## Lectures in Representation Theory

Last update: 27 August 2013

## Lecture 17

We saw last time that the Bratteli diagram describes the branching rules for restriction from irreducible modules over $B$ to irreducible modules over $A\text{.}$ Note that by Schur’s Lemma and Frobenius Reciprocity, $gλμ=dimℂ ( homA (Vλ,Wμ↓AB) ) =dimℂ ( homB (Vλ↑AB,Wμ) ) ,$ whence that ${V}^{\lambda }{↑}_{A}^{B}\simeq \sum _{\mu \in \stackrel{ˆ}{B}}{g}_{\lambda \mu }{W}^{\mu },$ so that the Young lattice also describes branching rules for induction of irreducible $A\text{-modules.}$

For $A$ semisimple, we could produce the irreducible modules ${V}^{\lambda }$ by noting that ${V}^{\lambda }=A{e}_{ii}^{\lambda },$ where the ${e}_{ij}^{\lambda }$ were the preimages of the ${E}_{ij}$ in the $\lambda \text{-th}$ block within $A\simeq \underset{\lambda \in \stackrel{ˆ}{A}}{⨁}{M}_{{d}_{\lambda }}\left(ℂ\right)\text{.}$

Conversely, given an idempotent $p$ in $A,$ we can write $p=\sum {p}_{i},$ where the ${p}_{i}$ are minimal idempotents, and $Ap\simeq ⨁A{p}_{i},$ with each $A{p}_{i}$ irreducible. For each minimal idempotent $p\in A,$ there exists a unique $\lambda \in A$ with $Ap\simeq {V}^{\lambda }\text{;}$ let us indicate this association by denoting our idempotents as ${p}_{\lambda }\text{.}$ Now if we set $Ap=V=\underset{\lambda \in \stackrel{ˆ}{A}}{⨁}{\left({V}^{\lambda }\right)}^{\oplus {m}_{\lambda }},$ one has that $p=∑λ∈Aˆ ∑i=1mλ pλi$ and further $Ap = A∑λ∈Aˆ ∑i=1mλ pλi ≃ ⨁λ∈Aˆ ∑i=1mλ Apλi$

Note that $Ap{↑}_{A}^{B}\simeq Bp,$ as $B\text{-modules.}$ (To prove this, one needs to produce a bijection $B{\otimes }_{A}Ap⇆Bp\text{.}$ The appropriate maps are the $C\text{-linear}$ extensions of $b\otimes ap↦bap$ and $bp↦bp\otimes p\text{.)}$

Now we know that ${V}^{\lambda }{↑}_{A}^{B}\simeq A{p}_{\lambda }^{A}{↑}_{A}^{B}\simeq B{p}_{\lambda }^{A},$ for ${p}_{\lambda }^{A}$ a minimal idempotent in $A$ (and naturally an idempotent of some sort in $B\text{).}$ The branching rule (the induction version) now tells us that ${V}^{\lambda }{↑}_{A}^{B}\simeq \sum _{\mu \in \stackrel{ˆ}{B}}{g}_{\lambda \mu }{W}^{\mu }\text{.}$ Thus if we express the idempotent ${p}_{\lambda }^{A}$ as a sum of minimal idempotents of $B,$ we will obtain ${p}_{\lambda }^{A}={\sum }_{\mu \in \stackrel{ˆ}{B}}{\sum }_{i=1}^{{g}_{\lambda \mu }}{q}_{\mu i}^{B},$ where the ${q}_{\mu i}^{B}$ are minimal idempotents in $B$ with $B{q}_{\mu i}^{B}\simeq {W}^{\mu }\text{.}$

