## Induction and Restriction

Last update: 6 November 2012

## Induction and Restriction

Let $A$ be a subalgebra of an algebra $B\text{.}$

Let $V$ be a representation of $B\text{.}$ The restriction $V$ ${↓}_{A}^{B}$ of $V$ to $A$ to be the representation of $A$ given by the action of $A$ on $V\text{.}$ Let $W$ be a representation of $A\text{.}$ Define $B{\otimes }_{A}W$ to be all formal linear combinations of elements $b\otimes w,$ where $b\in B,$ $w\in W$ with the relations

$(b1+b2)⊗w =(b1⊗w)+ (b2+w), b⊗(w1+w2) =(b⊗w1)+ (b⊗w2), (αb)⊗w=b⊗ (αw)=α (b⊗w), ba⊗w=b⊗aw, (5.1)$

for all $a\in A,$ $b,{b}_{1},{b}_{2}\in B,$ $w,{w}_{1},{w}_{2}\in W$ and $\alpha \in ℂ\text{.}$ The induced representation $W$ ${↑}_{A}^{B}$ is the representation of $B$ on $B{\otimes }_{A}W$ given by the action

$b(b′⊗w)= (bb′)⊗w, (5.2)$

for all $b,{b}^{\prime }\in B$ and $w\in W\text{.}$

(5.3) Proposition. Let $A\subset B\subset C$ be such that $A$ is a subalgebra of $B$ and $B$ is a subalgebra of $C\text{.}$ Let $V,{V}_{1},{V}_{2}$ be representations of $C$ and let $W,{W}_{1},{W}_{2}$ be representations of $C\text{.}$

$1) (V1⊕V2) ↓AC≅V1 ↓AC ⊕V2 ↓AC. 2) (V↓BC) ↓AB≅V ↓AC. 3) (V1⊕V2) ↑AB=V1 ↑AB⊕V2 ↑AB. 4) (V↑AB) ↑BC≅V ↑AC.$

 Proof. 1) and 2) are trivial consequences of the definition. he fact that the map $ϕ: B⊗A (V1⊕V2) ⟶ (B⊗AV1)⊕ (B⊗AV2) b⊗(v1,v2) ⟼ ( b⊗v1,b ⊗v2 ) .$ is a $B\text{-module}$ isomorphism gives 3). The map $ϕ1: C⊗B (B⊗AV) ⟶ (C⊗BB) ⊗AV c⊗(b⊗v) ⟼ (c⊗b)⊗v$ and the map $ϕ2: C⊗BB ⟶ C c⊗b ⟼ cb$ are both $C\text{-module}$ isomorphisms. So $C⊗B (B⊗AV)≅ (C⊗BB) ⊗AV≅C⊗A V,$ giving 4). $\square$

Note: Proving that these maps are isomorphisms is not a complete triviality. One must show that they are well defined (by showing that they preserve the bilinearity relations (5.1)) and that the inverse maps are also well defined. It is helpful to use the fact that the tensor product is a universal object as given in Ex. 1.

(5.4) Theorem. (Frobenius reciprocity) Let $A\subset B$ be algebras and ${V}_{\lambda }$ and ${W}_{\mu }$ be irreducible representations of $A$ and $B$ respectively. Then

$HomB ( Vλ ↑AB,, Wμ ) ≅ HomA ( Vλ,Wμ ↓AB ) .$

 Proof. The map $Ψ: HomB ( B⊗AVλ, Wμ ) ⟶ HomA ( Vλ,Wμ ↓AB ) ϕ ⟼ ϕ′ ,$ where $ϕ′(v)= ϕ(1⊗v),$ is an isomorphism. The inverse map is given by ${\Psi }^{-1}\left({\varphi }^{\prime }\right)=\varphi$ where $\varphi$ is given by $ϕ(b⊗v)=bϕ (1⊗v)=bϕ′ (v),$ so that $\varphi$ is a $B\text{-module}$ homomorphism. $\square$

