Last update: 12 August 2013

Continued proof. | |

Having shown that $V$ is an irreducible irreducible representation of ${M}_{d}\left(\u2102\right),$ we proceed to prove uniqueness of $V$ up to equivalence. Let $W$ be any simple module of ${M}_{d}\left(\u2102\right)$ and choose $w\in W\backslash \left\{0\right\}\text{.}$ Then $$0\ne w={I}_{d}w=\sum _{j=1}^{d}{E}_{j,j}w$$and so ${E}_{i,i}w\ne 0$ for some $i\text{.}$ The set ${M}_{d}\left(\u2102\right){E}_{i,i}w\subseteq W$ is a nonzero left submodule of $W$ and hence $W={M}_{d}\left(\u2102\right){E}_{i,i}w$ by the simplicity of $W\text{.}$ We claim that $V\cong {M}_{d}\left(\u2102\right){E}_{i,i}$ as left ${M}_{d}\left(\u2102\right)\text{-modules.}$ The map defined by $$\left(\begin{array}{c}{a}_{1}\\ {a}_{2}\\ \vdots \\ {a}_{d}\end{array}\right)\mapsto \left(\begin{array}{ccc}{\scriptscriptstyle 0}& {a}_{1}& {\scriptscriptstyle 0}\\ {a}_{2}\\ \vdots \\ {a}_{d}\end{array}\right)$$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 $$\varphi :V\cong {M}_{d}\left(\u2102\right){E}_{i,i}\u27f6{M}_{d}\left(\u2102\right){E}_{i,i}w=W$$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 $$\varphi \left({E}_{k,\ell}{e}_{j}\right)=\varphi \left({\delta}_{\ell ,j}{e}_{k}\right)={\delta}_{\ell ,j}{E}_{k,i}w={E}_{k,j}{E}_{j,i}w={E}_{k,\ell}\varphi \left({e}_{j}\right)$$and the module property follows by linearity. Since $\text{ker}\hspace{0.17em}\varphi \subseteq V$ is a submodule and ${e}_{i}\notin \text{ker}\hspace{0.17em}\varphi ,$ the simplicity of $V$ forces $\text{ker}\hspace{0.17em}\varphi =0\text{.}$ Similarly, $\text{im}\hspace{0.17em}\varphi $ is a nonzero submodule of the simple module $W,$ hence $\text{im}\hspace{0.17em}\varphi =W\text{.}$ The map $\varphi $ is therefore an isomorphism between $V$ and $W\text{.}$ $\square $ |

Next, let us fix some notation. As usual, let $\u2102$ be the complex numbers, $\mathbb{R}$ the reals, $\mathbb{Z}$ the integers. We depart from standard notation to write $\mathbb{N}=\{0,1,2,\dots \}$ for the set of nonnegative integers and let $\mathbb{P}=\{0,1,2,\dots \}$ denote the set of positive integers. We will use interval notation $[1,n]=\{1,2,\dots ,n\}$ 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 $A\cong {M}_{d}\left(\u2102\right)$ for some $d\text{.}$
*

**Definition 1.30** *We will say a finite dimensional algebra $A$ is* **semisimple**
*
provided $A\cong {\oplus}_{\lambda \in \stackrel{\u02c6}{A}}{M}_{{d}_{\lambda}}\left(\u2102\right),$
where $\stackrel{\u02c6}{A}$ is some finite index set, and ${d}_{\lambda}\in \mathbb{P}\text{.}$
*

We will often write ${A}_{\lambda}\cong {M}_{{d}_{\lambda}}\left(\u2102\right)$ for the simple component of $A$ indexed by $\lambda \in \stackrel{\u02c6}{A}\text{.}$ Thus a typical element $a\in A$ is of the form

$$a=\left(\begin{array}{cc}\begin{array}{c}{a}_{{\lambda}_{1}}\\ & {a}_{{\lambda}_{2}}\end{array}& {\scriptscriptstyle 0}\\ {\scriptscriptstyle 0}& \begin{array}{c}\ddots \\ & {a}_{{\lambda}_{k}}\end{array}\end{array}\right)$$where each ${a}_{{\lambda}_{j}}\in {A}_{{\lambda}_{j}}\cong {M}_{{d}_{{\lambda}_{j}}}\left(\u2102\right),$ which we regard as a ${d}_{{\lambda}_{j}}\times {d}_{{\lambda}_{j}}$ matrix.

Every finite dimensional semisimple algebra has a basis of **matrix units**
$\left\{{E}_{i,j}^{\lambda}\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le {d}_{\lambda},\lambda \in \stackrel{\u02c6}{A}\},$
which correspond to the usual matrix units in ${\oplus}_{\lambda \in \stackrel{\u02c6}{A}}{M}_{{d}_{\lambda}}\left(\u2102\right)\text{.}$
That is, the structure constants are given by

Define a trace ${\chi}^{\lambda}:A\u27f6\u2102$ 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 $\overrightarrow{t}$ be a trace on $A$ and let
${t}_{\lambda}=\overrightarrow{t}\left({E}_{1,1}^{\lambda}\right)\text{.}$ Then
*