Let $V$ be a representation of $A,$ and ${\chi }_{V}$ the associated character, with $V\simeq \sum _{\lambda \in \stackrel{ˆ}{A}}{m}_{\lambda }{V}^{\lambda }$ and ${\chi }_{V}=\sum _{\lambda \in \stackrel{ˆ}{A}}{m}_{\lambda }{\chi }_{A}^{\lambda }$ We would like to produce a formula for ${\chi }_{V{↑}_{A}^{B}}\text{.}$ As $B\text{-modules,}$ we know that $V↑AB ≃ ∑λ∈Aˆ mλVλ ↑AB ≃ ∑λ∈Aˆ mλ∑μ∈Bˆ gλμWμ,$ and by taking traces one then obtains the formula $χV↑AB= ∑λ∈Aˆ ∑μ∈Bˆ mλgλμ χBμ.$ This formula, while interesting, is not necessarily so nice for computational uses. With this in mind, let us define $z=∑λ∈Aˆ mλdλA zλA∈A,$ where (as usual) the ${z}_{\lambda }^{A}$ are the minimal central idempotents of $A,$ the ${d}_{\lambda }^{A}$ are the dimensions of the irreducible $A\text{-modules,}$ and the ${m}_{\lambda }$ are the multiplicities of these irreducibles. Let $\left\{a\right\}=𝒜$ be a basis for $A,$ and $\left\{{a}^{*}\right\}={𝒜}^{*}$ be its dual basis with respect to the trace of the regular representation of $A\text{.}$ Now recall that $zλA=∑a∈𝒜 dλAχAλ (a)a*,$ whence that $z = ∑λ∈Aˆ mλdλA zλA = ∑λ∈Aˆ mλdλA ∑a∈𝒜 dλAχAλ (a)a* = ∑λ∈Aˆ ( ∑a∈𝒜 mλχAλ (a) ) a* = ∑λ∈Aˆ χV(a)a*$ We find then that the values of ${\chi }_{V}$ are encoded in this distinguished element $z\text{.}$ Can we produce an element of $B$ within which the values of ${\chi }_{V{↑}_{A}^{B}}$ are similarly encoded?

Let $\left\{b\right\}=ℬ$ be a basis for $B,$ let $\stackrel{\to }{t}=\left({t}_{\lambda }\right)$ be any nondegenerate trace for $B,$ and let $\left\{{b}^{*}\right\}={ℬ}^{*}$ be the associated dual basis. Define for each $x\in B$ $[x]=∑b∈ℬ bxb*.$

Proposition 2.60 Given the above definitions, $\left[z\right]=\sum _{b\in ℬ}{\chi }_{V{↑}_{A}^{B}}\left(b\right){b}^{*}\text{.}$

 Proof. $[z] = ∑λ∈Aˆ mλdλA [zλA] = ∑λ∈Aˆ mλdλA ∑i=1dλ [pλiA] = ∑λ∈Aˆ mλdλA ∑i=1dλ ∑μ∈Bˆ ∑j=1gλμ [qμjB]$ (The second equality is expressing ${z}_{\lambda }^{A}$ in terms of minimal idempotents of $A,$ the third is expressing each of those minimal $A\text{-idempotents}$ in terms of the minimal idempotents of $B\text{.)}$ Note that if we take ${e}_{k,k}^{\mu }\in B$ and symmetrize it, we have $[ek,kμ] = ∑1≤i,j≤dνBν∈Bˆ eijν ekkμ (eijν)* = ∑1≤i,j≤dνBν∈Bˆ eijνekkμ ejiνtν = ∑1≤i≤dνBj=k ν=μ eijν ekkμ ejiν tν = ∑i=1dν eiiνtν =1tνzνB$ Thus we have that $[z] = ∑λ∈Aˆ mλdλA ∑i=1dλ ∑μ∈Bˆ ∑j=1gλμ [qμjB] = ∑λ∈Aˆ mλdλA ∑i=1dλ ∑μ∈Bˆ ∑j=1gλμ 1tμzμB = ∑λ∈Aˆ ∑μ∈Bˆ mλdλA dλAgλμ 1tμzμB = ∑λ∈Aˆ ∑μ∈Bˆ mλdλA gλμzμB = ∑λ∈Aˆ ∑μ∈Bˆ mλdλA gλμ ∑b∈ℬtμ χBμ(b)b* ⏟zμB = ∑b∈ℬ ∑λ∈Aˆ ∑μ∈Bˆ mλgλμ χBμ(b) ⏟χV↑AB(b) b* [z] = ∑b∈ℬ χV↑AB (b)b*$ $\square$

## Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.