## Lectures in Representation Theory

Last update: 12 August 2013

## Lecture 3

 Continued proof. Having shown that $V$ is an irreducible irreducible representation of ${M}_{d}\left(ℂ\right),$ we proceed to prove uniqueness of $V$ up to equivalence. Let $W$ be any simple module of ${M}_{d}\left(ℂ\right)$ and choose $w\in W\\left\{0\right\}\text{.}$ Then $0≠w=Idw=∑j=1d Ej,jw$ and so ${E}_{i,i}w\ne 0$ for some $i\text{.}$ The set ${M}_{d}\left(ℂ\right){E}_{i,i}w\subseteq W$ is a nonzero left submodule of $W$ and hence $W={M}_{d}\left(ℂ\right){E}_{i,i}w$ by the simplicity of $W\text{.}$ We claim that $V\cong {M}_{d}\left(ℂ\right){E}_{i,i}$ as left ${M}_{d}\left(ℂ\right)\text{-modules.}$ The map defined by $(a1a2⋮ad) ↦ ( 0 a1 0 a2 ⋮ ad )$ realizes this isomorphism, where the nonzero entries on the right appear in the $i\text{th}$ column (the proof is left as an exercise). We may then define a linear transformation $ϕ:V≅Md(ℂ) Ei,i⟶Md(ℂ) Ei,iw=W$ by $\varphi \left({e}_{j}\right)={E}_{j,i}{E}_{i,i}w={E}_{j,i}w\text{.}$ This map is a module homomorphism, for if we choose an arbitrary matrix unit ${E}_{k,\ell },$ then $ϕ(Ek,ℓej)= ϕ(δℓ,jek)= δℓ,jEk,iw= Ek,jEj,iw= Ek,ℓϕ(ej)$ and the module property follows by linearity. Since $\text{ker} \varphi \subseteq V$ is a submodule and ${e}_{i}\notin \text{ker} \varphi ,$ the simplicity of $V$ forces $\text{ker} \varphi =0\text{.}$ Similarly, $\text{im} \varphi$ is a nonzero submodule of the simple module $W,$ hence $\text{im} \varphi =W\text{.}$ The map $\varphi$ is therefore an isomorphism between $V$ and $W\text{.}$ $\square$

Next, let us fix some notation. As usual, let $ℂ$ be the complex numbers, $ℝ$ the reals, $ℤ$ the integers. We depart from standard notation to write $ℕ=\left\{0,1,2,\dots \right\}$ for the set of nonnegative integers and let $ℙ=\left\{0,1,2,\dots \right\}$ denote the set of positive integers. We will use interval notation $\left[1,n\right]=\left\{1,2,\dots ,n\right\}$ to denote the set of integers between two endpoints (inclusive).

Definition 1.29 A simple algebra $A$ is an algebra with no nonzero proper two-sided ideals. For this course, a simple algebra is any algebra $A$ such that $A\cong {M}_{d}\left(ℂ\right)$ for some $d\text{.}$

Definition 1.30 We will say a finite dimensional algebra $A$ is semisimple provided $A\cong {\oplus }_{\lambda \in \stackrel{ˆ}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right),$ where $\stackrel{ˆ}{A}$ is some finite index set, and ${d}_{\lambda }\in ℙ\text{.}$

We will often write ${A}_{\lambda }\cong {M}_{{d}_{\lambda }}\left(ℂ\right)$ for the simple component of $A$ indexed by $\lambda \in \stackrel{ˆ}{A}\text{.}$ Thus a typical element $a\in A$ is of the form

$a= ( aλ1 aλ2 0 0 ⋱ aλk )$

where each ${a}_{{\lambda }_{j}}\in {A}_{{\lambda }_{j}}\cong {M}_{{d}_{{\lambda }_{j}}}\left(ℂ\right),$ which we regard as a ${d}_{{\lambda }_{j}}×{d}_{{\lambda }_{j}}$ matrix.

Every finite dimensional semisimple algebra has a basis of matrix units $\left\{{E}_{i,j}^{\lambda } | 1\le i,j\le {d}_{\lambda },\lambda \in \stackrel{ˆ}{A}\right\},$ which correspond to the usual matrix units in ${\oplus }_{\lambda \in \stackrel{ˆ}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right)\text{.}$ That is, the structure constants are given by

$Ei,jλ Er,sμ= δλ,μ δj,r Ei,sλ.$

Define a trace ${\chi }^{\lambda }:A⟶ℂ$ by ${\chi }^{\lambda }\left(a\right)=\text{tr}\left({a}_{\lambda }\right)={\sum }_{i=1}^{{d}_{\lambda }}{a}_{i,i}^{\lambda }$ where $a=\sum _{i,j,\lambda }{a}_{i,j}^{\lambda }{E}_{i,j}^{\lambda }\text{.}$ That is ${\chi }^{\lambda }$ is the trace of the $\lambda$ block of $a\text{.}$