### Branching rules

Now suppose that $A$ is a subalgebra of $B$ and that both $A$ and $B$ are semisimple. Let $\stackrel{^}{A}$ and $\stackrel{^}{B}$ be index sets for the irreducible representations of $A$ and $B$ respectively. Let ${V}_{\lambda }$ and ${W}_{\mu }$ be the irreducible representations of $A$ and $B$ labelled by $\lambda \in \stackrel{^}{A}$ and $\mu \in \stackrel{^}{B}$ respectively. Let ${g}_{\lambda \mu }\in ℤ$ be such that

$Vλ ↑AB≅ ⊕μ∈B^ GλμWμ (5.5)$

for each pair $\left(\lambda ,\mu \right),$ $\lambda \in \stackrel{^}{A},$ $\mu \in \stackrel{^}{B}\text{.}$ Frobenius reciprocity implies that

$Wμ↓AB≅ ⊕λ∈A^ gλμVλ (5.5')$

for each $\mu \in \stackrel{^}{B}\text{.}$ An equation of the form (5.5) or (5.5') is called a branching rule between $A$ and $B\text{.}$

One can produce a visual representation of branching rules in the form of a graph. Construct a graph with two rows of vertices, the vertices in the first row labelled by the elements of $\stackrel{^}{A}$ and the vertices of the second row labelled by the elements of $\stackrel{^}{B}$ such that the vertex labelled by $\lambda \in \stackrel{^}{A}$ and the vertex labelled by $\mu \in \stackrel{^}{B}$ are connected by ${g}_{\lambda \mu }$ edges. This graph is the Bratteli diagram of $A\subset B\text{.}$

As an example, the following diagram is the Bratteli diagram of $ℂ{S}_{2}\subset ℂ{S}_{3},$ where ${S}_{n}$ denotes the symmetric group. Recall that the irreducible representations of ${S}_{2}$ and ${S}_{3}$ are indexed by partitions of 2 and of 3 respectively.

$S2 S3 (111) (21) (3) (2) (11)$

Note that in this example each ${g}_{\lambda \mu }$ is either 0 or 1; there are no multiple edges.

Let $p\in A$ and consider the representation of $A$ given by left multiplication on the space $Aa\text{.}$ Then

$(Ap) ↑AB≅ Bp. (5.6)$

To see this, informally, one notes that since $Ap\subset A$ we can move $Ap$ across the tensor product to give,

$(Ap)↑AB= B⊗AAp=BAp ⊗A1=Bp⊗A 1≅Bp.$

$BAp=Bp$ since $1\in A\text{.}$ More formally we should show that the map

$B⊗AAp ⟶ Bp b⊗ap ⟼ bap$

is well defined and has well defined inverse given by

$bop⟵bp.$

Now let ${p}_{\lambda }$ be a minimal idempotent of $A$ such that the action of $A$ by left multiplication on $A{p}_{\lambda }$ is a representation of $A$ isomorphic to the irreducible representation ${V}_{\lambda }$ of $A$ (3.6). Suppose that

$pλ=∑qi$

is a decomposition (§1 Ex. 7) of the minimal idempotent ${p}_{\lambda }$ of $A$ into minimal orthogonal idempotents of $B\text{.}$ Then $B{p}_{\lambda }=B\sum {q}_{i}=\sum B{q}_{i}$ gives a decomposition of $B{p}_{\lambda }$ into irreducible representations. So, by (5.6) and the branching rule (5.5), for exactly ${g}_{\lambda \mu }$ of the ${q}_{i}$ we will have that $B{q}_{i}$ is isomorphic to the irreducible representation ${W}_{\mu }$ of $B\text{.}$ We can write the decomposition of ${p}_{\lambda }$ as

$pλ= ∑μ∈B^ ∑i=1gλμ qμi (5.7)$

where each ${q}_{\mu i}$ is such that $B{q}_{\mu i}$ is isomorphic to the irreducible representation ${W}_{\mu }$ of $B\text{.}$

