# criterion for a Banach *-algebra representation to be irreducible

Theorem - Let $\mathcal{A}$ be a Banach *-algebra, $H$ an Hilbert space and $I$ the identity operator in $H$. A representation (http://planetmath.org/BanachAlgebraRepresentation) $\pi:\mathcal{A}\longrightarrow H$ is topologically irreducible if and only if $\pi(\mathcal{A})^{\prime}=\mathbb{C}I$, i.e. if and only if the commutant of $\pi(\mathcal{A})$ consists of scalar multiples of the identity operator.

Proof : $(\Longrightarrow)$

As $\pi(\mathcal{A})$ is selfadjoint, $\pi(\mathcal{A})^{\prime}$ is a von Neumann algebra.

Suppose $\pi(\mathcal{A})^{\prime}\neq\mathbb{C}I$. Then the dimension of $\pi(\mathcal{A})^{\prime}$ is greater than one.

It is known that von Neumann algebras of dimension greater than one contain non-trivial projections, so there is a projection $P\in\pi(\mathcal{A})^{\prime}$ such that $P\neq 0$ and $P\neq I$.

As $P\in\pi(\mathcal{A})^{\prime}$, $P$ commutes with every operator $T\in\pi(\mathcal{A})$, that is $PT=TP$.

Thus $Ran\;P$ is an invariant subspace of every $T\in\pi(\mathcal{A})$. Therefore $\pi$ is not an irreducible representation.

$(\Longleftarrow)$

Conversely, suppose that $\pi$ is not an irreducible representation. There exists a closed $\pi(\mathcal{A})$-invariant subspace different from $\{0\}$ and $H$.

Let $P$ be the projection onto that closed invariant subspace.

Invariance can be expressed as: $\pi(a)P=P\pi(a)P$ for every $a\in\mathcal{A}$. It follows that

 $P\pi(a)=(\pi(a)^{*}P)^{*}=(\pi(a^{*})P)^{*}=(P\pi(a^{*})P)^{*}=P\pi(a^{*})^{*}% P=P\pi(a)P=\pi(a)P$

for every $a\in\mathcal{A}$.

We conclude that $P$ commutes with every element of $\pi(\mathcal{A})$, i.e. $P\in\pi(\mathcal{A})^{\prime}$.

Thus $\pi(\mathcal{A})^{\prime}\neq\mathbb{C}I\;\;\square$

Title criterion for a Banach *-algebra representation to be irreducible CriterionForABanachalgebraRepresentationToBeIrreducible 2013-03-22 17:27:43 2013-03-22 17:27:43 asteroid (17536) asteroid (17536) 9 asteroid (17536) Theorem msc 46K10