Proposition 1.31 Let $\stackrel{\to }{t}$ be a trace on $A$ and let ${t}_{\lambda }=\stackrel{\to }{t}\left({E}_{1,1}^{\lambda }\right)\text{.}$ Then

$t→=∑λ∈Aˆ tλχλ.$

 Proof. By 1.23, tr is the unique trace on ${M}_{d}\left(ℂ\right),$ hence ${\chi }^{\lambda }$ restricted to ${A}_{\lambda }$ is the unique trace on ${A}_{\lambda }\text{.}$ Thus, for each $\lambda \in \stackrel{ˆ}{A},$ there exists ${t}_{\lambda }\in ℂ$ such that $\stackrel{\to }{t}\left({a}_{\lambda }\right)={t}_{\lambda }{\chi }^{\lambda }\left({a}_{\lambda }\right)$ for all ${a}_{\lambda }\in {A}_{\lambda }\text{.}$ In particular, $t→(E1,1λ) =tλχ(E1,1λ) =tλtr(E1,1) =tλ.$ The proposition follows by linearity of $\stackrel{\to }{t}\text{.}$ $\square$

Definition 1.32 Let $\stackrel{\to }{t}$ be a trace on $A\text{.}$ The vector ${\left({t}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}$ weight vector of $\stackrel{\to }{t}\text{.}$

The previous proposition shows that a trace $\stackrel{\to }{t}$ is completely determined by its weight vector. Conversely, one may define a trace by specifying its weight vector. Hence the trace functionals on $A$ are in one to one correspondence with the vectors in ${ℂ}^{|\stackrel{ˆ}{A}|}$ which explains the notation $\stackrel{\to }{t}\text{.}$

After classifying the traces on $A,$ we turn to the ideal structure. From the general theory of semisimple algebras, we know that $A$ possesses a finite collection of minimal two-sided ideals (i.e. nonzero ideals of $A$ containing no proper nonzero two-sided ideals). The algebra $A$ is the direct sum of these minimal ideals and, furthermore, every ideal of $A$ is a direct sum of some subset of these ideals.

Since ${M}_{d}\left(ℂ\right)$ is simple and $A\cong {\oplus }_{\lambda \in \stackrel{ˆ}{A}}{M}_{{d}_{\lambda }}\left(ℂ\right),$ evidently the minimal two-sided ideals of $A$ are the simple components ${A}_{\lambda }\cong {M}_{{d}_{\lambda }}\left(ℂ\right),$ $\lambda \in \stackrel{ˆ}{A}\text{.}$

Definitions 1.33 Two idempotents ${z}_{1}$ and ${z}_{2}$ are said to be orthogonal provided ${z}_{1}{z}_{2}={z}_{2}{z}_{1}=0\text{.}$ An idempotent $z$ is said to be minimal (or principal) provided $z$ cannot be written as a sum of orthogonal idempotents.

From the general theory, every minimal two-sided ideal ${A}_{\lambda }$ is generated by a central idempotent ${z}_{\lambda },$ i.e. ${A}_{\lambda }=A{z}_{\lambda }\text{.}$ Since ${z}_{\lambda }$ must be the identity of ${A}_{\lambda },$ necessarily ${z}_{\lambda }{I}_{{d}_{\lambda }}={\sum }_{i=1}^{{d}_{\lambda }}{E}_{i,i}^{\lambda }\text{.}$ Note that the ${z}_{\lambda }$ are orthogonal as they belong to different minimal ideals of $A\text{.}$

However, the minimality of the two-sided ideals ${A}_{\lambda }$ does not quite translate into minimality of the ${z}_{\lambda }$ in the sense of the above definition, since the ${E}_{i,i}^{\lambda }$ are orthogonal idempotents which sum to ${z}_{\lambda }\text{.}$ Yet ${z}_{\lambda }$ is minimal in the sense that it may not be written as the sum of central orthogonal idempotents:

Proposition 1.34 The minimal central idempotents of $A$ are $\left\{{z}_{\lambda } | \lambda \in \stackrel{ˆ}{A}\right\}\text{.}$ In particular, this set forms a basis for the center $Z\left(A\right)\text{.}$

