Lectures in Representation Theory

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

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, $$g_{\lambda \mu }={\text{dim}}_{ℂ}\hspace{0.17em}\left({\text{hom}}_A(V^{\lambda },W^{\mu }{\downarrow }_A^B)\right)={\text{dim}}_{ℂ}\hspace{0.17em}\left({\text{hom}}_B(V^{\lambda }{\uparrow }_A^B,W^{\mu })\right),$$ whence that $V^{\lambda }{\uparrow }_A^B\simeq {\displaystyle \sum _{\mu \in \hat{B}}}{g}_{\lambda \mu }{W}^{\mu },$ 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 $A\simeq {\displaystyle \underset{\lambda \in \hat{A}}{⨁}}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$

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 $Ap=V={\displaystyle \underset{\lambda \in \hat{A}}{⨁}}{\left(V^{\lambda }\right)}^{\oplus m_{\lambda }},$ one has that $$p=\sum _{\lambda \in \hat{A}}\sum _{i=1}^{m_{\lambda }}{p}_{\lambda i}$$ and further $$\begin{array}{ccc}Ap& =& A\sum _{\lambda \in \hat{A}}\sum _{i=1}^{m_{\lambda }}{p}_{\lambda i}\\ & \simeq & \underset{\lambda \in \hat{A}}{⨁}\sum _{i=1}^{m_{\lambda }}A{p}_{\lambda i}\end{array}$$

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

Now we know that $V^{\lambda }{\uparrow }_A^B\simeq A{p}_{\lambda }^A{\uparrow }_A^B\simeq B{p}_{\lambda }^A,$ 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 $V^{\lambda }{\uparrow }_A^B\simeq {\displaystyle \sum _{\mu \in \hat{B}}}{g}_{\lambda \mu }{W}^{\mu }\text{.}$ Thus if we express the idempotent $p_{\lambda }^A$ as a sum of minimal idempotents of $B,$ we will obtain $p_{\lambda }^A={\sum }_{\mu \in \hat{B}}{\sum }_{i=1}^{g_{\lambda \mu }}{q}_{\mu i}^B,$ 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 $V\simeq {\displaystyle \sum _{\lambda \in \hat{A}}}{m}_{\lambda }{V}^{\lambda }$ and ${\chi }_V={\displaystyle \sum _{\lambda \in \hat{A}}}{m}_{\lambda }{\chi }_A^{\lambda }$ We would like to produce a formula for ${\chi }_{V{\uparrow }_A^B}\text{.}$ As $B\text{-modules,}$ we know that $$\begin{array}{ccc}V{\uparrow }_A^B& \simeq & \sum _{\lambda \in \hat{A}}{m}_{\lambda }{V}^{\lambda }{\uparrow }_A^B\\ & \simeq & \sum _{\lambda \in \hat{A}}{m}_{\lambda }\sum _{\mu \in \hat{B}}{g}_{\lambda \mu }{W}^{\mu },\end{array}$$ and by taking traces one then obtains the formula $${\chi }_{V{\uparrow }_A^B}=\sum _{\lambda \in \hat{A}}\sum _{\mu \in \hat{B}}{m}_{\lambda }{g}_{\lambda \mu }{\chi }_B^{\mu }\text{.}$$ This formula, while interesting, is not necessarily so nice for computational uses. With this in mind, let us define $$z=\sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}{z}_{\lambda }^A\in A,$$ 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 $$z_{\lambda }^A=\sum _{a\in 𝒜}{d}_{\lambda }^A{\chi }_A^{\lambda }\left(a\right)a^{*},$$ whence that $$\begin{array}{ccc}z& =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}{z}_{\lambda }^A\\ & =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\sum _{a\in 𝒜}{d}_{\lambda }^A{\chi }_A^{\lambda }\left(a\right)a^{*}\\ & =& \sum _{\lambda \in \hat{A}}\left(\sum _{a\in 𝒜}{m}_{\lambda }{\chi }_A^{\lambda }\left(a\right)\right)a^{*}\\ & =& \sum _{\lambda \in \hat{A}}{\chi }_V\left(a\right)a^{*}\end{array}$$ 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 ${\chi }_{V{\uparrow }_A^B}$ are similarly encoded?

