a) ${\chi }_{N\uparrow },$ as defined in the statement, is a character, (b) $\sum _{b\in ℬ}{\chi }_{N\uparrow }\left(n\right){\chi }_P\left(b^{*}\right)=\sum _{a\in 𝒜}{\chi }_N\left(a\right){\chi }_P\left(a^{*}\right)$ for any $B\text{-module}$ $P\text{.}$ (a) $$\begin{array}{ccc}{\chi }_{N\uparrow }\left(b_1{b}_2\right)& =& \sum _{a\in 𝒜}{\chi }_N\left(a\right){⟨b_1{b}_2,\left[a^{*}\right]⟩}_B\\ & =& \sum _{a\in 𝒜}{\chi }_N\left(a\right){\text{tr}}_B\left(b_1{b}_2\sum _{b\in ℬ}b{a}^{*}{b}^{*}\right)\\ & =& \sum _{a\in 𝒜}{\chi }_N\left(a\right){\text{tr}}_B\left(b_1\left[a^{*}\right]b_2\right)\\ & =& \sum _{a\in 𝒜}{\chi }_N\left(a\right){\text{tr}}_B\left(\left[a^{*}\right]b_2{b}_1\right)\\ & =& \sum _{a\in 𝒜}{\chi }_N\left(a\right){⟨\left[a^{*}\right],b_2{b}_1⟩}_B\\ & =& {\chi }_{N\uparrow }\left(b_2{b}_1\right)\text{.}\end{array}$$ (b) $$\begin{array}{ccc}\sum _{a\in 𝒜}{\chi }_N\left(a\right){\chi }_P\left(a^{*}\right)& =& {\chi }_P\left(\sum _{a\in 𝒜}{\chi }_N\left(a\right)a^{*}\right)\\ & =& {\chi }_P\left(\sum _{a\in 𝒜}{\chi }_N\left(a\right)\left[a^{*}\right]\right)\\ & =& {\chi }_P\left(\sum _{a\in 𝒜}{\chi }_N\left(a\right)\sum _{b\in ℬ}⟨\left[a^{*}\right],b⟩b^{*}\right)\\ & =& \sum _{b\in ℬ}\left(\sum _{a\in 𝒜}{\chi }_N\left(a\right)⟨\left[a^{*}\right],b⟩\right){\chi }_P\left(b^{*}\right)\\ & =& \sum _{b\in ℬ}{\chi }_{N\uparrow }\left(b\right){\chi }_P\left(b^{*}\right)\end{array}$$ since, for any character $\chi$ of $B$ and any element $c\in B,$ $$\chi \left(\left[c\right]\right)=\chi \left(\sum _{b\in ℬ}bc{b}^{*}\right)=\chi \left(\sum _{b\in ℬ}c{b}^{*}b\right)=\chi \left(c\right)\text{.}$$ $\square$

Example 6. Suppose that $G$ is a finite group and $H$ is a subgroup. Then $G$ is a basis of $ℂG$ and $${\text{tr}}_G\left(g\right)=\{\begin{array}{cc}\left|G\right|,& \text{if}\hspace{0.17em}g=1,\\ 0,& \text{otherwise,}\end{array}$$ is the trace of the regular representation of $ℂG\text{.}$ The basis ${\{g^{-1}/\left|G\right|\}}_{g\in G}$ is the dual basis to $G$ with respect to ${⟨,⟩}_G\text{.}$ If $k\in G$ then $$\left[k\right]=\frac{1}{\left|G\right|}\sum _{g\in G}gk{g}^{-1},$$ and, for $k_1,$ $k_2\in G,$ $${⟨\left[k_1\right],k_2⟩}_G={⟨k_1,\left[k_2\right]⟩}_G=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}{k}_1\hspace{0.17em}\text{is conjugate to}\hspace{0.17em}{k}_2^{-1},\\ 0,& \text{otherwise,}\end{array}$$ Let $N$ be an $H\text{-module.}$ Then, by (???), the character ${\chi }_{N\uparrow }$ of ${\text{Ind}}_H^G\left(N\right)$ is given by $$\begin{array}{ccc}{\chi }_{N\uparrow }& =& \sum _{h\in H}{\chi }_N\left(h\right){⟨[h^{-1}/\left|H\right|],g⟩}_G\\ & =& \frac{1}{\left|H\right|}\sum _{h\in H}{\chi }_N\left(h\right){⟨h^{-1},\left[g\right]⟩}_G\\ & =& \frac{1}{\left|H\right|}\sum _{\underset{h\in {𝒞}_g}{h\in H}}{\chi }_N\left(h\right),\end{array}$$ where ${𝒞}_g$ is the conjugacy class of $g$ in $G\text{.}$

Example 7. Let $H$ be a subgroup of G. If $P$ is a $G\text{-module}$ then the subgroup $H$ acts on $P$ and so $P$ is an $H\text{-module.}$ This $H\text{-module}$ is denote ${\text{Res}}_H^G\left(P\right)\text{.}$ This provides a functor $$\begin{array}{cccc}{\text{Res}}_H^G:& \left\{G\text{-modules}\right\}& ⟶& \left\{H\text{-modules}\right\}\end{array}$$ and we define $$\begin{array}{cccc}{\text{Ind}}_H^G:& \left\{H\text{-modules}\right\}& ⟶& \left\{G\text{-modules}\right\}\end{array}$$ to be the left adjoint functor to ${\text{Res}}_H^G,$ i.e. $$\begin{array}{cc}{\text{Hom}}_H(M,{\text{Res}}_H^G\left(P\right))\cong {\text{Hom}}_G({\text{Ind}}_H^G\left(M\right),P)& \text{(}*\text{)}\end{array}$$ as vector spaces.

If $M$ is an $H\text{-module}$ then $${\text{Ind}}_H^G\left(M\right)=\{f:G\to M\hspace{0.17em}|\hspace{0.17em}f\left(hg\right)=hf\left(g\right),h\in H,g\in G\}$$ with $G\text{-action}$ given by $$\left(gf\right)\left(x\right)=f\left(xg\right),\text{for}\hspace{0.17em}g\in G,f\in {\text{Ind}}_H^G\left(M\right),x\in G\text{.}$$ If $f\in {\text{Ind}}_H^G\left(M\right)$ then the value of $f$ at any element of the coset $Hg$ is determined by the value of $f$ at $g\text{.}$ Thus, we sometimes view $f\in {\text{Ind}}_H^G\left(M\right)$ as a function $$f:H\backslash G⟶M\text{.}$$ For a proof of Frobenius reciprocity (i.e. $\text{(}*\text{))}$ 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 $A\text{.}$ Then $Ap$ is an $A\text{-module}$ and $${\text{Ind}}_A^B\left(Ap\right)=B{\otimes }_{A}Ap\cong Bp\text{.}$$

Let $p_1$ and $p_2$ be idempotents in $A\text{.}$ Then $$\begin{array}{cccc}\Psi :& p_{1}B{p}_2& \stackrel{\sim }{⟶}& {\text{Hom}}_B(B{p}_1,B{p}_2)\\ & x& ⟼& {\psi }_x\end{array}\text{where}\begin{array}{cccc}{\psi }_x:& B{p}_1& ⟶& B{p}_2\\ & b{p}_1& ⟼& b{p}_{1}x\end{array}\text{.}$$

		Proof.
	(a) If $x\in p_{1}B{p}_2$ then $x=p_1{b}_1{p}_2$ for some $b_1\in B\text{.}$ Then ${\psi }_x\left(b{p}_1\right)=b{p}_1{p}_1{b}_1{p}_2\in B{p}_2$ and so ${\psi }_x$ is well defined. (b) Since ${\psi }_x\left(b{p}_1\right)=b{p}_1{b}_1{p}_2=b{\psi }_x\left(p\right),$ for ${\psi }_x\in {\text{Hom}}_B(B{p}_1,B{p}_2),$ $\Psi$ is well defined. (c) If ${\psi }_x=0$ then ${\psi }_x\left(p_1\right)=p_{1}x=p_1{p}_1{b}_1{p}_2=p_1{b}_1{p}_2=0\text{.}$ So $x=0\text{.}$ So $\Psi$ is injective. (d) If $\psi \in {\text{Hom}}_B(B{p}_1,B{p}_2)$ then $\psi \left(p_1\right)=\psi \left(p_1{p}_1\right)=p_1\psi \left(p_1\right)=p_1\psi \left(p_1\right)p_2\in p_{1}B{p}_2,$ since $\psi \left(p_1\right)\in B{p}_2\text{.}$ Let $x=p_1\psi \left(p_1\right)p_2\text{.}$ Then $\psi \left(b{p}_1\right)=b\psi \left(p_1\right)=b{p}_1{p}_1\psi \left(p_1\right)p_2=b{p}_{1}x={\psi }_x\left(b{p}_1\right)\text{.}$ So $\psi ={\psi }_x\text{.}$ So $\Psi$ is surjective. $\square$

Example 9. If $X$ is a measure space with a positive measure $\mu$ then $$L^2\left(\mu \right)=\{f:X\to ℂ\hspace{0.17em}|\hspace{0.17em}{\int }_X{\left|f\left(x\right)\right|}^{2}d\mu \left(x\right)<\infty \}$$ is a Hilbert space with inner product $$⟨f_1,f_2⟩={\int }_X{f}_1\left(x\right)\overline{f_2\left(x\right)}d\mu \left(x\right)\text{.}$$ More generally, if $V$ is a Hilbert space with inner product ${⟨,⟩}_V$ then $$L^2\left(\mu \right)=\{f:X\to V\hspace{0.17em}|\hspace{0.17em}{\int }_X{\left|f\left(x\right)\right|}_V^{2}d\mu \left(x\right)<\infty \}$$ is a Hilbert space with inner product $$⟨f_1,f_2⟩={\int }_X{⟨f_1\left(x\right),f_2\left(x\right)⟩}_{V}d\mu \left(x\right)\text{.}$$ A unitary representation of a Lie group $G$ on a Hilbert space $V$ is an action of $G$ on $V$ such that

(a)	$⟨g{v}_1,g{v}_2⟩=⟨v_1,v_2⟩,$ for all $v_1,v_2\in V,$
(b)	The map $\begin{array}{ccc}G\times V& \to & V\\ (g,v)& \mapsto & gv\end{array}$ 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 $${\text{Ind}}_H^G=\{f:G\to V\hspace{0.17em}|\hspace{0.17em}f\left(hg\right)=hf\left(g\right),\hspace{0.17em}\text{for all}\hspace{0.17em}h\in H,\hspace{0.17em}\text{and}\hspace{0.17em}{\int }_G{\left|f\left(g\right)\right|}_V^{2}d\mu \left(g\right)<\infty \}$$ with $G\text{-action}$ given by $$\left(gf\right)\left(x\right)=f\left(xg\right),\text{for}\hspace{0.17em}x\in G,\hspace{0.17em}f\in {\text{Ind}}_H^G\left(V\right),\hspace{0.17em}g\in G\text{.}$$ Then the inner product on ${\text{Ind}}_H^G\left(V\right)$ given by $$\begin{array}{ccc}⟨f_1,f_2⟩& =& {\int }_G{⟨f_1\left(g\right),f_2\left(g\right)⟩}_{V}d\mu \left(g\right)\end{array}$$ satisfies $$\begin{array}{ccc}⟨g{f}_1,g{f}_2⟩& =& {\int }_G{⟨g{f}_1\left(x\right),g{f}_2\left(x\right)⟩}_{V}d\mu \left(x\right)\\ & =& {\int }_G{⟨f_1\left(xg\right),f_2\left(xg\right)⟩}_{V}d\mu \left(x\right)\\ & =& {\int }_G{⟨f_1\left(y\right),f_2\left(y\right)⟩}_{V}d\mu \left(y{g}^{-1}\right)\\ & =& {\int }_G{⟨f_1\left(y\right),f_2\left(y\right)⟩}_{V}d\mu \left(y\right)\\ & =& ⟨f_1,f_2⟩\end{array}$$ if the measure $\mu$ on $G$ is right invariant, i.e. if $\mu$ is a Haar measure on $G\text{.}$

Example 10. (Some “leftovers”) If $G$ is a finite group then $$\begin{array}{ccc}ℱ\left(G\right)& ⟶& ℂG\\ f& ⟼& {\sum }_{g\in G}f\left(g\right)g\end{array}$$ is an isomorphism of $G\text{-modules,}$ but, if $G$ is infinite then $ℱ\left(G\right)$ is much larger than $ℂG\text{.}$

Let $G$ be a finite group and let $H$ be a subgroup of $G\text{.}$ Let $\varphi :H\to {ℂ}^{*}$ be a one-dimensional representation of $H\text{.}$ Then $$p_{\varphi }=\frac{1}{\left|H\right|}\sum _{h\in H}\varphi \left(h^{-1}\right)h$$ is an element of $ℂG$ such that $$h{p}_{\varphi }\text{and}{p}_{\varphi }^2=p_{\varphi }\text{.}$$ Thus $ℂH{p}_{\varphi }=ℂp_{\varphi }\cong \varphi$ as $H\text{-modules.}$

Let $G$ be a finite group and let $B$ be a subgroup of $G\text{.}$ Let $1_B$ be the trivial representation of $B$ and let $p$ be the corresponding idempotent in $ℂB$ as defined in the previous paragraph. Then $$1_B^G={\text{Ind}}_B^G\left(1_B\right)\cong \left(ℂG\right)p\text{and}{\text{End}}_G\left(1_B^G\right)=p\left(ℂG\right)p$$ where $p\left(ℂG\right)p$ acts on $\left(ℂG\right)p$ by multiplication on the right.

Hecke algebras.

General Hecke algebras.

