# Finite dimensional irreducible representation

what's wrong about following proof? It's obviously wrong as we never used finite dimension of $\mathcal{H}$...

Lemma: Let $A$ be C*-algebra. Let $\pi: A \to \mathcal{B(H)}$ be irreducible *-representation and $\text{dim}\mathcal{H}=n\in \mathbb{N}$. Then $\pi$ is surjective.

(wrong) proof: The commutant $\pi(A)'$ of $\pi(A)$ is equal to $\mathbb{C}\cdot I$ (property of irreducible representations. $I$ is identity operator). Hence $\pi(A)''=\mathcal{B(H)}$. From Von neumann double commutant theorem we get $\overline{\pi(A)}=\pi(A)''=\mathcal{B(H)}$. But $\pi(A)$ is closed (as homomorphic image of *-algebra isomorphism) in $\mathcal{B(H)}$, hence $\pi(A)=\mathcal{B(H)}$.

• I'm not familiar with the theory anymore, but why is $\pi(A)$ closed? I don't get that part. On the other hand, I believe that in general any finite-dimensional subspace of a normed space is closed, that solves it. – Mathematician 42 Apr 27 '18 at 9:22
• @Mathematician42 This is because $\pi(A)=\tilde\pi(A/\ker\pi)$, where $\tilde\pi$ is the induced map on $A/\ker\pi$. But $\tilde\pi$ is injective, hence isometric, hence it's image is complete, and therefore closed. – Aweygan Apr 27 '18 at 10:43

In your second-to-last and last sentences, you're dealing with two different topologies of $\mathcal{B(H)}$. From the von Neumann double commutant theorem we obtain that $\overline{\pi(A)}^\text{SOT}=\pi(A)''=\mathcal{B(H)}$. Now since $\pi$ is a $*$-homomorphism, $\pi(A)$ is closed in the norm-topology. The way the argument is phrased, it seems to imply that closure in the norm topology implies closure in the strong operator topology, which is simply not true.