### Characters of induced representations

Let $V$ be a representation of $A$ where $A$ is a subalgebra of an algebra $B$ and both $A$ and $B$ are semisimple. Let ${\chi }_{V}$ be the character of $V$ and let ${\chi }_{V{↑}_{B}^{A}}$ be the character of $V{↑}_{A}^{B}\text{.}$ For each $A\in 𝒜$ let ${a}^{*}$ denote the element of the dual basis to $𝒜$ with respect to the trace, tr, of the regular representation of $A$ such that $\text{tr}\left(a{a}^{*}\right)=1\text{.}$

Let $ℬ$ be a basis of $B$ and let ${\stackrel{\to }{t}}_{B}=\left({t}_{\mu }^{B}\right)$ be a nondegenerate trace on $B\text{.}$ For each $b\in ℬ$ let ${b}^{*}$ denote the element of the dual basis to $ℬ$ with respect to the trace ${\stackrel{\to }{t}}_{B}$ such that $\stackrel{\to }{t}\left(b{b}^{*}\right)=1\text{.}$ For any element $x\in B$ we set (as in §3 EX.7)

$[x]=∑b∈ℬ bxb*.$

(5.8) Theorem.

$χV↑AB (b)=munde ∑a χV(a) ⟨ [b],a* ⟩ ,$

where $⟨{b}_{1},{b}_{2}⟩={\stackrel{\to }{t}}_{B}\left({b}_{1}{b}_{2}\right)\text{.}$

 Proof. In keeping with the notations of earlier sections, let $\stackrel{^}{A}$ and $\stackrel{^}{B}$ be index sets for the irreducible representations of $A$ and $B$ respectively and let ${\chi }_{A}^{\lambda },\lambda \in \stackrel{^}{A}$ and ${\chi }_{B}^{\mu },\mu \in \stackrel{^}{B}$ denote the irreducible characters of $A$ and $B$ respectively. Let ${z}_{\lambda }^{A},\lambda \in \stackrel{^}{A}$ and ${z}_{\mu }^{B},\mu \in \stackrel{^}{B}$ denote the minimal central idempotents of $A$ and $B$ respectively. Let ${d}_{\lambda }^{A}={\chi }_{A}^{\lambda }\left(1\right)$ so that ${d}_{\lambda }$ is the dimension of the irreducible representation of $A$ corresponding to $\lambda \in \stackrel{^}{A}\text{.}$ We have the following facts: (Theorem (3.10)) For each $\lambda \in \stackrel{^}{A},$ $\mu \in \stackrel{^}{B},$ $zλA = ∑a∈𝒜 tλA χAλ (a) a* ,and zμB = ∑b∈ℬ tμB χBμ (b) b* ,$ respectively. (§3 Ex. 5) The trace vector $\right)\left({t}_{\lambda }^{A}\right)$ of the trace of the regular representation of $A$ is given by ${t}_{\lambda }^{A}={d}_{\lambda }^{A}$ for all $\lambda \in \stackrel{^}{A}\text{.