Last update: 27 August 2013

The following theorem due to Burnside can be found in Burnside, Proc. London Math. Soc. (2), 3, 1905 p. 430. In his ”Classical Groups”, Weyl also directs the reader to Frobenius, Schur, Siztungsber Preuss. Akad. 1906, p. 209.

**Theorem 3.18 (Burnside)**
*
Let $A$ be an arbitrary algebra and let
$W:A\to {M}_{d}\left(\u2102\right)$
be an irreducible representation of $A\text{.}$ Then
$W\left(A\right)={M}_{d}\left(\u2102\right)\text{.}$
*

Proof. | |

Without loss, $W$ is faithful, since otherwise we may consider the representation $W:{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$\text{ker}\hspace{0.17em}W$}\right.}\to {M}_{d}\left(\u2102\right),$ which has the same image as $A\text{.}$ Choose a nonzero $a\in A\text{.}$ By faithfulness of $W,$ we may choose $w\in W$ such that $aw\ne 0\text{.}$ Since $W$ is an irreducible $A\text{-module,}$ $Aaw=W\text{.}$ Thus, there exists a $b\in A$ such that $baw=w,$ from which it follows that ${\left(ba\right)}^{k}w=w$ for all $k\text{.}$ In particular, $ba$ is not nilpotent, so $a$ is not in $\sqrt{A},$ the largest nil ideal. But $a$ was an arbitrary nonzero element, so $\sqrt{A}=0,$ hence $A$ is semisimple. If $A$ is not simple, we may write $A=I\oplus J$ as a direct sum of two-sided ideals. Since $W$ is faithful, we may choose $w\ne 0$ such that $Iw\ne 0$ Then $Iw$ is a submodule of $W$ and so irreducibility implies $Iw=W\text{.}$ But $JW=JIw=0,$ which contradicts the faithfulness of $W\text{.}$ Thus $A$ must be simple, so we must have $A\cong {M}_{n}\left(\u2102\right)\text{.}$ By 1.23, $A$ has a unique irreducible module, namely $W,$ so $n=d\text{.}$ $\square $ |

Last time (3.16) we showed that given an arbitrary algebra $A$ and a completely decomposable (finite dimensional) representation $V,$ whose decomposition is given by $V={\sum}_{\lambda \in \stackrel{\u02c6}{V}}{m}_{\lambda}{V}^{\lambda},$ then the algebra of the representation is given by $$V\left(A\right)=\underset{\lambda \in \stackrel{\u02c6}{V}}{\u2a01}{I}_{{m}_{\lambda}}\left({V}^{\lambda}\left(A\right)\right)\text{.}$$ From the above theorem, writing ${d}_{\lambda}=\text{dim}\hspace{0.17em}{V}^{\lambda},$ we have $$V\left(A\right)=\underset{\lambda \in \stackrel{\u02c6}{V}}{\u2a01}{I}_{{m}_{\lambda}}\left({M}_{d}\left(\u2102\right)\right)\text{.}$$ From proposition 3.17, recall that $$\stackrel{\u203e}{V\left(A\right)}\cong {\oplus}_{\lambda \in \stackrel{\u02c6}{V}}{M}_{{m}_{\lambda}}\left({I}_{{d}_{\lambda}}\left(\u2102\right)\right)\text{.}$$

Suppose $B$ is an arbitrary algebra which acts on $V$ such that $V\left(B\right)=\stackrel{\u203e}{V\left(A\right)}\text{.}$ Let $B\prime =\{b\in B\hspace{0.17em}|\hspace{0.17em}bV=0\}$ be the kernel of the representation of $B$ on $V\text{.}$ Then we have a representation $$\begin{array}{cccc}V:& \raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$B\prime $}\right.& \u27f6& {M}_{d}\left(\u2102\right)\\ & b+B\prime & \u27fc& V\left(b\right)\end{array}$$ which is, by definition, injective. Let $C={\displaystyle \raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$B\prime $}\right.}\text{;}$ by hypothesis, the map $V:C\to \stackrel{\u203e}{V\left(A\right)}$ is an isomorphism.

Hence, by proposition 3.17, $C$ is semisimple, and the irreducible representations of $C$ are indexed by $\lambda \in \stackrel{\u02c6}{V},$ and have dimensions ${m}_{\lambda}\text{.}$ Since $V\left(c\right)\in \stackrel{\u203e}{V\left(A\right)},$ by the above decomposition $$V\left(c\right)=\left(\begin{array}{c}{c}_{1}\otimes {I}_{{d}_{1}}\\ & {c}_{2}\otimes {I}_{{d}_{2}}\\ \multicolumn{2}{c}{}& \ddots \\ \multicolumn{3}{c}{}& {c}_{\left|\stackrel{\u02c6}{V}\right|}\otimes {I}_{{d}_{\left|\stackrel{\u02c6}{V}\right|}}\end{array}\right)$$ where ${c}_{\lambda}=\left({c}_{i,j}^{\lambda}\right)$ are ${m}_{\lambda}\times {m}_{\lambda}$ complex matrices.