Let $B$ be an algebra. The subspace $pBp\subseteq B$ is closed under multiplication and is an algebra with identity $p\text{.}$ The Hecke algebra of the pair $(B,p)$ is the algebra $pBp\text{.}$

Let $B$ be an algebra and let $p$ be an idempotent in $B\text{.}$ Then ${\text{End}}_B\left(Bp\right)=pBp,$ where $pBp$ acts on $Bp$ on the right. More precisely, $${\text{End}}_B\left(Bp\right)=\left\{{\varphi }_h\hspace{0.17em}\right|\hspace{0.17em}h\in pBp\},\text{where}\hspace{0.17em}{\varphi }_h\hspace{0.17em}\text{is given by}{\varphi }_h=bph,\text{for all}\hspace{0.17em}bp\in Bp\text{.}$$

		Proof.
	Let $\varphi \in {\text{End}}_B\left(Bp\right)$ and let $bp\in Bp$ be such that $\varphi \left(p\right)=bp\text{.}$ For all $b\prime p\in Bp,$ $\varphi \left(b\prime p\right)=\varphi \left(b\prime pp\right)=b\prime p\varphi \left(p\right)=b\prime pbp=\left(b\prime p\right)\left(pbp\right),$ and so $\varphi ={\varphi }_h$ with $h=pbp\text{.}$ The multiplication of the ${\varphi }_h\in {\text{End}}_B\left(Bp\right)$ corresponds to the action of $pBp$ on $Bp$ on the right since $({\varphi }_{h_1}\circ {\varphi }_{h_2})\left(bp\right)=\left(\left(bp{h}_2\right)h_1\right)=\left(bp\right)\left(h_2{h}_1\right),$ for all $h_1,h_2\in pBp$ and $bp\in Bp\text{.}$ $\square$

Recall that if $$A is a subalgebra of an algebra $B$ and $p$ is an idempotent in $A$ then the left ideal $Ap$ is an $A\text{-module}$ and $$\begin{array}{cc}{\text{Ind}}_A^B\left(Ap\right)\cong Bp\text{.}& \text{(3.2)}\end{array}$$

Haar measure on $G$

Let $G$ be a group. The vector space $ℱ$ of functions $f:G\to ℂ$ is a $G\text{-module}$ with $G$ action $$\left({}^{g}f\right)\left(h\right)=f\left(g^{-1}h\right),\text{for all}\hspace{0.17em}g,h\in G,\hspace{0.17em}f\in ℱ\text{.}$$ Fix a $G\text{-submodule}$ $C\left(G\right)$ of the space $ℱ$ of all functions on $G\text{.}$ A Haar measure on $G$ is a linear functional $\mu :C\left(G\right)\to ℂ$ such that

(a)	(continuity) $\mu$ is continuous with respect to the topology on $C\left(G\right)$ given by $${\Vert f\Vert }_{\infty }=\text{sup}\left\{\left|f\left(g\right)\right|\hspace{0.17em}\right|\hspace{0.17em}g\in G\},$$
(b)	(postivity) If $f$ is such that $f\left(g\right)\in {ℝ}_{\ge 0}$ for all $g\in G$ then $\mu \left(f\right)\in {ℝ}_{\ge 0}\text{.}$
(c)	(left invariance) For all $g\in G$ and $g\in C\left(G\right),$ $\mu \left({}^{g}f\right)=\mu \left(f\right)\text{.}$

Use the notation $$\mu \left(f\right)={\int }_{G}f\left(g\right)d\mu \left(g\right)\text{.}$$ The left invariance of $\mu$ means that $${\int }_{G}f\left(g\right)d\mu \left(g\right)=\mu \left(f\right)=\mu \left({}^{h}f\right)={\int }_{G}f\left(h^{-1}g\right)d\mu \left(g\right)={\int }_{G}f\left(k\right)d\mu \left(hk\right)\text{.}$$ In general it is not true that for a Haar measure $\mu ,$ $${\int }_{G}f\left(g\right)d\mu \left(g\right)={\int }_{G}f\left(g\right)d\mu \left(gh\right)\text{or}{\int }_{G}f\left(g^{-1}\right)d\mu \left(g\right)={\int }_{G}f\left(g\right)d\mu \left(g\right),$$ hold for all $f\in C\left(G\right)\text{.}$ A group $G$ is unimodular if one of the (equivalent) conditions in ??? hold.

The convolution of $f_1,f_2\in C\left(G\right)$ is the function $f_1*f_2$ on $G$ given by $$\begin{array}{cc}{f}_1*f_2\left(h\right)={\int }_G{f}_1\left(hg\right)f_2\left(g^{-1}\right)d\mu \left(g\right),\text{for all}\hspace{0.17em}h\in G\text{.}& \text{(3.3)}\end{array}$$

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

(a)	$C\left(G\right)$ is an associative algebra.
(b)	If $G$ is unimodular then the map $\overrightarrow{t}:C\left(G\right)\to ℂ$ given by $\overrightarrow{t}\left(f\right)=f\left(1\right)$ is a trace on $C\left(G\right)\text{.}$

		Proof.
	(a) Let $f_1,f_2,f_3\in C\left(G\right)\text{.}$ Then $$\begin{array}{ccc}((f_1*f_2)*f_3)\left(h\right)& =& {\int }_G(f_1*f_2)\left(h{g}_1\right)f_3\left(g_1^{-1}\right)d\mu \left(g_1\right)\\ & =& {\int }_G{\int }_G{f}_1\left(h{g}_1{g}_2\right)f_2\left(g_2^{-1}\right)f_3\left(g_1^{-1}\right)d\mu \left(g_2\right)d\mu \left(g_1\right)\end{array}$$ and $$\begin{array}{ccc}(f_1*(f_2*f_3))\left(h\right)& =& {\int }_G{f}_1\left(hk\right)f_2*f_3\left(k^{-1}\right)d\mu \left(k\right)\\ & =& {\int }_G{f}_1\left(hk\right){\int }_G{f}_2\left(k^{-1}{g}_1\right)f_3\left(g_1^{-1}\right)d\mu \left(g_1\right)d\mu \left(k\right)\\ & =& {\int }_G{\int }_G{f}_1\left(h{g}_1{g}_2\right){\int }_G{f}_2\left(g_2^{-1}\right)f_3\left(g_1^{-1}\right)d\mu \left(g_1\right)d\mu \left(g_1{g}_2\right)\\ & =& {\int }_G{\int }_G{f}_1\left(h{g}_1{g}_2\right)f_2{g}_2^{-1}{f}_3\left(g_1^{-1}\right)d\mu \left(g_1\right)d\mu \left(g_2\right),\end{array}$$ where the last equality is a consequence of the left invariance of $\mu \text{.}$ Thus, by Fubini’s theorem $(f_1*f_2)*f_3=f_1*(f_2*f_3)\text{.}$ (b) Since $$\begin{array}{ccccc}\overrightarrow{t}(f_1*f_2)& =& (f_1*f_2)\left(1\right)& =& {\int }_G{f}_1\left(p\right)f_2\left(p^{-1}\right)d\mu \left(p\right),\\ \overrightarrow{t}(f_2*f_1)& =& (f_2*f_1)\left(1\right)& =& {\int }_G{f}_2\left(p\right)f_1\left(p^{-1}\right)d\mu \left(p\right),\end{array}$$ and $G$ is unimodular, $\overrightarrow{t}(f_1*f_2)=\overrightarrow{t}(f_2*f_1)\text{.}$ $\square$

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

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

Let $H_1$ and $H_2$ be subgroups of $G$ and let $W_1$ and $W_2$ be representations of $H_1$ and $H_2,$ respectively. Define $$\mathscr{H}=\{f:G\to \text{Hom}(W_1,W_2)\hspace{0.17em}|\hspace{0.17em}f\left(h_{1}g{h}_2\right)=W_1\left(h_1\right)f\left(g\right)W_2\left(h_2\right),h_1\in H_1,h_2\in H_2\}\text{.}$$ Then $${\text{Hom}}_G({\text{Ind}}_{H_1}^G\left(W_1\right),{\text{Ind}}_{H_2}^G\left(W_2\right))\cong \mathscr{H}\text{.}$$

		Proof.
	$${\text{Hom}}_G({\text{Ind}}_{H_1}^G\left(W_1\right),{\text{Ind}}_{H_2}^G\left(W_2\right))\cong {\text{Hom}}_{H_1}(W_1,{\text{Res}}_{H_1}^G{\text{Ind}}_{H_2}^G\left(W_2\right))\text{.}$$ Then define $$\begin{array}{ccc}{\text{Hom}}_{H_1}(W_1,{\text{Ind}}_{H_2}^G\left(W_2\right))& ⟶& \mathscr{H}\\ \begin{array}{cccc}\psi :& W_1\lambda & \to & {\text{Ind}}_{H_2}^G\left(W_2\right)\\ & w_1& \mapsto & {\psi }_{w_1}\end{array}& ⟼& T:G\to \text{Hom}(W_1,W_2)\end{array}$$ by $${\psi }_{w_1}\left(g\right)=T\left(g\right)w_1\text{.}$$ Then $$\left(h_1{\psi }_{w_1}\right)\left(g\right)={\psi }_{w_1}\left(g{h}_1\right)=T\left(g{h}_1\right)w_1,\text{and}{\psi }_{h_1{w}_1}\left(g\right)=T\left(g\right)h_1{w}_1\text{.}$$ So $\varphi \in {\text{Hom}}_{H_1}(W_1,{\text{Ind}}_{H_2}^G\left(W_2\right))$ if and only if $T$ is right invariant under $H_1\text{.}$ Also $${\psi }_{w_1}\left(h_{2}g\right)=T\left(h_{2}g\right)w_1\text{and}{h}_2{\psi }_{w_1}\left(g\right)=h_{2}T\left(g\right)w_1,$$ so ${\psi }_{w_1}\in {\text{Ind}}_{H_2}^G\left(W_2\right)$ if and only if $T$ is left $H_2$ invariant. $\square$

The Hecke algebra $H(G,B,\chi )$

Fix a group $G$ and a subgroup $B$ and suppose that $\mu :C\left(G\right)\to ℂ$ is a Haar measure on $G\text{.}$ Let $\chi :B\to {ℂ}^{*}$ be a character of $B\text{.}$ Define $$\begin{array}{ccc}C(G/B)& =& \{f\in C\left(G\right)\hspace{0.17em}|\hspace{0.17em}f\left(gb\right)=f\left(g\right)\chi \left(b\right)\hspace{0.17em}\text{for all}\hspace{0.17em}b\in B,\hspace{0.17em}g\in G\},\\ C(B\backslash G/B)& =& \{f\in C\left(G\right)\hspace{0.17em}|\hspace{0.17em}f\left(b_{1}g{b}_2\right)=\chi \left(b_1\right)f\left(g\right)\chi \left(b_2\right)\hspace{0.17em}\text{for all}\hspace{0.17em}{b}_1,b_2\in B,\hspace{0.17em}g\in G\}\text{.}\end{array}$$ The following proposition puts these objects into the context of Section ??? It shows that $C(B\backslash G/B)$ is an algebra under convolution and that $C(G/B)$ is a right $C(B\backslash G/B)\text{-module.}$ The Hecke algebra of the pair $(G,B)$ is the algebra $C(B\backslash G/B)\text{.}$

Define a function $p:G\to ℂ$ by $$p\left(g\right)=\{\begin{array}{cc}\frac{\chi \left(g\right)}{\mu \left(B\right)},& \text{if}\hspace{0.17em}g\in B,\\ 0,& \text{otherwise.}\end{array}$$

(a)	$p$ is an idempotent in $C\left(G\right)\text{.}$
(b)	$C(G/B)=C\left(G\right)p,$
(c)	$C(B\backslash G/B)=pC\left(G\right)p,$

		Proof.
	(a) If $h\in G,$ $$(p*p)\left(h\right)={\int }_{G}p\left(hk\right)p\left(k^{-1}\right)d\mu \left(k\right)=\{\begin{array}{cc}0& \text{if}\hspace{0.17em}h\notin B,\\ \left(\frac{\chi \left(h\right)}{\mu {\left(B\right)}^2}\right)\mu \left(B\right),& \text{if}\hspace{0.17em}h\in B\end{array}=p\left(h\right),$$ since $\mu \left(B\right)=1\text{.}$ (b) Let $f\in C\left(G\right)$ and let $h\in G,$ $b\in B\text{.}$ Then, by the left invariance of $\mu ,$ $$\begin{array}{ccc}(f*p)\left(hb\right)& =& {\int }_{G}f\left(hbk\right)p\left(k^{-1}\right)d\mu \left(k\right)={\int }_{G}f\left(hg\right)p\left(g^{-1}b\right)d\mu \left(b^{-1}g\right)\\ & =& \chi \left(b\right){\int }_{G}f\left(hg\right)p\left(g^{-1}\right)d\mu \left(g\right)=(f*p)\left(h\right)\chi \left(b\right)\text{.}\end{array}$$ So $f*p\in C(G/B)$ and thus $C\left(G\right)*p\subseteq C(G/B)\text{.}$ Let $f\in C(G/B)$ and $h\in G\text{.}$ Then $$(f*p)\left(h\right)={\int }_{G}f\left(hk\right)p\left(k^{-1}\right)d\mu \left(k\right)={\int }_{G}f\left(h\right)\chi \left(k\right)\chi \left(k^{-1}\right)d\mu \left(k\right)=f\left(h\right)\mu \left(B\right)=f\left(h\right)\text{.}$$ So $f=f*p\in C\left(G\right)*p\text{.}$ Thus $C(G/B)\subseteq C\left(G\right)*p\text{.}$ The proof of (c) is similar to the proof of (b). $\square$

Assume that

(a)	Each function in $C(G/B)$ is supported on only a finite number of cosets of $B,$
(b)	Each function in $C(B\backslash G/B)$ is supported on only a finite number of cosets $BwB$ in $B\backslash G/B,$
(c)	For each $BwB\in B\backslash G/B$ the number of $B$ cosets in $BwB$ is finite.