Let $\left\{b\right\}=ℬ$ be a basis for $B,$ let $\overrightarrow{t}=\left(t_{\lambda }\right)$ be any nondegenerate trace for $B,$ and let $\left\{b^{*}\right\}={ℬ}^{*}$ be the associated dual basis. Define for each $x\in B$ $$\left[x\right]=\sum _{b\in ℬ}bx{b}^{*}\text{.}$$

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

		Proof.
	$$\begin{array}{ccc}\left[z\right]& =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\left[z_{\lambda }^A\right]\\ & =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\sum _{i=1}^{d_{\lambda }}\left[p_{\lambda i}^A\right]\\ & =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\sum _{i=1}^{d_{\lambda }}\sum _{\mu \in \hat{B}}\sum _{j=1}^{g_{\lambda \mu }}\left[q_{\mu j}^B\right]\end{array}$$ (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 $$\begin{array}{ccc}\left[e_{k,k}^{\mu }\right]& =& \sum _{\underset{\nu \in \hat{B}}{1\le i,j\le d_{\nu }^B}}{e}_{ij}^{\nu }{e}_{kk}^{\mu }{\left(e_{ij}^{\nu }\right)}^{*}\\ & =& \sum _{\underset{\nu \in \hat{B}}{1\le i,j\le d_{\nu }^B}}{e}_{ij}^{\nu }{e}_{kk}^{\mu }\frac{e_{ji}^{\nu }}{t_{\nu }}\\ & =& \sum _{\underset{j=k\hspace{0.17em}\nu =\mu }{1\le i\le d_{\nu }^B}}\frac{e_{ij}^{\nu }{e}_{kk}^{\mu }{e}_{ji}^{\nu }}{t_{\nu }}\\ & =& \sum _{i=1}^{d_{\nu }}\frac{e_{ii}^{\nu }}{t_{\nu }}=\frac{1}{t_{\nu }}{z}_{\nu }^B\end{array}$$ Thus we have that $$\begin{array}{ccc}\left[z\right]& =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\sum _{i=1}^{d_{\lambda }}\sum _{\mu \in \hat{B}}\sum _{j=1}^{g_{\lambda \mu }}\left[q_{\mu j}^B\right]\\ & =& \sum _{\lambda \in \hat{A}}\frac{m_{\lambda }}{d_{\lambda }^A}\sum _{i=1}^{d_{\lambda }}\sum _{\mu \in \hat{B}}\sum _{j=1}^{g_{\lambda \mu }}\frac{1}{t_{\mu }}{z}_{\mu }^B\\ & =& \sum _{\lambda \in \hat{A}}\sum _{\mu \in \hat{B}}\frac{m_{\lambda }}{d_{\lambda }^A}{d}_{\lambda }^A{g}_{\lambda \mu }\frac{1}{t_{\mu }}{z}_{\mu }^B\\ & =& \sum _{\lambda \in \hat{A}}\sum _{\mu \in \hat{B}}\frac{m_{\lambda }}{d_{\lambda }^A}{g}_{\lambda \mu }{z}_{\mu }^B\\ & =& \sum _{\lambda \in \hat{A}}\sum _{\mu \in \hat{B}}\frac{m_{\lambda }}{d_{\lambda }^A}{g}_{\lambda \mu }\underset{\underset{z_{\mu }^B}{⏟}}{\sum _{b\in ℬ}{t}_{\mu }{\chi }_B^{\mu }\left(b\right)b^{*}}\\ & =& \sum _{b\in ℬ}\underset{\underset{{\chi }_{V{\uparrow }_A^B}\left(b\right)}{⏟}}{\sum _{\lambda \in \hat{A}}\sum _{\mu \in \hat{B}}{m}_{\lambda }{g}_{\lambda \mu }{\chi }_B^{\mu }\left(b\right)}\hspace{0.17em}{b}^{*}\\ \left[z\right]& =& \sum _{b\in ℬ}{\chi }_{V{\uparrow }_A^B}\left(b\right)b^{*}\end{array}$$ $\square$

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.

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