}$ Suppose that $V\cong {\oplus }_{\lambda \in \stackrel{^}{A}}{V}_{\lambda }^{\oplus {m}_{\lambda }}$ gives the decomposition of $V$ into irreducible representations of $A\text{.}$ Then $χV(a)= ∑λ∈A^ mλχAλ (a),$ for all $a\in A\text{.}$ The branching rule (5.5) for $A\subset B$ gives that $χV↑AB (b)= ∑λ∈A^mλ ∑μ∈B^ gλμχBμ (b),$ for all $b\in B\text{.}$ For each $\lambda \in \stackrel{^}{A}$ let $zλA= ∑i=1dλA pλiA$ be a decomposition of ${z}_{\lambda }^{A}$ into minimal orthogonal idempotents of $A\text{.}$ For each $\lambda \in \stackrel{^}{A}$ and $1\le i\le {d}_{\lambda }^{A}$ let $pλiA= ∑μ∈B^ ∑j=1gλμ qμjB$ be a decomposition (5.7) of ${p}_{\lambda i}^{A}$ into minimal orthogonal idempotents of $B\text{.}$ ${q}_{\mu j}$ denotes a minimal idempotent in the minimal ideal of $B$ corresponding to $\mu \in \stackrel{^}{b},$ i.e., a minimal idempotent such that the representation $B{q}_{\mu }j$ of $B$ is isomorphic to the irreducible representation of $B$ corresponding to $\mu \in \stackrel{^}{B}\text{.}$ Then, by (3.12), $[qμjB]= (1/tμB) zμB,$ for each minimal idempotent ${q}_{\mu j}^{B},$ since for each $\nu \in \stackrel{^}{B},$ ${\chi }_{\nu }\left({q}_{\mu j}^{B}\right)={\delta }_{\mu \nu }\text{.}$ Let ${b}_{1},{b}_{2}\in B\text{.}$ Using the trace property, $⟨ [b1], b2 ⟩ = t→B ( ∑b∈ℬ bb1b*b2 ) = t→B ( ∑b∈ℬ b1b*b2 b ) = ⟨ b1,[b2] ⟩ .$ Now, define $z=∑λ∈A^ (mλ/dλA) zλA.$ Then, using 1), 2) and 3), $z = ∑λ∈A^ mλ∑a (tλA/dλA) χAλ(a)a* = ∑aχV (a)a*,$ and, by 5), 1) and 4), $[z] = ∑λ (mλ/dλA) [zλA] = ∑λ (mλ/dλA) ∑i=1dλA [pλiA] = ∑λ (mλ/dλA) ∑i=1dλA ∑μ ∑j=1gλμ [qμjB] = ∑λ (mλ/dλA) ∑i=1dλA ∑μ ∑j=1gλμ (1/tμB) zμB = ∑λ∑μ (mλ/dλA) dλAgλμ (1/tuB) ∑btμB χBμ(b)b* = ∑bχV↑AB (b)b*.$ Combining these and using 6) we get $χV↑AB (b) = ⟨ [z],b ⟩ = ⟨ [ ∑aχV (a)a* ] ,b ⟩ = ∑aχV(a) ⟨ [a*],b ⟩ = ∑aχV(a) ⟨ a*,[b] ⟩ ,$ as desired. $\square$