 Proof. One may verify by using the basis of matrix units that $Z\left({M}_{d}\left(ℂ\right)\right)=ℂ{I}_{d}\text{.}$ Since $A$ is a direct sum of matrix algebras, the second statement of the proposition follows from the definition of the ${z}_{\lambda }\text{.}$ Suppose $z\in Z\left(A\right)$ is any central idempotent of $A\text{.}$ If we express $z={\sum }_{\lambda \in \stackrel{ˆ}{A}}{c}_{\lambda }{z}_{\lambda },$ ${c}_{\lambda }\in ℂ,$ in terms of the new basis for the center, we obtain $zλ=zλ2= (∑λ∈Aˆcλzλ)2 =∑λ∈Aˆ cλ2zλ$ using the orthogonality of the ${z}_{\lambda }\text{.}$ By uniqueness of expression, we have ${c}_{\lambda }^{2}={c}_{\lambda }$ and hence ${c}_{\lambda }\in \left\{0,1\right\}\text{.}$ Thus, any central idempotent of $A$ is a sum of one or more ${z}_{\lambda }\text{.}$ The result then follows from the independence of the ${z}_{\lambda }\text{.}$ $\square$

Homework Problem 1.35 Show that every element of the form

$Ei,iλ+ ∑j≠i ci,j Ei,jλ,$

where each ${c}_{i,j}\in \left\{0,1\right\}$ is an idempotent which generates a minimal left ideal of $A\text{.}$ Conclude that these elements are minimal idempotents of $A\text{.}$

Similarly, show that the idempotents

$Ei,iλ+ ∑j≠i ci,jEj,iλ$

generate minimal right ideals of $A$ and hence are minimal idempotents.

Finally, we turn to the irreducible representations and simple modules of $A\text{.}$ Let ${W}^{\mu }:A⟶{M}_{{d}_{\mu }}\left(ℂ\right),$ $\mu \in \stackrel{ˆ}{A}$ be defined by ${W}^{\mu }\left(a\right)={a}_{\mu }$ where ${a}_{\mu }={\sum }_{i,j=1}^{{d}_{\mu }}{a}_{i,j}^{\mu }{E}_{i,j}^{\mu }\in {A}_{\mu }$ is the $\mu \text{-block}$ of the element $a\text{.}$

Proposition 1.36 The representations ${W}^{\mu },$ $\mu \in \stackrel{ˆ}{A}$ are the complete set of irreducible representations of $A$ up to equivalence.

 Proof. The restriction ${W}^{\mu }{↓}_{{A}_{\mu }}:{A}_{\mu }⟶{M}_{{d}_{\mu }}\left(ℂ\right)$ is the unique irreducible representation of ${A}_{\mu }\text{.}$ Any $A\text{-subrepresentation}$ $\varphi :A⟶{M}_{{d}_{\mu }}\left(ℂ\right)$ of ${W}^{\mu }$ is also a ${A}_{\mu }\text{-subrepresentation}$ of ${W}^{\mu },$ hence is zero on ${A}_{\mu }\text{.}$ However, $\text{ker} {W}^{\mu }={\sum }_{\lambda \ne \mu }{A}_{\lambda }\text{.}$ Hence, $\varphi =0$ on $A$ and ${W}^{\mu }$ is an irreducible representation of $A\text{.}$ If $W:A⟶{M}_{d}\left(ℂ\right)$ is any irreducible representation of $A,$ then $\text{ker} W$ is an ideal of $A,$ and hence is the direct sum of one or more of the ${A}_{\lambda }\text{.}$ If ${A}_{\mu }$ is in the complement of $\text{ker} W,$ then ${W}^{\mu }$ is a summand of $W\text{.}$ Irreducibility of $W$ then implies that $W$ is equivalent to exactly one of the ${W}^{\mu }\text{.}$ $\square$

Turning to simple modules, denote the space ${ℂ}^{{d}_{\mu }}$ by ${W}^{\mu },$ and let $A$ act on ${W}^{\mu }$ by $a·w={a}_{\mu }w\text{.}$ The proof of the following proposition is left as an exercise.

Proposition 1.37 The set ${W}^{\mu }$ form a complete set of simple modules of $A$ up to isomorphism.

Observe that $A{E}_{i,j}^{\lambda }\cong {W}^{\lambda }\text{.}$

Observe also that the irreducible representation ${W}^{\lambda }:A⟶{M}_{{d}_{\lambda }}\left(ℂ\right)$ defined by a $a↦{W}^{\lambda }\left(a\right)={a}_{\lambda }$ has character ${\chi }_{{W}^{\lambda }}:A⟶ℂ$ defined by

$χWλ(a)=tr (Wλ(a))=tr (aλ)=χλ.$

Hence the irreducible characters of $A$ are the traces ${\chi }^{\lambda },$ $\lambda \in \stackrel{ˆ}{A}\text{.}$

## 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.