Showing representation on $C^*$-algebra is equivalent to multiple of identity.

Question

I'm stuck at the proof of corollary 1 in the text by Arveson (Invitation to $C^*$-algebras):

(here $\mathcal{A}$ is a $C^*$-subalgebra of $B_0(\mathcal{H})$, the compact operators on the Hilbert space $\mathcal{H}$).

We have a representation $\mathcal{A} \to B(\mathcal{K})$. If we apply the theorem on $\pi$, shouldn't we get orthogonal subrepresentations $\pi_i:\mathcal{A} \to B(\mathcal{K}_i)$ for some subspace $\mathcal{K}_i \leq \mathcal{K}$? In the proof we have subrepresentations $\mathcal{A} \to B(\mathcal{H}_i)$ where $\mathcal{H}_i\leq \mathcal{H}$. What am I missing?

Answer 1

Firstly, in the first sentence of the proof for the corollary, is a typo: $\mathcal A$ should be $\mathcal C(\mathcal H)$ (or $B_0(\mathcal H)$ as you notate it). This proof fails for arbitrary $C^*$-algebras of compact operators.

Secondly, each $\mathcal H_i$ is a subspace of $\mathcal K$, not a subspace of $\mathcal H$. I believe the confusing part is just bad notation.