Fix a set $R$ of coset representatives of $G/B$ and a subset $W\subseteq R$ of coset representatives of $B\backslash G/B\text{.}$ For each $x\in R$ and each $w\in W$ define functions $p_x\in C(G/B)$ and $T_w\in C(B\backslash G/B)$ by $$p_x\left(g\right)=\{\begin{array}{cc}\frac{\chi \left(b\right)}{\mu \left(B\right)},& \text{if}\hspace{0.17em}g=xb\in xB,\\ 0,& \text{otherwise,}\end{array}\text{and}{T}_w\left(g\right)=\{\begin{array}{cc}\frac{\chi \left(b_1\right)\chi \left(b_2\right)}{\mu \left(B\right)},& \text{if}\hspace{0.17em}g=b_{1}w{b}_2\in BwB,\\ 0,& \text{otherwise.}\end{array}$$ Then $$C\left(G/B\right)\text{has basis}\left\{p_x\hspace{0.17em}\right|\hspace{0.17em}x\in R\},\text{and}C(B\backslash G/B)\text{has basis}\left\{T_w\hspace{0.17em}\right|\hspace{0.17em}w\in W\}\text{.}$$ The following proposition completely determines the structure of the Hecke algebra $C(B\backslash G/B)$ and its action on $C(G/B)$ in terms of the combinatorics of cosets.

(a)	If $x\in R$ and $w\in W$ then $$p_x*T_w=\sum _{\underset{yB\in xBwB}{yB\in G/B}}\mu \left(B\right)T_w\left(x^{-1}y\right)p_y\text{.}$$
(b)	If $v,w\in W$ then $$T_u*T_v=\sum _{w\in W}{c}_{u,v}^w{T}_w,\text{where}{c}_{u,v}^w=\mu {\left(B\right)}^2\sum _{xB\in BuB\cap wB{v}^{-1}B}{T}_u\left(x\right)T_v\left(x^{-1}w\right)\text{.}$$

		Proof.
	(a) Let $y\in R\text{.}$ Then $$\begin{array}{ccc}(p_x*T_w)\left(y\right)& =& {\int }_G{p}_x\left(g\right)T_w\left(g^{-1}y\right)d\mu \left(g\right)={\int }_B{p}_x\left(xb\right)T_w\left(b^{-1}{x}^{-1}y\right)d\mu \left(xb\right)\\ & =& \{\begin{array}{cc}{p}_x\left(x\right)T_w\left(x^{-1}y\right)\mu \left(B\right),& \text{if}\hspace{0.17em}{x}^{-1}y\in BwB,\\ 0,& \text{otherwise,}\end{array}\\ & =& \{\begin{array}{cc}\mu \left(B\right)T_w\left(x^{-1}y\right)p_y\left(y\right),& \text{if}\hspace{0.17em}yB\in xBwB,\\ 0,& \text{otherwise,}\end{array}\end{array}$$ (b) Let $w\in W\text{.}$ Then $$\begin{array}{ccc}(T_u*T_v)\left(w\right)& =& {\int }_G{T}_u\left(x\right)T_v\left(x^{-1}w\right)d\mu \left(x\right)=\mu \left(B\right)\sum _{xB\in wB{v}^{-1}B\cap BuB}{T}_u\left(x\right)T_v\left(x^{-1}w\right)\\ & =& \mu {\left(B\right)}^2\left(\sum _{xB\in wB{v}^{-1}B\cap BuB}{T}_u\left(x\right)T_v\left(x^{-1}w\right)\right)T_w\left(w\right)\text{.}\end{array}$$ By ???, the map $\overrightarrow{t}:C\left(G\right)\to C\left(G\right)$ given by $$\overrightarrow{t}\left(h\right)=h\left(1\right),h\in C\left(G\right),$$ is a trace on $C\left(G\right)$ and, by restriction, $\overrightarrow{t}$ is a trace on $C(B\backslash G/B)\text{.}$ Let $⟨,⟩$ be the bilinear form on $C(B\backslash G/B)$ given by $$⟨h_1,h_2⟩=\overrightarrow{t}\left(h_1{h}_2\right),h_1,h_2\in C(B\backslash G/B)\text{.}$$ $\square$

Let $W$ be a set of coset representatives of the cosets in $B\backslash G/B$ and define $$\text{ind}\left(w\right)=\mu \left(BwB\right)=\text{(}\text{number of cosets of}\hspace{0.17em}B\hspace{0.17em}\text{in}\hspace{0.17em}BwB\text{),}$$ for each $w\in W\text{.}$ Then

(a)	${\left\{\frac{T_{w^{-1}}}{\text{ind}\left(w\right)}\right\}}_{w\in W}$ is the dual basis to ${\left\{T_w\right\}}_{w\in W}$ with respect to $⟨,⟩\text{.}$
(b)	If $|G/B|$ is finite then $\overrightarrow{t}=\frac{1}{|G/B|}\text{tr},$ where $\text{tr}$ is the trace of the action of $C(B\backslash G/B)$ on $C(G/B)\text{.}$

		Proof.
	(a) If $u,v\in W$ then $$\overrightarrow{t}\left(T_u{T}_v\right)=\sum _{w\in W}{c}_{u,v}^w\overrightarrow{t}\left(T_w\right)=c_{u,v}^1=\{\begin{array}{cc}\text{ind}\left(u\right),& \text{if}\hspace{0.17em}BuB=B{v}^{-1}B,\\ 0,& \text{otherwise.}\end{array}$$ (b) If $w\in W,$ $$\begin{array}{ccc}\text{tr}\left(T_w\right)& =& \sum _{xB\in G/B}(p_x*T_w){|}_{p_x}=\sum _{xB\in G/B}\sum _{yB\in xBwB}{T}_w\left(x^{-1}y\right)p_y{|}_{p_x}\\ & =& \{\begin{array}{cc}\sum _{xB\in G/B}1,& \text{if}\hspace{0.17em}w=1,\\ 0,& \text{otherwise,}\end{array}=\{\begin{array}{cc}|G/B|,& \text{if}\hspace{0.17em}w=1,\\ 0,& \text{otherwise.}\end{array}\end{array}$$ $\square$

Examples.

1. Let $G$ be a finite group. Then the delta functions ${\delta }_g,$ $g\in G,$ given by $${\delta }_g\left(h\right)=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}g=h,\\ 0,& \text{otherwise,}\end{array}\text{for all}\hspace{0.17em}h\in G,$$ form a basis of the space $C\left(G\right)$ of all functions on $G\text{.}$ If $\mu$ is a Haar measure on $C\left(G\right)$ then $$\mu \left(f\right)=\mu \left(\sum _{g\in G}f\left(g\right){\delta }_g\right)=\sum _{g\in G}f\left(g\right)\mu \left(g{\delta }_1\right)=\mu \left({\delta }_1\right)\sum _{g\in G}f\left(g\right),$$ and so there is, up to multiplication by constants, a unique Haar measure on the space of all functions on $G\text{.}$ Choosing $\mu \left({\delta }_1\right)=1/\left|G\right|$ normalizes $\mu$ so that $\mu \left(G\right)=1\text{.}$ The vector space $C\left(G\right)$ is closed under convolution with respect to $\mu$ and the associative algebra $ℱ$ is isomorphic to the group algebra $ℂG$ of $G$ via the map $$\begin{array}{ccc}C\left(G\right)& ⟶& ℂG\\ f& ⟼& \mu \left({\delta }_1\right)\sum _{g\in G}f\left(g\right)g\text{.}\end{array}$$

2. Let $$\begin{array}{ccc}G& =& {GL}_2{\left(\mathbb{Q}\right)}^{+}=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a,b,c,d\in \mathbb{Q},ad-bc>0\},\\ B& =& {SL}_2\left(\mathbb{Z}\right)=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a,b,c,d\in \mathbb{Z},ad-bc=1\}\text{.}\end{array}$$ Then the matrices in $$W=\left\{\left(\begin{array}{cc}{d}_1& 0\\ 0& d_2\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}{d}_1,d_2\in \mathbb{Q},d_1/d_2\in {\mathbb{Z}}_{>0}\}$$ are a set of coset representatives of the cosets in $B\backslash G/B\text{.}$ 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 ${𝔽}_q$ be a finite field with $q$ elements and let $$\begin{array}{ccc}G& =& {GL}_n\left({𝔽}_q\right)=\{A\in M_n\left({𝔽}_q\right)\hspace{0.17em}|\hspace{0.17em}\text{det}\left(A\right)\ne 0\},\\ W& =& \left\{\text{upper triangular matrices in}\hspace{0.17em}{GL}_n\left({𝔽}_q\right)\right\}\text{.}\end{array}$$ Then the set $$W=S_n=\left\{\text{permutation matrices in}\hspace{0.17em}{GL}_n\left({𝔽}_q\right)\right\}$$ is a set of coset representatives of the cosets in $B\backslash G/B$ and the Hecke algebra of the pair $(G,B)$ is a deformation of the group algebra of the symmetric group. If $s_i=(i,i+1)$ is the transposition switching $i$ and $i+1$ in the symmetric group and $T_i=T_{s_i}={\chi }_{B{s}_{i}B}$ denotes the corresponding basis element of the Hecke algebra then $$\begin{array}{ccc}{T}_i{T}_j& =& T_j{T}_i,\text{if}\hspace{0.17em}|I-j|>1,\\ T_i{T}_{i+1}{T}_i& =& T_{i+1}{T}_i{T}_{i+1},1\le i\le n-1,\\ T_i^2& =& (q-1)T_i+q,1\le i\le n\text{.}\end{array}$$ 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 $G\text{.}$ Then the cosets in $B\backslash G/B$ 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 $W\text{.}$ These algebras were introduced by Iwahori in [Iw].

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

Clifford Theory

Let $R$ be an algebra over $ℂ,$ $K$ a finite group, and fix $\beta :K\times K\to R$ and automorphisms $c_k:R\to R,$ $k\in K,$ such that the semidirect product algebra $R⋊{\left(ℂK\right)}_{\beta }$ is associative, where $$R⋊{\left(ℂK\right)}_{\beta }=\left\{\sum _{k\in K}{r}_k{t}_k\hspace{0.17em}\right|\hspace{0.17em}{r}_k\in R\},$$ with multiplication determined by the multiplication in $R$ and the relations $$t_{k_1}{t}_{k_2}=\beta (k_1,k_2)t_{k_1{k}_2},t_{k}r=c_k\left(r\right)t_k,\text{and}{t}_1=1,$$ for $k,k_1,k_2\in K$ and $r\in R\text{.}$

Let $N$ be an $R$ module. Define an new $R\text{-module}$ ${}^{k}N$ to have the same underlying vector space $N$ but with $R\text{-action}$ given by $$c_k\left(r\right)\circ n=rn,\text{for all}\hspace{0.17em}r\in R,n\in N\text{.}$$ If $P\subseteq N$ is an $R\text{-submodule}$ of $N$ then ${}^{k}P$ is an $R\text{-submodule}$ of ${}^{k}N,$ and so, $N$ is a simple $R\text{-module}$ if and only if ${}^{k}N$ is a simple $R\text{-module.}$ In this way $K$ acts on the (isomorphism classes of) simple $R\text{-modules.}$

The inertia group of a simple $R\text{-module}$ $R^{\lambda }$ is $$H=\{h\in K\hspace{0.17em}|\hspace{0.17em}{}^h{R}^{\lambda }\cong R^{\lambda }\}\text{.}$$ For each $h\in H$ fix an $R\text{-module}$ isomorphism ${\varphi }_h:{}^h{R}^{\lambda }\to R^{\lambda }\text{.}$ The condition that ${\varphi }_h$ is an $R\text{-module}$ homomorphism is the same as saying that, as linear transformations on the vector space $R^{\lambda },$ $$c_h\left(r\right){\varphi }_h={\varphi }_{h}r,\text{for all}\hspace{0.17em}r\in R,\hspace{0.17em}h\in H\text{.}$$ (Proof: $\left(c_h\left(r\right){\varphi }_{h}m\right)={\varphi }_h(c_h\left(r\right)\circ m)={\varphi }_{h}rm\text{.)}$ Thus $$\beta {(h_1,h_2)}^{-1}{\varphi }_{h_1}{\varphi }_{h_2}r=\beta {(h_1,h_2)}^{-1}{c}_{h_1}\left(c_{h_2}\left(r\right)\right){\varphi }_{h_1}{\varphi }_{h_2}=c_{h_1{h}_2}\left(r\right)\beta {(h_1,h_2)}^{-1}{\varphi }_{h_1}{\varphi }_{h_2}\text{.}$$ By Schur’s lemma ${\varphi }_h$ is unique up to constant multiples and so $$\beta {(h_1,h_2)}^{-1}{\varphi }_{h_1}{\phi }_{h_2}=\alpha (h_1,h_2){\varphi }_{h_1{h}_2},\text{for some}\hspace{0.17em}\alpha (h_1,h_2)\in ℂ\text{.}$$ Let ${\left(ℂH\right)}_{1/\alpha }$ be the algebra over $ℂ$ with basis $\left\{b_h\hspace{0.17em}\right|\hspace{0.17em}h\in H\}$ and multiplication given by $$\begin{array}{cc}{b}_{h_1}{b}_{h_2}=\alpha {(h_1,h_2)}^{-1}{b}_{h_1{h}_2},h_1,h_2\in H\text{.}& \text{(4.1)}\end{array}$$ If $H^{\mu }$ is a ${\left(ℂH\right)}_{1/\alpha }\text{-module}$ then $R^{\lambda }\otimes H^{\mu }$ is an $R⋊{\left(ℂH\right)}_{\beta }$ module with action given by $$r{t}_h(m\otimes n)=r{\varphi }_{h}m\otimes b_{h}n,$$ for all $r\in R,$ $h\in H,$ $m\in R^{\lambda }$ and $n\in H^{\mu }\text{.}$ This is an $R⋊{\left(ℂH\right)}_{\beta }$ action since $$\begin{array}{ccc}{t}_{h}r(m\otimes n)& =& {\varphi }_{h}rm\otimes b_{h}n=c_h\left(r\right){\varphi }_{h}m\otimes b_{h}n,\\ t_{h_1}{t}_{h_2}(m\otimes n)& =& {\varphi }_{h_1}{\varphi }_{h_2}m\otimes b_{h_1}{b}_{h_2}n=\beta (h_1,h_2)\alpha (h_1,h_2)\alpha {(h_1,h_2)}^{-1}{\varphi }_{h_1{h}_2}m\otimes b_{h_1{h}_2}n\text{.}\end{array}$$