Define maps ${C}^{\lambda}:C\to {M}_{{m}_{\lambda}}\left(\u2102\right)$ by ${C}^{\lambda}\left(c\right)={c}_{\lambda}\text{.}$ These are the irreducible representations of $C,$ but they extend to irreducible representations of $B$ via the composition $B\to {\displaystyle \raisebox{1ex}{$B$}\!\left/ \!\raisebox{-1ex}{$B\prime $}\right.}=C\stackrel{{C}^{\lambda}}{\to}{M}_{{m}_{\lambda}}\left(\u2102\right)$

Suppose $V$ and $W$ are vector spaces with bases ${\left\{{v}_{i}\right\}}_{1\le i\le m}$ and ${\left\{{w}_{j}\right\}}_{1\le j\le n}\text{.}$ Then $V\otimes W$ is a vector space with basis $\{{v}_{i}\otimes {v}_{j}\hspace{0.17em}|\hspace{0.17em}1\le i\le m,1\le j\le n\}\text{.}$

If $A$ and $B$ any are algebras, then $A\otimes B$ is a vector space which becomes an algebra under the multiplication $$({a}_{1}\otimes {b}_{1})({a}_{2}\otimes {b}_{2})=({a}_{1}{a}_{2}\otimes {b}_{1}{b}_{2})\text{.}$$

If $V$ and $W$ are modules for $A$ and $B,$ respectively, then $V\otimes W$ is an $A\otimes B$ module with the action defined by $$(a\otimes b)(v\otimes w)=av\otimes bw$$

**Problem 3.19**
*
Check that the multiplication in $A\otimes B$ is well-defined and that
$A\otimes B$ satisfies the axiom of an algebra over $\u2102\text{.}$
Check that the action of $A\otimes B$ on
$V\otimes W$ is a well-defined module action.
*

If $V\left(a\right)={\left({a}_{i,j}\right)}_{1\le i,j\le m}$ and $W\left(b\right)={\left({b}_{k,\ell}\right)}_{1\le k,\ell \le n}$ with respect to bases $\left\{{v}_{i}\right\}$ and $\left\{{w}_{k}\right\}$ of $V$ and $W,$ respectively, then $$(V\otimes W)(a\otimes b){v}_{j}\otimes {w}_{\ell}=\sum _{i=1}^{m}{a}_{i,j}{v}_{i}\otimes \sum _{j=1}^{n}{b}_{k,\ell}{w}_{k}=\sum _{i,k}{a}_{i,j}{b}_{k,\ell}{v}_{i}\otimes {w}_{k}\text{.}$$ If we order the basis $\{{v}_{i}\otimes {w}_{j}\}$ lexicographically $(1,1),$ $\dots ,$ $(1,n),$ $(2,1),$ $\dots ,$ then the $(i,j)\otimes (k,\ell )$ entry of the matrix of $(V\otimes W)(a\otimes b)$ is ${a}_{i,j}{b}_{k,\ell}\text{.}$

On the other hand, the matrix $V\left(a\right)\otimes W\left(b\right)$ is also a $mn\times mn$ matrix, so acts on $V\otimes W\text{.}$ If we consider the block structure given by $V\left(a\right)\otimes W\left(b\right)={\sum}_{i,j=1}^{m}{a}_{i,j}{E}_{i,j}\otimes W\left(b\right),$ and label the rows and columns in the form $(x,y)$ where $x$ indicates the number of the $n\times n$ block row or block column and $y$ indicates the row or column within that block, then the $(i,j)\times (k,\ell )$ entry is the $(k,\ell )$ entry of ${a}_{i,j}W\left(b\right),$ namely ${a}_{i,j}{b}_{k,\ell}\text{.}$

**Proposition 3.20**
*
If $A$ and $B$ are algebras with representations $V$ and $W,$
respectively, then $V\otimes W$ is an $A\otimes B\text{-module,}$
and the matrix of the representation is given by
$(V\otimes W)(a\otimes b)=V\left(a\right)\otimes W\left(b\right)\text{.}$
*

Apply these tensor product constructions to the previous situation where $A$ is an arbitrary algebra with completely reducible module $V\cong {\sum}_{\lambda}{m}_{\lambda}{V}^{\lambda},$ and where $B$ is an arbitrary algebra with $V\left(B\right)=\stackrel{\u203e}{V\left(A\right)}\text{.}$

The ${V}^{\lambda}$ represent a subset of the isomorphism classes of the irreducible $A\text{-modules.}$ The representations ${C}^{\lambda}:B\to {M}_{{m}_{\lambda}}\left(\u2102\right)$ are, similarly, a subset of the irreducible representations of $B\text{.}$

The algebras $A$ and $B$ both act on $V\text{.}$ Define an action of $B\otimes A$ on $V$ by $(b\otimes a)v=bav\text{.}$ This is well-defined on $B\times A$ since it is the linearization of a bilinear map on $B\times A\text{.}$ Moreover, since $A$ and $B$ commute on $V,$ we have $$({b}_{1}\otimes {a}_{1})({b}_{2}\otimes {a}_{2})v={b}_{1}{a}_{1}{b}_{2}{a}_{2}v={b}_{1}{b}_{2}{a}_{1}{a}_{2}v=({v}_{1}{v}_{2}\otimes {a}_{1}{a}_{2})v\text{.}$$ Hence, this represents a well-defined module action. This brings us to the most important theorem of the course.