### Centralizers

Let $A$ be a subalgebra of an algebra $B,$ and let $B$ be a representation of $B\text{.}$ Let $\stackrel{‾}{A}$ and $\stackrel{‾}{B}$ be the centralizers of $V\left(A\right)$ and $V\left(B\right)$ respectively. Then $\stackrel{‾}{B}$ is a subalgebra of $\stackrel{‾}{A};$ $A\subset B$ and $\stackrel{‾}{A}\supset \stackrel{‾}{B}\text{.}$

(5.9) Theorem. Suppose that

$Wμ ↓AB ≅ ∑λgμλ Vλ, V‾λ ↓B‾A‾ ≅ ∑μgλμ′ W‾μ$

are the branching rules for $A\subset B$ and $\stackrel{‾}{B}\subset \stackrel{‾}{A}$ respectively. Then for all $\lambda ,\mu$

$gμλ= gλμ′.$

 Proof. We know, Theorem (4.11), that, as $A\otimes \stackrel{‾}{A}$ representations, $V≅⊕λVλ⊗ V‾λ,$ and as $B\otimes \stackrel{‾}{B}$ representations, $V≅⊕μWμ ⊗W‾μ,$ where ${V}_{\lambda },$ ${\stackrel{‾}{V}}_{\lambda },$ ${W}_{\mu },$ and ${\stackrel{‾}{W}}_{\mu }$ are irreducible representations of $A,\stackrel{‾}{A},B,$ and $\stackrel{‾}{B}$ respectively. $A\otimes \stackrel{‾}{B}$ is a subalgebra of both $A\otimes \stackrel{‾}{A}$ and $B\otimes \stackrel{‾}{B}\text{.}$ We have that as $A\otimes \stackrel{‾}{B}$ representations $V≅V ↓ A⊗B‾ A⊗A‾ ≅ ⊕λVλ⊗ ( ⊕μgλμ′ W‾μ ) ≅ ⊕λ,μ gλμ′ Vλ⊗W‾μ.$ On the other hand as $A\otimes \stackrel{‾}{B}$ representations $V≅V ↓ A⊗B‾ B⊗B‾ ≅ ⊕μ ( ⊕λ gμλ Vλ ) ⊗W‾μ ≅ ⊕λ,μ gμλVλ ⊗W‾μ.$ $\square$

### Examples

1. Let $A,B$ and $C$ be vector spaces. A map $f:A×B\to C$ is bilinear if $f a 1 + a 2 b =f a 1 b +f a 2 b ,$$f a b 1 + b 2 =f a b 1 +f a b 2 ,$$f αa b =f a αb =αf ab ,$ for all $a,{a}_{1},{a}_{2}\in A,b,{b}_{1},{b}_{2}\in B,\alpha \in ℂ.$
2. The tensor product is given by a vector space $A\otimes B$ and a map $i:A×B\to A\otimes B$ such that for every bilinear map $f:A×B\to C$ there exists a linear map $\stackrel{-}{f}:A\otimes B\to C$ such that the following diagram commutes:

One constructs the tensor product $A\otimes B$ as the vector space of elements $a\otimes b,a\in A,b\in B,$ with relations $a 1 + a 2 ⊗b= a 1 ⊗b+ a 2 ⊗b,$$a⊗ b 1 + b 2 =a⊗ b 1 +a⊗ b 2 ,$$αa b=a⊗ αb =α a⊗b ,$ for all $a,{a}_{1},{a}_{2}\in A,b,{b}_{1},{b}_{2}\in B$ and $\alpha \in ℂ.$ The map $i:A×B\to A\otimes B$ is given by $i\left(a,b\right)=a\otimes b.$ Using the above universal mapping property one gets easily that the tensor product is unique in the sense that any two tensor products of $A$ and $B$ are isomorphic.

If $R$ is an algebra and $A$ is a right $R$-module (a vector space that affords an antirepresentation of $R$) and $B$ is a left $R$-module them one forms the vector space $A{\otimes }_{R}B$ as above except that we require a bilinear map $f:A×B\to C$ to satisfy the additional condition $f ar b =f a rb$ for all $r\in R.$ Then the tensor product $A{\otimes }_{R}B$ once again is constructed by using the vector space of elements $a\otimes b,a\in A,b\in B,$ with the relations above and the additional relation $ar⊗b=a⊗rb,$ for all $r\in R.$

3. Let $A\subseteq B$ be semisimple algebras such that $A$ is a subalgebra of $B$ Let $\stackrel{~}{A}$ and $\stackrel{~}{B}$ be index sets of the irreducible representations of $A$ and $B$ respectively, and suppose that $\left\{{f}_{ij}^{\mu }\right\},\mu \in \stackrel{~}{A},$ is a complete set of matrix units of $A$

[Bra1972] There exists a complete set of matrix units $\left\{{e}_{rs}^{\lambda }\right\},\lambda \in \stackrel{~}{B},$ of $B$ that is a refinement of the ${f}_{ij}^{\mu }$ in the sense that for each $\mu \in \stackrel{~}{A}$ and each $i$, $f ii μ =∑ e rr λ ,$ for some set of ${e}_{rr}^{\lambda }$.