(Clifford Theory)

(a)

Let $M$ be a finite dimensional simple $R⋊{\left(ℂK\right)}_{\beta }\text{-module.}$ Then $$M\cong {\text{Ind}}_{R⋊{\left(ℂH\right)}_{\beta }}^{R⋊{\left(ℂK\right)}_{\beta }}(R^{\lambda }\otimes H^{\mu }),$$ where $R^{\lambda }$ is a simple $R$ submodule of $M,$ $H=\{h\in K\hspace{0.17em}|\hspace{0.17em}{}^h{R}^{\lambda }\cong R^{\lambda }\hspace{0.17em}\text{as}\hspace{0.17em}R\text{-modules}\},$ $\alpha :H\times H\to ℂ$ is determined by choices of $R$ module isomorphisms ${\varphi }_h:t_h{R}^{\lambda }\to R^{\lambda },$ and $H^{\mu }$ is the simple ${\left(ℂH\right)}_{1/\alpha }\text{-module}$ given by $H^{\mu }={\text{Hom}}_R(R^{\lambda },M)$ with $$\begin{array}{cc}\left(b_h\psi \right)\left(m\right)=\alpha {(h,h^{-1})}^{-1}{t}_h\psi \left({\varphi }_{h^{-1}}\beta {(h,h^{-1})}^{-1}m\right),h\in H,\psi \in H^{\mu },m\in R^{\lambda },& \text{(4.3)}\end{array}$$ where $\left\{b_h\hspace{0.17em}\right|\hspace{0.17em}h\in H\}$ is the basis of ${\left(ℂH\right)}_{1/\alpha }$ in (???).

		Proof.
	Let $M$ be a simple $R⋊{\left(ℂK\right)}_{\beta }$ module. Let $R^{\lambda }$ be a simple $R\text{-submodule}$ of $M\text{.}$ Then $t_k{R}^{\lambda }$ is an $R\text{-submodule}$ of $M$ isomorphic to ${}^k{R}^{\lambda }\text{.}$ The sum ${\sum }_{k\in K}{t}_k{R}^{\lambda }$ is an $R⋊{\left(ℂK\right)}_{\beta }$ submodule of $M,$ and, since $M$ is simple, $$M=\sum _{k\in K}{t}_k{R}^{\lambda }=\sum _{k_i\in K/H}{t}_{k_i}N,\text{where}N=\sum _{h\in H}{t}_h{R}^{\lambda }$$ and the second sum is over a set of coset representatives of the cosets in $K/H\text{.}$ So $$M\cong {\text{Ind}}_{R⋊{\left(ℂH\right)}_{\beta }}^{R⋊{\left(ℂK\right)}_{\beta }}\left(N\right),$$ as $R⋊{\left({ℂ}_K\right)}_{\beta }\text{-modules.}$ We shall define a ${\left(ℂH\right)}_{1/\alpha }$ action on $H^{\mu }={\text{Hom}}_R(R^{\lambda },N)={\text{Hom}}_R(R^{\lambda },M)$ so that $$\begin{array}{cccc}\Theta :& R^{\lambda }\otimes {\text{Hom}}_R(R^{\lambda },N)& ⟶& N\\ & m\otimes \psi & ⟼& \psi \left(m\right)\end{array}$$ is an $R⋊{\left({ℂ}_K\right)}_{\beta }$ module isomorphism. The condition that $\Theta$ is an $R⋊{\left({ℂ}_H\right)}_{\beta }\text{-module}$ homomorphism is that $$t_h\left(\Theta (m\otimes \psi )\right)=\Theta \left(t_h(m\otimes \psi )\right),\text{for all}\hspace{0.17em}m\in M,\hspace{0.17em}\psi \in H^{\mu },\hspace{0.17em}h\in H,$$ and so $$t_h\psi \left(m\right)=\Theta ({\varphi }_{h}m\otimes b_h\psi )=\left(b_h\psi \right)\left({\varphi }_{h}m\right),\text{for all}\hspace{0.17em}m\in M,\hspace{0.17em}h\in H,\hspace{0.17em}\psi \in H^{\mu }\text{.}$$ Thus the appropriate formula for the action of ${\left(ℂH\right)}_{1/\alpha }$ on $H^{\mu }$ must be $$\left(b_h\psi \right)\left(m\right)=t_h\psi \left({\varphi }_h^{-1}m\right)=\alpha {(h,h^{-1})}^{-1}{t}_h\psi \left({\varphi }_{h^{-1}}\beta {(h,h^{-1})}^{-1}m\right)\text{.}$$ Given these formulas, it is straightforward to check that $H^{\mu }$ is a well defined ${\left(ℂH\right)}_{1/\alpha }$ module and $\Theta$ is an $R⋊{\left({ℂ}_H\right)}_{\beta }$ module isomorphism. The $R⋊{\left({ℂ}_H\right)}_{\beta }$ module $N$ is simple, since if $P$ is an $R⋊{\left({ℂ}_H\right)}_{\beta }$ submodule of $N$ then ${\text{Ind}}_{R⋊{\left({ℂ}_H\right)}_{\beta }}^{R⋊{\left({ℂ}_K\right)}_{\beta }}\left(P\right)$ is and $R⋊{\left({ℂ}_K\right)}_{\beta }$ submodule of $M\text{.}$ Since $M$ is simple $P$ must be equal to $N\text{.}$ Since $N$ is simple the map $\Phi$ is surjective. If $\Phi :R^{\lambda }\to N$ is a nonzero $R$ module homomorphism then it must be injective, since $R^{\lambda }$ is simple. So $\Phi \left(m\right)\ne 0$ for all nonzero $m\in R^{\lambda }\text{.}$ So the map $\Phi$ is injective. Thus $$R^{\lambda }\otimes {\text{Hom}}_R(R^{\lambda },N)\cong N\text{.}$$ It follows that ${\text{Hom}}_R(R^{\lambda },N)$ is a simple ${\left(ℂH\right)}_{1/\alpha }$ modules since any submodule $P$ would yield a submodule $\Phi (R^{\lambda }\otimes P)$ of $N\text{.}$ $\square$

Remark. Note that the map $$\begin{array}{ccc}{}^k{R}^{\lambda }& ⟶& t_k{R}^{\lambda }\\ n& ⟼& t_{k}n\end{array}$$ is an isomorphism.

(a)	The simple $R⋊{\left({ℂ}_K\right)}_{\beta }$ modules $R{K}^{(\lambda ,\mu )}$ are indexed by pairs $(\lambda ,\mu )$ with $\lambda \in \hat{R}/K,$ $\mu \in {\left(ℂH\right)}_{a/\alpha }^{ˆ},$ where $\hat{R}/K$ is a set of representatives of the $K$ orbits on $\hat{R},$ and ${\left(ℂH\right)}_{1/\alpha }^{ˆ}$ is an index set for the simple ${\left(ℂH\right)}_{1/\alpha }$ modules.
(b)	$\text{dim}\left(R{K}^{(\lambda /\mu )}\right)=\text{dim}\left(R^{\lambda }\right)\text{dim}\left(H^{\mu }\right)\text{Card}(K/H)\text{.}$
(c)	The irreducible $R⋊{\left({ℂ}_K\right)}_{\beta }$ module $R{K}^{(\lambda ,\mu )}$ is given by $$R{K}^{(\lambda ,\mu )}\cong {\text{Ind}}_{R⋊{\left({ℂ}_H\right)}_{\beta }}^{R⋊{\left({ℂ}_K\right)}_{\beta }}(R^{\lambda }\otimes H^{\mu })\text{.}$$

		Proof.
	Let $R^{\lambda }$ be a simple $R$ module. If the inertia group of $R^{\lambda }$ is $H$ then the inertia group of ${}^k{R}^{\lambda }$ is $kH{k}^{-1}\cong H\text{.}$ If $h\in H$ and $$\begin{array}{c}{\varphi }_h:{}^h{R}^{\lambda }\to R^{\lambda }\text{is an isomorphism, then}\\ \begin{array}{cccc}{\psi }_{gh{g}^{-1}}:& {}^{gh}{R}^{\lambda }& ⟶& {}^g{R}^{\lambda }\\ & m& ⟼& {\varphi }_h\left(m\right)\end{array}\text{is an}\hspace{0.17em}R\hspace{0.17em}\text{module isomorphism,}\end{array}$$ since $$\begin{array}{ccc}r\circ {\psi }_{gh{g}^{-1}}\left(m\right)& =& r\circ {\varphi }_h\left(m\right)=g^{-1}\left(r\right){\varphi }_h\left(m\right)\\ & =& {\varphi }_h(g^{-1}\left(r\right)\circ m)={\varphi }_h\left(h^{-1}{g}^{-1}\left(r\right)m\right)\\ & =& {\psi }_{gh{g}^{-1}}(r\circ m)\text{.}\end{array}$$ So, in fact we may choose ${\varphi }_{gh{g}^{-1}}={\varphi }_h\text{.}$ Then the factor set for $gH{g}^{-1}$ is $\alpha :H\times H\to ℂ$ and clearly ${\left({ℂ}^{g}H\right)}_{1/\alpha }\cong {\left(ℂH\right)}_{1/\alpha }$ and $$\begin{array}{ccc}{R}^{\lambda }\otimes H^{\mu }& ⟶& {}^g{R}^{\lambda }\otimes {}^g{H}^{\mu }\\ r\otimes m& ⟼& r\otimes m\end{array}$$ is an $R\text{-module}$ isomorphism. A different choice $${\stackrel{\sim }{\varphi }}_h:{}^h{R}^{\lambda }\to R^{\lambda }$$ may yield a different factor set $\stackrel{\sim }{\alpha }:H\times H\to ℂ,$ $${\stackrel{\sim }{\varphi }}_{h_1}{\stackrel{\sim }{\varphi }}_{h_2}=\stackrel{\sim }{\alpha }(h_1,h_2){\stackrel{\sim }{\varphi }}_{h_1{h}_2}\text{.}$$ By Schur’s lemma $${\stackrel{\sim }{\varphi }}_h=\gamma \left(h\right){\varphi }_h,\text{for some constant}\hspace{0.17em}\gamma \left(h\right)\in {ℂ}^{*}\text{.}$$ So $${\stackrel{\sim }{\varphi }}_{h_1}{\stackrel{\sim }{\varphi }}_{h_2}=\gamma \left(h_1\right)\gamma \left(h_2\right){\varphi }_{h_1}{\varphi }_{h_2}=\gamma \left(h_1\right)\gamma \left(h_2\right)\alpha (h_1,h_2){\varphi }_{h_1{h}_2}$$ implies $$\stackrel{\sim }{\alpha }(h_1,h_2)=\frac{\gamma \left(h_1\right)\gamma \left(h_2\right)}{\gamma \left(h_1{h}_2\right)}\alpha (h_1,h_2)\text{.}$$ Then the algebra ${\left(ℂH\right)}_{1/\stackrel{\sim }{\alpha }}$ given by $${\left(ℂH\right)}_{1/\stackrel{\sim }{\alpha }}=\text{span}\left\{d_h\hspace{0.17em}\right|\hspace{0.17em}h\in H\},\text{with}{d}_{h_1}{d}_{h_2}=\stackrel{\sim }{\alpha }{(h_1,h_2)}^{-1}{d}_{h_1{h}_2},$$ is isomorphic to ${\left(ℂH\right)}_{1/\alpha }$ via the isomorphism $$\begin{array}{ccc}{\left(ℂH\right)}_{1/\stackrel{\sim }{\alpha }}& \stackrel{\Phi }{⟶}& {\left(ℂH\right)}_{1/\alpha }\\ d_h& ⟼& c_h\gamma {\left(h\right)}^{-1}\text{.}\end{array}$$ Just to check: $$\Phi \left(d_{h_1}{d}_{h_2}\right)=c_{h_1}\gamma {\left(h_1\right)}^{-1}{c}_{h_2}\gamma {\left(h_2\right)}^{-1}=\gamma \left(h_1^{-1}\right)\gamma \left(h_2^{-1}\right)c_{h_1{h}_2}\alpha {(h_1,h_2)}^{-1}=\stackrel{\sim }{\alpha }{(h_1,h_2)}^{-1}\Phi \left(d_{h_1{h}_2}\right)\text{.}$$ $\square$

Example 1. Let $G$ be a group and let $N$ be a normal subgroup of $G\text{.}$ Let $K$ be the quotient group $N\backslash G$ and choose a set $\left\{t_k\hspace{0.17em}\right|\hspace{0.17em}k\in K\}$ of the representatives of the cosets in $N\backslash G\text{.}$ If the map $\beta :K\times K\to N$ and automorphisms $c_k:N\to N$ are $$t_{k_1}{t}_{k_2}=\beta (k_1,k_2)t_{k_1{k}_2}\text{and}{c}_k\left(n\right)=t_{k}n{t}_k^{-1},\text{then}ℂG\cong ℂN⋊{\left(ℂK\right)}_{\beta }\text{.}$$

