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

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

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

Note that ${𝒰}^{\prime }$ is closed with regard to the weak topology (see this entry (http://planetmath.org/CommutantIsAWeakOperatorClosedSubalgebra)). So ${𝒰}^{\prime }$ 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 $𝒰$ be a $\ast$-algebra and let $\pi$ be a $\ast$-representation of $𝒰$ on the Hilbert space $\mathscr{H}$. Then $\pi$ is topologically irreducible iff $\pi {(𝒰)}^{\prime }=ℂI$.

We can now prove the result.

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

Proof.

By the Lemma it follows that $\pi {(𝒰)}^{\prime }=ℂI$. Hence $\pi {(𝒰)}^{\prime \prime }=ℒ(\mathscr{H})$. By the double commutant theorem (http://planetmath.org/VonNeumannDoubleCommutantTheorem) every operator in $ℒ{(\mathscr{H})}_1$ (the unit ball in the set of bounded operators $ℒ(\mathscr{H})$) belongs to the strong operator closure of $\pi {(𝒰)}_1$ (the unit ball in $\pi (𝒰)$).

To show the algebraical irreducibility of $\pi (𝒰)$ it is enough to find for two given vectors $x,y\in \mathscr{H},x\ne 0$ an element $T\in 𝒰$ such that $\pi (T)x=y$ holds. Indeed, it is enough to consider the case $\parallel x\parallel =\parallel y\parallel =1$.

Now construct the rank one approximation ${\tilde{T}}_1:=y\otimes x$ ($\iff {\tilde{T}}_{1}z=⟨x,z⟩y,z\in \mathscr{H}\Rightarrow {\tilde{T}}_{1}x=\parallel x\parallel y=y$) with a corresponding $T_1\in 𝒰,\pi (T_1)\in \pi {(𝒰)}_1$, so that $\parallel y-\pi (T_1)x\parallel =\parallel {\tilde{T}}_{1}x-\pi (T_1)x\parallel \le \frac{1}{2}$.

Approximate further ${\tilde{T}}_2:=(y-\pi (T_1)x)\otimes x\in \frac{1}{2}ℒ{(\mathscr{H})}_1$ and choose $\pi (T_2)\in \frac{1}{2}\pi {(𝒰)}_1$ with $\parallel y-\pi (T_1)x-\pi (T_2)x\parallel =\parallel {\tilde{T}}_{2}x-\pi (T_2)x\parallel \le \frac{1}{2^2}$.

Proceed by induction with ${\tilde{T}}_n:=(y-{\sum }_{j=1}^{n-1}\pi (T_j)x)\otimes x\in 2^{-j}ℒ{(\mathscr{H})}_1$. Choose $\pi (T_n)\in 2^{-n}\pi {(𝒰)}_1$ with $\parallel y-{\sum }_{j=1}^n\pi (T_j)x\parallel =\parallel {\tilde{T}}_{n}x-\pi (T_n)x\parallel \le 2^{-n}$. Then we have $\pi (T):={\sum }_{j=1}^n\pi (T_n)$ in $𝒰$ and $\pi (T)x=y$ which completes the proof.∎