Proof. | |

By 1.23, tr is the unique trace on ${M}_{d}\left(\u2102\right),$ hence ${\chi}^{\lambda}$ restricted to ${A}_{\lambda}$ is the unique trace on ${A}_{\lambda}\text{.}$ Thus, for each $\lambda \in \stackrel{\u02c6}{A},$ there exists ${t}_{\lambda}\in \u2102$ such that $\overrightarrow{t}\left({a}_{\lambda}\right)={t}_{\lambda}{\chi}^{\lambda}\left({a}_{\lambda}\right)$ for all ${a}_{\lambda}\in {A}_{\lambda}\text{.}$ In particular, $$\overrightarrow{t}\left({E}_{1,1}^{\lambda}\right)={t}_{\lambda}\chi \left({E}_{1,1}^{\lambda}\right)={t}_{\lambda}\text{tr}\left({E}_{1,1}\right)={t}_{\lambda}\text{.}$$The proposition follows by linearity of $\overrightarrow{t}\text{.}$ $\square $ |

**Definition 1.32**
*
Let $\overrightarrow{t}$ be a trace on $A\text{.}$ The vector
${\left({t}_{\lambda}\right)}_{\lambda \in \stackrel{\u02c6}{A}}$
*
**weight vector**
*
of $\overrightarrow{t}\text{.}$
*

The previous proposition shows that a trace $\overrightarrow{t}$ 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 ${\u2102}^{\left|\stackrel{\u02c6}{A}\right|}$ which explains the notation $\overrightarrow{t}\text{.}$

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 ${M}_{d}\left(\u2102\right)$ is simple and $A\cong {\oplus}_{\lambda \in \stackrel{\u02c6}{A}}{M}_{{d}_{\lambda}}\left(\u2102\right),$ evidently the minimal two-sided ideals of $A$ are the simple components ${A}_{\lambda}\cong {M}_{{d}_{\lambda}}\left(\u2102\right),$ $\lambda \in \stackrel{\u02c6}{A}\text{.}$

**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 $\left\{{z}_{\lambda}\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \stackrel{\u02c6}{A}\}\text{.}$
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 $Z\left({M}_{d}\left(\u2102\right)\right)=\u2102{I}_{d}\text{.}$ 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 $z={\sum}_{\lambda \in \stackrel{\u02c6}{A}}{c}_{\lambda}{z}_{\lambda},$ ${c}_{\lambda}\in \u2102,$ in terms of the new basis for the center, we obtain $${z}_{\lambda}={z}_{\lambda}^{2}={\left(\sum _{\lambda \in \stackrel{\u02c6}{A}}{c}_{\lambda}{z}_{\lambda}\right)}^{2}=\sum _{\lambda \in \stackrel{\u02c6}{A}}{c}_{\lambda}^{2}{z}_{\lambda}$$using the orthogonality of the ${z}_{\lambda}\text{.}$ By uniqueness of expression, we have ${c}_{\lambda}^{2}={c}_{\lambda}$ and hence ${c}_{\lambda}\in \{0,1\}\text{.}$ 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*

*
where each ${c}_{i,j}\in \{0,1\}$
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
*

*
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 ${W}^{\mu}:A\u27f6{M}_{{d}_{\mu}}\left(\u2102\right),$ $\mu \in \stackrel{\u02c6}{A}$ 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 ${W}^{\mu},$
$\mu \in \stackrel{\u02c6}{A}$
are the complete set of irreducible representations of $A$ up to equivalence.
*

Proof. | |

The restriction ${W}^{\mu}{\downarrow}_{{A}_{\mu}}:{A}_{\mu}\u27f6{M}_{{d}_{\mu}}\left(\u2102\right)$ is the unique irreducible representation of ${A}_{\mu}\text{.}$ Any $A\text{-subrepresentation}$ $\varphi :A\u27f6{M}_{{d}_{\mu}}\left(\u2102\right)$ of ${W}^{\mu}$ is also a ${A}_{\mu}\text{-subrepresentation}$ of ${W}^{\mu},$ hence is zero on ${A}_{\mu}\text{.}$ However, $\text{ker}\hspace{0.17em}{W}^{\mu}={\sum}_{\lambda \ne \mu}{A}_{\lambda}\text{.}$ Hence, $\varphi =0$ on $A$ and ${W}^{\mu}$ is an irreducible representation of $A\text{.}$ If $W:A\u27f6{M}_{d}\left(\u2102\right)$ is any irreducible representation of $A,$ then $\text{ker}\hspace{0.17em}W$ 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 $\text{ker}\hspace{0.17em}W,$ 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 ${\u2102}^{{d}_{\mu}}$ by ${W}^{\mu},$ and let $A$ act on ${W}^{\mu}$ by $a\xb7w={a}_{\mu}w\text{.}$ 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 ${W}^{\lambda}:A\u27f6{M}_{{d}_{\lambda}}\left(\u2102\right)$ defined by a $a\mapsto {W}^{\lambda}\left(a\right)={a}_{\lambda}$ has character ${\chi}_{{W}^{\lambda}}:A\u27f6\u2102$ defined by

$${\chi}_{{W}^{\lambda}}\left(a\right)=\text{tr}\left({W}^{\lambda}\left(a\right)\right)=\text{tr}\left({a}_{\lambda}\right)={\chi}^{\lambda}\text{.}$$Hence the irreducible characters of $A$ are the traces ${\chi}^{\lambda},$ $\lambda \in \stackrel{\u02c6}{A}\text{.}$

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