Example 2. $G(r,1,n)={(\mathbb{Z}/r\mathbb{Z})}^n⋉S_n\text{.}$

Recall that $S_n$ acts on ${(\mathbb{Z}/r\mathbb{Z})}^n$ by permuting the factors. The simple $\mathbb{Z}/r\mathbb{Z}$ representations are indexed by $\lambda =0,1,2,\dots ,r-1$ and are given by $$\begin{array}{cccc}{x}^{\lambda }:& \mathbb{Z}/r\mathbb{Z}& ⟶& ℂ\\ & e^{2\pi ik/r}& ⟼& e^{2\pi i\lambda k/r}\end{array}$$ The simple ${(\mathbb{Z}/r\mathbb{Z})}^n$ modules are indexed by $n\text{-tuples}$ $\lambda =({\lambda }_1,\dots ,{\lambda }_n)$ with ${\lambda }_i\in \{0,1,2,\dots ,r-1\}\text{.}$ The group $S_n$ acts on the simple ${(\mathbb{Z}/r\mathbb{Z})}^n\text{-modules}$ and on the indexes $\lambda =({\lambda }_1,\dots ,{\lambda }_n)$ by permuting the factors. Each orbit under this action has a unique representative of the form $$\lambda =(\underset{\underset{m_0\hspace{0.17em}\text{times}}{⏟}}{0,0,\dots ,0},\underset{\underset{m_1\hspace{0.17em}\text{times}}{⏟}}{1,1,\dots ,1}\dots ,\underset{\underset{m_{r-1}\hspace{0.17em}\text{times}}{⏟}}{r-1,r-1,\dots ,r-1})\text{.}$$ The inertia group of this representation is $$S_{\overrightarrow{m}}=S_{m_0}\times S_{m_1}\times \cdots \times S_{m_{r-1}}$$ and ${(\mathbb{Z}/r\mathbb{Z})}^n$ module isomorphisms ${\varphi }_h:{}^h{R}^{\lambda }\to R^{\lambda }$ can be fixed to be ${\varphi }_h={\text{Id}}_{R^{\lambda }}$ for all $h\in H\text{.}$ So the factor set is trivial in this case.

The simple $S_{m_0}\times S_{m_1}\times \cdots \times S_{m_{r-1}}$ modules are indexed by $r\text{-tuples}$ of partitions $$\mu =({\mu }^{\left(0\right)},\dots ,{\mu }^{(r-1)})\text{with}{\mu }^{\left(i\right)}⊢m_i\text{.}$$ So the simple $G(r,1,n)\text{-modules}$ are indexed by pairs $$(\lambda ,\mu ),\text{with}\hspace{0.17em}\lambda \hspace{0.17em}\text{as in ??? and}\hspace{0.17em}\mu \hspace{0.17em}\text{as in ???.}$$ Given $\mu =({\mu }^{\left(0\right)},\dots ,{\mu }^{(r-1)})$ the information in $\lambda$ is redundant and so we may index the simple $G(r,1,n)$ modules by $r\text{-tuples}$ of partitions $\mu =({\mu }^{\left(0\right)},\dots ,{\mu }^{(r-1)})$ with $n$ boxes total.

The same type of analysis can be carried through for the affine symmetric group ${\stackrel{\sim }{S}}_n={\mathbb{Z}}^n⋉S_n\text{.}$ The simple $\mathbb{Z}\text{-modules}$ are given by $$\begin{array}{cccc}{x}^{\lambda }:& \mathbb{Z}& ⟶& ℂ\\ & 1& ⟶& \lambda \end{array}\lambda \in {ℂ}^{*}\text{.}$$ Thus the representations of ${\stackrel{\sim }{S}}_n$ are indexed by ${ℂ}^{*}\text{-tuples}$ of partitions $\mu ={\left({\mu }^{\left(\lambda \right)}\right)}_{\lambda \in {ℂ}^{*}}$ with $n$ boxes total. Alternatively the simple ${\stackrel{\sim }{S}}_n$ modules are indexed by functions from ${ℂ}^{*}$ to the set $𝒫$ of partitions $$\begin{array}{cccc}\mu :& {ℂ}^{*}& ⟶& 𝒫\\ & \lambda & ⟼& {\mu }^{\left(\lambda \right)}\end{array},\text{such that}n=\sum _{\lambda \in {ℂ}^{*}}\left|{\mu }^{\left(\lambda \right)}\right|\text{.}$$

The same analysis works for $G^n⋉S_n=G\wr S_n$ where $G$ is any finite group. The general statement is that the simple $G\wr S_n$ modules are indexed by functions $$\begin{array}{cccc}\mu :& \hat{G}& ⟶& 𝒫\\ & \lambda & ⟼& {\mu }^{\left(\lambda \right)}\end{array}\text{such that}n=\sum _{\lambda \in \hat{G}}\left|{\mu }^{\left(\lambda \right)}\right|,$$ where $\hat{G}$ is an index set for the simple $G\text{-modules.}$

Monomial groups

A composition series of $G$ is a sequence $\left\{1\right\}=G_0\subseteq G_1\subseteq \cdots \subseteq G_n=G$ with $G_{i-1}$ normal in $G_i\text{.}$

(a)	A group is solvable if there exists composition series of $G$ with $G_i/G_{i+1}$ abelian.
(b)	A group is supersolvable if there exists a composition series with $G_i$ normal in $G$ and $G_i/G_{i-1}$ cyclic.
(c)	A group is nilpotent if there exists a composition series of $G$ with $G_i/G_{i-1}\subseteq Z(G/G_{i-1})\text{.}$ $$\text{nilpotent}⟹\text{supersolvable}⟹\text{solvable.}$$
(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 $G\text{-module.}$ Case 1. Suppose that $K=\text{ker}\hspace{0.17em}M\ne \left\{1\right\}\text{.}$ Then $M$ is a $G/K\text{-module.}$ $$M={\text{Ind}}_{K⋊{(G/K)}_{\beta }}^{K⋊{(G/K)}_{\beta }}(1_K\otimes M)=1_K\otimes M\text{.}$$ Since $|G/K|<\left|G\right|,$ by induction, $M={\text{Ind}}_{H/K}^{G/K}\left(1_{\xi }\right),$ for some one dimensional representation $1_{\xi }$ of a subgroup $H/K\text{.}$ Case 2. If $K=\text{ker}\hspace{0.17em}M=\left\{1\right\}$ then $M$ is a faithful representation of $G\text{.}$ Since $G$ is supersolvable, $G/Z\left(G\right)$ is supersolvable and has a composition series in which the first nontrivial term ${(G/Z\left(G\right))}_1$ is a cyclic subgroup of $G/Z\left(G\right)\text{.}$ The inverse image of ${(G/Z\left(G\right))}_1$ in $G$ is a normal abelian subgroup $A$ of $G$ which is not contained in the center of $G\text{.}$ Then $$M={\text{Ind}}_{A⋊{(I/A)}_{\beta }}^{A⋊{(G/A)}_{\beta }}(1_{\xi }\otimes M\prime )={\text{Ind}}_I^G(1_{\xi }\otimes M\prime ),$$ where $I$ is the inertia group of $M$ and $1_{\xi }\otimes M$ is a simple $I$ module. Since $A$ is not contained in $Z\left(G\right)$ and $M$ is a faithful representation of $G$ the group $A$ does not act on $M$ by scalars, but $A$ does act on $1_{\xi }\otimes M$ by scalars. So $1_{\xi }\otimes M\prime \ne M$ and therefore $I\ne G\text{.}$ Thus, by induction, there is a one dimensional representation $1_{\eta }$ of a subgroup $H$ such that $$1_{\xi }\otimes M\prime ={\text{Ind}}_H^I\left(1_{\eta }\right)\text{.}\text{So}M={\text{Ind}}_I^G\left({\text{Ind}}_H^I\left(1_{\eta }\right)\right)={\text{Ind}}_H^G\left(1_{\eta }\right)\text{.}$$ $\square$

Nilpotent groups are monomial.

		Proof.
	Let $M$ be an irreducible representation of $G\text{.}$ 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 $\rho :G\to GL\left(M\right)$ be the hommorphism determined by $M\text{.}$ Let $K=\text{ker}\hspace{0.17em}\rho \text{.}$ Then $G/K$ acts on $M$ and $M$ is an irreducible representation of $G/K\text{.}$ 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 $$M\cong {\text{Ind}}_{H/K}^{G/K}\left(1_{\xi }\right),\text{as a}\hspace{0.17em}G/K\hspace{0.17em}\text{module.}$$ But then $$M\cong {\text{Ind}}_H^G\left(1_{\xi }\right),$$ since ${\text{Ind}}_H^G\left(1_{\xi }\right)$ and ${\text{Ind}}_{H/K}^{G/K}\left(1_{\xi }\right)$ can both be viewed as functions on $G/H\cong (G/K)(H/K)\text{.}$ So the theorem is proved for $K\ne \left\{1\right\}\text{.}$ Now assume that $K=\left\{1\right\},$ i.e. $\rho$ is injective. The last nontrivial term ${𝒞}^k\left(G\right)$ of the lower central series of $G$ is $Z\left(G\right)$ and ${𝒞}^{k-1}$ is a subgroup of $G$ containing $Z\left(G\right)\text{.}$ Let $x\in {𝒞}^{k-1}\left(G\right)$ be such that $x\notin Z\left(G\right),$ $x\in Z({𝒞}^{k-1}\left(G\right)/Z\left(G\right))\text{.}$ Then let $$\begin{array}{ccc}A& =& \left\{x^{k}z\hspace{0.17em}\right|\hspace{0.17em}k\in {\mathbb{Z}}_{\ge 0},z\in Z\left(G\right)\},\text{(finite case)}\\ A& =& \left\{e^{tX}z\hspace{0.17em}\right|\hspace{0.17em}t\in ℝ,z\in Z\left(G\right)\},\text{(Lie case)}\end{array}$$ so that $A$ is cyclic (finite case), $A$ is one dimensional (Lie case). Then $A$ is abelian since $$x^{k_1}{z}_1{x}^{k_2}{z}_2=z^{k_1+k_2}{z}_1{z}_2=x^{k_2}{z}_2{x}^{k_1}{z}_1,\text{for}\hspace{0.17em}{k}_i\in {\mathbb{Z}}_{\ge 0},\hspace{0.17em}{z}_i\in Z\left(G\right)\text{.}$$ Let $L$ be an irreducible $A\text{-submodule}$ of $M\text{.}$ Then $\text{dim}\hspace{0.17em}L=1$ (since $A$ is abelian). If $g\in G$ then $gL$ is a representation of $A$ that looks just like $L$ except $$a\left(g\ell \right)=g(g^{-1}a\mathrm{g)},\text{for}\hspace{0.17em}\ell \in L\text{.}$$ Since $L$ is a simple $A$ module so is $gL\text{.}$ Now $$\sum _{g\in G}gL\text{is a}\hspace{0.17em}G\text{-submodule of}\hspace{0.17em}M\text{.}$$ Since $M$ is simple $M={\sum }_{g\in G}gL\text{.}$ Let $$\stackrel{\sim }{A}=\{g\in G\hspace{0.17em}|\hspace{0.17em}gL\cong L\hspace{0.17em}\text{as}\hspace{0.17em}A\hspace{0.17em}\text{modules}\}\text{.}$$ Then $$\stackrel{\sim }{L}=\sum _{\stackrel{\sim }{a}\in \stackrel{\sim }{A}}\stackrel{\sim }{a}L$$ is an $\stackrel{\sim }{A}\text{-module.}$ Then $$M=\sum _{g\in G}gL=\sum _{g_i\in G/\stackrel{\sim }{A}}{g}_i\stackrel{\sim }{L},$$ where $g_i$ runs over a set of coset representatives of $G/\stackrel{\sim }{A}\text{.}$ This is a decomposition of $M$ as an $A\text{-module}$ such that (a) every irreducible $A$ submodule of $g_i\stackrel{\sim }{L}$ is isomorphic, (b) The irreducible $A$ submodules of $g_i\stackrel{\sim }{L}$ and $g_j\stackrel{\sim }{L}$ are not isomorphic. Since $A$ acts on $L$ by scalars, $A$ acts on $\stackrel{\sim }{L}$ by scalars. So $\stackrel{\sim }{L}\ne M,$ since $\rho$ is injective and $A\subseteq Z\left(G\right),$ $A\ne Z\left(G\right)\text{.}$ So $\stackrel{\sim }{A}\ne G$ and $M={\text{Ind}}_{\stackrel{\sim }{A}}^G\left(\stackrel{\sim }{L}\right)\text{.}$ Since $\stackrel{\sim }{A}$ is smaller than $G$ (since $G$ is not abelian) $$\stackrel{\sim }{L}\cong {\text{Ind}}_H^{\stackrel{\sim }{A}}\left(1_{\xi }\right)\text{.}$$ So $$M={\text{Ind}}_{\stackrel{\sim }{A}}^G\left({\text{Ind}}_H^{\stackrel{\sim }{A}}\left(1_{\xi }\right)\right)\cong {\text{Ind}}_H^G\left(1_{\xi }\right)\text{.}$$ The theorem now follows from the fact that all representations of an abelian group are 1 dimensional. $\square$

$$\left\{\text{supersolvable groups}\right\}\subseteq \left\{\text{monomial groups}\right\}\subseteq \left\{\text{solvable groups}\right\}\text{.}$$

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

A Lie group is exponential if the map $𝔤\stackrel{\text{exp}}{⟶}G$ is a diffeomorphism. $$\left\{\text{nilpotent Lie groups}\right\}\subseteq \left\{\text{exponential Lie groups}\right\}\subseteq \left\{\text{solvable Lie groups}\right\},$$ and the inclusions are strict.

