Lecture 2

Lectures in Representation Theory

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

Last update: 12 August 2013

Homework Problem 1.16 Study $L^{2} (G)$ for large groups (the algebra of $L^{2}$ functions on $G).$ Go find out about compact groups and representation theory.

If $V$ is a (finite dimensional) $A -module$ of dimension $d,$ then there is an associated representation, which we shall also denote by $V,$ such that $V : A ⟶ M_{d} (ℂ) .$

Definition 1.17 Let $A$ be an algebra, and $V_{1}$ and $V_{2}$ be $A -modules.$ The direct sum $V_{1} \oplus V_{2}$ is the $A -module$ of all pairs $(v_{1}, v_{2}),$ where $v_{i} \in V_{i} .$ Addition is defined componentwise; the action of $A$ is $a \cdot (v_{1}, v_{2}) = (a v_{1}, a v_{2}) .$

We can make a parallel definition for representations.

Definition 1.18 Let $A$ be an algebra, and $V_{1} : A \to M_{d_{1}} (ℂ)$ and $V_{2} : A \to M_{d_{2}} (ℂ)$ representations of $A .$ Then the direct sum $(V_{1} \oplus V_{2}) : A \to M_{d_{1} + d_{2}} (ℂ)$ is the representation given by

(V_{1} \oplus V_{2}) (a) = (\begin{matrix} V_{1} (a) & 0 \\ 0 & V_{2} (a) \end{matrix}) .

Definition 1.19 An $A -module$ $V$ is completely decomposable if it is isomorphic (as an $A -module)$ to a direct sum of simple $A -modules.$

Problem 1.20 Let $A$ be an algebra.

1.	What are the irreducible representations of $A ?$
2.	Are the other representations of $A$ completely decomposable?

Definitions 1.21 Given an algebra $A .$ A trace on $A$ is a linear functional $\vec{t} : A \to ℂ$ such that $\vec{t} (a b) = \vec{t} (b a),$ for all $a, b \in A .$ The center of $A$ is the subalgebra $Z (A) = {t \in A | t a = a t \forall a \in A} .$ An idempotent is an element $p \in A$ such that $p \neq 0$ and $p^{2} = p .$ A central idempotent is an idempotent $z$ such that $z \in Z (A) .$ An ideal of $A$ is an ideal in the ring theory sense.

We will use the words “trace” and “character” interchangeably; some would have us call our traces “virtual characters”.

Definition 1.22 Let $V : A \to M_{d} (ℂ)$ be a representation. Then the character of the representation $V$ is given by $χ_{V} : A \to ℂ : a \mapsto tr (V (a)),$ where $tr$ denotes the vanilla trace of high school matrix theory. Any such character is a trace in the sense defined above.

The following propositions can be nicely summarized as follows:

Theorem 1.23 There exists a unique nonzero

{\begin{matrix} trace \\ element of the center \\ central idempotent \\ ideal \\ irreducible representation \end{matrix}} in M_{d} (ℂ) {\begin{matrix} up to scalar multiples \\ up to scalar multiples \\ up to equivalence \end{matrix}} .

Proposition 1.24 There is a unique (up to scalar multiples) nonzero trace on $M_{d} (ℂ) .$

Proof.

Let $\vec{t}$ be a trace, and let $a \in M_{d} (ℂ) .$

\begin{matrix} \vec{t} (a) & = & \vec{t} (\sum_{i j} a_{i j} E_{i j}) = \sum_{i j} a_{i j} \vec{t} (E_{i j}) = \sum_{i j} a_{i j} \vec{t} (E_{i 1} E_{1 j}) \\ = & \sum_{i j} a_{i j} \vec{t} (E_{1 j} E_{i 1}) = \sum_{i j} a_{i j} δ_{i j} \vec{t} (E_{11}) = \vec{t} (E_{11}) tr (a) \end{matrix}

Thus any such $\vec{t}$ is a scalar multiple of the ordinary trace functional.

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Proposition 1.25 There is a unique nonzero element in $Z (M_{d} (ℂ))$ (up to scalar multiples).

Proof.

Let $z \in Z (M_{d} (ℂ)),$ $z \neq 0 .$ Now $z = \sum_{i j} z_{i j} E_{i j},$ and $z \neq 0$ implies that for some $m$ and $n$ the coefficient $z_{m n} \neq 0,$ Fix these $m$ and $n .$

Let $k \neq s .$ Since $z$ is central, $E_{m k} z E_{s n} = E_{m k} E_{s n} z = 0 .$ On the other hand,

\begin{matrix} E_{m k} z E_{s n} & = & E_{m k} (\sum_{i j} z_{i j} E_{i j}) E_{s n} \\ = & E_{m k} z_{k s} E_{k s} E_{s n} \\ = & z_{k s} E_{m n} \end{matrix}

Thus if $k \neq s,$ then $z_{k s} = 0 .$ Thus $z$ central implies that $z$ is a diagonal matrix.

To show that $z$ central implies $z$ is scalar, fix $1 \leq i \leq d,$ and consider $E_{i i} z .$ For any $j,$ $E_{i i} z = E_{i j} E_{j i} z = E_{i j} z E_{j i} = z_{j j} E_{i i} .$ Since this holds for all $j,$ it follows that $z_{j j} = z_{11}$ for all $j .$

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Proposition 1.26 There is a unique nonzero ideal in $M_{d} (ℂ) .$

Proof.

Exercise.

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Proposition 1.27 There is a unique central idempotent in $M_{d} (ℂ) .$

Proof.

Let $z \in Z (M_{d} (ℂ))$ with $z^{2} = z .$ $z$ central implies that $z = z_{11} I_{d} .$ Now $z^{2} = {(z_{11} I_{d})}^{2} = {z_{11}}^{2} I_{d};$ thus $z_{11}$ is a complex number satisfying ${z_{11}}^{2} = z_{11},$ so either $z_{11} = 0$ or $z_{11} = 1 .$ This is one place where these developments will not go through so nicely in other characteristics.

$□$

Proposition 1.28 There is a unique (nonzero) irreducible representation of $M_{d} (ℂ)$ (up to equivalence).

Proof.

First we shall prove that such a representation exists.

Let $V$ be the space of all column vectors of length $d,$ $ℂ^{d} .$ $M_{d} (ℂ)$ acts on $V$ by matrix multiplication.

Claim: $V$ is irreducible.

Let $e_{1}, \dots, e_{d}$ be the standard basis for $V .$ Let $W$ be a subrepresentation, with $W \neq 0 .$ Choose $v \in W$ with $v \neq 0,$ $v = \sum v_{i} e_{i},$ and fix an $i$ such that $v_{i} \neq 0 .$

Claim: $M_{d} (ℂ) v = V$ (whence $W = V,$ proving the previous claim.)

\begin{matrix} \frac{1}{v_{i}} E_{j i} v & = & \frac{1}{v_{i}} E_{j i} \sum_{k} v_{k} e_{k} \\ = & \frac{1}{v_{i}} \sum_{k} v_{k} E_{j i} e_{k} \\ = & \frac{1}{v_{i}} \sum_{k} v_{k} δ_{i k} e_{j} \\ = & \frac{1}{v_{i}} v_{i} e_{j} = e_{j} \end{matrix}

Thus for every $j,$ $e_{j} \in M_{d} (ℂ) v .$ Thus $M_{d} (ℂ) v = V .$

Continued in next lecture.

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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