# proof of topologically irreducible representations are algebraically irreducible for $C^{*}$-algebras

Denote by $\mathcal{H}$ an arbitrary Hilbert space. To fix notation let $\mathcal{U}\subset\mathcal{L}(\mathcal{H})$ be a $C^{\ast}$ subalgebra of $\mathcal{L}(\mathcal{H})$. We then define the commutator of $\mathcal{U}$ by

 $\displaystyle\mathcal{U}^{\prime}$ $\displaystyle:=\{T\in\mathcal{L}(\mathcal{H}):TU=UT\ \forall U\in\mathcal{U}\}$

Note that $\mathcal{U}^{\prime}$ is closed with regard to the weak topology (see this entry (http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra)). So $\mathcal{U}^{\prime}$ is always a von Neumann algebra.

As an immediate consequence of Schur’s Lemma for group representations on a Hilbert space we obtain the following result.

Lemma. Let $\mathcal{U}$ be a $\ast$-algebra and let $\pi$ be a $\ast$-representation of $\mathcal{U}$ on the Hilbert space $\mathcal{H}$. Then $\pi$ is topologically irreducible iff $\pi(\mathcal{U})^{\prime}=\mathbb{C}I$.

We can now prove the result.

Theorem. Let $\mathcal{U}$ be a $C^{\ast}$ algebra. Assume the $\ast$-representation $\pi$ of $\mathcal{U}$ on the Hilbert space $\mathcal{H}$ is topologically irreducible. Then $\pi$ is algebraically irreducible.

###### Proof.

By the Lemma it follows that $\pi(\mathcal{U})^{\prime}=\mathbb{C}I$. Hence $\pi(\mathcal{U})^{\prime\prime}=\mathcal{L}(\mathcal{H})$. By the double commutant theorem (http://planetmath.org/VonNeumannDoubleCommutantTheorem) every operator in $\mathcal{L}(\mathcal{H})_{1}$ (the unit ball in the set of bounded operators $\mathcal{L}(\mathcal{H})$) belongs to the strong operator closure of $\pi(\mathcal{U})_{1}$ (the unit ball in $\pi(\mathcal{U})$).

To show the algebraical irreducibility of $\pi(\mathcal{U})$ it is enough to find for two given vectors $x,y\in\mathcal{H},x\not=0$ an element $T\in\mathcal{U}$ such that $\pi(T)x=y$ holds. Indeed, it is enough to consider the case $\|x\|=\|y\|=1$.

Now construct the rank one approximation $\tilde{T}_{1}:=y\otimes x$ ($\Leftrightarrow\tilde{T}_{1}z=\langle x,z\rangle y,z\in\mathcal{H}\Rightarrow% \tilde{T}_{1}x=\|x\|y=y$) with a corresponding $T_{1}\in\mathcal{U},\pi(T_{1})\in\pi(\mathcal{U})_{1}$, so that $\|y-\pi(T_{1})x\|=\|\tilde{T}_{1}x-\pi(T_{1})x\|\leq\frac{1}{2}$.

Approximate further $\tilde{T}_{2}:=(y-\pi(T_{1})x)\otimes x\in\frac{1}{2}\mathcal{L}(\mathcal{H})_% {1}$ and choose $\pi(T_{2})\in\frac{1}{2}\pi(\mathcal{U})_{1}$ with $\|y-\pi(T_{1})x-\pi(T_{2})x\|=\|\tilde{T}_{2}x-\pi(T_{2})x\|\leq\frac{1}{2^{2}}$.

Proceed by induction with $\tilde{T}_{n}:=(y-\sum_{j=1}^{n-1}\pi(T_{j})x)\otimes x\in 2^{-j}\mathcal{L}(% \mathcal{H})_{1}$. Choose $\pi(T_{n})\in 2^{-n}\pi(\mathcal{U})_{1}$ with $\|y-\sum_{j=1}^{n}\pi(T_{j})x\|=\|\tilde{T}_{n}x-\pi(T_{n})x\|\leq 2^{-n}$. Then we have $\pi(T):=\sum_{j=1}^{n}\pi(T_{n})$ in $\mathcal{U}$ and $\pi(T)x=y$ which completes the proof. ∎

Title proof of topologically irreducible representations are algebraically irreducible for $C^{*}$-algebras ProofOfTopologicallyIrreducibleRepresentationsAreAlgebraicallyIrreducibleForCalgebras 2013-03-22 19:04:12 2013-03-22 19:04:12 karstenb (16623) karstenb (16623) 8 karstenb (16623) Proof msc 46L05