A nilpotent Lie group is exponential.

		Proof.
	First note that if $X\in 𝔤$ then $e^X=1+X+\frac{X^2}{2!}+\frac{X^3}{3!}+\cdots$ is a finite sum, since $X^n=0$ for sufficiently large $n\text{.}$ Further $$X=\text{log}(1+(e^X-1))=(e^X-1)-\frac{{(e^X-1)}^2}{2}+\frac{{(e^X-1)}^3}{3}-\cdots$$ is also finite, and so the map exp is invertible if $𝔤\subseteq 𝔫\subseteq {𝔤𝔩}_n$ for some $n\text{.}$ $\square$

Note that the proof of this theorem uses the fact that any nilpotent Lie algebra can be imbedded in the nilpotent Lie algebra ${𝔫}_n$ of strictly upper triangular matrices for some $n\text{.}$ 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 $G,G\subseteq S_n$ for some $n,$ and (b) for an algebraic group $G,G\subseteq {GL}_n,$ for some $n\text{.}$ 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 $G\text{-module.}$ Let $\rho :G\to \text{End}\left(M\right)$ be the corresponding homomorphism. If $g\in G$ then $\rho \left(g\right)\in {\text{End}}_G\left(M\right)$ and so, by Schur’s lemma, $\rho \left(g\right)=\alpha {\text{Id}}_M$ for some $\alpha \in ℂ\text{.}$ So every element $g\in G$ acts on $M$ by scalars. Thus, if $m\in M$ then $ℂm\subseteq M$ is a submodule of $M\text{.}$ Since $M$ is irreducible, $ℂm\subseteq M,$ and therefore $M$ is one dimensional. $\square$

Kirillov’s classification

Let $G$ be a simply connected nilpotent Lie group. The coadjoint representation of $G$ is the action of $G$ on ${𝔤}^{*}=\text{Hom}(𝔤,ℝ)$ given by $$\left(g\varphi \right)\left(x\right)=\varphi \left({\text{Ad}}_{g^{-1}}\left(x\right)\right),\text{for}\hspace{0.17em}g\in G,\hspace{0.17em}x\in 𝔤\hspace{0.17em}\text{and}\hspace{0.17em}\varphi \in {𝔤}^{*}\text{.}$$ A coadjoint orbit is a set $G\varphi ,$ $\varphi \in {𝔤}^{*}\text{.}$ Then $$\left\{\text{coadjoint orbits}\right\}\leftrightarrow \left\{\text{irreducible unitary representations of}\hspace{0.17em}G\right\}$$

Let $\Omega$ be a coadjoint orbit $(\Omega \subseteq {𝔤}^{*})\text{.}$ Let $\varphi \in \Omega$ so that $\Omega =G\varphi \text{.}$ A Lie subalgebra $𝔥\subseteq 𝔤$ is subordinate to $\varphi$ if $$\varphi {|}_{[𝔥,𝔥]}=0,\text{or, equivalently,}\varphi \left([x,y]\right)=0\text{for all}\hspace{0.17em}x,y\in 𝔥\text{.}$$ A subalgebra $𝔥$ is subordinate to $𝔤$ if and only if $$\begin{array}{ccc}H& ⟶& ℂ\\ e^X& ⟼& e^{2\pi i\varphi \left(X\right)}\end{array}$$ defines a representation of $H=\text{exp}\left(𝔥\right)\text{.}$

Choose a maximal dimensional subalgebra $𝔥\subseteq 𝔤$ which is subordinate to $\varphi \text{.}$ (Choosing a maximal subalgebra $𝔥$ with respect to inclusion turns out to be the same thing.) Let $H=\text{exp}\left(𝔥\right)$ and define a one dimensional representation of $H$ by $$\begin{array}{cccc}{1}_{\varphi }:& H& ⟶& {ℂ}^{*}\\ & e^x& ⟼& e^{2\pi i\varphi \left(x\right)}\end{array}$$ for $x\in 𝔥\text{.}$ Then $$V^{\Omega }={\text{Ind}}_H^G\left(1_{\varphi }\right)$$ is the irreducible representation of $G$ associated to $\Omega \text{.}$ This says that every irreducible unitary representation of $G$ is obtained by inducing from a one dimensional representation of some subgroup $H\text{.}$ We will want to show that

(a)	$V$ is an irreducible representation of $G,$
(b)	$V$ does not depend on the choice of $\varphi$ and $𝔥$ (only on the orbit $\Omega \text{).}$

and answer the questions

(c)	Why is it sufficient to define $W^{\varphi }$ by its values on $e^X$ for $X\in 𝔥\text{?}$
(d)	What is ${\text{Ind}}_H^G\left(W\right)\text{?}$

Let $𝔥\subseteq \stackrel{\sim }{𝔥}$ be Lie subalgebras of $𝔤$ and let $\varphi \in {𝔤}^{*}$ be such that $𝔥$ and $\stackrel{\sim }{𝔥}$ are both subordinate to $\varphi \text{.}$ Then, since $\stackrel{\sim }{H}\supseteq H$ there are fewer functions in ${\text{Ind}}_{\stackrel{\sim }{H}}^G\left(W^{\varphi }\right)$ than in ${\text{Ind}}_H^G\left(W^{\varphi }\right)\text{.}$ The $G\text{-actions}$ on the two spaces are defined in the same way so ${\text{Ind}}_{\stackrel{\sim }{H}}^G\left(W^{\varphi }\right)$ is a submodule of ${\text{Ind}}_H^G\left(W^{\varphi }\right)\text{.}$ So $V^{\Omega }$ is not irreducible unless $𝔥$ is maximal.

Suppose $\Omega$ is a coadjoint orbit, $\varphi \in \Omega$ and $𝔥$ is a maximal dimensional subalgebra of $𝔤$ subordinate to $\varphi \text{.}$ Suppose $\stackrel{\sim }{\varphi }$ is another element of $\Omega \text{.}$ Then there is a $g\in G$ such that $$\stackrel{\sim }{\varphi }=g\varphi -{\text{Ad}}_g^{*}\varphi \text{.}\text{Let}\stackrel{\sim }{𝔥}=g𝔥={\text{Ad}}_g𝔥\text{.}$$ Then, for $gX\in g𝔥,$ $$\begin{array}{ccc}\stackrel{\sim }{\varphi }\left(gX\right)& =& \left(g\varphi \right)\left(gX\right)=\varphi \left(g^{-1}gX\right)=\varphi \left(X\right),\text{and since}\\ g[X,Y]& =& {\text{Ad}}_g\left([X,Y]\right)=R_g[X,Y]R_{g^{-1}}\\ & =& R_g(XY-YX)R_{g^{-1}}\\ & =& \left(R_{g}X{R}_{g^{-1}}\right)\left(R_{g}Y{R}_{g^{-1}}\right)-\left(R_{g}Y{R}_{g^{-1}}\right)\left(R_{g}X{R}_{g^{-1}}\right)\\ & =& [gX,gY],\end{array}$$ it follows that $\stackrel{\sim }{𝔥}$ is subordinate to $\stackrel{\sim }{\varphi }$ if and only if $𝔥$ is subordinate to $\varphi \text{.}$ If $H=\text{exp}\left(𝔥\right)$ and $\stackrel{\sim }{H}=\text{exp}\left(\stackrel{\sim }{𝔥}\right)$ then $\stackrel{\sim }{H}=gH{g}^{-1},$ since ${\text{Ad}}_g$ is the differential of ${\text{Int}}_g\text{.}$ So we get one dimensional representations $$\begin{array}{cccc}{W}^{\varphi }:& H& ⟶& ℂ\\ & h& ⟼& e^{2\pi i\varphi \left(X\right)}\end{array}\text{and}\begin{array}{cccc}{W}^{\stackrel{\sim }{\varphi }}:& gH{g}^{-1}& ⟶& ℂ\\ & gh{g}^{-1}& ⟼& e^{2\pi i\stackrel{\sim }{\varphi }\left(gX\right)}& =& e^{2\pi i\varphi \left(X\right)}\end{array}$$ if $h=e^X\text{.}$ Define a map $$\begin{array}{ccc}{\text{Ind}}_H^G\left(W^{\varphi }\right)& \stackrel{\Phi }{⟶}& {\text{Ind}}_{\stackrel{\sim }{H}}^G\left(W^{\stackrel{\sim }{\varphi }}\right)\\ f& ⟼& \stackrel{\sim }{f}\end{array}\text{by}\stackrel{\sim }{f}\left(t\right)=\widetilde{f\left(g^{-1}t\right)}\text{.}$$ Then $$\stackrel{\sim }{f}\left(gh{g}^{-1}t\right)=\widetilde{f\left(h{g}^{-1}t\right)}=\widetilde{hf\left(g^{-1}t\right)}=\stackrel{\sim }{h}\stackrel{\sim }{f}\left(t\right)$$ and $$\widetilde{xf}\left(t\right)=\widetilde{xf\left(g^{-1}t\right)}=\widetilde{f\left(g^{-1}tx\right)}=\stackrel{\sim }{f}\left(tx\right)=\left(x\stackrel{\sim }{f}\right)\left(t\right)\text{.}$$ So $\Phi$ is a well defined $G\text{-module}$ 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 $V\text{.}$ $$f=(0\subseteq V_1\subseteq V_2\subseteq \cdots \subseteq V_n=V),\text{dim}\hspace{0.17em}{V}_i=i\text{.}$$ Let $$W(f,⟨,⟩)=\sum _{i=0}^n{V}_i\cap V_i^{\perp },$$ Then

(a)	$W(f,⟨,⟩)$ is a maximal isotropic subspace of $V,$
(b)	If $⟨x,y⟩=⟨\varphi ,[x,y]⟩$ for some $\varphi \in {𝔤}^{*}$ then $𝔥$ is a subalgebra of $𝔤\text{.}$

Let $𝔤$ be a nilpotent Lie algebra such that $\text{dim}\left(Z\left(𝔤\right)\right)=1\text{.}$ Then $$𝔤=ℝx\oplus ℝy\oplus ℝz\oplus W,$$ where $ℝz=Z\left(𝔤\right),$ $[x,y]=z,$ and $[y,w]=0$ for $w\in W\text{.}$

		Proof.
	Let $y\in 𝔤$ such that the image of $y$ in $𝔤/Z\left(𝔤\right)$ is a nonzero element of $Z(𝔤/Z\left(𝔤\right))\text{.}$ Then $$\begin{array}{cccc}{\text{ad}}_y:& 𝔤& ⟶& Z\left(𝔤\right)\\ & t& ⟶& [y,t],\end{array}$$ has $\text{im}\left({\text{ad}}_y\right)=Z\left(𝔤\right),$ and $\text{ker}\left({\text{ad}}_y\right)=Z_g\left(y\right)\text{.}$ Thus $\text{dim}\left(\text{ker}\left({\text{ad}}_y\right)\right)=\text{dim}\left(𝔤\right)-1$ and $\text{ker}\hspace{0.17em}{\text{ad}}_y$ is an ideal of $𝔤$ since $${\text{ad}}_y\left([x,t]\right)=[y,[x,t]]=-[t,[y,x]]-[x,[t,y]]=0,\text{for}\hspace{0.17em}t\in \text{ker}\hspace{0.17em}{\text{ad}}_y\hspace{0.17em}\text{and}\hspace{0.17em}x\in 𝔤\text{.}$$ So $$Z_{𝔤}\left(y\right)=ℝy\oplus ℝz\oplus W,\text{where}\hspace{0.17em}[y,w]=0\hspace{0.17em}\text{for}\hspace{0.17em}w\in W,$$ is an ideal of $𝔤\text{.}$ Let $x\in 𝔤$ such that $[x,y]\ne 0$ and then normalize $x$ so that $[x,y]=z\text{.}$ $\square$

Note that, in the previous lemma, $ℝx+ℝy+ℝz$ is a subalgebra isomorphic to a Heisenberg algebra: $$𝔤=\left\{\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}x,y,z\in ℝ\}\text{.}$$ and let $$x=\left(\begin{array}{ccc}0& 1& 0\\ & 0& 0\\ & & 0\end{array}\right),y=\left(\begin{array}{ccc}0& 0& 0\\ & 0& 1\\ & & 0\end{array}\right),z=\left(\begin{array}{ccc}0& 0& 1\\ & 0& 0\\ & & 0\end{array}\right)\text{.}$$ Then $𝔤=ℝx+ℝy+ℝz,$ $[x,y]=z$ and $ℝz=Z\left(𝔤\right)\text{.}$ So, if $G$ is a nilpotent Lie group with one dimensional center the $$G_0=\text{exp}\left(Z_{𝔤}\left(y\right)\right),\text{where}\hspace{0.17em}{Z}_{𝔤}\left(y\right)=ℝy\oplus ℝz\oplus W,$$ and $G_0$ is normal since $e^{tx}{e}^h{e}^{-tx}=e^{t[x,h]+\cdots },$ where $t[x,h]+\cdots \in Z_{𝔤}\left(y\right)$ since only involves brackets of $x$ and $h$ and $Z_{𝔤}\left(y\right)$ is an ideal. So $$G=\left\{g_0{e}^{tx}\hspace{0.17em}\right|\hspace{0.17em}{g}_0\in G_0,t\in ℝ\}\text{.}$$ Since $Z_{𝔤}\left(y\right)$ is an ideal of $𝔤,$ $$G_0=\text{exp}\left(Z_{𝔤}\left(y\right)\right)\text{is a normal subgroup of}\hspace{0.17em}G,$$ and, in fact, every element of $G$ can be written uniquely in the form $$g_0{e}^{tx},g_0\in G_0,t\in ℝ\text{.}$$ Since $G_0$ is normal in $G,$ the group $G$ acts on $G_0$ be automorphisms. Let $V_0$ be an irreducible representation of $G_0\text{.}$ Define ${}^g{V}_0$ to be the $G_0$ module with the same vector space ${}^g{V}_0=V$ and with $G_0$ action $$g_0·v=\left(g^{-1}{g}_{0}g\right)v,\text{for}\hspace{0.17em}{g}_0\in G_0\hspace{0.17em}\text{and}\hspace{0.17em}v\in V_0\text{.}$$ Then ${}^g{V}_0$ is a new $G_0$ representation and ${}^g{V}_0$ is irreducible if and only if $V_0$ is (since, if $P\subseteq V_0$ is a submodule of $V_0$ then ${}^{g}P\subseteq {}^g{V}_0$ is a submodule of ${}^g{V}_0\text{).}$ So $G$ acts on the irreducible representations of $G_0\text{.}$