**Theorem 3.21**
*
Let $A,$ $B$ and $V$ be as above. Then as
$B\otimes A$ representations,
*
$$V\cong \underset{\lambda \in \stackrel{\u02c6}{V}}{\u2a01}{C}^{\lambda}\otimes {V}^{\lambda}$$
*
where ${C}^{\lambda}$ are irreducible representations of $B$ and
${V}^{\lambda}$ are irreducible representations of $A\text{.}$
*

Proof. | |

Let $n=\left|\stackrel{\u02c6}{V}\right|\text{.}$ From the decompositions $V\left(A\right)={\sum}_{\lambda}{I}_{{m}_{\lambda}}\otimes {V}^{\lambda}$ and $\stackrel{\u203e}{V\left(A\right)}\cong {\sum}_{\lambda}{M}_{{m}_{\lambda}}\left({I}_{{d}_{\lambda}}\left(\u2102\right)\right)$ we have shown that the representations of $A$ and $B$ on $V$ decompose into irreducible representations as $$\begin{array}{ccc}V\left(a\right)& =& \left(\begin{array}{c}{I}_{{m}_{1}}\otimes {V}^{1}\left(a\right)\\ & \ddots \\ \multicolumn{2}{c}{}& {I}_{{m}_{n}}\otimes {V}^{n}\left(a\right)\end{array}\right)\\ V\left(b\right)& =& \left(\begin{array}{c}{C}^{1}\left(b\right)\otimes {I}_{{d}_{1}}\\ & \ddots \\ \multicolumn{2}{c}{}& {C}^{n}\left(b\right)\otimes {I}_{{d}_{n}}\end{array}\right)\end{array}$$ Note that each of the matrices above are block diagonal matrices, with block sizes of ${m}_{\lambda}{d}_{\lambda}\times {m}_{\lambda}{d}_{\lambda},$ as $\lambda $ runs over $\stackrel{\u02c6}{V}\text{.}$ Thus, the $B\otimes A$ action on $V$ defined by $(a\otimes b)v=bav$ is represented by the matrices of the form $$V(b\otimes a)=V\left(b\right)V\left(a\right)=\left(\begin{array}{cccc}{C}^{1}\left(b\right)\otimes {V}^{1}\left(a\right)& & & \\ & {C}^{2}\left(b\right)\otimes {V}^{2}\left(a\right)& & \\ & & \ddots & \\ & & & {C}^{n}\left(b\right)\otimes {V}^{n}\left(a\right)\end{array}\right)\text{.}$$ Thus the $B\otimes A$ representation decomposes as $V={\oplus}_{\lambda \in \stackrel{\u02c6}{V}}{C}^{\lambda}\otimes {V}^{\lambda}\text{.}$ $\square $ |

**Remarks.** Note that $V\left(A\right)$ has the same center as
$\stackrel{\u203e}{V\left(A\right)},$
since both are block diagonal with the same block sizes.

Note also that if we suppose $A$ and $B$ are semisimple, and that ${V}^{\lambda},\lambda \in \stackrel{\u02c6}{A},$ and ${V}^{\mu},\mu \in \stackrel{\u02c6}{B}$ are representatives of the isomorphism classes of irreducible $A$ and $B$ modules, respectively, then $A\otimes B$ is semisimple and the set of all ${V}^{\lambda}\otimes {V}^{\mu},$ $\lambda \in \stackrel{\u02c6}{A},$ $\mu \in \stackrel{\u02c6}{B},$ are a complete set of irreducible modules for $A\otimes B\text{.}$

However, even if $V\left(A\right)$ is semisimple (as always, $\stackrel{\u203e}{V\left(A\right)}$ is semisimple), the decomposition of $V$ into $\stackrel{\u203e}{V\left(A\right)}\otimes V\left(A\right)$ modules contains only those irreducibles of the form ${C}^{\lambda}\otimes {V}^{\lambda}\text{.}$ There are no summands isomorphic to the irreducible $\stackrel{\u203e}{V\left(A\right)}\otimes V\left(A\right)$ modules of the form ${C}^{\mu}\otimes {V}^{\lambda}$ for $\lambda \ne \mu \text{.}$

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