Takesaki: Lemma about enveloping von Neumann algebra

Question

Consider the following lemma with proof from Takesaki's book "Theory of operator algebra I" (p121):

It appears to me that Takesaki claims at the end of the proof that $\pi(A)_1$ is $\sigma$-weakly dense in $(\pi(A)'')_1$. Of course, this should somehow follow from the Kaplansky density theorem. However, don't we need that the representation $\pi$ is non-degenerate for this? In that case, $\pi(A)$ is a non-degenerate $*$-algebra and thus its bicommutant is its $\sigma$-weak closure, so $\pi(A)$ is $\sigma$-weakly dense in $\pi(A)''$ and then $\pi(A)_1$ is also $\sigma$-weakly dense in $(\pi(A)'')_1$ by the Kaplansky density theorem. So concretely my question is: do we need to assume that $\pi: A \to B(H)$ is non-degenerate, i.e. that $\pi(A)H$ is dense in $H$? Thanks in advance for any help/comment/remark!

Answer 1

"Yes".

Degenerate representations of $C^*$-algebras have a simple form: each $\pi(a)$ restricts to $H_1 := [\pi(A)(H)]$ and acts as $0$ on $H_1^\perp$. Thus we obtain a non-degenerate representation $\pi_1$ on $H_1$, and the $0$ representation on $H_1^\perp$.

A calculation shows that $\tilde\pi:A^{**}\rightarrow B(H)$ will have the same decomposition. However, $\pi(A)''$ will be $\{ T+\alpha 1_{H_1^\perp} : T\in \pi_1(A)'', \alpha\in\mathbb C\}$, compare with Theorem 3.9 in the previous chapter of Takesaki. So we can never hope to have that $\tilde\pi$ maps onto $\mathcal M(\pi) = \pi(A)''$, unless $\pi$ is non-degenerate.

I think there are some other statements in this section of Takesaki which need "non-degenerate". However, we've just seen that degenerate representations are in a sense trivial, and so really we can just read "representation" as always meaning "non-degenerate representation".

(Notice there is another small typo: surely $\mathcal M$ should be $\mathcal M(\pi)$ in the statement of (iii). This is just a fact of life: authors are human, and we have to expect the occasional slip.)