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 $Z\left(G\right)$ acts on $M$ by scalars. Let $z_1,\dots ,z_k$ be a basis of $Z\left(𝔤\right)\text{.}$ Let $\varphi :Z\left(𝔤\right)\to ℂ$ be the map such that $$e^{2\pi i\varphi \left(z\right)}=\rho \left(e^z\right),\text{for}\hspace{0.17em}z\in Z\left(𝔤\right),\hspace{0.17em}\text{and let}{Z}_0=\{z\in Z\left(𝔤\right)\hspace{0.17em}|\hspace{0.17em}\varphi \left(z\right)=0\}\text{.}$$ So $z\in Z_0$ if ${\sum }_{i=1}^k{c}_i\varphi \left(z_i\right)=0\text{.}$ So $\text{dim}\left(z_0\right)\ge \text{dim}\hspace{0.17em}Z-1$ and $\text{exp}\left(Z_0\right)$ acts trivially on $M\text{.}$ Let $\rho :G\to GL\left(M\right)$ has a nontrivial kernel unless $\text{dim}\hspace{0.17em}Z\left(𝔤\right)=1\text{.}$ In the case when $\text{dim}\hspace{0.17em}Z\left(𝔤\right)=1$ we can use Kirillov’s lemma to obtain the normal subgroup $G_0=\text{exp}\left(Z_{𝔤}\left(y\right)\right)\text{.}$

The Heisenberg group

The Heisenberg group $$G=\left\{\left(\begin{array}{ccc}1& x& z\\ & 1& y\\ & & 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}x,y,z\in ℝ\}\text{.}$$ has the Lie algebra $$𝔤=\left\{\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}x,y,z\in ℝ\},$$ where we exhibit $𝔤$ as a Lie subalgebra of $M_n\left(ℝ\right)$ since $H$ is a subgroup of ${GL}_n\left(ℝ\right)\text{.}$ Setting $$X=\left(\begin{array}{ccc}0& 1& 0\\ & 0& 0\\ & & 0\end{array}\right),Y=\left(\begin{array}{ccc}0& 0& 0\\ & 0& 1\\ & & 0\end{array}\right),Z=\left(\begin{array}{ccc}0& 0& 1\\ & 0& 0\\ & & 0\end{array}\right)\text{.}$$ $𝔥=\text{span-}\{X,Y,Z\}$ with bracket determined by $$[X,Y]=Z,[X,Z]=0,[Y,Z]=0\text{.}$$

Since $[𝔤,[𝔤,𝔤]]=0,$ the lower central series of $𝔤$ is $$𝔤\supseteq [𝔤,𝔤]\supseteq 0,\text{where}[𝔤,𝔤]=\left\{\left(\begin{array}{ccc}0& 0& z\\ & 0& 0\\ & & 0\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}z\in ℝ\},$$ is the center of $𝔤\text{.}$ The exponential map is given by $$\begin{array}{ccc}𝔤& ⟶& G\\ \left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)& ⟼& \left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)\end{array},$$ since $$\text{exp}\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)=1+\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)+\frac{1}{2}=\left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)\text{.}$$ Let $$X=\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)\in 𝔤,g=e^X=\left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)\in G,\text{and}Y=\left(\begin{array}{ccc}0& a& c\\ & 0& b\\ & & 0\end{array}\right)\in 𝔤\text{.}$$ Then $${\text{Ad}}_{𝔤}Y=gY{g}^{-1}=\left(\begin{array}{ccc}0& a& c+bx\\ & 0& b\\ & & 0\end{array}\right)\left(\begin{array}{ccc}1& -x& -z+xy/2\\ & 1& -y\\ & & 1\end{array}\right)=\left(\begin{array}{ccc}0& a& -ay+bx+c\\ & 0& b\\ & & 0\end{array}\right)$$ describes the adjoint representation of $G\text{.}$ (Alternatively one can use the formula $\left({\text{Ad}}_{e^X}\right)\left(Y\right)=\left(e^{{\text{ad}}_X}\right)\left(Y\right)$ to do this calculation.)

To compute the coadjoint representation we need a good way of looking at ${𝔤}^{*}=\text{Hom}(𝔤,ℝ)\text{.}$ Use the trace on $M_n\left(ℝ\right)$ $$\begin{array}{cccc}\text{Tr}:& M_n\left(ℝ\right)& ⟶& ℝ\\ & X& ⟼& \text{Tr}\left(X\right)\end{array}$$ to define a symmetric bilinear form $⟨,⟩:M_n\left(ℝ\right)\times M_n\left(ℝ\right)\to ℝ$ by $$⟨X,Y⟩=\text{Tr}\left(XY\right),\text{for}\hspace{0.17em}X,Y\in 𝔤\text{.}$$ The form $⟨,⟩$ provides a vector space isomorphism $$\begin{array}{ccc}{M}_n\left(ℝ\right)& ⟼& M_n{\left(ℝ\right)}^{*}\\ X& ⟼& ⟨X,·⟩\end{array}\text{where}\begin{array}{cccc}⟨X,·⟩:& M_n\left(ℝ\right)& ⟶& ℝ\\ & Y& ⟼& ⟨X,Y⟩\end{array},\text{for}\hspace{0.17em}X\in M_n\left(ℝ\right)\text{.}$$ Use this isomorphism to identify $M_n\left(ℝ\right)$ and $M_n{\left(ℝ\right)}^{*}\text{.}$ Our favourite basis of $M_n\left(ℝ\right)$ is $\left\{E_{ij}\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le n\},$ where $E_{ij}$ denotes the matrix with a $1$ in the $(i,j)\text{th}$ entry and $0$ everywhere else. Then $\{E_{ij}^{*}=E_{ji}\hspace{0.17em}|\hspace{0.17em}1\le i,j\le n\}$ is the dual basis with respect to $⟨,⟩\text{.}$ Since $𝔤=\text{span-}\{E_{12},E_{13},E_{23}\},$ $${𝔤}^{*}=\text{span-}\{E_{12}^{*},E_{13}^{*},E_{23}^{*}\}=\left\{\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ \gamma & \beta & 0\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}\alpha ,\beta ,\gamma \in ℝ\}\text{.}$$ Let $$g=e^X=\left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)\in G,\varphi =\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ \gamma & \beta & 0\end{array}\right)\in {𝔤}^{*},Y=\left(\begin{array}{ccc}0& a& c\\ & 0& b\\ & & 0\end{array}\right)\in 𝔤\text{.}$$ Then $$\begin{array}{ccc}\left({\text{Ad}}_g^{*}\right)\left(Y\right)& =& \varphi \left({\text{Ad}}_{g^{-1}}Y\right)=\varphi \left(g^{-1}Yg\right)=⟨\varphi ,g^{-1}Yg⟩\\ & =& ⟨\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ \gamma & \beta & 0\end{array}\right),\left(\begin{array}{ccc}0& a& -ay+bx+c\\ & 0& b\\ & & 0\end{array}\right)⟩\\ & =& \alpha a+\beta b+\gamma ay-\gamma bx+\gamma c\\ & =& ⟨\left(\begin{array}{ccc}0& & \\ \alpha +\gamma y& 0& \\ \gamma & \beta -\gamma x& 0\end{array}\right),\left(\begin{array}{ccc}0& a& c\\ & 0& b\\ & & 0\end{array}\right)⟩\text{.}\end{array}$$ Thus $${\text{Ad}}_g^{*}(\alpha ,\beta ,\gamma )=(\alpha +\gamma y,\beta -\gamma x,\gamma ),$$ where we use the more concise notation $(\alpha ,\beta ,\gamma )$ for the matrix $\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ \gamma & \beta & 0\end{array}\right)\in {𝔤}^{*}\text{.}$

To compute the coadjoint orbits note that

(a)	If $\gamma \ne 0$ then we can choose $x,y\in ℝ$ so that ${\text{Ad}}_g^{*}(\alpha ,\beta ,\gamma )=(0,0,\gamma )\text{.}$ So $(0,0,\gamma )$ is in the orbit of $(\alpha ,\beta ,\gamma )\text{.}$
(b)	If $\gamma =0$ then ${\text{Ad}}_g^{*}(\alpha ,\beta ,\gamma )=(0,0,\gamma ),$ for all $g\in G\text{.}$ So $(\alpha ,\beta ,0)$ is in an orbit all by itself.

Thus there are two kinds of coadjoint orbits $${\Omega }_{\gamma }=\left\{(\alpha ,\beta ,\gamma )\hspace{0.17em}\right|\hspace{0.17em}\alpha ,\beta \in ℝ\},\gamma \in ℝ\backslash \left\{0\right\},\text{and}{\Omega }_{\alpha ,\beta }=\left\{(\alpha ,\beta ,0)\right\},\alpha ,\beta \in ℝ,$$ with $\text{dim}\hspace{0.17em}{\Omega }_{\gamma }=2$ and $\text{dim}\hspace{0.17em}{\Omega }_{\alpha ,\beta }=0\text{.}$

Note that $\text{dim}\hspace{0.17em}𝔤=3$ and $$\left\{\left(\begin{array}{ccc}0& 0& c\\ & 0& b\\ & 0& \end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}b,c\in ℝ\}$$ is a two dimensional subalgerba of $𝔤$ such that $[𝔥,𝔥]=0\text{.}$ So $𝔥$ is a subalgebra of $𝔤$ which is subordinate to any $\varphi \in {𝔤}^{*},$ $$\varphi \left([x,y]\right)=\varphi \left(0\right)=0,\text{for all}\hspace{0.17em}x,y\in 𝔥\text{.}$$ So, if $\varphi \in {𝔤}^{*}$ then either (a) there is a 3 dimensional subalgebra subordinate to $\varphi$ (which must be all of $𝔤\text{),}$ or (b) $𝔥$ is a maximal subordinate subalgebra to $\varphi \text{.}$

(a) If $\varphi =(\alpha ,\beta ,0)=\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ 0& \beta & 0\end{array}\right)\in {\Omega }_{\alpha ,\beta }$ then $𝔤$ is subordinate to $\varphi$ and $$\begin{array}{ccc}G& ⟶& {ℂ}^{*}\\ e^X& ⟼& e^{2\pi i⟨\varphi ,X⟩}\end{array}$$ is the representation $V^{{\Omega }_{\alpha ,\beta }}$ associated to the orbit ${\Omega }_{\alpha ,\beta }\text{.}$ More precisely, $V^{{\Omega }_{\alpha ,\beta }}=\text{span-}\left\{v\right\}$ and $$\left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)v=e^{2\pi i(\alpha x+\beta y)}v,\text{for}\hspace{0.17em}x,y,z\in ℝ,$$ since $$⟨\varphi ,X⟩=⟨\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ 0& \beta & 0\end{array}\right),\left(\begin{array}{ccc}0& x& z\\ \alpha & 0& y\\ & & 0\end{array}\right)⟩=\alpha x+\beta y\text{.}$$

(b) If $$\varphi =(\alpha ,\beta ,\gamma )=\left(\begin{array}{ccc}0& 0& 0\\ \alpha & 0& 0\\ \gamma & \beta & 0\end{array}\right)\in {\Omega }_{\gamma },\gamma \ne 0,$$ then $𝔤$ is not subordinate to $\varphi$ and so $𝔥$ is a maximal subordinate subalgebra for $\varphi \text{.}$ The one dimensional representation of $H=\text{exp}\left(𝔥\right)$ is given by $$\begin{array}{cccc}{W}^{\varphi }:& H& ⟶& {ℂ}^{*}\\ & \left(\begin{array}{ccc}1& 0& z\\ & 1& y\\ & & 1\end{array}\right)& ⟼& e^{2\pi i(y\beta +\gamma z)}\end{array}$$ since $$\text{exp}\left(\left(\begin{array}{ccc}0& 0& z\\ & 0& y\\ & & 0\end{array}\right)\right)=\left(\begin{array}{ccc}1& 0& z\\ & 1& y\\ & & 1\end{array}\right)\text{and}⟨\left(\begin{array}{ccc}0& 0& z\\ & 0& y\\ & & 0\end{array}\right),\left(\begin{array}{ccc}0& & \\ \alpha & 0& \\ \gamma & \beta & 0\end{array}\right)⟩=y\beta +\gamma z\text{.}$$ The representation of $G$ corresponding to the orbit ${\Omega }_{\gamma }$ is $$V^{{\Omega }_{\gamma }}={\text{Ind}}_H^G\left(W^{\varphi }\right),$$ and we can view elements of ${\text{Ind}}_H^G\left(W^{\varphi }\right)$ as functions on $H\backslash G\text{.}$ Since $$H=\left\{\left(\begin{array}{ccc}1& 0& z\\ & 1& y\\ & & 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}y,z\in ℝ\}\text{and}G=\left\{\left(\begin{array}{ccc}1& x& z\\ & 1& y\\ & & 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}y,z\in ℝ\},$$ the elements $$\left(\begin{array}{ccc}1& t& 0\\ & 1& 0\\ & & 1\end{array}\right),t\in ℝ,$$ are coset representatives of the cosets in $H\backslash G\text{.}$ So we may view elements of $V^{{\Omega }_{\gamma }}$ as functions $$\begin{array}{cccccc}f:& ℝ& ⟶& ℂ& =& W^{\varphi }\\ & t& ⟼& f\left(t\right)\end{array}\text{.}$$ If $X=\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)$ then $$\left(e^{X}f\right)\left(t\right)=f\left(t{e}^X\right)=f\left(\left(\begin{array}{ccc}1& t& 0\\ & 1& 0\\ & & 1\end{array}\right)\left(\begin{array}{ccc}1& x& z+xy/2\\ & 1& y\\ & & 1\end{array}\right)\right)=f\left(\left(\begin{array}{ccc}1& 0& a_2\\ & 1& a_1\\ & & 1\end{array}\right)\left(\begin{array}{ccc}1& s& 0\\ & 1& 0\\ & & 1\end{array}\right)\right),$$ where $a_1=y,$ $s=t+x,$ and $a_2=z+\left(xy\right)/2+ty\text{.}$ So $$\begin{array}{ccc}\left(e^{X}f\right)\left(t\right)& =& f\left(\left(\begin{array}{ccc}1& x& z+xy/2+ty\\ & 1& y\\ & & 1\end{array}\right)\left(\begin{array}{ccc}1& t+x& 0\\ & 1& 0\\ & & 1\end{array}\right)\right)\\ & =& \left(\begin{array}{ccc}1& x& z+xy/2+ty\\ & 1& y\\ & & 1\end{array}\right)f\left(\left(\begin{array}{ccc}1& t+x& 0\\ & 1& 0\\ & & 1\end{array}\right)\right)\\ & =& e^{2\pi i(y\beta +\gamma z+\gamma \left(xy\right)/2+\gamma ty)}f(t+x),\end{array}$$ and this formula describes the action of $G$ on $V^{{\Omega }_{\gamma }}\text{.}$