 Proof. Suppose that $B\cong {\oplus }_{\lambda \in \stackrel{~}{B}}{M}_{{d}_{\lambda }}\left(ℂ\right)$. Let ${z}_{\lambda }^{B}$ be the minimal central idempotent of $B$ such that ${I}_{\lambda }={B}_{{z}_{\lambda }}$ is the minimal ideal corresponding to the $\lambda$ block of matrices in ${\oplus }_{\lambda }{M}_{{d}_{\lambda }}\left(ℂ\right).$ For each $\mu \in \stackrel{~}{A}$ and each $i$ decompose ${f}_{ii}^{\mu }$ into minimal orthogonal idempotents of $B$ (Section 1, Ex 7), ${f}_{ii}^{\mu }=\sum {p}_{j}.$ Label each ${p}_{j}$ appearing in this sum by the element $\lambda \in \stackrel{~}{B}$ which indexes the minimal ideal ${I}_{\lambda }=B{p}_{j}B$ of $B$. Then $1= ∑ μ,i f ii μ = ∑ λ∈ B ~ ∑ j=1 d λ p j λ .$ Now $B=1.B.1= ∑ λ,μ∈ B ~ ∑ 1≤i≤ d λ ,1≤j≤ d μ p i λ B p j μ .$ If $\lambda \ne \mu$ then the space ${p}_{i}^{\lambda }B{p}_{j}^{\mu }={p}_{i}^{\lambda }B\left({z}_{\mu }^{B}{p}_{j}^{\mu }\right)={p}_{i}^{\lambda }{z}_{\mu }^{B}B{p}_{j}^{\mu }=0$ for all $i,j.$ Since ${p}_{i}^{\lambda }={p}_{i}^{\lambda }.1.{p}_{i}^{\lambda }\in {p}_{i}^{\lambda }I{p}_{i}^{\lambda }$ and ${p}_{i}^{\lambda }B{p}_{j}^{\lambda }{p}_{j}^{\lambda }B{p}_{i}^{\lambda }={p}_{i}^{\lambda }{I}_{\lambda }{p}_{i}^{\lambda }\ne 0$, we know that ${p}_{i}^{\lambda }B{p}_{j}^{\lambda }$ is not zero for any $1\le i,j\le {d}_{\lambda }.$ Furthermore, since the dimension of $B$ is ${\sum }_{\lambda }{d}_{\lambda }^{2}$ each of the spaces ${p}_{i}^{\lambda }B{p}_{j}^{\lambda }$ is one dimensional. For each ${p}_{i}^{\lambda }$ define ${e}_{ii}^{\lambda }={p}_{i}^{\lambda }.$ For each $\lambda$ and each $1\le i let ${e}_{ii}^{\lambda }$ be some element of ${p}_{i}^{\lambda }B{p}_{j}^{\lambda }.$ Then choose ${e}_{ii}^{\lambda }\in {p}_{j}^{\lambda }B{p}_{i}^{\lambda }$ such that ${e}_{ij}^{\lambda }{e}_{ji}^{\lambda }={e}_{ii}^{\lambda }.$ This defines a complete set of matrix units of $B.\square$

4. Let $G$ be a finite group and let $H$ be a subgroup of $G.$ Let $R=\left\{{g}_{i}\right\}$ be a set of representatives for the left cosets $gH$ of $H$ in $G.$ The action of $G$ on the cosets of $H$ in $G$ by left multiplication defines a representation ${\pi }_{H}$ of $H$ in $G.$ This representation is a permutation representation of $G.$ Let $g\in G.$ The entries ${\pi }_{H}{\left(g\right)}_{i\text{'}i}$ of the matrix ${\pi }_{H}\left(g\right)$ are given by ${\pi }_{H}{\left(g\right)}_{i\text{'}i}={\delta }_{i\text{'}k}$ where $k$ is such that $g{g}_{i}\in {g}_{k}H.$

Let $V$ be a representation of $H.$ Let $B=\left\{{v}_{j}\right\}$ be a basis of $V.$ Then the elements $g\otimes {v}_{j}$ where $g\in G,{v}_{j}\in B$ span $ℂG{\otimes }_{ℂH}V.$ The fourth relation in 5.1 gives that the set $\left\{{g}_{i}\otimes {v}_{j}\right\},{g}_{i}\in R,{v}_{j}\in B$ forms a basis of $ℂG{\otimes }_{ℂH}V.$

