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

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 $w\in W\backslash \left\{0\right\}\text{.}$ 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 $M_d\left(ℂ\right)E_{i,i}w\subseteq W$ 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 $$\left(\begin{array}{c}{a}_1\\ a_2\\ ⋮\\ a_d\end{array}\right)\mapsto \left(\begin{array}{ccc}0& a_1& 0\\ a_2\\ ⋮\\ a_d\end{array}\right)$$ 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 $$\varphi :V\cong M_d\left(ℂ\right)E_{i,i}⟶M_d\left(ℂ\right)E_{i,i}w=W$$ 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 $ℂ$ be the complex numbers, $ℝ$ 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 $A\cong M_d\left(ℂ\right)$ for some $d\text{.}$

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

We will often write $A_{\lambda }\cong M_{d_{\lambda }}\left(ℂ\right)$ for the simple component of $A$ indexed by $\lambda \in \hat{A}\text{.}$ Thus a typical element $a\in A$ is of the form

$$a=\left(\begin{array}{cc}\begin{array}{c}{a}_{{\lambda }_1}\\ & a_{{\lambda }_2}\end{array}& 0\\ 0& \begin{array}{c}\ddots \\ & a_{{\lambda }_k}\end{array}\end{array}\right)$$

where each $a_{{\lambda }_j}\in A_{{\lambda }_j}\cong M_{d_{{\lambda }_j}}\left(ℂ\right),$ which we regard as a $d_{{\lambda }_j}\times d_{{\lambda }_j}$ matrix.

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

$$E_{i,j}^{\lambda }{E}_{r,s}^{\mu }={\delta }_{\lambda ,\mu }{\delta }_{j,r}{E}_{i,s}^{\lambda }\text{.}$$

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 $\overrightarrow{t}$ be a trace on $A$ and let $t_{\lambda }=\overrightarrow{t}\left(E_{1,1}^{\lambda }\right)\text{.}$ Then

$$\overrightarrow{t}=\sum _{\lambda \in \hat{A}}{t}_{\lambda }{\chi }^{\lambda }\text{.}$$

		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 $\lambda \in \hat{A},$ there exists $t_{\lambda }\in ℂ$ 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 ${\left(t_{\lambda }\right)}_{\lambda \in \hat{A}}$ 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 ${ℂ}^{\left|\hat{A}\right|}$ 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 $M_d\left(ℂ\right)$ is simple and $A\cong {\oplus }_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right),$ evidently the minimal two-sided ideals of $A$ are the simple components $A_{\lambda }\cong M_{d_{\lambda }}\left(ℂ\right),$ $\lambda \in \hat{A}\text{.}$

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 $\left\{z_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A}\}\text{.}$ 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 $Z\left(M_d\left(ℂ\right)\right)=ℂI_d\text{.}$ 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 $z={\sum }_{\lambda \in \hat{A}}{c}_{\lambda }{z}_{\lambda },$ $c_{\lambda }\in ℂ,$ in terms of the new basis for the center, we obtain $$z_{\lambda }=z_{\lambda }^2={\left(\sum _{\lambda \in \hat{A}}{c}_{\lambda }{z}_{\lambda }\right)}^2=\sum _{\lambda \in \hat{A}}{c}_{\lambda }^2{z}_{\lambda }$$ 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

$$E_{i,i}^{\lambda }+\sum _{j\ne i}{c}_{i,j}{E}_{i,j}^{\lambda },$$

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

$$E_{i,i}^{\lambda }+\sum _{j\ne i}{c}_{i,j}{E}_{j,i}^{\lambda }$$

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 $W^{\mu }:A⟶M_{d_{\mu }}\left(ℂ\right),$ $\mu \in \hat{A}$ 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 $W^{\mu },$ $\mu \in \hat{A}$ are the complete set of irreducible representations of $A$ up to equivalence.

		Proof.
	The restriction $W^{\mu }{\downarrow }_{A_{\mu }}:A_{\mu }⟶M_{d_{\mu }}\left(ℂ\right)$ 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, $\text{ker}\hspace{0.17em}{W}^{\mu }={\sum }_{\lambda \ne \mu }{A}_{\lambda }\text{.}$ Hence, $\varphi =0$ on $A$ and $W^{\mu }$ is an irreducible representation of $A\text{.}$ If $W:A⟶M_d\left(ℂ\right)$ 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 ${ℂ}^{d_{\mu }}$ 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 $a\mapsto W^{\lambda }\left(a\right)=a_{\lambda }$ has character ${\chi }_{W^{\lambda }}:A⟶ℂ$ 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 ${\chi }^{\lambda },$ $\lambda \in \hat{A}\text{.}$

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.

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