Summary: For each orbit ${\Omega }_{\alpha ,\beta },$ $\alpha ,\beta \in RR,$ we get a one dimensional representation $V^{{\Omega }_{\alpha ,\beta }}=ℂv$ with action $$e^{X}v=e^{2\pi i(\alpha x+\beta y)}v,\text{if}X=\left(\begin{array}{ccc}0& x& z\\ & 0& y\\ & & 0\end{array}\right)\in 𝔤,$$ and for each orbit ${\Omega }_y,$ $\gamma \ne 0,$ we get an action of $G$ on functions $$W^{{\Omega }_{\gamma }}=\{f:ℝ\to ℂ\}\text{given by}\left(e^{X}f\right)\left(t\right)=e^{2\pi i(y\beta +\gamma z+\gamma \left(xy\right)/2+\gamma ty)}f(t+x)\text{.}$$

Let us show that the representations $V^{\gamma }$ are irreducible by computing $${\text{Hom}}_G(V^{{\Omega }_g},V^{{\Omega }_g})={\text{Hom}}_G({\text{Ind}}_H^G\left(W^{\varphi }\right),{\text{Ind}}_H^G\left(W^{\varphi }\right))\cong \{T:G\to \text{Hom}(W^{\varphi },W^{\varphi })\hspace{0.17em}|\hspace{0.17em}T\left(h_{1}g{h}_2\right)=T\left(g\right)\}\text{.}$$ Then $$\begin{array}{ccc}T\left(\begin{array}{ccc}1& x& z\\ & 1& y\\ & & 1\end{array}\right)& =& T\left(\left(\begin{array}{ccc}1& 0& z\\ & 1& y\\ & 1& \end{array}\right)\left(\begin{array}{ccc}1& x& 0\\ & 1& 0\\ & & 1\end{array}\right)\right)=e^{2\pi i(\beta y+\gamma z)}T\left(x\right)\\ & =& T\left(\left(\begin{array}{ccc}1& x& 0\\ & 1& 0\\ & 1& \end{array}\right)\left(\begin{array}{ccc}1& 0& z-xy\\ & 1& y\\ & & 1\end{array}\right)\right)=e^{2\pi i(\beta y+\gamma z-\gamma xy)}T\left(x\right)\text{.}\end{array}$$ So $T\left(x\right)=0$ unless $e^{2\pi i(-\gamma xy)}=1$ for all $y\in ℝ\text{.}$ So $T\left(x\right)=0$ unless $x=0\text{.}$ So $T$ is determined by its value at the identity matrix. So $\text{dim}\hspace{0.17em}{\text{Hom}}_G(V^{{\Omega }_g},V^{{\Omega }_g})=1\text{.}$ So $V^{{\Omega }_g}$ is irreducible.

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

Let $\varphi \in {𝔤}^{*}$ and let $𝔥$ be a maximal subalgebra of $G$ subordinate to $\varphi \text{.}$ Then (if $V^{\varphi }$ is unitary) $$V^{\varphi }={\text{Ind}}_H^G\left(W^{\varphi }\right)\text{is irreducible.}$$

		Proof.
	Let $x$ be a coset representative for a coset in $H\backslash G\text{.}$ Let $t:G\to \text{Hom}(W^{\varphi },W^{\varphi })$ be such that $$T\left(h_{1}x{h}_2\right)=h_{1}T\left(x\right)h_2,\text{for all}\hspace{0.17em}{h}_1,h_2\in H\text{.}$$ Let $\eta \in 𝔥\text{.}$ Then $$e^{2\pi i\varphi \left(\eta \right)}T\left(x\right)=T\left(e^{\eta }x\right)=T\left(x{x}^{-1}{e}^{\eta }x\right)=T\left(x{e}^{{\text{Ad}}_{x^{-1}}\eta }\right)=T\left(x\right)e^{2\pi i\varphi \left({\text{Ad}}_{x^{-1}}\eta \right)}\text{.}$$ So $T\left(x\right)=0$ unless $\varphi \left({\text{Ad}}_{x^{-1}}\eta \right)=\varphi \left(\eta \right)$ for all $\eta \in 𝔥\text{.}$ So $T\left(x\right)=0$ unless $\varphi \left(e^{\text{ad}(-X)}\eta \right)=\varphi \left(\eta \right)\text{.}$ So $T\left(x\right)=0$ unless $\varphi \left((e^{\text{ad}(-X)}-1)\eta \right)=0\text{.}$ Now $$\text{ad}(-X)=\text{log}(1+(e^{\text{ad}(-X)}-1))=(e^{\text{ad}(-X)}-1)-\frac{{(e^{\text{ad}(-X)}-1)}^2}{2}+\cdots \text{.}$$ So $\varphi \left(\text{ad}(-X)\eta \right)=0$ for all $\eta \in 𝔥\text{.}$ So $\varphi (-[X,\eta ])=0$ for all $\eta \in 𝔥\text{.}$ Let $\stackrel{\sim }{𝔥}=ℂX\oplus 𝔥\text{.}$ We know $X\notin 𝔥$ since $x$ is a coset representative of $H\backslash G$ (not $1\text{).}$ Then $\stackrel{\sim }{𝔥}$ is a subalgebra of $𝔤$ subordinate to $\varphi \text{.}$ This is a contradiction to the maximality of $𝔥\text{.}$ So $T\left(x\right)=0$ unless $x=1\text{.}$ So $T$ is determined by its value at $e^0=1\text{.}$ So $\text{dim}\hspace{0.17em}{\text{Hom}}_G(V^{\Omega },V^{\Omega })=1\text{.}$ $\square$

Remark 1. Let $\mathbb{Z}$ be the subgroup of $G$ given by $$\mathbb{Z}=\left\{\left(\begin{array}{ccc}1& 0& n\\ & 1& 0\\ & & 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}n\in \mathbb{Z}\}\text{.}$$ The group $G/\mathbb{Z}$ is often called the Heisenberg group. This group has a subgroup $$S_1=\left\{\left(\begin{array}{ccc}1& 0& t\\ & 1& 0\\ & & 1\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}t\in ℝ\}\cong \{z\in ℂ\hspace{0.17em}|\hspace{0.17em}\left|z\right|=1\}\cong ℝ/\mathbb{Z},$$ which is contained in $Z(G/\mathbb{Z})=[G/\mathbb{Z},G/\mathbb{Z}],$ and this can be used to show that $G/\mathbb{Z}$ does not have a faithful finite dimensional representation. The Lie group $G$ is the simply connected cover of $G/\mathbb{Z}\text{.}$ There is an exact sequence $$1⟶S^1⟶G/\mathbb{Z}⟶R^2⟶1\text{and if}⟨\xi ,\eta ⟩={\xi }_1{\eta }_2-{\xi }_2{\eta }_1,$$ then $$G/\mathbb{Z}=\left\{u\hspace{0.17em}\text{exp}\left(\xi \right)\hspace{0.17em}\right|\hspace{0.17em}u\in S^1,\xi \in {ℝ}^2\},\text{with}u\hspace{0.17em}\text{exp}\left(\xi \right)·v\hspace{0.17em}\text{exp}\left(\eta \right)=uv{e}^{i⟨\xi ,\eta ⟩}\text{exp}(\xi +\eta )\text{.}$$

Let $P,Q$ be such that $PQ-QP=-i\text{.}$ Then $$e^{iaP}{e}^{ibQ}=-e^{iab}{e}^{ibQ}{e}^{iaP}$$ and $G/\mathbb{Z}$ is the group generated by $e^{iaP}$ and $e^{ibQ}\text{.}$ Let $T_a,$ $M_b$ and $U_c$ be the operators on $L^2\left(ℝ\right)$ given by $$\left(T_{a}f\right)\left(x\right)=f(x-a),M_{b}f=e^{2\pi i\lambda x}f,U_{c}f=e^{2\pi i\lambda c}f,\text{for fixed}\hspace{0.17em}\lambda \in {ℝ}^{*}\text{.}$$ Then this is a unitary representation of $G/\mathbb{Z}$ for each $\lambda$ and $G/\mathbb{Z}\cong \left\{T_a{M}_b{U}_c\hspace{0.17em}\right|\hspace{0.17em}a,b,c\in ℝ\}\text{.}$

Remark 2. If $P,Q$ are such that $PQ-QP=-i$ and if $$a=\frac{1}{\sqrt{2}}(P+iQ)\text{and}{a}^{*}=\frac{1}{\sqrt{2}}(P-iQ),\text{then}[a,a^{*}]=1\text{.}$$ Then, on $L^2\left(ℝ\right),$ $$a^{*}\hspace{0.17em}\text{acts by}\frac{-i}{\sqrt{2}}(\frac{d}{dx}+x)\text{and}{a}^{*}\Omega =0\text{for}\hspace{0.17em}\Omega =e^{-\frac{1}{2}{x}^2}\text{.}$$ So $\Omega$ is a lowest weight vector and ${\left\{a^n\Omega \right\}}_{n\ge 0}$ is a basis of $L^2\left(ℝ\right)\text{.}$ $${𝔰𝔩}_2\left(ℝ\right)\cong \left(\text{Lie algebra generated by}\hspace{0.17em}\{(i/2)P^2,(i/2)(PQ+QP),(i/2)Q^2\}\right)\subseteq 𝔤\text{.}$$

Remark 3. If $$P=\left(\begin{array}{ccc}0& 1& 0\\ & 0& 0\\ & & 0\end{array}\right),Q=\left(\begin{array}{ccc}0& 0& 0\\ & 0& 1\\ & & 0\end{array}\right),Z=\left(\begin{array}{ccc}0& 0& 1\\ & 0& 0\\ & & 0\end{array}\right),$$ then $[P,Q]=Z$ and these act on functions $$V^{{\Omega }_{\gamma }}=\{f:ℝ\to ℂ\},$$ via the differential of the representation $V^{{\Omega }_{\gamma }}$ of $G\text{.}$ Then $$\left(Zf\right)\left(t\right)=2\pi i\gamma f\left(t\right)\text{.}$$ In particular, by considering the action of $P,Q$ on $V^{{\Omega }_{\gamma }}$ when $\gamma =-1/2\pi ,$ we have operators $P,Q$ on functions $f:ℝ\to ℂ$ which satisfy $$[P,Q]=PQ-QP=-i\text{.}$$ This solves a basic problem in quantum mechanics, see [Dirac] and [Heis].

Remark 4. Define an action of $${SL}_2\left(ℝ\right)=\left\{\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}ad-bc=1\}$$ on $𝔤$ by $$\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)P=aP+cQ,\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)Q=bP+dQ,\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)Z=Z\text{.}$$ Then $$[\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)P,\left(\begin{array}{cc}a& b\\ c& c\end{array}\right)Q]=[aP+cQ,bP+dQ]=ad[P,Q]+cb[Q,P]=(ad-bc)[P,Q]=[P,Q],$$ and so ${SL}_2\left(ℝ\right)$ acts on $𝔤$ by automorphisms. Let $m\in {SL}_2\left(ℝ\right)$ and let $\varphi \in {𝔤}^{*}\text{.}$ Let $𝔥$ be subordinate to $\varphi \text{.}$ Say $\varphi =(0,0,\gamma )\text{.}$ Then $m𝔥$ is subordinate to $\varphi$ (since $m𝔥$ is, after all, abelian). Then let us make the isomorphism $$\begin{array}{ccc}{\text{Ind}}_H^G\left(W^{\varphi }\right)& ⟶& {\text{Ind}}_{\stackrel{\sim }{H}}^G\left(W^{\varphi }\right)\\ f& ⟼& \stackrel{\sim }{f}\end{array}$$ where $\stackrel{\sim }{H}=\text{exp}\left(m𝔥\right)\text{.}\hspace{0.17em}\text{.}\hspace{0.17em}\text{.}\hspace{0.17em}\text{.}$

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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