Let $g\in G$ and suppose that $g{g}_{i}={g}_{k}h,$ where $h\in H$ and ${g}_{k}\in R.$ Then $g g i ⊗ v j = g k h⊗ v j = g k ⊗h v j = ∑ j g k ⊗ v j' V h j'j = ∑ i',j' g i' ⊗ v j' V h j'j δ i'k = ∑ i',j' g i' ⊗ v j' V h j'j π H g i'i .$ Then $χ V ↑ H G g = ∑ g i ∈R, v j ∈B g g i ⊗ v j | g i ⊗ v j = ∑ g i , v j ,g g i ∈ g i H V g i -1 g g i jj .$

Since characters are constant on conjugacy classes we have that $χ V ↑ H G g = 1 H ∑ h∈H ∑ g i ; h -1 g i -1 g g i h∈H χ V h -1 g i -1 g g i h = 1 H ∑ a∈H,a∈ C g χ V a ,$ where ${C}_{g}$ denotes the conjugacy class of $g.$ This is an alternate proof of Theorem 5.8 for the special case of inducing from a subgroup $H$ of a group $G$ to the group $G.$

5. Define $ℂG{\otimes }_{d}ℂG$ to be the subalgebra of the algebra $ℂG\otimes ℂG$ consisting of the span of the elements $g\otimes g$, $g\in G.$ Then $ℂG\cong ℂG{\otimes }_{d}ℂG$ as algebras.

Let ${V}_{1}$ and ${V}_{2}$ be representations of $G.$ Then the restriction of the $ℂG\otimes ℂG$ representation $V={V}_{1}\otimes {V}_{2}$ to the algebra $ℂG{\otimes }_{d}ℂG$ is the Kronecker product (Section 4, Ex 1) $V 1 ⊗ d V 2 = V 1 ⊗ V 2 ↓ ℂG⊗ℂG ℂG ⊗ d ℂG$ of ${V}_{1}$ and ${V}_{2}.$ Since $ℂG\cong ℂG{\otimes }_{d}ℂG$ we can view ${V}_{1}{\otimes }_{d}{V}_{2}$ as a representation of $G.$

Let ${V}_{\lambda }$ and ${V}_{\mu }$ be irreducible representations of $G$ such that ${V}_{\lambda }\otimes {v}_{\mu }$ appears as an irreducible component of the $ℂG\otimes ℂG$ representation ${V}_{1}\otimes {V}_{2}$. The decomposition of the Kronecker product $V λ ⊗ d V μ = V 1 ⊗ V 2 ↓ ℂG⊗ℂG ℂG ⊗ d ℂG ≅ ⊕ ν g λμ ν V ν$ into irreducible representations ${V}_{\nu }$ of $G$ is given by the branching rule for $ℂG\otimes ℂG\supset ℂG{\otimes }_{d}ℂG.$ Let ${C}_{1}$ and ${C}_{2}$ be the centralisers of the representations ${V}_{1}$ and ${V}_{2}$ respectively. Let $C$ be the centraliser of the $ℂG\otimes ℂG$ representation $V={v}_{1}\otimes {V}_{2}.$ Applying Theorem 5.9 to $V$ where $A=ℂG\otimes ℂG$ and $ℂG{\otimes }_{d}ℂG=B\cong G$ shows that the ${g}_{\lambda \mu }^{\nu }$ are also given by the branching rule for ${C}_{1}\otimes {C}_{2}\subset C.$

### Notes and References

The main result, Theorem (5.8), of this section is a generalization of the formula for the induced character for finite groups, see [Ser1977] §7.2. I have been unable to find any similar result in previous literature.

## Notes and References

This is an excerpt from the unpublished first chapter of Arun Ram's dissertation entitled Representation Theory, written July 4, 1990.